i
Advancing in
Mathematics
Student’s Book for Form 3
Kinyua Mugo
Lucy Maina
Jared Odera
ii
Published by
Longhorn Publishers (K) Ltd
Funzi Road, Industrial Area,
P.O. Box 18033
Nairobi, Kenya
Longhorn Publishers (U) Ltd
Plot M220, Ntinda Industrial Area
Jinja Road, P.O. Box 24745
Kampala, Uganda
Longhorn Publishers (T) Ltd
Msasani Village, Block F, House No. 664,
Old Bagamoyo Rd.
P.O. Box 1237,
Dar es Salaam, Tanzania
© Kinyua Mugo, Lucy Maina, Jared Odera, 2004
All rights reserved. No part of this publication may be reproduced,
stored in a retrieval system or transmitted in any form or by any means,
electronic, mechanical, photocopying, recording or otherwise without
the prior written permission of the publisher.
First published 2004
Reprinted 2006, 2007
Revised 2008
ISBN 9966 49 584 3
Typsetting, design and graphics by Justice Mogaki
Printed by Printpoint Ltd., Changamwe Road, Industrial Area,
P. O. Box 30975 - 00100 Nairobi, Kenya.
iii
Contents
1. Quadratic expressions and
equations (2).................................... 1
Factorisation of quadratic
expressions............................................. 1
Perfect squares ....................................... 2
Solving quadratic equations................... 3
Completing the square ........................... 4
The quadratic formula............................ 6
Tables of values and graphs of quadratic
expressions............................................. 8
Graphical solution of quadratic
equations ................................................ 11
Simultaneous equations: One linear,
one quadratic.......................................... 14
Application of quadratic equations to
real life situations................................... 16
2. Approximations and errors ........... 19
Computing using calculators ................. 19
Estimation and approximation............... 21
Counting and measuring errors.............. 22
Rounding off, truncating and signicant
gures .................................................... 23
Absolute error ........................................ 23
Relative and percentage error ................ 24
Propagation of error............................... 25
Error in multiplication ........................... 28
Error in division..................................... 29
3. Trigonometry .................................. 32
Introduction............................................ 32
The unit circle ........................................ 32
Trigonometric ratios of angles between
0° and 360° ............................................ 33
Trigonometric ratios of negative
angles ..................................................... 36
Angles greater than 360° or less
than
–
360°............................................... 36
Use of trigonometric tables.................... 37
Radian measure...................................... 39
Trigonometric graphs............................. 40
Graph of cos
θ
........................................ 42
Graph of tan
θ
........................................ 44
The sine rule .......................................... 45
The cosine rule....................................... 47
Application of sine and cosine rules to
real life situations................................... 49
4. Surds ............................................... 52
Rational and irrational numbers............. 52
Rational and irrational square roots ...... 52
Surds ...................................................... 54
Simplication of surds........................... 54
Operations on surds ............................... 55
Multiplication of surds........................... 55
Division of surds: rationalising the
denominator ........................................... 57
5. Further logarithms ........................ 59
Logarithmic notation ............................. 59
Laws of logarithms ................................ 59
Simplication and evaluation of
logarithmic expressions ......................... 60
Equations involving logarithms............. 61
Further computations using logarithmic
laws ........................................................ 62
Revision exercises 1 .................... 64
Revision exercise 1.1 ............................. 64
Revision exercise 1.2 ............................. 65
Revision exercise 1.3 ............................. 66
6. Commercial arithmetic .................. 68
Principal, rate, time and interest ............ 68
Simple interest ....................................... 68
Compound interest................................. 70
Compound interest formula ................... 71
Appreciation and depreciation............... 74
Hire purchase ......................................... 76
Taxation ................................................. 77
Income tax ............................................. 77
Pay as you earn (PAYE)......................... 79
Benets .................................................. 79
7. Circles — Chords and tangents .... 82
Arc of circle ........................................... 82
Properties of chords ............................... 83
iv
Length of a chord................................... 84
Parallel chords and equal chords ........... 84
Intersecting chords................................. 86
Tangent to a circle.................................. 89
Construction of tangents to circles ........ 91
Common tangents to two circles............ 92
Contact of circles ................................... 95
Angles in alternate segment................... 97
Inscribed circles ..................................... 99
Circumscribed circles ............................ 99
Escribed circles...................................... 100
Centroid and orthocentre of a triangle ...100
Application of arcs, chords and
tangents to real life situations ................102
8. Matrices...........................................105
Matrix ....................................................105
Order of a matrix ...................................105
Addition and subtraction of matrices.....106
Scalar multiplication of matrices ...........108
Multiplication of matrices......................109
General matrix multiplication................110
Compatibility in multiplication.............. 112
Identity and zero matrices...................... 114
Determinant and inverse of a 2 × 2
matrix..................................................... 115
Inverse of a 2 × 2 matrix........................115
Simultaneous linear equations in two
unknowns ..............................................117
9. Formulae and variation .................121
Formulae ................................................121
The subject of a formula ........................122
Variation.................................................123
Direct variation ......................................124
Non-linear direct variation.....................125
Inverse variation ....................................126
Joint variation ........................................127
Partial variation......................................128
10. Sequences and series......................131
Number patterns.....................................131
Sequences ..............................................132
Arithmetic sequences............................. 134
Geometric sequence............................... 135
Series......................................................137
Arithmetic progression (A.P.)
[Arithmetic series] .................................137
Geometric progression (G.P.)
(Geometric series)..................................138
Application of A.P. and G.P. to real life
situations ................................................140
Revision exercises 2 .....................142
Revision exercise 2.1 .............................142
Revision exercise 2.2 .............................143
Revision exercise 2.3 .............................144
11. Vectors .............................................146
Coordinates in two and three
dimensions .............................................146
Column vectors and position vectors in
three dimensions ....................................146
The unit vectors i, j and k .................... 148
Combination of vectors..........................150
Parallel vectors and collinear points ......151
Proportional division of a line ...............152
The ratio theorem...................................153
Further problems on ratio theorem ........155
Application of vector methods in
geometry ................................................156
12. Binomial expansion ........................158
Binomial expansion ...............................158
Pascal’s triangle .....................................158
Finding coefcients of individual terms 160
Applying binomial expansions ..............161
13. Probability....................................... 162
What is probability?...............................162
Experimental probability .......................163
Range of probability measure................ 165
Theoretical probability...........................165
The possibility or sample space............. 167
Finite and innite sample space.............168
Discrete probability space......................170
Continuous possibility space .................171
Combined events ...................................172
Independent events ................................174
Tree diagram ..........................................175
v
14. Compound proportion, mixtures
and rates of work............................177
Proportion ..............................................177
Proportional parts...................................177
Ratios and weighting .............................178
Continued proportion............................. 179
Derived proportion.................................179
Compound proportions and rates of
work .......................................................181
Mixtures................................................. 183
15. Graphical methods .........................186
Tables of values and graphs of given
relations..................................................186
Graphs of cubic relations .......................188
Exercise 15.2..........................................190
Solving cubic equations......................... 190
Average rate of change ..........................192
Rate of change at an instant ...................193
Empirical data and their graphs .............194
The equation of a circle .........................198
Revision exercises 3...............................202
Revision exercise 3.1 .............................202
Revision exercise 3.2 .............................203
Revision exercise 3.3 .......................... 204
Mathematical formulae and tables....
0207
vi
1
1
QUADRATIC EXPRESSIONS AND
EQUATIONS (2)
Factorisation of quadratic expressions
In Form 2, we learnt that the product of two
rst-degree (i.e. linear) expressions in x is a
quadratic expression in x. For example,
(2x + 15)(x – 1) = 2x (x – 1) + 15 (x – 1)
= 2x
2
– 2x + 15x – 15
= 2x
2
+ 13x – 15
To factorise a quadratic expression, we
express it in such a way that the process of
expansion can be worked backwards. This
requires that the term in x be replaced by two
terms chosen in such a way that the grouping
method of factorisation can be used.
Example 1.1
Factorise 3x
2
+ 13x + 4
Solution
Replace 13x by two terms whose product is
equal to 3x
2
× 4 = 12x
2
, i.e. the product of the
two terms.
Since the product, 12x
2
, is positive, the terms
must have the same sign and since the sum, 13x,
is positive, both terms must be positive.
12x
2
= 12x × x and 12x + x = 13x
∴3x
2
+ 13x + 4 = 3x
2
+ 12x + x + 4
= 3x(x + 4) + 1(x + 4)
= (x + 4)(3x + 1)
Example 1.2
Factorise 2y
2
– 11y + 12
Solution
Replace
–
11y by two terms whose product is
equal to 2y
2
× 12 = 24y
2
.
Since the product, 24y
2
, is positive, the terms
must have the same sign, but since the sum,
–
11y,
is negative both terms must be negative.
24y
2
= 8y × 3y and
–
8y – 3y =
–
11y
∴2y
2
– 11y + 12 = 2y
2
– 8y – 3y + 12
= 2y(y – 4) – 3(y – 4)
= (y – 4)(2y – 3)
Example 1.3
Factorise 2n
2
– 7n – 15
Solution
Replace
–
7n by two terms whose product is
2n
2
× (–15) =
–
30n
2
.
Since the product is negative, the terms
must have opposite signs and since the sum
is negative, the numerically larger term is
negative.
–
30n
2
= 3n × (
–
10n) and 3n – 10n =
–
7n.
∴2n
2
– 7n – 15 = 2n
2
+ 3n – 10n –15
= n (2n + 3) – 5(2n + 3)
= (2n + 3)(n – 5)
Note:
If the expression to be factorised has a common
factor in three terms, the process becomes
simpler if we start by factoring it out.
Example 1.4
Factorise 2x
2
– 14x + 20
Solution
First factor out 2 to get
2x
2
– 14x + 20 = 2(x
2
– 7x + 10)
Now factorise x
2
–7x + 10 by replacing
–
7x
by two terms whose product is x
2
× 10 = 10x
2
.
These two terms are
–
2x and
–
5x.
Hence
2x
2
– 14x + 20 = 2[x
2
– 2x – 5x + 10]
= 2[x(x – 2) – 5(x – 2)]
= 2(x – 2)(x –5).
2
Exercise 1.1
1. Expand the following products.
(a) (x + 4)(x – 6) (b) (2x – 9)(x + 8)
(c) (5 – y)(12 – 5y) (d) (2x + 3)(x + 4)
(e) (3x + 2y)(2x – 5y) (f) (1 – 2t)(2 + 5t)
2. Factorise the following quadratic expressions.
(a) x
2
+ 6x + 8 (b) 3x
2
+ 5x + 2
(c) 3p
2
+ 8p + 4 (d) y
2
– 7y + 12
(e) n
2
– 5n – 14 (f) z
2
+ 3z – 70
(g) 4 – 3x – x
2
(h) 12q
2
+ 33q – 9
(i) 2y
2
– 22y + 60 (j) x
2
– 13x + 40
(k)
–
20x
2
+ 14x + 24 (l) 6b
2
– 13b + 6
(m) 24x
2
+ 2xy – y
2
(n) m
2
n
2
– 7mn + 10
Perfect squares
Recall that
1. (a + b)
2
= a
2
+ b
2
+ 2ab, and
2. (a – b)
2
= a
2
+ b
2
– 2ab
The right hand sides of 1 and 2 are the
expansions of the squares on the left hand sides.
Such expressions are called perfect squares.
They may be written in words as:
1. The square of the sum of two numbers
is equal to the sum of their squares plus
twice their product.
2. The square of the difference of two
numbers is equal to the sum of their
squares minus twice their product.
Example 1.5
Write down the square of
(a) 4x + y (b) 2x – 3y
Solution
(a) (4x + y)
2
= (4x)
2
+ (y)
2
+ 2(4x)(y)
= 16x
2
+ y
2
+ 8xy
(b) (2x – 3y)
2
= (2x)
2
+ (3y)
2
– 2(2x)(3y)
= 4x
2
+ 9y
2
– 12xy
After a little practice, it should be easy to work
out the middle step mentally.
Example 1.6
Is 4x
2
– 40xy + 25y
2
a perfect square?
Solution
4x
2
= (2x)
2
and 25y
2
= (5y)
2
.
Thus, if the expression is a perfect square it
must be the square of 2x – 5y.
But (2x – 5y)
2
= 4x
2
+ 25y
2
– 20xy.
∴ 4x
2
– 40xy + 25y
2
is not a perfect square.
Note:
Instead of expanding (2x – 5y)
2
in order to
compare the middle term with that of the given
expression, we could alternatively rst check
if
–
40xy is equal to 2(2x)(
–
5y). In this case, it is
not. This indicates that 4x
2
– 40xy + 25y
2
is not
a perfect square.
Example 1.7
Complete 9p
2
– 12pq + 4q
2
= ( )
2
.
Solution
Factorising:
9p
2
– 12pq + 4q
2
= 9p
2
– 6pq – 6pq + 4q
2
= 3p(3p – 2q) – 2q(3p – 2q)
= (3p – 2q)(3p – 2q)
= (3p – 2q)
2
Alternative method
Since 9p
2
– 12pq + 4q
2
is a perfect square, we
realize that
9p
2
= (3p)
2
, 4q
2
= (2q)
2
and
2 × (3p) × (2q) = 12pq
∴ 9p
2
– 12pq + 4q
2
= (3p – 2q)
2
Notice that the second method is quicker and
shorter. Thus, given an expression that is a
perfect square, we need not use the grouping
method when factorising it.
3
Exercise 1.2
1. Write down the squares of the following.
(a) x + 5 (b) x – 6 (c) 3 – x
(d)
–
y – 4 (e) 2x + 9 (f) 2y – 7
(g) 4x – y (h) 5y – 4z (i) x +
2. State whether each of the following
expressions is a perfect square. If it is,
write it in the form (a + b)
2
or (a – b)
2
.
(a) x
2
+ 6x + 9 (b) x
2
– 9x + 9
(c) x
2
– 8x +16 (d) x
2
– 2x + 1
(e) y
2
– 10y – 25 (f) 4 – 12m – 9m
2
(g) a
2
+ 3a + (h) t
2
+ 2t +
(i) 16a
2
– 20ab + 25b
2
(j) 25p
2
– 70pq + 49q
2
Solving quadratic equations
Recall that when solving quadratic equations,
we use the following fact:
If the product of two numbers is equal to zero,
then one of the numbers or both of them must
be zero, i.e.
if ab = 0 ,
either a = 0 or b = 0 or a = b = 0
Example 1.8
Solve the equation 2x
2
– 22x + 60 = 0
Solution

Thus, 2x
2
– 22x + 60 = 2x
2
– 12x – 10x + 60
= 2x (x – 6) – 10(x – 6)
= (x – 6)(2x – 10)
 ∴ (x – 6)(2x – 10) = 0
If (x – 6)(2x – 10) = 0
either x – 6 = 0 or 2x – 10 = 0
i.e. x = 6 or x = 5.
Alternatively
2x
2
–22x + 60 = 0
⇒ 2(x
2
– 11 + 30) = 0
⇒ 2(x – 6)(x – 5) = 0
There are three factors but only (x– 6) and
(x – 5) give the same answer as before.
If the expression on the LHS of an equation is a
perfect square, we use the fact that:
Every positive number has two square roots.
For example, if n
2
= 36, then n =
+
6 or
–
6
since (
+
6)
2
= 36 and (
–
6)
2
= 36. This is usually
written as n = ±6.
Example 1.9
Solve the equation (2x – 5)
2
= 9.
Solution
If (2x – 5)
2
= 9,
then 2x – 5 = ±3
∴ 2x = 5 ± 3
i.e. 2x = 8 or 2 (since 5 + 3 = 8 and 5 – 3 = 2)
⇒ x = 4 or 1
Note that you could also solve this equation by
rst rearranging the equation in the form
ax
2
+ bx + c = 0 and then factorising the LHS.
Example 1.10
Solve the equation p
2
– 10p + 25 = 8
Solution
p
2
– 10p + 25 = 8
(p – 5)
2

⇒ p – 5 = ±√8
⇒ p = 5 ± √8
Thus, p = 5 + √8 or p = 5 – √8
A result like this may be expressed approximately
in decimals by putting ±√8 = ±2.828. Such
roots are said to be non-rational. (Rational
and non-rational numbers will be dealt with in
Chapter 4).
1
x
–
1
4
–
9
4
–
4
1
4
–
7
9
–
16
9
––
7
9
–
1
4
–
7
9
–
9
4
–
Exercise 1.3
Solve the following equations. If an equation
has non-rational roots, leave the answer in the
form given in Example 1.10.
1. (a) (x + 4)(x + 5) = 0
(b) (x + 3)(2x – 5) = 0
2. (a) (2x – 3)(2x + 7) = 0
(b) (3x – 4)(4x – 1) = 0
3. (a) x
2
+ 6x + 8 = 0 (b) a
2
– 7a + 12 = 0
4. (a) 15 – 2y – y
2
= 0 (b) 4x
2
+ 16x + 15 = 0
5. (a) 6 + 5y – 6y
2
= 0 (b) 2x
2
– 7x – 15 = 0
6. (a) (4x + 3)
2
= 16 (b) (2p + 9)
2
= 9
7. (a) (4y – 15)
2
= 25 (b) (3 – b)
2
= 7
8. (a) (6 – x)
2
= 2 (b) (x – 2)
2
= 6
9. (a) x
2
+ 6x + 9 = (b) x
2
– 8x + 16 =
–
3
10. (a) y
2
+ 3y + = 1 (b) h
2
– 6h + 9 = 14
11. (a) 4x
2
+ 12x + 9 = (b) 1 – 6x + 9x
2
= 2
Completing the square
In some cases, it is not possible to solve
a quadratic equation by the method of
factorisation because the LHS does not
factorise. In such a case, we rst rearrange the
equation to make the LHS a perfect square, i.e.
we complete the square on the LHS.
Example 1.11
What must be added to x
2
+ 8x to make the
result a perfect square?
Solution
Suppose x
2
+ 8x +k is a perfect square and that
it is equal to (x + a)
2
, i.e.
x
2
+ 8x + k = (x + a)
2
.
We know that
(x + a)
2
= x
2
+ 2ax + a
2
∴ x
2
+ 8x + k = x
2
+ 2ax + a
2
.

can only be equal if the corresponding terms
are equal.

2a = 8
∴ a = 4.
Comparing the constant terms:
k = a
2
= 4
2
= 16.
∴ 16 must be added to the expression x
2
+ 8x.
Then x
2
+ 8x + 16 = (x + 4)
2
.
Check by opening the bracket on the RHS!
Example 1.12
Find the term that must be added to p
2
– 6p to
make the expression a perfect square.
Solution
Suppose p
2
– 6p + k is a perfect square.
Then, p
2
– 6p + k = (p – a)
2
i.e. p
2
– 6p + k = p
2
– 2ap + a
2

–
2a =
–
6
 ∴a = 3.
Comparing the constant terms:
k = a
2
= 3
2
= 9
∴ 9 must be added to the expression.
Then, p
2
– 6p + 9 = (p – 3)
2
.
In general:
The quantity to be added is
[
1
2
(coecient of x)]
i.e. to make x
2
± bx a perfect square, we add
(
±b
2
)
2
to get x
2
+ bx +
(
±b
2
)
2
=
(
x ±
b
2
)
2
,
else x
2
– bx +
(
–b
2
)
2
=
(
x –
b
2
)
2
Check:
In Example 1.11, the coecient of x is 8; half
of 8 is 4; and the square of 4 is 16. Hence, 16
must be added.
5
–
5
2
––
–
5
2
––
–
5
2
––
25
4
––
8 + 25
4
–––––
In Example 1.12, the coecient of p is
–
6; half
of
–
6 is
–
3; and the square of
–
3 is 9. Hence, 9
must be added.
Example 1.13
Solve the equation x
2
+ 8x + 9 = 0.
Solution

method of completing the square.
x
2
+ 8x + 9 = 0
x
2
+ 8x =
–
9 (subtracting 9 from both
sides)
x
2
+ 8x + =
–
9 + (adding to
both sides)
x
2
+ 8x + 4
2
= 7
⇒(x + 4)
2
= 7
 ⇒ x + 4 = ±√7
i.e. x =
–
4 ± √7.
Thus, x =
–
4 + √7 or
–
4 – √7.
Example 1.14
Solve the equation q
2
– 5q – 2 = 0.
Solution

method of completing the square.
q
2
– 5q – 2 = 0
q
2
– 5q = 2
Add the square of to both sides
q
2
– 5q +
( )
2
= 2 +
( )
2
= 2 +
=

∴
(
q –
)
2
=
⇒ q – = ±
√
⇒q = ±
√
= ±
=
= or
Exercise 1.4
1. What term must be added to each of the
expressions below to make the expression a
perfect square? Write each expression as
(a + b)
2
or (a – b)
2
.
(a) y
2
+ 10y (b) x
2
– 8x
(c) p
2
+ 6pq (d) d
2
+ 5d
(e) a
2
– 12ab (f) q
2
– 7q
(g) n
2
+ n (h) m
2
– mn
2. Solve the equations below by completing
the square. Do not put the answer in
decimal form.
(a) x
2
– 4x – 21 = 0 (b) y
2
+ 3y – 10 = 0
(c) x
2
+ 3x – 11 = 0 (d) d
2
+ 4d – 4 = 0
(e) p
2
+ 3p – 2 = 0 (f) 25 – 10x + x
2
= 0
(g) n
2
– 14n + 2 = 0 (h) t
2
– 15t – 4 = 0
Completing the square when the
coecient of x
2
is not equal to one
If the quadratic equation is of the form
ax
2
+ bx + c = 0, where a ≠ 1 and the LHS
does not factorise, rst divide both sides by
a to make the coecient of x
2
one and then
complete the square.
Example 1.15
Solve the equation 2x
2
+ 14x + 9 = 0, giving
your answer correct to 3 d.p.
Solution
2x
2
+ 14x + 9 = 0

2
one by dividing
both sides by 2.
The equation becomes
x
2
+ 7x + 4
1
2
= 0
⇒x
2
+ 7x =
–
4
1
2
.
8
2
–
( )
8
2
–
( )
8
2
–
( )
5
2
–
33
4
––
5
2
–

2
–––
5 – √33
2
––––––
5 + √33
2
––––––
5
2
–
33
4
––
5
2
–
33
4
––
–––––––
5 ± √33
2
6
Completing the square gives
x
2
+ 7x +
( )
2
=
–
9
2
+
( )
2

⇒
(
x +
)
2
=
–
9
2
+
= +
=
=
∴ x + = ±

⇒ x =
–
±
=
=
= or
=
–
0.716 or
–
6.284
Exercise 1.5
Solve the following equations by factorisation
if possible, otherwise by completing the square.
Give your answers to 2 d.p. where necessary.
1. (a) 2x
2
– 3x = 9 (b) 2y
2
– 4y + 1 = 0
2. (a) 2b
2
+ b + 1 =0 (b) 4x
2
– 8x = – 1
3. (a) 4p
2
– 8p + 3 = 0 (b) 4p
2
= 8p + 3
4. (a) 3e
2
= 9e – 2 (b) 3a
2
– 2 = 12a
5. (a) 5x
2
= – 1 – 15x (b) 2z
2
+ 10z + 5 = 0
The quadratic formula
The general form of a quadratic equation is
ax
2
+ bx + c = 0. Its roots can be found by
the method of completing the square as follows:
ax
2
+ bx + c = 0
Making the coecient of x
2
unity:
x
2
+ x +
= 0
∴ x
2
+
x =
.
Completing the square on the LHS:
x
2
+ x +
( )
2
= +
( )
2
= +
= =
–
18 + 49
4
–––––––
7
2
–
7
2
–
49
4
––
31
4
––
7
2
–

2
–––
–

2
–––––––
–
7 ± 5.568
2
––––––––
–
1.432
2
–––––
–
12.568
2
––––––
31
4
––
7
2
–
–
9
2
––
49
4
––
b
a
–
b
2a
––
–
c
a
––
b
2a
––
b
2
4a
2
–––
–
c
a
––
b
2a
––
b
2
– 4ac
4a
2
–––––––
b
2
– 4ac
4a
2
–––––––
b
2a
––
√b
2
– 4ac
2a
––––––––
√b
2
– 4ac
2a
––––––––
–
b ± √b
2
– 4ac
2a
––––––––––––
–

2
– 4ac
2a
––––––––––––
Factorising the LHS:
(
x +
)
2
=
∴ x + = ±
√
= ±
∴ x =
±
i.e. x =
This is called the quadratic formula.
The value b
2
– 4ac is called the discriminant. It
determines whether or not a quadratic equation
has roots, as follows:
1. If b
2
– 4ac > 0, the equation has two
distinct (i.e. unequal) roots.
2. If b
2
– 4ac = 0, the equation has a repeated
root (i.e. two roots that are equal).
3. If b
2
– 4ac < 0, the equation has no real
roots since the square root of a negative
number is not real.
Example 1.16
Find the roots of the equation 2x
2
+ x – 12 = 0 ,
giving your answer correct to 2 d.p.
Solution
Comparing 2x
2
+ x –12 = 0 with ax
2
+ bx + c
gives a = 2, b = 1 and c =
–
12
Now, x =
∴ x =
=
=
=
= or
= 2.21 or
–
2.71
7
2
–
b
a
–
–
c
a
––
b
a
–
c
a
–
–
b
2a
–––
–

2
– 4 × (2) × (
–
12)
2 × (2)
––––––––––––––––––––––
–
4ac + b
2
4a
2
–––––––
b
2
– 4ac
4a
2
–––––––
–
1 ± 9.849
4
–––––––––
8.849
4
–––––
–
10.849
4
––––––
–

4
–––––––
–

4
––––––––––
7
Exercise 1.6
1. Determine whether or not each of the
following equations has roots. If it does,
state the nature of the roots.
(a) 2x
2
+ 4x – 3 = 0 (b) 2x
2
+ 3x + 4 = 0
(c) 3x
2
– 6x + 3 = 0 (d) 2x
2
+ 4x + 3 = 0
(e) 3x
2
– 2x – 4 = 0 (f) 4x
2
– 20x + 25 = 0
(g) x
2
– 5x + 2 = 0 (h) 16x
2
+ 24x + 9 = 0
2. Use the quadratic formula to solve the
following equations, giving your answers
correct to 2 d.p. where necessary. Check the
results of parts (a) to (f) by factorisation.
(a) x
2
+ x – 12 = 0 (b) x
2
– 8x + 12 = 0
(c) 3x
2
+ x – 10 = 0 (d) x
2
– 6x + 9 = 0
(e) x
2
– x + 1 = 0 (f) x
2
+ 12x + 36 = 0
(g) x
2
+ 5x + 3 = 0 (h) x
2
– 3x – 4 = 0
(i) 5x
2
+ 5x – 3 = 0 (j) 5x
2
– 10x + 4 = 0
(k) 2 – 3x = (l)
= – 5
Problems leading to quadratic
equations
There are many problems that require us to
form and solve quadratic equations. At times
we may be given roots and be required to form
the quadratic equation. At other times, we may
be given word problems from which we form
and solve quadratic equations. Here are some
examples.
Example 1.17
Find the quadratic equation whose roots are 3
and
–
2.
Solution

The root 3 means that
x = 3 ⇒ x – 3 = 0
The root
–
2 means that
x =
–
2 ⇒ x + 2 = 0
Thus, the two factors of the quadratic equation
are x – 3 and x + 2.
∴ the equation is (x – 3)(x + 2) = 0
⇒ x(x + 2) – 3(x + 2) = 0
⇒ x
2
+ 2x – 3x – 6 = 0
⇒ x
2
– x – 6 = 0
Example 1.18
A rectangular strip of land has a perimeter of
124 m and an area of 385 m
2
. Find its possible
dimensions.
Solution

Then, width = (124 – 2x) m
= (62 – x) m
∴area = x(62 – x) m
= (62x – x
2
) m
⇒ 62x – x
2
= 385
⇒ x
2
– 62x + 385 = 0
a = 1, b =
–
62, c = 385
∴ x =
=
=
=
= = 55 or = 7
1
2x + 1
–––––
2
2x – 1
–––––
–
4
x
––
5
2
–
7
2
–
110
2
–––
1
2
–

2
–––––––––––––––––

2
––––––––––
62 ± 48
2
––––––
× 1 × 385
2
–––––––––––––––––––––
14
2
––
x
x
124 – 2x
2
8
–24
5
–––
4
5
–
4
5
–
1
2
–
1
3
–
3
4
–
1
5
–
1
4
–
1
2
–
If length = 55 m, width = 62 – 55 = 7 m.
If length = 7 m, width = 62 – 7 = 55 m.
In both cases, the strip of land measures
55 m × 7 m.
Check: Perimeter = 2(55 + 7) m = 124 m.
Area = 55 m × 7 m = 385 m
2
.
Example 1.19

years ago, the product of their ages was 324.
Find their present ages.
Solution


Six years ago, their respective ages were (x – 6)
and (5x – 6) years.
∴ (x – 6)(5x – 6) = 324
⇒ 5x
2
– 36x + 36 = 324
⇒ 5x
2
– 36x – 288 = 0
⇒ 5x
2
– 60x + 24x – 288 = 0
⇒ 5x(x – 12) + 24(x – 12) = 0
⇒ (x – 12)(5x + 24) = 0
∴ x = 12 or x = =
–
4
The value
–
4 is unsuitable since a person can
not have a negative age.
∴ 
 
Check: Six years ago, their ages were 6 and 54,
and 6 × 54 = 324.
Exercise 1.7
1. Write down the quadratic equation whose
roots are
(a) 4 and 7 (b)
–
3 and 2
(c) 2 and (d)
–
5 and
–
4
(e)
–
3 and
–
4 (f) 4 and
–
3
2. Find a whole number such that twice the
square of the number subtracted from seven
times the number gives 3.
3. A number is such that three times its square
added to ve times the number gives 12.
Find two possible values of the number.
4. The base of a triangle is 2 cm less than the
altitude, and the area of the triangle is
60 cm
2
. Calculate the altitude.
5. The hypotenuse of a right-angled triangle
is 25 cm long. If the the other two sides
are such that the length of one side is 5 cm
more than the other, nd the length of the
two sides.
6. The parallel sides of a trapezium are such
that one is 4 cm longer than the other. The
perpendicular distance between the sides
is equal to the length of the shorter of the
parallel sides. If the area of the trapezium is
40 cm
2
, nd the length of the longer of the
parallel sides correct to 3 s.f.
7. The formula S = n(n + 1) gives the sum S
of the counting numbers 1, 2, 3, …, n. Find
n if S = 105.
8. The hypotenuse of a right-angled triangle
is 13 cm long, and the perimeter is 30 cm.
Find the length of the other two sides of the
triangle.
Tables of values and graphs of
quadratic equations
As we have already seen, a quadratic expression
in x is one in which the highest power of x is
2. For example, 2x
2
– 3x + 4 is a quadratic
expression.
The following are examples illustrating how to
construct the tables of values from the equation,
and drawing the graph of the quadratic equation.
Example 1.20
Draw the graph of y = x
2
+ 2x – 2 and use it to
 
(b) the value of x when y =
–
1,
(c) the least value of y.
9
–
4
–
3
–
2 1 2
–
1
x
–
1
6
–
2
5
–
3
4
3
2
1
y
0
y = x
2
+ 2x – 2
×
×
×
×
×
×
×
Solution
Table 1.1 is the required table of values. Values
of y are obtained by adding the values of x
2
, 2x
and
–
2. Note that breaking down the expression
into such components makes the working much
easier.
x
–
4
–
3
–
2
–
1 0 1 2
x
2
16 9 4 1 0 1 4
2x
–
8
–
6
–
4
–
2 0 2 4
–
2
–
2
–
2
–
2
–
2
–
2
–
2
–
2
y 6 1
–
2
–
3
–
2 1 6
Table 1.1
Choose a suitable scale and plot the values of y
against the corresponding values of x. Join the
various points with a continuous smooth curve
to obtain a graph like that of Fig. 1.1.
(b) when y =
–
1, x =
–
2.4 or 0.4,
(c) the least value of y is
–
3.
Note:
The graph of a quadratic expression is a curve.
The shape of the curve is called a parabola.
This curve is symmetrical about a line parallel
to the y-axis, passing through its lowest or
highest point.
Example 1.21
Draw the graph of y = 2 + 2x – x
2
for values of
x from
–

(a) the maximum value of 2 + 2x – x
2
,
(b) the value of x for which y is greatest,
(c) the range of values of x for which y is
positive.
Solution
Table 1.2 is the required table of values.
x
–
2
–
1 0 1 2 3 4
2 2 2 2 2 2 2 2
2x
–
4
–
2 0 2 4 6 8
–
x
2
–
4
–
1 0
–
1
–
4
–
9
–
16
y
–
6
–
1 2 3 2
–
1
–
6
Table 1.2
Fig. 1.2 on the following page shows the graph
of y = 2 + 2x – x
2
.
From the graph,
(a) the maximum value of 2 + 2x – x
2
is 3,
(b) the value of x for which y is greatest is 1,
(c) y is positive for all parts of the curve above
the x-axis, i.e. when x >
–
0.8 and x < 2.8.
In short, y is positive over the range
–
0.8 < x < 2.7.
Notice that the parabola in Fig. 1.2 is upside
down compared to Fig. 1.1.
This is always the case when the coecient of
x
2
is negative.
Fig. 1.1
From Fig. 1.1,
(a) when x = 1.5, y = 3,
10
–
2
–
1 0 3 4
1
x
–
4
3
–
5
2
–
6
1
2
–
1
–
2
y
–
3
y = 2 + 2x – x
2
×
×
×
×
×
×
×
Fig. 1.2
Example 1.22
Draw the graph of y = x
2
– 4x + 5 for
–

5. Where does the curve cut the x-axis?
Solution
The required table of values is as in Table 1.3.
x
–
1 0 1 2 3 4 5
x
2
1 0 1 4 9 16 25
–
4x 4 0
–
4
–
8
–
12
–
16
–
20
5 5 5 5 5 5 5 5
y 10 5 2 1 2 5 10
Table 1.3
Fig. 1.3 shows the required graph. The graph
does not cut the x-axis anywhere.
This means the equation x
2
– 4x + 5 = 0 has no
solution
Fig. 1.3
From Examples 1.20 to 1.22, notice that:
The graph of y = ax
2
+ bx + c is ∪-shaped if
a is positive; it is ∩-shaped if a is negative.
Exercise 1.8
1. Use Fig. 1.1 to nd
(a) the value of y when x =
–
1.5,
(b) the value of x when y = 3,
(c) the range of values of x for which y is
negative.
2. Use Fig. 1.3 to nd
(a) the value of y when x =
–
0.5,
(b) the values of x when y = 6.2,
(c) the least value of y,
–
1
0 1 4 5
2
x
2
9
1
8
7
6
5
4
3
10
3
×
×
×
×
×
×
y = x
2
– 4x + 5
×
y
11
(d) the value of x when y is least,
(e) State the line of symmetry of the curve.
3. Table 1.4 is the table of values of y = 3x – x
2
.
x
–
1 0 1 2 3 4
y
–
4 0 2 2 0
–
4
Table 1.4
(a) Use a scale of 1 cm to 1 unit on both
axes to draw the graph of y = 3x – x
2
.
(b) From your graph what is the maximum
value of y?
(c) What is the value of x for which y is
maximum?
(d) Find the values of x when y =
–
3.
4. Table 1.5 is the table of values of
y = x
2
– 3x + 2.
x
–
1 0 1 2 3 4
y 6 2 0 0 2 6
Table 1.5
Using a suitable scale, draw the graph of
y = x
2
– 3x + 2 and use it to answer the
following.
(a) State the range of values of x for which
x
2
– 3x + 2 is negative.
(b) What is the minimum value of
x
2
– 3x + 2?
(c) For what value of x is y least?
(d) State the range of values of x for which
x
2
– 3x + 2 < 3.
5. (a) Copy and complete Table 1.6 for
y = 2x
2
– 3x – 7.
x
–
2
–
1 0 1 2 3 4
2x
2
8 2 0 2 __ __ __
–
3x 6 3 0 __ __
–
9
–
12
–
7
–
7
–
7
–
7
–
7
–
7
–
7
–
7
y 7
–
2
–
7 __ __ __ __
Table 1.6
(b) Use a scale of 1 cm to 1 unit on the
x-axis and of 1 cm to 2 units on the
y-axis to draw the graph of
y = 2x
2
– 3x – 7.
(c) Use your graph to nd
(i) the value of y when x = 3.4,
(ii) the values of x when y =
–
7,
(iii) the minimum value of y.
6. (a) Draw the graph of y = x
2
– 3x for values
x from
–
1 to 4.
(i) What is the main dierence
between this graph and that of
Question 3?
(ii) What is the minimum value of y?
(iii) What transformation maps this
graph onto that of Question 3?
(b) Draw the graph of y = x
2
– 3x – 1 for
values of x from
–
1 to 4.
(i) What is the dierence between this
graph and that of Question 4?
(ii) For what value of x is y minimum
and what is the minimum value
of y?
(iii) What transformation maps this
graph onto that of Question 4?
Graphical solution of quadratic
equations
Consider the equation x
2
– x – 2 = 0.
x
2
– x – 2 is a quadratic expression in x.
Let y = x
2
– x – 2.
Table 1.7 is the table of values for y = x
2
– x – 2.
x
–
2
–
1 0 1 2 3
y 4 0
–
2
–
2 0 4
Table 1.7
Fig. 1.4 is the graph of y = x
2
– x – 2.
To solve the equation x
2
– x – 2 = 0 means that
we nd values of x for which y (i.e. x
2
– x – 2)
is equal to zero.
12
Fig. 1.4
From the graph
The values are
–
1 and 2.
These are the required solutions.
Check: By factorisation
x
2
– x – 2 = 0
(x + 1)(x – 2) = 0
∴x =
–
1 or x = 2.
Thus to obtain the solution of the quadratic
equation ax
2
+ bx + c = 0 graphically:
1. Draw the graph of y = ax
2
+ bx + c, and
2. Read the values of x at the points where
the graph cuts the x-axis.
Example 1.23

equation x
2
+ 2x – 4 = 0
Solution
x
2
+ 2x – 4 does not factorise, so we may use


2
+ 2x – 4 and construct a table of
values (Table 1.8).
x
–
4
–
3
–
2
–
1 0 1 2
y 4
–
1
–
4
–
5
–
4
–
1 4
Table 1.8
Fig 1.5 is the graph of y = x
2
+ 2x – 4
Fig. 1.5
From the graph, y = 0 when x =
–
3.2 and 1.2.
Hence, the approximate roots of the equation
x
2
+ 2x – 4 = 0 are
–
3.2 and 1.2.
Note:
The accuracy of the results depends on
(i) the appropriateness of the scale used. For
example, a scale of 2 cm : 1 unit leads to
greater accuracy than a scale of 1 cm : 1 unit
(ii) the smoothness of the curve.
Example 1.24
 
roots of the equation
4x
2
– 12x + 9 = 0
–
5
y = x
2
+ 2x – 4
–
4
–
3
–
2 1 2
–
1
x
–
3
4
–
4
3
2
1
0
–
1
y
–
2
×
×
×
×
×
×
×
x
y
4
3
2
1
0
–
1
–
3
–
3
–
2
–
1
1 2 3 4
–
2
×
×
×
×
×
×
y = x
2
– x – 2
13
3
2
–
Check: By factorisation:
4x
2
– 12x + 9 = 0
(2x – 3) (2x – 3) = 0
i.e. (2x – 3)
2
= 0
∴x =
= 1.5 (twice).
When the curve just touches the x-axis, the
equation has a repeated root, and the roots are
said to be coincident.
As we have already seen, the curve of a
quadratic expression can be in one of three
positions relative to the x-axis (Fig 1.7).
(b) (c)
(a)
Fig 1.7
In Fig. 1.7(a), the curve cuts the x-axis at two
distinct points. The related equation has two
distinct roots.
In Fig. 1.7(b), the curve touches the x-axis at
only one point. The corresponding equation
has one repeated root.
In Fig. 1.7(c), the curve does not cut, nor
touch, the x-axis. The related equation has
no real roots.
Note that the curves could also be ∩-shaped but
similarly positioned. The interpretation remains
the same.
Exercise 1.9
1. Use Fig. 1.1 to write down the roots of the
equation x
2
+ 2x – 2 = 0. Check your results
using the quadratic formula.
2. Use Fig. 1.2 to state the roots of the
equation 2 + 2x – x
2
= 0. Check your result
using the quadratic formula.
Solution

2
– 12x + 9 and construct a table of
values (Table 1.9).
x
–
1 0 1 2 3 4
4x
2
4 0 4 16 36 64
–
12x 12 0
–
12
–
24
–
36
–
48
9 9 9 9 9 9 9
y 25 9 1 1 9 25
Table 1.9
Fig. 1.6
Fig. 1.6 is the graph of y = 4x
2
– 12x + 9.
The curve does not cut the x-axis.
It only touches it where x = 1.5.
∴ the root of the equation 4x
2
– 12x + 9 = 0 is
1.5 (twice).
–
2
x
y
–
1 0 1 2 3 4
25
20
15
10
5
×
××
×
×
×
14
3. Using Fig. 1.3, state the roots of the
equation x
2
– 4x + 5 = 0. Check your result
using the quadratic formula.
4. Table 1.10 is a table of values for a certain
quadratic expression. Using the table,
draw a graph and state the solutions of the
corresponding quadratic equation.
x
–
7
–
6
–
4
–
2
–
1
y
–
9
–
6
–
4
–
6
–
9
Table 1.10
5. Table 1.11 is a table of values for
y = 5x
2
– 10x + 4. Use the table to solve the
equation 5x
2
– 10x + 4 = 0.
x
–
1 0 1 2 3
y 19 4
–
1 4 19
Table 1.11
6. (a) Copy and complete Table 1.12 for
y = 4x
2
– 20x + 25.
x 0 1 2 3 4 5
4x
2
0 4 16 — — —
–
20x 0
–
20 — — — —
25 25 25 25 25 25 25
y 25 9 — — — —
Table 1.12
(b) Hence, solve the equation
4x
2
– 20x + 25 = 0 graphically.
7. Draw the graph of y = 2x
2
+ x – 2 from
x =
–
3 to x = 2. Hence, nd the approximate
values of the roots of the equation
2x
2
+ x – 2 = 0.
8. Draw the graph of the expression
2 – 3x – 2x
2
from x =
–
3 to x = 3.
Hence, nd the approximate values of the
roots of the equation 2 – 3x – 2x
2
= 0.
Simultaneous equations: One linear,
one quadratic
Suppose we are given a linear equation and
a quadratic one and asked to solve them
simultaneously. We could do this as illustrated
in Example 1.25.
Example 1.25
Solve the simultaneous equations
y = x
2
– x – 2, and
y = 3 – 2x.
Give your answers correct to 2 s.f.
Solution
Substitution method
Substitute y = 3 – 2x into y = x
2
– x – 2 and
solve the resulting quadratic equation in x.
Thus, 3 – 2x = x
2
– x –2
⇒x
2
+ x – 5 = 0
Using the quadratic formula,
a = 1, b = 1, c =
–
5
∴ x =
–

2
– 4 × 1 × (
–
5)
2 × 1
=
–

2
=
–
1 ± 4.5826
2
=
3.5826
or
–
5.5826
2 2
∴ 
–
2.8 (2 s.f.)
×
–
0.6

–
2.8, y = 3 – 2(
–

i.e. the solutions are
x = 1.8, y =
–
0.6 or x =
–
2.8, y = 8.6.
Graphical method
On the same axes, draw the graphs of both
equations.
Table 1.13 gives the values for drawing the
graphs.
(a) y = x
2
– x – 2
x
–
3
–
2
–
1 0 1 2 3
y 10 4 0
–
2
–
2 0 4
15
x
y
10
8
6
4
2
0
–
2
–
4
–
3
–
2
–
1
1
2
–
3
+
B
A
×
×
×
×
×
×
×
×
•
•
+
(b) y = 3 – 2x
x
–
3 0 3
y 9 3
–
3
Table 1.13
Graphical method as follows:
Draw the graphs of y = ax
2
+ bx + c and
y = mx + n on the same axes and read the
coordinates of the points of intersection of
the two graphs.
Note that there are two pairs of solutions as
there are two points of intersection between the
line and the curve.
Example 1.26
Draw the graph of y = 2x
2
+ x – 2 and use it to
solve the equation
(a) 2x
2
+ x – 2 = 5 (b) 2x
2
+ x – 5 = 0
(c) 2x
2
+ 2x – 3 = 0.
Solution
Table 1.14 is the required table of values.
x
–
3
–
2
–
1 0 1 2
y 13 4
–
1
–
2 1 8
Table 1.14
Fig. 1.9 shows the required graph.
Fig. 1.8
Fig. 1.8 shows both graphs drawn on the same
axes.
At A, y = 8.6 when x =
–
2.8.
At B, y =
–
0.6 when x = 1.8.
Thus, the approximate solutions of the given
simultaneous equations are:
x = 1.8, y =
–
0.6
or x =
–
2.8, y = 8.6.
We notice that given two simultaneous
equations, one linear and one quadratic, the
method of solution is similar to that of solving
two linear simultaneous equations. Thus, given
the simultaneous equations y = ax
2
+ bx + c and
y = mx + n, we use either:
Substitution method as follows:
Substitute mx + n for y in y = ax
2
+ bx + c
and solve the resulting quadratic equation in
x, or
Fig. 1.9
–
3
–
2
–
1
0 1 2 3
x
y
14
12
10
8
6
4
2
–
2
– – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – –
(c)
(a)
(b)
y =
–
x + 1
y = 3
y = 5
×
×
×
×
×
y = 2x
2
+ x – 2
y = x
2
– x – 2
y = 3 – 2x
×
16
x
3
––
2
3
––
To solve each equation, we transform it so that
we have two simultaneous equations, one linear
and the other quadratic. The quadratic one in
each case equation of the graph. In this case,
the graph equation is y = 2x
2
+ x – 2.
(a) 2x
2
+ x – 2 = 5 at the points where the line
y = 5 cuts the curve (construction (a) on
Figure 1.9).
At these points, x =
–
2.1 or x = 1.6
(b) 2x
2
+ x – 5 = 0.
Adding 3 to both sides gives
2x
2
+ x – 2 = 3.
2x
2
+ x – 2 = 3 at the points where the line
y = 3 cuts the curve ( construction (b) on
Figure 1.9)
At these points
x =
–
1.9 or x = 1.3 (1 d.p.)
Note: The line y = 3 could also have been
obtained as follows:
y = 2x
2
+ x – 2
–
0 = 2x
2
+ x – 5
y = 3
(c) 2x
2
+ 2x – 3 = 0
We can write this as 2x
2
=
–
2x + 3.
We then add x – 2 to both sides to get
2x
2
+ x – 2 =
–
2x + 3 – x – 2.
⇒ 2x
2
+ x – 2 =
–
x + 1.
 
form.
Now, 2x
2
+ x – 2 =
–
x + 1 at the points
where the line y =
–
x + 1 cuts the curve
[construction (c) on Fig. 1.9].
At these points
x =
–
1.8 or x = 0.8 (1 d.p.).
Note: The line y =
–
x + 1 could also be
obtained as follows:
y = 2x
2
+ x – 2
–
0 = 2x
2
+ 2x – 3
y =
–
x + 1
Exercise 1.10
1. Use Fig. 1.8 to answer the following.
(a) What is the minimum value of x
2
– x – 2?
(b) For what range of values of x is
x
2
– x – 2 < 1?
(c) What are the roots of the equation
x
2
– x – 3 = 0?
(d) What are the solutions of the
simultaneous equations
y = x
2
– x – 3, and y = – ?
2. Use Figure 1.3 to solve
(a) the equation x
2
– 4x + 1 = 0
(b) the simultaneous equations
y = x
2
– 4x + 5 and y = 8 – 2x.
3. Use substitution method to solve the
simultaneous equations
(a) y = 2x
2
– x + 2 and y = 4x + 2
(b) y = 4x
2
– x + 3 and y = 5x + 2.
4. Taking values of x from
–
3 to 3, draw the
graph of y = 3x
2
– x – 7. From your graph
read o the roots of the equation
(a) 3x
2
– x – 7 = 0 (b) 3x
2
= x + 2
(c) 3x
2
– x – 18 = 0 (d) 3x
2
= x + 24.
5. Draw the graph of y = 2x
2
– 3x – 7, taking
values of x from
–
2 to 4. Use your graph
to solve
(a) the equation 2x
2
– 3x – 7 = 0
(b) the equation 2x
2
– 3x – 12 = 0
(c) the simultaneous equations
y = 2x
2
– 3x – 7 and y = 2x – 1.
Application of quadratic equations to
real life situations.
There are many real life situations that require
the use of knowledge of quadratic equations.
Examples 1.18 and 1.19 present such situations.
This section provides yet another example and
more exercises in such applications.
17
20
18
–––
20
18
–––
1
10
––
Example 1.27

using either 16 large square tiles or 25 small
square tiles. If the side of one small tile is 2

length of a side of the large tile.
Solution

Then, the length of the side of the small tile is
(x – 2) cm.
Area of one large tile = x
2
cm
2
Area of one small tile = (x – 2)
2
cm
2
16 large tiles may be used. They have a total
area of 16x
2
.
.
Alternatively, 25 small tiles may be used, and
their total area is 25(x – 2)
2
Thus,
25(x – 2)
2
= 16x
2
⇒ 25(x
2
– 4x + 4) = 16x
2
⇒ 25x
2
– 100x + 100 = 16x
2
⇒ 25x
2
– 16x
2
– 100x +100 = 0
⇒ 9x
2
– 100x + 100 = 0
Using the quadratic formula;
x =
–
(
–

–
100)
2
– 4 × 9 × 100
2 × 9
=

18
=
100 ± 80
18
= 10 cm or cm.
cm is an invalid length as it is too small and
at the same time it would mean that the length
of the side of a small tile is negative. Thus, the
length of the side of a large tile is 10 cm.
Exercise 1.11
1. The width of a page of a book is 6 cm less
than the length. If the area of the page is
432 cm
2
, nd its width.
2. A rectangle is formed by bending a piece of
wire whose length is 56 cm. If the rectangle
has an area of 171 cm
2
, nd its dimensions.
3. A rectanglular lawn measures 12 m by 8 m.
It is surrounded by a owerbed whose area
is 52 m
2
. If the width of the owerbed is
uniform all round the lawn, nd this width
giving your answer to 1 d.p.
4. A square carpet is laid on the oor of a
room so that one of its sides is against a side
of the room. It leaves strips of uncovered
oor 1 m wide along the two opposite sides
and 2 m wide along the remaining side.
If the area of the room is 64 m
2
, nd the
dimensions of the carpet.
5. A circular lawn is surrounded by a pavement
1 m wide. If the area of the pavement is
that of the lawn, nd the radius of the lawn
in metres, correct to 3 s.f.
6. A stone is thrown vertically upwards from
the top of a 60 m tall tower. The height h of
the stone above the top of the tower after t
seconds is given by the formula h = 20t – 5t
2
.
Copy and complete Table 1.15 for this
motion.
t 0 1 2 3 4 5 6
20t 0 20 40 60
__ __ __
–
5t
2
0
–
5
__ __ __ __ __
h 0 15
__ __ __ __ __
Table 1.15
Draw a graph showing this motion and use
it to answer the following questions:
(a) For what values of t is the stone below
the top of the tower?
(b) What is the maximum height, above the
ground, that the stone attains? After
how long does it attain it?
(c) How long after being thrown, is the
stone at a height of 68 m above the
18
ground? Why are there two answers
and what do they mean?
(d) After how long does the stone pass the
top of the tower, coming down?
(e) Solve the equation 14 + 20t – 5t
2
= 0.
(f) After how long does the stone hit the
ground?
(g) Is there a value of t for which
20t – 5t
2
= 23? Explain what your
answer means.
19
2
APPROXIMATIONS AND ERRORS
Consider the product of 560 002 × 820 004.
By long multiplication, the product is
459203880008
= 4.59203880008 × 10
11
.
On an 11 digit calculator, this answer would be
displayed as 4.5920388
11
which means the same
as 4.5920388 × 10
11
.
Examples 2.1 and 2.2 illustrate the use of a
scientic calculator in computation.
Example 2.1
Use a calculator to evaluate the following
(a) 81.4 (b) 163
2
726 + 5 192 17 – 9.11
(c) 174 × 76.1
8.96
Solution
(a) 81.4 = 81.4 × 1
726 + 5 192 726 + 5192
Here we can
 
 
 
numerator.
The procedure is as follows:
Key in 726, press + , key in 5 192, press
= , press x
–1
, press × , key in 81.4, press
= .
The answer you obtain is 0.013754646 (on
the 11-digit calculator).
Alternatively, we can use the following
procedure.
Key in 81.4, press ÷ , press
(
, key in
726 , press + , key in 5 192, press
)
,
press = .
 
√
3
( )
In numerical computations we often work with
approximations of numbers. It is therefore
not surprising that the results obtained in
computations are often subject to error. In this
chapter, we shall nd out how approximation
aects our calculations.
Computing using calculators
In Form 2, we learnt how to use mathematical
tables as an aid to ease numerical computations.
In this section we shall further ease numerical
computation by using a calculator. A scientic
calculator is recommended for use in this book.
It has more operations and functions which
are required in this and other chapters. Some
of the commonly used keys are as shown in
Table 2.1.
AC/ON switch calculator on
or
x
–1
reciprocal of x
+ add
x
y
raise x to power y
– subtract log nd common

× multiply
logarithm of
10
x
nd number whose
÷ divide
logarithm to base 10
is x
= equal to
( ) brackets
x
2
square of x
± change sign
√
x
square root of x OFF switch o power
x
3
cube of x
SHIFT or INV
change
cube root of x
function
sin / sin
–1
nd sine / angle whose sine is
tan / tan
–1
nd tangent / angle whose tangent is
cos / cos
–1
nd cosine / angle whose cosine is
Table 2.1
Most calculators give answers to an accuracy
of eight or more digits. For example on an
8-digit screen, the smallest and largest positive
numbers that can be displayed are 0.00000001
and 99999999 respectively. If the answer to a
calculation lies outside this range it is displayed
in standard form as illustrated below.
3
√
x
1
/
x
20
(b)
163
2
17 – 9.11
As in part (a), there are two possible
alternative procedures.
Either: Key in 17, press – , key in 9.11,
press = , press , press × , key
in 163, press
= , press x
2
.
or Key in 163, press ÷ , press
(
,
key in 17, press – , key in 9.11,
press
)
, press = , press x
2
.
They both give 426.796 847 as the answer.
(c) To evaluate
3
174 × 76.1
,
work as follows:
8.96
Key in 174, press × , key in 76.1, press
÷
,
key in 8.96, press = , press
, press
. You should obtain 11.390 478 41.
As with mathematical tables, we can use a
calculator to perform computations involving
logarithms and antilogarithms. In Form 2, we
learnt that to nd the number whose logarithm
is given means reversing the process of nding
logarithms. We can do the same on a calculator
by using either 10
x
, i.e. INV log , or key in 10,
press x
y
and key in the given value.
Example 2.2

(a) the logarithm of (i) 28.95 (ii) 0.008 615
(b) the number whose logarithm is 0.2678
Solution

the sequence:
Press log , key in 28.95, press = .
Answer: 1.461 648 568.
(ii) Press
log , key in 0.008 615, press = .
Answer:
–
2.064 744 718.
Note that with regard to logarithm of
a number less than 1, the calculator
does not separate the characteristic
from the mantissa as we do when using
logarithm tables. However, should it
( )
√
1
/x
SHIFT
3
√
–
3
3
3
3
5
–
√
( )
3
3
be necessary to do so, you should be
able to convert this logarithm to bar
form.
For example, given log x =
–
2.06, we
convert it as follows:
log x =
–
2 +
–
0.06
=
–
1 +
–
2 +
–
0.06 + 1 (to have
a positive mantissa)
= 3.94

be x.
log
10
x = 0.2678
∴ x = 10
0.2678
Key in 10, press x
y
, key in 0.267 8.
press = .
The answer is 1.852 678 823 7.
Important point:
You may have found out that with some
calculators, the sequence in Example 2.2 does
not work. This happens because dierent
calculators are programmed dierently. It is
important, therefore, that you study the manual
of your calculator so that you know how it
works.
Exercise 2.1
Use a calculator to evaluate the following:
1. (a) 0.68 × 0.422 (b) 0.5276 ÷ 3.842
(c) 0.75 × 38.74 (d) 14.67 × π
16.91 22.35
(e) 0.734
3
(f) 0.062 1
2
(g) √0.846 3 (h) √0.00328
(i) 2π
76.4
(j)
22.67 – 11.43
32.2 22.67 + 11.43
(k)
√0.843 2 –
√0.752 6
√0.843 2 + √0.752 6
(l)
√11 × 14.23
2
(m)
0.671
3
× 0.042
2
39.67
√0.061
(n)
1 + √0.072
(o) 0.9 (p) 2.718
1.6
1 – √0.072
21
2. Given that u = 22.3 and v = 31.4, evaluate
(a)
1
+
1
(b)
1
–
1
u v u v
3. Find u given that
1
+
1
=
1
.
u
17.24 16.41
4. If x
3
= 0.746, nd x.
5. If x = 0.7 + log 0.7, nd the value of x.
6. Evaluate
log 0.8
.
log 0.7
7. Find x given that
(a) log x = log 4 + log 17.
(b) x = log 0.24 + log 2.31.
(c) log x = 4 log 2 + 2 log 3.
Estimation and approximation
Estimation
A friendly match between the Harambee Stars
and the Uganda Cranes at the Nyayo Stadium
was attended by 55 298 people. A television
commentator told his viewers that 55 000
people attended the match. Do you think he
lied to the viewers?
The commentator did not lie. His interest
was to inform the viewers of the number of
spectators correct to the nearest 1 000. He
was stating the number correct to 2 signicant
gures (2 s.f.). The commentator would still
have not lied had he given the number as
60 000 or 55 300.
Thus 55 298 ≈ 60 000 correct to the nearest
10 000 (or 1 s.f.)
≈ 55 000 correct to the nearest
1 000 (or 2 s.f.)
≈ 55 300 correct to the nearest
100 (or 3 s.f. or 4 s.f.)
We use signicant gures when estimating
numbers or amounts and when we are doing
rough calculations or estimates.
We use estimates to help us make decisions.
Estimates are only meant to give rough
guidelines.
In order to estimate the answer to a calculation,
we round o each number to its highest place
value and then compute.
Example 2.3
Find a rough estimate of the sum of
1 497 + 2 565.
Solution
Number Estimate (1 s.f)
1 497
+
1 000
+
2 565 3 000
4 062 4 000
In this case, the sum 4 000 is an estimate of
1 497 + 2 565, i.e. 4 062.
Example 2.4
Find a rough estimate of
8.76 × 0.495
.
17.8
Solution

8.76 × 0.495

9 × 0.5
17.8 20
= 0.225.
Approximation
Consider the following:
(a) Calculate the following exactly:
(i) 7.2 × 1.2 (ii) 0.64 × 0.914
(iii) 0.75 ÷ 1.25
(iv) Area of a square of sides 11.2 cm
(b) Measure the width of your mathematics
text book using a ruler and compare your
answer with those of other members of
your class.
In (a), your answers will agree with those of
other members of your class, i.e.
(i) 8.64 (ii) 0.58496 (iii) 0.6
(iv) 125.44 cm
2
22
In (b), some answers may agree with yours,
while others may be quite dierent. So, whose
answer is exactly correct? None of the answers
are exact, but all may be correct depending on
the degree of accuracy stated.
All measurements, no matter how sophisticated
the measuring instruments, are always
approximated. When we say, for example, that
the length of a line segment AB = 7.4 cm (1 d.p.),
we mean that the length of AB can be anywhere
between 7.35 cm and 7.45 cm.
So, 7.35 ≤ 7.4 < 7.45, which means that the true
length must lie between 7.35 inclusive and 7.45
exclusive.
ACTIVITY 2.1
1. Use your ruler to measure the length of your
mathematics text book correct to (i) 1 d.p.
(ii) the nearest cm (iii) 2 s.f.
Compare your answers with those of other
members of your class. Determine which
of the answers give the best approximation
and why.
2. Do you think it is more accurate to use a pair
of dividers and a ruler to measure lengths
than to use only a ruler? Why? Do you
think there are any limitations to the use of
dividers? Discuss.
Example 2.5
A lorry driver is contracted to transport
20 000 building blocks to a site. The lorry has
a capacity to hold 5 120 blocks. Estimate the
number of lorry loads necessary to transport
the blocks.
Solution

is 5.
So, 5 120 may be estimated as 5 000 (1 s.f.)
Total number of blocks = 20 000
∴ Number of lorry loads is
20 000
= 4.
5000
We see that approximately 4 lorry loads are
required to transport 20 000 blocks.
This is an example of an estimate necessary to
help make a decision and plan.
Exercise 2.2
1. Work out rough estimates of the following.
(a) 275 × 992 (b) 8 806 × 0.845
(c) 67.4 × 0.001 2 (d) 25.04 ÷ 0.45
(e) 375 × 569 ÷ 42.3 (f) 0.006 43 ÷ 5.81
2. Estimate how much it would cost to buy
a copy of this text book for every pupil in
your school given that a copy costs sh 480.
3. You wish to paint two surfaces measuring
6.4 m by 4.6 m and 10.8 m by 6.7 m. Each
tin of paint is expected to cover 12 m
2
and
costs sh 1 250. Estimate the number of tins
of paint you should buy. Roughly, how
much do you expect the paint to cost?
4. Estimate the total area of paper, in square
metres, used to produce this text book.
Discuss.
5. A metal rod is measured as 3 m long (1 s.f.)
and another as 4 m long (1 s.f.). Do you
think it is true to say that the total length of
the rods is 7 m (1 s.f.)? Discuss.
6. I measured the length of my ower garden
as 15 m (2 s.f.). State (a) the minimum length
and (b) the maximum length of the garden.
Counting and measuring errors
In counting we get exact values. However, as
we are already familiar, measurements of length,
distance, mass, volume, etc. are obtained only
to a certain degree of accuracy. No matter how
carefully we measure, we can never know what
the true or exact measurement is though it exists.
The dierence between this true value and the
one obtained by measurement is called absolute
error.
23
1
10
––
Errors can never be completely eliminated
in measurements. Only their sizes can be
reduced by using more accurate instruments.
For example, in measuring time, a stop watch is
more accurate than a wall clock, an electronic
balance measures mass more accurately than an
ordinary balance, and so on.
Rounding o, truncating and
signicant gures
In real life situations, exact answers to most
questions are rare. For example, the accuracy
of a measurement depends on its purpose or the
method employed in obtaining it. Usually, we
indicate the degree of accuracy by means of:
(a) Rounding o to the required (i) number of
decimal places, (ii) number of signicant
gures, (iii) number of truncated gures.
(b) Stating the smallest unit of measurement
used.
In Form 1, we learnt how to round o numbers
to the nearest place value and to a given number
of decimal places or signicant gures. It would
be a good idea for you, at this point, to remind
yourself of how to carry out this process.
If a number is  or truncated to a
desired number of decimal places or signicant
gures, an error is introduced as long as the new
number is dierent from the original one. Such
an error is called a round o or truncation error.
For example, 43.48 is rounded o to 43.5 (1 d.p.).
The round o error is 43.5 – 43.48 = 0.02.
Similarly, 43.48 can be truncated to 1 d.p. as
43.4. In truncation, we simply take the number
of decimal places required, and ignore the digits
thereafter. In this case, the truncation error is
43.48 – 43.4 = 0.08.
If the length of a line segment is stated as 5.2 cm,
we understand that it is measured correct to one
decimal place or two signicant gures or to the
nearest 1 mm.
Thus 0.1 cm or 1 mm is the smallest unit of
measurement.
Exercise 2.3
1. Express each of the following correct to
(i) one decimal place (ii) one signicant
gure.
(a) 7.82 (b) 12.19 (c) 50.701
(d) 38.09 (e) 4.98 (f) 8.056
2. Write each of the following correct to the
number of signicant gures indicated in
brackets.
(a) 3.1416 (2) (b) 0.589 63 (3)
(c) 19.189 8 (4) (d) 0.006 193 (2)
(e) 10.046 (3) (f) 4.078 6 (4)
3. Express 67 348.07 m to the nearest (a) m,
(b) 10 m, (c) 100 m, (d) 1 000 m.
In each case, state the smallest unit of
measurement.
4. For each of the following numbers,
(i) round o to two signicant gures.
(ii) truncate to two signicant gures.
(iii) nd the round o and truncation error
in each case.
(a) 0.237 (b) 4.05
(c) 7.218 (d) 1.086
5. A factory worker making precision
instruments is told that the diameter of a
piston he is making is to be 6.7 cm with
a 0.01 cm tolerance. State (a) the least
possible diameter, (b) the largest possible
diameter.
Absolute error
If the length of a line segment is stated as
3.4 cm, we understand that the true length is a
value less than 3.45 cm and greater than
3.35 cm. It may lie anywhere between 3.35 cm
inclusive and 3.45 cm exclusive.
24
The dierence between 3.4 and the maximum
value 3.45 is 0.05. The dierence between
3.4 and the minimum value 3.35 is 0.05. The
value 0.05 is the maximum possible dierence
and is known as the absolute error. The term
“absolute” means that we are only interested in
the magnitude of the error and not whether the
error is positive or negative (see also page 22).
The measurement 3.4 cm is stated to the nearest
0.1 cm or correct to 1 d.p. Thus the least unit of
measurement is 0.1 cm.
Note: The maximum absolute error is equal to
half the least unit of measurement.
Relative and percentage error
The ratio
absolute error
is called the relative
true value
error. This ratio is more important than the
error itself, since it gives the magnitude of error
in relation to the quantity being measured.
If the true value is not available, we use the
measured value, sometimes called working
value or observed value.
If the relative error is multiplied by 100, the
result is called percentage error.
From above, we note that:
Absolute error = True value – measured
value (or estimate), or
Absolute error = the least unit of
measurement
Relative error =
Absolute error
True value
Percentage error =
Absolute error 
× 100
True value
Consider a volume of 2.24 l
The least unit of measurement is 0.01 l
∴ the absolute error =
0.01
= 0.005 l.
2
This means that the actual volumes lies within
the limits 2.235 l and 2.245 l, where 2.235 l is
called the lower limit of the volume and 2.245 l
the upper limit of the volume.
Relative error =
0.005
(since we do not know
2.24
the true value, we use the working value 2.24 l
).
Percentage error =
0.005
× 100
2.24
= 0.22% (2 d.p.)
Example 2.6
A line segment which is actually 3.5 cm long
is measured as 3.53. Calculate the percentage
error in this measurement.
Solution
True value is 3.5 cm
Measured value is 3.53 cm
∴ Absolute error is 3.53 – 3.5 = 0.03 cm
Relative error is
0.03
3.5
∴ Percentage error =
0.03
× 100
3.5
= 0.008571 × 100
= 0.9% (1 s.f)
Suppose 12.3 cm is measured as 12.35 cm.
Then, the absolute error is 0.05 cm and
relative error =
0.05
12.3
= 0.004 (1 s.f.)
Though the actual error is greater in this case
than in the above example, the relative error is
much smaller. Consequently, the error is of less
signicance in this case than in the above example.
In geometrical constructions, therefore, it is
advisable to draw large diagrams since an error
of, say, 1 mm in a length of, say, 7.6 cm will not
be as serious as the same error in a length of,
say, 3.8 cm.
1
2
–
25
12
13
––
13
12
––
2
3
–
Exercise 2.4
1. Which of the following measures are exact
and which are approximates?
(a) The mass of a sack of maize, in
kilograms.
(b) The number of days in the month of
June.
(c) The distance from Mombasa to
Kisumu, in kilometres.
(d) The number of pupils in a class.
(e) The number of subjects being taken by
each member of a class.
(f) The distance round the football eld,
in metres.
(g) Length and breadth of your
mathematics text book, in centimetres.
2. State the least unit of measurement in each
of the following. Hence state the maximum
absolute error in each case.
(a) 18.32 cm
(b) 6.7 m
(c) 9.217 kg
(d) 5.092 6 km
3. If the following measurements are given
correct to 3 s.f., nd their maximum relative
errors, giving your answers correct to 1 s.f.
(a) 24.23 m (b) 4.827 km
(c) 107.4 l (d) 0.032 78 kg
In question 4 to 8, nd (a) absolute error,
(b) relative error, (c) percentage error.
4. 2.54 is read as 2.53.
5. 22.4 cm is used instead of 22.2 cm.
6. is taken instead of .
7. is taken as 0.6.
8. 0.416 kg is to the nearest 10 g.
9. Find the limits within which the following
measurements lie.
(a) 2.7 cm (b) 62.07 m
(c) 12.01 cm. (c) 3.14 kg
10. State the greatest possible error in each of
the following measurements.
(a) 3.142 m (b) 2.025 kg
(c) 1215 g (d) 7.26 l
Propagation of error
As mentioned earlier, all measurements involve
some error, say due to inaccuracy in measuring,
due to faults in the measuring instruments,
etc. Thus, when we are doing numerical
computations involving measurements, errors
will be carried through the various stages of the
computations, all the way to the nal answer.
This is known as propagation of error.
Let us now consider the propagation of error in
the basic arithmetic operations.
Addition
The following example illustrates propagation
of error during addition.
Example 2.7
Consider the lengths 68.07 m and 13.03
m. Each is given to the nearest cm (2 d.p.).
Between what two values does the true sum lie?
Solution
In each length, there is a maximum possible
error of cm, i.e. 0.005 m. The greatest
possible value of 68.07 is just less than 68.075,
and the least possible value is 68.065. Thus
 
Similarly,
 
Thus, the true sum must lie between
68.065 + 13.025 and 68.075 + 13.035,
i.e. between 81.09 and 81.11.
1
2
–
26
Note:
1. The value 68.065 + 13.025 = 81.09 is
called the lower limit (minimum) of the
sum of the two measurements in Example
2.7, and 68.075 + 13.035 = 81.11 is the
upper limit (maximum). The working
sum is 68.07 + 13.03 = 81.1. Thus:
Maximum sum – working sum
= 81.11 – 81.1 = 0.01
Working sum – minimum sum
= 81.1 – 81.09 = 0.01
∴ True sum = working sum ± 0.01
= 81.1 ± 0.01.
The error in each of the given lengths is
±0.005 and the propagated error, in the
sum, is ±0.01.
2. If expressed to 1 d.p., the maximum and
minimum sums are equal. If we give the
values correct to 4 s.f., the answer will be
81.11 in one case and 81.09 in the other,
which are dierent and, therefore, not
suitable. Though the individual lengths
were given correct to 4 s.f., the sum is best
given to 3 s.f., as 81.1.
Sometimes, the two dierent answers may
work out to a higher number of signicant
gures than with the given measurements.
This is a greater degree of accuracy than is
warranted by the given data, which must
not be allowed. We, therefore, state the
answer to 1 s.f. less than in the given data.
3. If three lengths were given, the propagated
error would have been 3 × 0.005. Likewise
if the number of measurements was n, the
propagated error would have been 0.005n.
Example 2.8
Find the error in the sum of 7.32 m and 5.51 m.
Solution
The least unit of measurement in each case is
0.01 m or 1 cm.
1
2
–
∴ Maximum possible error in each
measurement is 0.005 m or cm.
∴ Upper limit of sum is
7.32 + 0.005 7.325
5.51 + 0.005
+
i.e. 5.515
+
12.83 + 0.01 12.840
 
7.32 – 0.005 7.315
5.51 – 0.005
+
i.e. 5.505
+
12.83 – 0.01 12.820
Observed (i.e. working) sum is
7.32
+
5.51
12.83
The sum lies between 12.820 and 12.840.
This sum may be written as 12.83 ± 0.01.
∴ the error in the sum is ± 0.01 m.
If x and y are measurements such that the
maximum possible error in each is ± e, then
the real value of the sum must be less than
(x + e) + (y + e), i.e. (x + y) + 2e, and greater
than (x – e) + (y – e), i.e. (x + y) – 2e.
Thus the limits of error in the sum are ± 2e.
Maximum error = max. sum – min. sum
2
or
=
(max. sum – working sum) + (working sum – min. sum)
2
Subtraction
Consider the measurements 35.3 g and 14.8 g.
The upper limit for 35.3 g is 35.35 g, and the
lower limit is 35.25 g.
The upper limit for 14.8 g is 14.85 g, and the
lower limit is 14.75 g.
The greatest possible dierence between 35.3 g
and 14.8 g must be,
the upper limit for 35.3 g – the lower limit for 14.8 g,
i.e. maximum dierence = 35.35 – 14.75
= 20.6 g.
27
The minimum dierence must be,
minimum limit for 35.3 g – upper limit for 14.8 g,
i.e. minimum dierence = 35.25 – 14.85
= 20.4 g.
The observed dierence is 35.3 – 14.8
= 20.5 g.
∴ the dierence may be written as 20.5 ± 0.1 g
∴ the error in the dierence is ± 0.1 g.
If x and y are measurements such that the
maximum possible error in each is ± e, then
the real value of the dierence must be less
than (x + e) – (y – e), i.e. (x – y) + 2e and
greater than (x – e) – (y + e), i.e. (x – y) – 2e.
Thus, the limits of error in the dierence
are ± 2e.
Example 2.9

each given to 4 s.f.
Solution
The possible error in each case is 0.005.

= (14.57 – 13.29) + 0.005 × 2 = 1.29.

= (14.57 – 13.29) – 0.005 × 2 = 1.27.


The limits of error of the given measurements
need not always be equal.
Example 2.10

13.32 and 28.5.
Solution
(a) The possible error in 13.32 is 0.005 and in
28.5 it is 0.05.
13.32 ± 0.005
+
The sum is 28.5 ± 0.05
41.82 ± 0.055
∴The sum is 41.82 ± 0.055
This means that the limiting values of the


suggest an accuracy of 5 s.f., while the
most accurate of the given measurements
was given correct to 4 s.f.
∴ the sum may be given as 42 (2 s.f.) since
the upper and the lower limits of the
sum are equal correct to 2 s.f.
 
28.5 + 0.05
–
28.5 – 0.05
–
13.32 – 0.005 13.32 + 0.005
15.18 + 0.055 15.18 – 0.055

–
13.32
15.18
∴
The limiting values are 15.235 and
15.125
Let us now consider the general case:
Suppose x and y are measurements which have
to be added, and suppose the possible errors
are e
1
and e
2
respectively. It is unlikely that
the errors will cancel each other out when we
add. Thus, we must think of the maximum and
minimum possible sums. If x′ and y
′ are the
true values of x and y, then
x′ = x ± e
1
and y
′ = y ± e
2
, and
Maximum possible sum = (x + e
1
) + (y + e
2
)
= (x + y) + (e
1
+ e
2
)
Minimum possible sum = (x – e
1
) + (y – e
2
)
= (x + y) – (e
1
+ e
2
)
Thus,
(x′ – y
′) = (x + y) ± (e
1
+ e
2
)
In each case, the maximum possible error is
e
1
+ e
2
.
28
We note that:
The maximum possible error in the sum of
two measurements is the sum of the two
errors involved.
Now suppose we were to subtract y
′ from x′. Then:
Maximum dierence = (x + e
1
) – (y – e
2
)
= (x – y) + (e
1
–
–
e
2
)
= (x – y) + (e
1
+ e
2
)
Minimum dierence = (x – e
1
) – (y + e
2
)
= (x – y) – e
1
– e
2
= (x – y) – (e
1
+ e
2
)
Thus,
x′ – y
′ = (x – y) ± (e
1
+ e
2
).
Like in the case of addition:
The maximum possible error in the dierence
of two measurements is the sum of the two
errors involved.
Note: In approximations the answers must not
have more signicant gures than the least
number of s.f. in the given data.
Exercise 2.5
1. Find the minimum and maximum possible
(i) sum (ii) dierence of the following
measurements.
(a) 30.14 m and 0.96 m.
(b) 62.3 cm and 45 cm.
(c) 2.61 l and 4.83 l.
(d) 2.089 km and 208.4 km.
2. Find the sums of the following measurements,
giving your answers in the form x ± e.
(a) 25.5 + 31.7 + 12.3
(b) 8.35 + 6.45 + 2.03
(c) 157 + 725 + 536
(d) 13.32 + 27.85 + 173.3
3. Find the dierences in the following
measurements, give your answers in the
form x ± e.
(a) 38.7 – 14.6 (b) 9.324 – 8.558
(c) 5.86 – 2.97 (d) 373 – 89.7
4. A rectangle has sides of length 25 cm and
5.7 cm. Within what limits does its
perimeter lie?
5. My watch is subject to a possible error of 5
minutes and it takes me 12 min ± 3 min to
get to school.
(a) By my watch, when must I leave in
order to reach school by 8.40 a.m.?
(b) What is the earliest time I am likely to
reach school if I set o at 8.00 a.m. by
my watch?
Error in multiplication
The dimensions of a rectangle are given as
65.7 m by 42.3 m to the nearest m.
The upper limits of the dimensions are 65.75 m
and 42.35 m. The lower limits are 65.65 m and
42.25 m.
∴ The area of the rectangle lies between
65.75 × 42.35 m
2
and 65.65 × 42.25 m
2
,
i.e. between 2 785.5125 and 2 773.7125.
The working area is 65.7 × 42.3 = 2 779 m
2
.
This area can be stated as 2 779.11 ± 5.4 m
2
.
∴ area = 2 800 m
2
(see note below).
Given two measurements x and y whose
limits of errors are ±e:
Maximum product
= (x + e)(y + e)
= xy + (x + y)e (ignore e
2
)
Minimum product
= (x – e)(y – e)
= xy – (x + y)e (ignore e
2
).
The limits of error are ±(x + y)e.
1
10
––
29
If the limits of error in x are ±e
1
, and those in y
are ±e
2
, then:
Maximum product = (x + e
1
)(y + e
2
)
= xy + xe
2
+ ye
1
+ e
1
e
2
≈ xy + (xe
2
+ ye
1
) — ignore
e
1
e
2
as it should be too
small under normal
conditions.
Minimum product = (x – e
1
)(y – e
2
)
= xy – xe
2
– ye
1
+ e
1
e
2
≈ xy – (xe
2
+ ye
1
) — ignore
e
1
e
2
.
Hence:
The true product is xy ± (xe
2
+ ye
1
), i.e. the
limits of error in the product are ±(xe
2
+ ye
1
).
Example 2.11
Find the product of 17.3 and 13.8. Find also
the percentage error in the product.
Solution

17.35 × 13.85 = 240.2975 (upper limit), and
17.25 × 13.75 = 237.1875 (lower limit).
Working product is
17.3 × 13.8 = 238.74.

max. product – working product
= 240.2975 – 238.74 = 1.5575 and
working product – min. product
= 238.74 – 237.1875 = 1.5525
∴ maximum error =
1.5575 + 1.5525
2
=
3.11
= 1.55
2
∴ the product can be written as
238.74 ± 1.555
Maximum error = 1.555
Relative error =
1.555
238.74
Percentage error =
1.55 
× 100
238.7
= 0.6513 × 100
Alternatively, you may shorten your working by
using the formula;
True product = xy ± (xe
2
+ ye
1
).
Thus, true product
= 17.3 × 13.8 ± (17.3 × 0.05 + 13.8 × 0.05).

Using the results of Example 2.9, conrm that
the sum of the maximum relative errors of
the two given numbers gives the maximum
relative error of the product correct to 2 s.f.
Error in division
A rectangle has an area of 87.4 cm
2
and a length
of 15.3 cm, correct to 3 s.f. To what degree of
accuracy must the width be measured?
The greatest possible width would be found
by using the maximum possible area and the
minimum possible length.
The limits of area are 87.35 cm
2
and 87.45 cm
2
.
The limits of length are 15.25 cm and 15.35 cm.
∴ Maximum breadth =
max. area
min. length
= 87.45
15.25
= 5.7344 cm (4 s.f.)
∴ Minimum breadth =
min. area
max. length
=
87.35
15.35
= 5.691 cm (4 s.f.)
Working breadth =
87.4
15.3
= 5.712 cm (4 s.f.)
Now, max breadth – working breath
= 5.734 – 5.712
= 0.022 cm and
working breadth – min breadth
= 5.712 – 5.691
= 0.021 cm.
30
∴ limits of error are ± 0.0215 cm.
∴ breadth is 5.712 ± 0.0215 cm.
If the limits of error of x and y are ±e, then
the value of the quotient
x
lies between
y
x – e
(min) and
x + e
(max).
y + e y – e
Suppose the limits of error in x and y are ±e
1
and ±e
2
respectively. Then,
maximum quotient =
x + e
1
,
and
y – e
2
minimum quotient =
x – e
1
.
y + e
2
Maximum error = max. quotient – min. quotient
2
=
x + e
1
– x – e
1
÷ 2
y – e
2
y + e
2
=
(x + e
1
)(y + e
2
) – (x – e
1
)(y – e
2
)
2(y
2
– e
2
2
)
=
(xy + xe
2
+ ye
1
+ e
1
e
2
) – (xy – xe
2
– ye
1
+ e
1
e
2
)
2(y
2
– e
2
2
)
=
2xe
2
+ 2ye
1
2(y
2
– e
2
2
)
=
xe
2
+ ye
1
y
2
– e
2
2
≈
xe
2
+ ye
1
(ignoring e
2
2
as it is very small)
y
2
True quotient =
x
±
xe
2
+ ye
1
y
y
2
Relative error =
xe
2
+ ye
1
÷
x
y
2
y
=
y xe
2
+ ye
1
x y
2
=
e
1
+
e
2
x y
Note: Just as in addition and subtraction, if
the measurements involved in multiplication or
division are correct to n s.f., the result should
be given correct to (n – 1) s.f.
Example 2.12
Evaluate 40.3 ÷ 8.71.
Find (a) the limits of error in the quotient.
(b) the relative error
(c) the percentage error
Solution
(a) Maximum quotient =
40.35
= 4.635 (4 s.f.)
8.705
Minimum quotient =
40.25
= 4.618 (4 s.f.)
8.715
Working quotient =
40.3
= 4.627 (4 s.f.)
8.71
Max. quotient – working quotient
= 4.635 – 4.627 = 0.008
Working quotient – min. quotient
= 4.627 – 4.618 = 0.009

∴
0.008 + 0.009
2
= ± 0.008 5
∴Quotient = 4.627 ± 0.008 5.
(b) Relative error =
0.008 5
= 0.002
4.627
(c) Percentage error =
0.008 5
× 100
4.627
= 0.2%.
Note:
1. Using the formulas:
True quotient =
x
±
xe
2
+ ye
1
y y
2
Relative error =
e
1
+ e
2
x y
shortens your working. Thus;
True quotient
=
40.3
±
40.3 × 0.005 + 8.71 × 0.05
8.71 8.71
2
= 4.627 ± 0.008 4
Relative error =
0.05
+
0.005
40.3 8.71
= 0.002
( )
( )
(
)
31
2. The maximum relative error in
multiplication or division is equal to the
sum of the relative errors of the separate
numbers.
Exercise 2.6
1. Work out the following products in the
form x ± e.
(a) 5.3 × 6.4
(b) 0.35 × 6.4
(c) 3.7 × 0.85
(d) 823 × 14.6
2. Find the values of the following quotients
in the form x ± e.
(a) 8.4 ÷ 4.3 (b) 35 ÷ 3.1
(c) 5.9 ÷ 4.7 (d) 18.6 ÷ 31.5
3. The volume of a cuboid is 49.72 cm
3
and
the base area is 8.43 cm
2
(2 d.p.).
Calculate the absolute error in the height of
the cuboid.
4. Find the limits of error in the volume of a
block 9.7 cm by 8.4 cm by 7.6 cm (2 s.f.).
5. A square has sides of length 3.6 cm (2 s.f.).
Between what two values does the area of
the square lie?
6. The measurements a = 6.3, b = 15.8,
c = 14.2 and d = 0.017 3 have maximum
possible errors of 1%, 2%, 3% and
4% respectively. Find the maximum
possible percentage errors in the following
quantities.
(a) a + b (b) c + d (c) b – c
(d) ad (e)
a
(f) abcd
d
(g)
ab
(h)
ad
(i) cd
3
cd bc
7. The edges of a cuboid are measured to the
nearest millimetre. If the cross-section is
a square of side 9.3 cm and the length of
the cuboid is 28.7 cm, calculate percentage
error in
(a) the area of the cross-section,
(b) the total surface area,
(c) the volume.
8. Amina is using the formula s = gt
2
(where
s is distance in metres, g is acceleration due
to gravity in m/s
2
and t is time in seconds)
to determine the height of a vertical cli.
She drops a stone from the top of the cli
and nds that it takes 3.2 ± 0.2 s to fall to
the base of the cli. Given that g = 9.8 m/s
2
,
calculate the height of the cli.
1
2
–
32
Fig. 3.3
In the unit circle (Fig. 3.3), the sector labelled
A is called the rst quadrant, B is called the
second quadrant, C is the third quadrant and
D is the fourth quadrant.
Angles are measured from OX (the positive
direction of x-axis) in either anti-clockwise
(positive) or clockwise (negative) direction.
Fig. 3.4(a) shows angles AOA
1
, AOA
2
, AOA
3
and AOA
4
in the 1st, 2nd, 3rd and 4th quadrants
respectively, measured in the anticlockwise
(positive) direction.
Fig. 3.4(b) shows the same angles now
measured in the clockwise (negative) direction.
Angles in Fig. 3.4(a) are positive while those in
Fig. 3.4(b) are negative.
Introduction
In Form 2, we used a right-angled triangle to
dene the trigonometric ratios sine, cosine and
tangent for angles between 0° and 90°.
Fig. 3.1
From Fig. 3.1 we have
sin
θ
=
opposite side
=
AC ,
hypotenuse AB
cos
θ
=
adjacent side
=
BC ,
hypotenuse AB
tan
θ
=
opposite side
=
AC .
adjacent side BC
In this chapter, we shall consider trigonometric
ratios for angles greater than 90° and also for
negative angles.
Angles measured in the anti-clockwise direction
from a xed horizontal line [Fig. 3.2(a)] are
taken to be positive, while angles measured in
the clockwise direction [Fig. 3.2(b)] are taken
to be negative.
Fig. 3.2
Thus, if
θ
= 30°, then ∠PQR = 30° [Fig. 3.2
(a)], while ∠STU =
–
30° [Fig. 3.2 (b)].
The unit circle
A circle centre (0 , 0) and radius one unit is
called a unit circle (Fig. 3.3).
3
TRIGONOMETRY
θ
C
B
A
θ
R
Q
P
(a)
T
U
S
(b)
θ
y
x
–
1
0
1
1
–
1
D
C
A
B
33
y
x
–
1
0 1
1
–
1
A
A
4
A
1
A
2
A
3
y
x
–
1
0
1
1
–
1
A
A
4
A
1
A
2
A
3
Fig. 3.6
y
x
–
1
O
1
1
–
1
A
A
4
A
1
A
2
A
3
Trigonometric ratios from the unit
circle
In this section, we are going
to nd trigonometric
ratios using the unit
circle.
∆A
1
ON in the
unit circle in
Fig. 3.5 is a
right-angled
triangle with
OA
1
= 1 unit.
∴ sin
θ
= = A
1
N = y-coordinate
cos
θ
= = ON = x-coordinate
tan
θ
= =
y-coordinate
x-coordinate
In general, in a unit circle as in Fig. 3.5,
sin
θ
= y-coordinate of A
cos
θ
= x-coordinate of A
tan
θ
=
y-coordinate of A
=
sin
θ
x-coordinate of A cos
θ
We can also use the unit circle to nd the
trigonometric ratios of angles greater than 90°.
Examine points A
1
, A
2
, A
3
and A
4
on the unit
circle in Fig. 3.6.
(a)
Fig. 3.4
(a) What are the coordinates of these points?
(b) What are the sines, cosines and tangents of
the angles formed between OA and OA
1
,
OA
2
, OA
3
and OA
4
shown in Fig. 3.6?
(c) Measure the angles.
Examination shows that:
(a) The coordinates are: A
1
(0.86 , 0.5),
A
2
(
–
0.5 , 0.86), A
3
(
–
0.5 ,
–
0.86) and
A
4
(0.7 ,
–
0.7).
(b)
A
1
N
OA
1
––––
ON
OA
1
––––
A
1
N
ON
––––
–
1
A
1
(x
1
, y
1
)
–
1
1
1
y
x
– – – – – – –
θ
A
O
Fig. 3.5
N
34
(b) sin ∠AOA
1
= 0.5, cos ∠AOA
1
= 0.86,
tan ∠AOA
1
= 0.58
sin ∠AOA
2
= 0.86, cos AOA
2
=
–
0.5,
tan ∠AOA
2
=
–
1.72
sin ∠AOA
3
=
–
0.86, cos AOA
3
=
–
0.5,
tan ∠AOA
3
= 1.72
sin ∠AOA
4
=
–
0.7, cos ∠AOA
4
= 0.7,
tan ∠AOA
4
=
–
1.0
(c) ∠AOA
1
= 30°, ∠AOA
2
= 120°, 
∠AOA
3
= 240°, ∠AOA
4
= 315°.
You should have noticed that:
(a) A
1
is in the rst quadrant; all the
trigonometric ratios are positive.
(b) A
2
is in the second quadrant; the sine is
positive, while the cosine and tangent are
negative.
(c) A
3
is in the third quadrant; the tangent
is positive, while the sine and cosine are
negative.
(d) A
4
is in the fourth quadrant; the cosine is
positive, while the sine and tangent are
negative.
In general:
(a) All trigonometric ratios of angles in the
rst quadrant are positive.
(b) Only the sines of angles in the second
quadrant are positive.
(c) Only the tangents of angles in the third
quadrant are positive.
(d) Only the cosines of angles in the fourth
quadrant are positive.
You might nd it easier
to remember this using
Fig. 3.7 and reading in a
clockwise direction from
the rst quadrant.
All students Take Chapati
(neutral statement)
Alternatively, you could read in the
anticlockwise direction starting from the rst
quadrant and recall “All Sinners Take Care”.
Fig. 3.8 shows the signs, in the four quadrants
for the three basic trigonometric ratios.
Fig. 3.8
ACTIVITY 3.1
1. Draw a unit circle. Copy and complete
Table 3.1 using the unit circle. What do
you notice about the values of the sines of
these angles?
θ
20° 160° 200° 340°
sin
θ
Table 3.1
Repeat the procedure for the following sets
of angles.
(a) 30°, 150°, 210°, 330°
(b) 45°, 135°, 225°, 315°
(c) 60°, 120°, 240°, 300°
2. Copy and complete Table 3.2 using the unit
circle you drew in 1 above. What do you
notice about the values of cosines of these
angles?
θ
20° 160° 200° 340°
cos
θ
Table 3.2
Repeat the procedure for the following sets
of angles.
(a) 30°, 150°, 210°, 330°.
(b) 45°, 135°, 225°, 315°.
(c) 60°, 120°, 240°, 300°.
3. Copy and complete Table 3.3 using the
values obtained in Tables 3.1 and 3.2.
+
––
+
–
+
–
+
–
–
+
+
A
C
T
S
Fig. 3.7
Sine Cosine Tangent
35
What do you notice about the values of the
tangents of these angles?
θ
20° 160° 200° 340°
tan
θ
=
sin
θ
cos
θ
Table 3.3
Repeat the procedure for the following
sets of angles
(a) 30°, 150°, 210°, 330°
(b) 45°, 135°, 225°, 315°
(c) 60°, 120°, 240°, 300°
In part 1 of Activity 3.1, you are expected to
have noticed the following:
2nd quadrant:
sin 160° = sin (180° – 160°) = sin 20°.
3rd quadrant:
sin 200° =
–
sin (200° – 180°) =
–
sin 20°.
4th quadrant:
sin 340° =
–
sin (360° – 340°) =
–
sin 20°.
You are expected to have made similar
observations for (a), (b) and (c).
In part (2) you are expected to have noticed the
following:
2nd quadrant:
cos 160° =
–
cos (180° – 160°) =
–
cos 20°.
3rd quadrant:
cos 200° =
–
cos (200° – 180°) =
–
cos 20°.
4th quadrant:
cos 340° = cos (360° – 340°) = cos 20°.
In part (3), you are expected to have noticed the
following:
2nd quadrant:
tan 160° =
–
tan (180° – 160°) =
–
tan 20°.
3rd quadrant:
tan 200° = tan (200° – 180°) = tan 20°.
4th quadrant:
tan 340° =
–
tan (360° – 340°) =
–
tan 20°.
In general, for any angle
θ
such that
90° <
θ
< 360°,
1. For 90° <
θ
< 180°
sin
θ
= sin (180° –
θ
),
cos
θ
=
–
cos (180° –
θ
) and
tan
θ
=
–
tan (180° –
θ
) for 90° <
θ
< 180°.
2. For 180° <
θ
< 270°
sin
θ
=
–
sin (
θ
– 180°),
cos
θ
=
–
cos (
θ
– 180°) and
tan
θ
= tan (
θ
– 180°) for 180° <
θ
< 270°.
3. For 270° <
θ
< 360°
sin
θ
=
–
sin (360° –
θ
),
cos
θ
= cos (360° –
θ
) and
tan
θ
=
–
tan (360° –
θ
) for 270° <
θ
< 360°.
Exercise 3.1
1. Draw a unit circle and use it to nd the
(a) sines, (b) cosines (c) tangents
of the following angles.
(i) 25° (ii) 55° (iii) 105°
(iv) 165° (v) 220° (vi) 285°
(vii) 310° (viii) 350°
2. Express the following sines in terms of
sines of acute angles.
(a) sin 115° (b) sin 175°
(c) sin 205° (d) sin 290°
(e) sin 305° (f) sin 349°
3. Express the following cosines in terms of
cosines of acute angles.
(a) cos 100° (b) cos 163°
(c) cos 199° (d) cos 283°
(e) cos 309° (f) cos 337°
4. Express the following tangents in terms of
tangents of acute angles.
(a) tan 100° (b) tan 170°
(c) tan 225° (d) tan 250°
(e) tan 300° (f) tan 330°
36
Fig. 3.9
ACTIVITY 3.2
Draw a similar unit circle. Copy and
complete Table 3.4 using the unit circle.

θ
30°
–
30° 45°
–
45° 60°
–
60°
sin
θ
 0.5
–
0.5
cos
θ
tan
θ
Table 3.4
What do you notice about the values of the
trigonometric ratios of 30° and
–
30°, 45° and
–
45°, 60° and
–
60°?
You should have noticed that
cos (30°) = cos
–
30°, tan (
–
30°) =
–
tan 30°.
y
x
–
1
0 1
1
–
1
30°
O
A
1
A
In general, for any angle
θ
:
sin (
–
θ
) =
–
sin
θ
, cos (
–
θ
) = cos
θ
,
tan (
–
θ
) =
–
tan
θ
Note: It is not a must that you memorize these
formulae. They can easily be derived from a
sketch of the unit circle whenever one needs them.
Example 3.1

trigonometric ratios for
θ
=
–
250°.
Solution

∠AOA
1
= 250° measured in the clockwise
direction. (Fig. 3.10)
Fig. 3.10
Then, by reading the coordinates,
sin (
–
250° 
cos (
–
250°
–
0.35, and
tan (
–
250°) =
0.94

–
2.69
–
0.35
Angles greater than 360° or less
than
–
360°
Trigonometric ratios of angles greater than 360°
or less than
–
360° can also be found using the
unit circle.
Trigonometric ratios of negative angles
A unit circle can also be used to obtain
trigonometric ratios of negative angles.
Angle AOA
1
, in the unit circle (Fig. 3.9), is
measured in a clockwise direction from OA.
∠AOA
1
=
–
30°
Sin
–
30 =
–
0.5
But sin ∠30 = 0.5
∴ sin
–
30 =
–
sin 30°.
y
x
–
1
0
1
1
–
1
–
250°
O
A
1
A
37
(a) (b) (c)
Fig. 3.11
750°
y
x
–
1
0 1
1
–
1
y
x
–
1
0 1
1
–
1
390°
y
x
–
1
0 1
1
–
1
30°
Consider Fig. 3.11. Read and state the values of
the trigonometric ratios of each of the marked
angles.
All the angles are in the rst quadrant, i.e. the
nal position is the same. Thus, beyond 360°,
we are just repeating angles. For example,
390° ≡390° – 360° = 30° in terms of position.
Therefore, 390° just describes the rst 30° of a
second revolution.
From this, we see that the respective
trigonometric ratios of these angles are the
same, i.e.
(i) sin 30° = sin 390° = sin 750°.
(ii) cos 30° = cos 390° = cos 750°.
(iii) tan 30° = tan 390° = tan 750°.
Note that sin 390° = sin (390° – 360°) and
sin 750° = sin (750° – 360° × 2) : Similarly for
cosine and tangent.
In general, if
θ
> 360° then
sin
θ
= sin (
θ
– 360°n),
cos
θ
= cos (
θ
– 360°n),
tan
θ
= tan (
θ
– 360°n) where n is the
number of complete revolutions.
Similarly, if
θ
<
–
360°, then
sin
θ
= sin (
θ
+ 360°n),
cos
θ
= cos (
θ
+ 360°n),
tan
θ
= tan (
θ
+ 360°n) where n is the
number of complete revolutions.
Exercise 3.2
1. Use the unit circle to nd the basic
trigonometric ratios of
(a)
–
25° (b)
–
65° (c)
–
130° (d)
–
160°
(e)
–
220° (f)
–
265° (g)
–
310° (h)
–
345°
2. Use the unit circle to nd the basic
trigonometric ratios of
(a) 370° (b) 420° (c) 480° (d) 510°
(e) 630° (f) 660° (g) 720° (h) 850°
Use of trigonometric tables
Trigonometric ratios of any angle can be
obtained using a unit circle as shown in the
preceding sections. Remember that values of
trigonometric ratios of acute angles are obtained
from Mathematical Tables, as we saw in Form 2.
We shall now see how to obtain trigonometric
ratios of angles greater than 90° using tables.
Example 3.2
Find the trigonometric ratios of
(a) 105° (b) 223° (c) 304°.
Solution
Fig. 3.12 is a sketch of a unit circle showing the
position of each of the angles.
38
(a) 105° is in the 2nd quadrant.
∴ sin 105° = sin (180° – 105°) = sin 75°
= 0.965 9 (from tables).
cos 105° =
–
cos (180° – 105°) =
–
cos 75°
=
–
0.258 8 (reading cos 75°
from tables and making it
negative).
tan 105° =
–
tan (180° – 105°) =
–
tan75°
=
–
3.732 1.
(b) sin 223° =
–
sin (223° – 180°) =
–
sin 43°
=
–
0.682 0.
cos 223° =
–
cos (223° – 180°) =
–
cos 43°
=
–
0.731 4.
tan 223° = tan (223° – 180°) = tan 43°
= 0.932 5.
(c) sin 304° =
–
sin (360° – 304°) =
–
sin 56°
=
–
0.829 0.
cos 304° = cos (360° – 304°) = cos 56°
=
–
0.559 2.
tan 304° =
–
tan (360° – 304°) =
–
tan 56°
=
–
1.482 6.
Example 3.3
Find angles between 0° and 360° whose
(a) sin is
–
0.573 6 (b) cosine is 0.309 0.
(c) tangent is 1.191 8.
Solution

θ
.
So, sin
θ
=
–
0.573 6.
Since sin
θ
is negative,
θ
must be in the 3rd
or 4th quadrant.
Now, from tables, sin 35° = 0.573 6.
So, the acute angle involved must be 35°.
Hence, the required angles are
180° + 35° = 215° and 360° – 35° = 325°.

θ
.
So, cos
θ
= 0.309 0.
∴the acute angle is 72°.
Since cosine is positive,
θ
lies in the 1st or
4th quadrant, so the required angles are
72° and (360° – 72°) = 288°.
 
θ
.
So, tan
θ
= 1.191 8 ⇒ 
θ
= 50°.
Tangent is positive in the 1st and 3rd
quadrants.
 ∴ required angles are 50° and
(180° + 50°) = 230°.
Example 3.4
Find the trigonometric ratios of
(a) 520° (b)
–
850°.
Solution
(a) sin 520° = sin (520° – 360°) = sin 160°
= sin 20° = 0.342 0.
(a) (b) (c)
Fig. 3.12
y
x
–
1
0 1
1
–
1
304°
y
x
–
1
0 1
1
–
1
223°
y
x
–
1
0 1
1
–
1
105°
39
cos 520° = cos (520° – 360°) = cos 160°
=
–
cos 20° =
–
0.939 7.
tan 520°
= tan (520° – 360°) = tan 160°
=
–
tan 20° =
–
0.364 0.
(b) sin (
–
850°) = sin (
–
850° + 720°)
= sin (
–
130°) =
–
sin 50°
=
–
0.766 0.
cos (
–
850°) = cos (
–
850° + 720°)
= cos (
–
130°) =
–
cos 50°
=
–
0.642 8.
tan (
–
850°) = tan (
–
850° + 720°)
= tan (
–
130°) = tan 50°
= 1.191 8.
Use of calculators
For any angle, we can nd any trigonometric
ratio by carrying out the following steps:
Press trigonometric ratio button,
Key in angle size
Press = .
For example, we get tan (–125°) as follows:
Press tan , key in 125 , press
+
/
–
,
press = ; or for some calculators
Key in 125 , press
+
/
–
, press tan , or
Press tan , key in 125 , press = .
This gives tan (
–
125°) as 1.428 148 006 7 (on
an 11-digit calculator).
Exercise 3.3
1. Find the values of the trigonometric ratios
for 25°. Hence state the value of
(a) sin 155° (b) cos 205°
(c) tan 335°.
2. Find the values of the trigonometric ratios
for 115°. Hence state the value of
(a) sin 295° (b) cos 245°
(c) tan 65°.
3. Find the values of the trigonometric ratios
for 78°. Hence state the values of
(a) sin 258° (b) cos 102° (c) tan 282°.
1
r
–
4. Find the values of the trigonometric ratios
for
(a) 0° (b) 45° (c) 90°
(d) 270° (e) 360°.
5. Find the values of the trigonometric ratios
for
(a)
–
27° (b)
–
127° (c)
–
227°
(d)
–
327°
6. Find the values of the trigonometric ratios
for
(a) 628° (b) 1 094° (c) 4 509°
(d) 5 648°
Radian measure
In this section, a new unit for measuring angles
is introduced.
Let the radius of a circle,
centre O, be r and let
the length of the arc
AB be r (Fig. 3.13).
The magnitude of 
∠AOB is said to be
1 radian.
Fig. 3.13
Thus, one radian is defined as the angle
subtended at the centre of a circle by an arc
equal in length to the radius of the circle.
Denotation: 1 radian = 1 rad = 1
c
or 
θ
radians = 
θ
rad = 
θ

c
We know that the circumference of a circle is 2πr.
Since an arc of length r subtends an angle of 1
c
at the centre of a circle, then the circumference
2πr subtends an angle of ( × 2πr)

= 2π
c
.
But we know that the angle at a point is 360°.
∴2π
c
= 360°.
It follows that π
c
= 180°, = 90°, = 60°,
= 45°, etc.
π
c
4
––
π
c
3
––
π
c
2
––
r
r
r
O
B
A
40
Note that 2π
c
= 360° ⇔ 1
c
=
°
=
°
Taking π ≈ 3.142, then 1
c
=
°
≈ 57.30°
(4 s.f.).
Example 3.5
Covert
(a) 129° into radians, (b) 1.45
c
into degrees,
(c) 3 π
c
into degrees.
Solution
(a) 360° = 2π
c
1° =
∴129° = ×2π
c
= 2.252
c
.
(b) 2π
c
= 360°
∴ 1.45
c
= × 360
°
= 83.07°.
(c) 2π
c
= 360°
∴3 π
c
= × 360
°
= 570°.
Exercise 3.4
1. Convert the following angles into radians.
State your answers in terms of π in the
simplest form.
(a) 180° (b) 720° (c) 450° (d) 135°
(e) 225° (f) 30° (g) 15° (h) 20°
(i) 50° (j) 150° (k) 345° (l) 650°
2. Write each of the following angles in degrees.
(a) (b) π
c
(c) π
c
(d) π
c
(e) (f)
(g) π
c
(h) 14π
c
(i) π
c
(j) 5π
c
(k) 7 π
c
(l) 11 π
c
3. Convert the following into degrees.
(a) 4
c
(b) 3.5
c
(c) 1.2
c
(d) 0.8
c
(e) 0.25
c
(f) 0.48
c
(g) 0.65
c
(h) 2.3
c
(i) 0.05
c
(j) 10.5
c
(k) 6.8
c
(l) 3.9
c
( )
5
4
––
2π
c
360°
––––
( )
1.45
2π
––––
( )
129
360
–––
π
c
3
––
π
c
12
––
3
4
––
5
3
––
7
8
––
π
c
10
––
7
6
––
7
3
––
1
2
–
1
6
–
y
x
–
1
0
1
1
–
1
θ
1 unit
A
1
(x
1
, y
1
)
A
O
4. Convert the following into radians.
(a) 35° (b) 44.5° (c) 145°
(d) 258° (e) 369° (f) 400°
5. Evaluate
(a) sin π
c
(b) sin (2.56
c
)
(c) cos (d) cos (0.48
c
)
(e) tan π
c
(f) tan (0.3
c
)
(g) tan (1.78
c
) (h) cos (3.62
c
)
Trigonometric graphs
Graph of sin
θ
Taking a point A on the unit circle (Fig. 3.14).
We see that
(i) sin
θ
= y-coordinate,
(ii) cos
θ
= x-coordinate and
(iii) tan
θ
=
y-coordinate
=
sin
θ
.
x-coordinate
cos
θ
Fig. 3.14
Table 3.5 shows values of
θ
and the
corresponding values of sin
θ
. Note that
the values are approximated to 2 d.p. for the
purpose of graphing. When you read them from
tables, you are required to record them as they
are and approximate only when graphing.
180
π
–––
( )
360
2π
–––
( )
–––
180
π
(
)
1
6
–
1
6
–
1
6
–
3 π
c
2π
c
––––
( )
( )
π
c
9
––
( )
2
3
––
41

θ
° 0 20 40 60 80 100 120
sin
θ
0 0.34 0.64 0.87 0.98 0.98 0.87
140 160 180 200 220 240 260
0.64 0.34 0
–
0.34
–
0.64
–
0.87
–
0.98
280 300 320 340 360
–
0.98
–
0.87
–
0.64
–
0.34 0
Table 3.5
Example 3.6
 °
(b) solve for
θ
in the equation 4 sin
θ
= 3.
Solution
(a) From Fig. 3.15, sin 290° =
–
0.9.
(b) 4 sin
θ
= 3 ⇒ sin
θ
= = 0.75
⇒
θ
= 50° or 130°.
Example 3.7
Using mathematical tables or a calculator,
solve the following equations for values of
θ
between 0° and 180°.
(a) 4 sin
θ
= 3 (b) 4 sin 3
θ
= 3
Solution
(a) 4 sin
θ
= 3
sin
θ
=
∴
θ
= 48.59° or 180° – 48.59° = 131.4°
(b) 4 sin 3
θ
= 3
⇒ sin (3
θ
) =
3
4
–
Plotting values of sin
θ
against
θ
and joining
the points with a smooth curve, we obtain the
graph in Fig. 3.15.
This graph is called the graph of y = sin
θ
for
0° ≤
θ
≤ 360°.
This graph could be extended to include
negative values of
θ
.
Fig. 3.16 shows the graph of y = sin
θ
for values
of
θ
such that
–
2π
c
≤
θ
≤ 2π
c
, which is the
equivalent of
–
360° ≤
θ
≤ 360°.
The graph of y = sin
θ
is also called the sine
curve. Since the curve is in the form of a wave,
it is also referred to as a sine wave.
–
2π
π
––
–
3π
2
–
π
2
–
–
π
2π
––
3π
2
π
2
–
π
×
+
+
×
× ×
+
+
×
×
+
+
×
×
××
×
+
+
– – – – – – – – –
3
4
–
3
4
–
Fig. 3.16
sin
θ
θ
0
–
1
1
Fig. 3.15
sin
θ
40 80 120 160 200 240 280 320
360
180
1
–
1
0
θ
42
3π
4
––
π
5
––
π
8
––
5π
3
––
3
2
–
Fig. 3.17
 ∴
3
θ
= 48.59°(∠a in Fig. 3.17)


∴
θ
= 16.20°
 or

3
θ
= 131.41°(∠b in Fig. 3.17)


∴
θ
= 16.20°
 or

3
θ
= 408.59°(∠c in Fig. 3.17)


∴
θ
= 136.2°
 or

3
θ
= 360°+ 131.41° (2nd quadrant)
= 491.41°


∴
θ
= 163.8°
Exercise 3.5
1. Use the graph of y = sin
θ
to solve the
following for 0° ≤
θ
≤ 360°.
(a) sin
θ
= 0.35 (b) sin
θ
= 0.58
(c) sin
θ
= 0.72 (d) sin
θ
= 1.0
(e) sin
θ
=
–
0.2 (f) sin
θ
=
–
0.6
(g) sin
θ
=
–
0.85 (h) sin
θ
=
–
1.0
2. Draw the graph of y = sin
θ
for
–
180° ≤
θ
≤ 180°. Use the graph to solve
the following.
(a) sin
θ
= 0 (b) sin
θ
= 0.5
(c) sin
θ
= 0.8 (d) sin
θ
=
–
0.35
(e) sin
θ
=
–
1.02 (f) sin
θ
=
–
0.1
3. Use the graph of y = sin
θ
for 0° ≤
θ
≤ 360°
to solve the following.
(a) 6 sin
θ
= 5 (b) 2 sin
θ
+ 1 =
0
y
x
a
b
c
(c) 25 sin
θ
– 4 = 0 (d) 8 sin
θ
+ 7 = 0
4. Use the graph of y = sin
θ
for 0° ≤
θ
≤ 360°
to nd the following.
(a) sin 25° (b) sin 78° (c) sin 118°
(d) sin 235° (e) sin 295° (f) sin 350°
(g) sin (
–
48°) (h) sin (
–
140°)
5. Using mathematical tables or a calculator,
nd values of
θ
between 0° and 360° for
each of the following.
(a) sin
θ
= 0.643 0 (b) sin
θ
= 0.777 5
(c) sin
θ
= 0.292 4 (d) sin (
θ
) =
–
0.425
(e) 3 sin 4
θ
= 0.845 (f) 2 sin
θ
=
–
0.174 5
6. Express the following sines in terms of
sines of acute angles. For example,
sin 340° =
–
sin 20°
(a) sin 130° (b) sin 215° (c) sin 265°
(d) sin 295° (e) sin 305° (f) sin 355°
7. Using Fig. 3.16, state the
(a) angle whose sines is (i) 0.25,
(ii) 0.75, (iii)
–
0.3, (iv)
–
0.8.
(b) sines of the following angles
(i)
–
(ii)
–
(iii) (iv) .
Graph of cos
θ
ACTIVITY 3.3
Make a table of values of
θ
and cos
θ
similar
to Table 3.5. Plot values of cos
θ
against
θ
.
The graph obtained is called the cosine graph
or cosine curve. Compare your graph with that
on Fig. 3.15. What do you notice?
Draw another cosine graph for
–
2π
c
≤
θ
≤ 2π
c
.
Compare it with Fig. 3.16. What do you notice?
You should have obtained graphs of y = cos
θ
(Fig. 3.18(a) and (b)) similar to the ones shown.
43
Fig. 3.18
You should have noticed that:
1. All values of sin
θ
and cos
θ
lie between
–
1
and
+
1.
2. The sine curve and the cosine curve have
the same shape. The cosine curve be
obtained by translating the sine curve
through (=90°) to the right.
3. The curves have peaks (high points) and
troughs (low points).
Example 3.8
Use your graph of y = cos
θ
to
°,
(b) solve 5 cos
θ
= 2.
Solution
(a) From the graph, cos 125° =
–
0.57.
(b) 5 cos
θ
= 2 ⇒ cos
θ
= = 0.4
⇒
θ
= 66° or 294°.
π
c
2
––
2
5
–
–
1
θ
1
cos
θ
0
–
2π
––
–
3π
2
–
π
2
–
–
π
2π
––
3π
2
π
2
–
π
(b)
Exercise 3.6
1. Use the graph of y = cos
θ
for 0° ≤
θ
≤ 360°
to solve the following.
(a) cos
θ
= 0.25 (b) cos
θ
= 0.84
(c) cos
θ
= 0.75 (d) cos
θ
=
–
0.20
(e) cos
θ
=
–
0.62 (f) cos
θ
=
–
0.9
2. Draw the graph of y = cos
θ
for
–
180° ≤
θ
≤ 180° and use it to solve the
following.
(a) cos
θ
=
–
0.35 (b) cos
θ
= 0.5
(c) cos
θ
= 0.78 (d) cos
θ
=
–
0.50
(e) cos
θ
=
–
0.87 (f) cos
θ
=
–
1.0
3. Use the graph of y = cos
θ
for 0° ≤
θ
≤ 360°
to nd the following.
(a) cos 15° (b) cos 53° (c) cos 98°
(d) cos 155° (e) cos 245° (f) cos 290°
(g) cos 330° (h) cos 300°
4. Use the graph of y = cos
θ
for 0° ≤
θ
≤
360° to solve the following.
(a) 3 cos
θ
=
–
1 (b) cos (3
θ
) = 0.5
(c) 13 cos ( 
θ
) – 3 = 0
(d) cos
θ
– 0.14 = 0
5. Express the following cosines in terms of
cosines of acute angles.
(a) cos 120° (b) cos 170° (c) cos 225°
(d) cos 285° (e) cos 300° (f) cos 330°
(g) cos 350° (h) cos 276°
6. Find values of
θ
between 0° and 360° for
parts (a) to (f) and between 0° and 180° for
part (g).
(a) cos
θ
= 0.415 0 (b) cos
θ
= 0.555 5
(c) cos
θ
= 0.25 (d) 4 cos
θ
= 0.8
(e) 2 cos (2
θ
) = 0.60 (f) cos ( 
θ
) = 0 . 7 5
(g) 90 cos (3
θ
) =
–
55.
1
2
–
1
4
–
1
2
–
θ
(a)
–
1
90° 270° 360°
1
0
cos
θ
180°
44
Graph of tan
θ
ACTIVITY 3.4
Table 3.6 shows values of
θ
and the
corresponding values of tan
θ
(approximated
to 2 d.p. for the purpose of graphing

θ
° 0 20 40 60 80 100 120
tan
θ
° 0 0.36 0.84 1.73 5.67
–
5.67
–
1.73
 140 160 180 200 220 240 260

–
0.84
–
0.36 0 0.36 0.84 1.73 5.67
 280 300 320 340 360

–
5.67
–
1.73
–
0.84
–
0.36 0
Table 3.6
Plot the values of tan
θ
against
θ
.
Join the points in the interval 0°≤
θ
< 90°,
90° <
θ
< 270° and 270° <
θ
≤ 360° using
smooth curves.
1. Compare the shape of the graph of y = tan
θ
to that of y = sin
θ
and y = cos
θ
.
2. What do you notice about the behaviour
of the curves near the lines
θ
= 90° and 
θ
= 270°.
You should have obtained a graph similar to the
one in Fig. 3.19.
Fig. 3.19
Note that:
1. The graph of y = tan
θ
is very dierent
from that of y = sin
θ
or that of y = cos
θ
.
In these graphs, every value of
θ
has a
corresponding value of sin
θ
and cos
θ
.
sin
θ
cos
θ
––––
sin
θ
0
––––
90 180 270 3600
θ
y
2. As
θ
gets closer to 90° from the left, the
value of tan
θ
becomes larger and larger and
positive. Also, as
θ
gets closer to 90° from
the right, the value of tan
θ
becomes larger
and larger but negative. Thus, the portions
of the curve do not meet, yet they get closer
and closer to the line
θ
= 90°. Such line as
θ
= 90° is called an asymptote.
3. For all angles,
θ
, whose cosine is zero,
such as
θ
= 90°,
θ
= 270°, etc.,
tan
θ
= = , which is undened.
Thus, all the lines
θ
= 90°,
θ
= 270°, etc. are
asymptotes.
Example 3.9
(a) Use the graph you drew in Activity 3.4 to
°
(ii)
θ
, given that 3 tan
θ
– 6 = 0.
(b) Using mathematical tables or a calculator,
solve the equation
2 + tan 2
θ
= 5, for 0°
θ
°.
Solution
(a) (i) By graph, tan 125° =
–
1.45
(ii) 3 tan
θ
– 6 = 0
⇒ tan
θ
= = 2
By graph,
θ
= 63° or 243°.
(b) 2 + tan 2
θ
= 5.
tan 2
θ
= 3
∴ 2
θ
= 71.57°
⇒
θ
= 35.78°
or 2
θ
= 71.57° + 180° = 251.57°
⇒
θ
= 125.8°
or 2
θ
= 251.57° + 180° = 431.57°
⇒
θ
= 215.8°
or 2
θ
= 431.57° + 180° = 611.57°
⇒
θ
= 305.8°
6
3
––
45
θ
2
–
Exercise 3.7
1. Use the graph of y = tan
θ
, for 0° ≤
θ
≤
360°, to nd the following.
(a) tan 15° (b) tan 40° (c) tan 60°
(d) tan 80° (e) tan 100° (f) tan 160°
(g) tan 225° (h) tan 290° (i) tan 305°
(j) tan 350°
2. Use the graph of y = tan
θ
, for
–
90° <
θ
< 90°
 to solve the following.
(a) tan
θ
= 0 (b) tan
θ
= 0.70
(c) tan
θ
= 2.75 (d) tan
θ
=
–
0.58
(e) tan
θ
=
–
2 (f) tan
θ
=
–
3
3. Use the graph of y = tan
θ
, for 0°≤
θ
≤ 360°
to solve the following.
(a) tan
θ
= 0.5 (b) tan
θ
= 1.5
(c) tan
θ
=
–
1 (d) tan
θ
= 4.5
(e) tan
θ
=
–
2.1 (f) tan
θ
=
–
4
4. Find the value of
θ
between 0° and 360° for
the given equations.
(a) tan
θ
= 1 (b) tan
θ
= 2.75
(c) tan 2
θ
= 0.455 (d) 5 tan
θ
+ 4 = 0
(e) tan
θ
– 2 = 0 (f) tan (
) =
–
8
The sine rule
In Fig. 3.20, ∆ABC has sides of lengths a, b, c.
AD and CE are the altitudes from A to BC and
C to AB respectively.
Using ∆ADC: = sin C ⇒ AD = AC sin C
i.e. AD = b sin C.
1
3
–
AD
AC
–––
– – – – – – – – – – – – – – –
– – – – – – – – – – – – – –
C
A
E
B
D
a
b
c
Fig. 3.20
1
2
–
1
2
–
CE
BC
–––
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
Area of ∆ABC = BC
.
AD
= ab sin C ……………(i)
Using ∆CBE: = sin B ⇒ CE = BC sin B
i.e. CE = a sin B
∴ Area of ∆ABC = AB
.
CE
= ac sin B ……………(ii)
It can similarly be shown that area of ∆ABC
= bc sin A ……………(iii)
Combining (i) and (ii), we obtain
Area of ∆ABC = ab sin C = ac sin B
Hence ab sin C = ac sin B
⇔b sin C = c sin B
Thus = ………………(iv)
Combining (ii) and (iii), we obtain
Area of ∆ABC = ac sin B = bc sin A
Hence ac sin B = bc sin A
⇔a sin B = b sin A
Thus = ………………(v)
Combining (iv) and (v), we obtain
= =
which is known as the sine rule.
Alternatively
Consider ∆ABC which is circumscribed by a
circle, centre O, radius R.
(Fig. 3.21)
Let BC = a, AC = b,
and AB = c. Let C′
be a point on the circle
such that BC′ is a
diameter of the circle.
Join AC′.
Using ∆ABC′, sin C′ = ⇒ BC′ = .
(Why?)
b
sin B
–––––
c
sin C
–––––
1
2
–
a
sin A
–––––
b
sin B
–––––
c
sin C
–––––
a
sin A
–––––
b
sin B
–––––
AB
BC′
–––
AB
sin C′
–––––
Fig. 3.21
– – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
•
C′
O
A
B
C
46
Fig. 3.23
The two possible areas are 8.258 cm
2
and
1.670 cm
2
.
Note:
We use the sine rule when given
1. two sides and an angle opposite one of
the given sides, or
– – – – – – – – – – – – – – –
19°
A
1
4 cm
5.7 cm
A
B
C
4 cm
a
sin A
–––––
c
sin C
–––––
b
sin B
–––––
⇒2R = =
. (Why?)
⇒2R =
Similar construction lines through A and C
give us
sin B = i.e. 2R = =
and sin A = i.e. 2R = =
Hence
2R = = = — the
sine rule.
Example 3.10
   
QR = 5.12 cm, ∠Q = 43°
and ∠R = 74° (Fig. 3.22)

Solution
To solve a triangle
angles of the triangle, which are not known,
and the area of triangle.
Since ∠Q = 43° and ∠R = 74°, then 
∠P = 180° – (43° + 74°
Using sine rule: = 
⇒ =
⇒PQ = 5.12 = 5.524 cm
= ⇒ =
⇒ PR = 5.12 = 3.919 cm
× QR sin 43°
= × 5.524 × 5.12 sin 43°
= 9.644 cm
2
.
4
sin 19°
––––––
c
sin 8.6°
––––––
1.8 cm
8.9 cm
43°
Q
74°
5.12 cm
P
R
Fig. 3.22
1
2
–
1
2
–
PQ
sin ∠R
––––––
PQ
sin 74°
––––––
QR
sin ∠P
––––––
QR
sin 63°
––––––
sin 74°
sin 63°
––––––
PR
sin ∠Q
––––––
QR
sin ∠P
––––––
5.12
sin 63°
––––––
PR
sin 43°
––––––
sin 43°
sin 63°
––––––
5.7
sin A
––––
4
sin 19°
––––––
c
sin 133.4°
––––––––
sin 133.4°
sin 19°
––––––––
4
sin 19°
––––––
Example 3.11

and ∠B = 19°

Solution
Using Sine Rule: = 
⇒ sin A = 5.7 = 0.464 0
∴A = 27.65° or 152.35°°
(i) When A = 27.65°, then
C = 180° – (19° + 27.65°)
= 133.35°°.
Then =
⇒c = 4
= 8.927 cm
= 8.9 cm (1 d.p.)
(ii) When A = 152.4°, then
C = 180° – (19° + 152.4°) = 8.6°
Then = 
⇒c = 4
= 1.837 cm

 
1
BC as the
possible triangles.
sin 19°
4
––––––
sin 8.6°
sin 19°
––––––
AB
sin C′
––––––
AB
sin C
–––––
c
sin C
–––––
AC
2R
–––
b
sin B
––––
BC
2R
–––
BC
sin A
––––
AC
sin B
––––
a
sin A
––––
47
2. two sides and any two angles, or
3. two angles and a side.
Exercise 3.8
In this exercise, state your answers correct to
3 s.f.
1. Solve ∆ABC where
(a) a = 2.2 m, c = 4 m, C = 36°
(b) b = 5 cm, A = 45°, C = 25°
(c) a = 11 cm, A = 120°, B = 24°
(d) b = 4 cm, c = 7 cm, B = 28°
(e) a = 6 cm, b = 6 cm, A = 50°
(f) a = 5 cm, A = 80°, B = 30°
2. In ∆PQR, p =10 cm, q = 5 cm, P = 30°.
Find Q and r.
3. In ∆XYZ, X = 50°, Y = 60°, z = 20 cm.
Find x and y.
4. Solve ∆PQR, where
(a) p = 5 cm, P = 70°, Q = 30°.
(b) P = 105°, Q = 36°, q = 4.17 cm.
5. In ∆PQR, p = 10 cm, Q = 15°, R = 45°.
Find r.
6. In ∆KLM, k = 15 cm, L = 25°, M = 120°.
Find m.
7. In ∆XYZ, y = 10 cm, Z = 84°, Y = 20°.
Calculate x.
8. At one instant, the distance from the earth
to the sun is 149 million kilometres. The
distance from the sun to Venus is 107
million kilometres. The line from earth to
Venus and the line from the earth to the sun
make an angle of 37°.
How far is Venus
from the earth at
that instant?
9. In Fig. 3.24, nd
the length of SR.
S
Q
25°
P
R
80°
55°
30°
5 cm
Fig. 3.24
P
58°
12 cm
10 cm
Q
R
10. ∆PQR has sides PR = 30 cm,
∠QPR = 52° 30′, ∠PQR = 66° 10′.
Find PQ and QR.
The cosine rule
Consider ∆PQR (Fig. 3.25)
in which PQ = 10 cm,
QR = 12 cm and 
∠Q = 58°. Can you
solve it?
Note that ∆PQR is not
right-angled and the
sine rule does not help one to solve it. So
another relationship between the sides and
angles is needed.
Consider Fig. 3.26 in which  
∠C is an acute angle
and AD ⊥ BC.
In ∆ABD,
AD
2
= AB
2
– BD
2
= c
2
– (a – x)
2
……………………(i)
(Pythagoras theorem).
In ∆ACD, AD
2
= AC
2
– DC
2
= b
2
– x
2
………………(ii)
(Pythagoras theorem).
∴ From (i) and (ii) we obtain
c
2
– (a – x)
2
= b
2
– x
2
⇒ c
2
– (a
2
– 2ax + x
2
) = b
2
– x
2
⇒ c
2
– a
2
+ 2ax – x
2
= b
2
– x
2
   ⇒ c
2
= a
2
+ b
2
– 2ax
In ∆ACD: x = b cos C
 ∴ c
2
= a
2
+ b
2
– 2ab cos C.
Consider Fig. 3.27 in which  
∠ACB is an obtuse angle.
AD ⊥ BC produced.
Using Pythagoras theorem
in ∆ABD,
⇒ AD
2
= AB
2
– BD
2
= AB
2
– (BC + CD)
2
= c
2
– (a + x)
2
= c
2
– (a
2
+ 2ax + x
2
)
= c
2
– a
2
– 2ax – x
2
………………(i)
Fig. 3.25
C
c
B
A
b
a
– – – – – – – – –
D
x
Fig. 3.26
– – – – – –
D
c
B
A
b
a
– – – – – – – – – – – – –
C
x
Fig. 3.27
48
Using Pythagoras theorem in ∆ACD,
 AD
2
= AC
2
– CD
2
= b
2
– x
2
…………(ii)
From (i) and (ii) we obtain
c
2
– a
2
– 2ax – x
2
= b
2
– x
2
  ⇒ c
2
= a
2
+ b
2
+ 2ax
But in ∆ACD, x = b cos ∠ACD
= b cos (180° – C)
=
–
b cos C.
 ∴ c
2
= a
2
+ b
2
– 2ab cos C.
So in either case c
2
= a
2
+ b
2
– 2ab cos C.
It can similarly be shown that
a
2
= b
2
+ c
2
– 2bc cos A
and b
2
= a
2
+ c
2
– 2ac cos B.
The expression a
2
= b
2
+ c
2
– 2bc cos A
or b
2
= a
2
+ c
2
– 2ac cos B
or c
2
= a
2
+ b
2
– 2ab cos C
is known as the cosine rule.
To nd the angles of a triangle given the lengths
of the three sides, we need to rearrange the
cosine rule. Thus,
a
2
= b
2
+ c
2
– 2bc cos A becomes
cos A =
b
2
+ c
2
– a
2
2bc
Similarly,
cos B =
a
2
+ c
2
– b
2
, and
2ac
cos C =
a
2
+ b
2
– c
2
.
2ab
We now just require to evaluate the right hand
side of each of the formulas and then nd the
angle whose cosine has been calculated.
Example 3.12
Find the angles of a triangle with sides of
lengths 3 cm, 5 cm and 7 cm.
Solution
Fig. 3.28 is a sketch of the given triangle.
cos A =
7
2
+ 3
2
– 5
2
=
33
= 0.785 7 (4 s.f.)
2 × 7 × 3 42
Fig. 3.28
⇒ A = 38.21° (4 s.f.)
C must be an obtuse angle (i.e. an angle in the
second quadrant of the unit circle) due to the
negative value of the cosine.
cos C =
5
2
+ 3
2
– 7
2
2 × 5 × 3
= 0.5
Now, cos
–
1
–0.5 = 60°.
∴C = 180° – 60° = 120°
cos B =
5
2
+ 3
2
– 7
2
2 × 5 × 3
=
65
70
= 0.928 6 (4 s.f.)
⇒ B = 21.79° (4 s.f.)
Check: A + B + C = 38.21° + 120° + 21.79°
= 180°.
Example 3.13

and ∠Q = 58°.
Solution
The triangle is as shown
in Fig. 3.29.
q
2
= p
2
+ r
2
– 2pr cos Q
q
2
= 12
2
+ 10
2
– 2 × 12 × 10 × cos 58°
= 144 + 100 – 240 × 0.530
= 116.8
q = 10.81 cm.
Using the cosine rule,
r
2
= p
2
+ q
2
– 2pq cos R
10
2
= 12
2
+ 10.81
2
– 2(12)(10.81) cos R
100 = 144 + 116.856 1 – 259.44 cos R
100 = 260.856 1 – 259.44 cos R
–
160.856 1 =
–
259.44 cos R
 ⇒ cos R =
cos R = 0.620
R = 51.68°.
A
7 cm
C
B
3 cm
5 cm
P
58°
12 cm
10 cm
Q
R
q
160.856 1
259.44
––––––––
Fig. 3.29
49
4 cm
B
A
D
C
5 cm
5 cm
8 cm
5. Solve ∆ABC where a = 5 cm, b = 3 cm,
and B = 30°.
6. The sides of a triangle are 7 cm, 9 cm and
14 cm. What is the length of the shortest
median?
7. The sides of a triangle are 11 cm, 8 cm
and 9 cm. What are the lengths of the two
shorter medians?
8. In Fig. 3.32, AB = 4 cm, BC = 5 cm,
AC = 8 cm and CD = 5 cm. Find AD.
Fig.3.32
9. Find the lengths of the diagonals of a
parallelogram whose sides are 3.6 cm and
4.8 cm and one angle of which is 54.2°.
10. Three vectors a, b and c are such that
a + b + c = 0. If |a| =10, |b| = 14 and
|c| = 17, nd size of the angles between the
vectors.
Application of sine and cosine rules to
real life situations
There are many real life situations which
require the use of sine and cosine rules. In such
cases, a sketch diagram of the situation will be
very useful. Example 3.14 illustrates this.
Example 3.14
The bearing of Kisumu from Thika is 285°. The
two towns are 400 km apart on a straight line.
Embu is 140 km from Thika on a bearing of
025°.
(a) Find the distance, on a straight line,
between Kisumu and Embu.
(b) What is the bearing of Kisumu from Embu?
Note
We use the cosine rule when given
(i) two sides and the included angle, or
(ii) all the three sides.
Exercise 3.9
In this exercise, give answers correct to 3 s.f.
1. Solve ∆ABC where
(a) a = 5 cm, b = 8 cm, c = 7 cm
(b) b = 6 cm, c = 14.5 cm, A = 95°
(c) a = 17 cm, c = 12 cm, B = 80°
(d) a = 4.1 cm, b = 8.5 cm, c = 5.9 cm,
(e) a = 6 cm, b = 6 cm, c = 50°
(f) a = 3.49 cm, b = 4.62 cm, c = 6.93 cm.
2. In a triangle, two sides are 2.8 cm and
12 cm long, and the angle between them is
60°. Find the length of the third side.
3. Find the value
of x in Fig. 3.30.
4. Find ∠ABC in Fig. 3.31.
56°
3 cm
B
x
130°
A
D
C
2.1 cm
2.3 cm
Fig. 3.30
5 cm
B
A
D
C
10 cm
11 cm
5 cm
7 cm
Fig. 3.31
50
25°
25°
N
1
N
S
T
285°
100°
E
t
K
400 km
140 km
Solution
Sketch: (Fig. 3.33).
Exercise 3.10
1. Two boats A and B leave a port at 0700
h. Boat A travels at 25 km/h on a bearing
of 037°. Boat B travels at 15 km/h on a
bearing of 140°. After 3 hours, how far is
A from B?
2. At one instant, the distance from the Earth
to the Sun is 149 million kilometres and the
distance from Mars to the Sun is 225 million
kilometres. The line from the earth to Mars
and the line from the Earth to the Sun form
an angle of 132.5°. How far is Mars from
the Earth?
3. An aeroplane ies 120 km in the direction
113°, then turns and ies 160 km in the
direction 156°. Find its distance from the
starting point.
4. The distance from the Earth to the Sun is
149 million kilometres. The distance from
the Earth to Venus is 160 million kilometres
and the distance from the Sun to Venus is
107 million kilometres. What is the angle
between the line joining the sun to Venus
and the line joining the Earth to Venus?
5. Two aeroplanes start from an airport at the
same time. One plane ies West at 400 km
per hour while the other ies at 500 km
per hour on a bearing of 040°. What is the
distance between the two planes after 15
minutes?
6. In a soccer game, players A and B are 15 m
apart. Player C has the ball and wants to
pass it either to A or B, whoever is nearest
to him. If the angle CAB = 45.6° and the
angle ABC = 37.9°, nd by calculation,
between A and B, is nearer to C?
7. Mary was driving along a straight level
road in the direction 053°. She saw a
billboard on a bearing 037°. After covering
a distance of 800 m, the billboard was on a
Fig. 3.33
Since K is on a bearing of 285° from T,
∠KTN = 360° – 285° = 75°.
∠ETN = 25°
∴ ∠KTE = ∠KTN + ∠NTE
= 75° + 25° = 100°.
(a) Two sides and an included angle are
available. So we use the cosine rule. Thus,
t
2
= 400
2
+ 140
2
– 2 × 400 × 140 × cos 100°
= 160 000 + 19 600 – 112 000 × (
–
cos 80°)
= 179 600 + 112 000 × 0.173 6
= 199 048.595 9
⇒
(b) To get the bearing of Kisumu from Embu,
 ∠KET.
Using sine rule,
=
⇒ sin E =
=
= 0.883 2
∴°
∠N
1
ES = 180° (a straight line).
∠TES = 25° (alternate to ∠NTS)
∴ The bearing of Kisumu from Embu is
180° + 25° + ∠E
= 180° + 25° + 62°
= 267°
* Do you remember working out this example,
by scale drawing, in Form 1?
400 × sin 100°
t
––––––––––––
400 × sin 100°
446
––––––––––––
400
sin E
––––
t
sin 100°
––––––
51
bearing of 296°. How far is the billboard
from the road?
8. A ranger cadet was walking towards a
tower in a village and noticed that the angle
of elevation of the top of the tower was
10°. After walking a distance of 20 m, she
noticed that the angle of elevation was 15°.
What is the height of the tower?
9. Mulei walked from a point A along a
straight path that meets a straight road at
point B. At point B he turned right and
walked 300 m along the road to point C.
He was then 400 m from A. If at point B
he had turned left and walked 300 m, he
would have been 700 m from A.
(a) What is the distance from point A to
point B?
(b) What is the size of ∠ABC?
52
4
SURDS
a
b
–
1
3
–
1
2
–
2
3
–
1
2
–
2
3
–
1
3
–
2
3
–
p
q
–
a
b
–
Rational and irrational numbers
Numbers which can be expressed as exact
fractions, i.e. in the form where a and b
are integers and b ≠ 0, are called rational
numbers. Those which cannot be written in
this form are called irrational numbers.
In Form 1, we learnt that both terminating
decimals and recurring decimals can be
expressed as exact fractions, and so they are
rational numbers. Are all integers rational
numbers? Is zero a rational number?
Some decimals neither terminate nor recur. An
example of this type is 0.20200 2000 20000 2…
This decimal cannot be written as exact fraction
and so it is irrational. 0.404404440… is another
example of an irrational number. πis another
3.141592664......
Write down some more irrational numbers of
the same type.
Exercise 4.1
1. By nding the mean of the two numbers
in each of the following cases, obtain a
rational number between them.
(a) and (b) and 1
(c) and (d) and
2. Given that Q represents the group of all
rational numbers, Z all integers, T all
terminating decimals, and P all recurring
decimals, which of the following
statements are true?
(a) Z is wholly contained in Q.
(b) Z, P and T put together make Q.
(c) There are no numbers that are in T as
well as in P.
3. Express the following recurring decimals in
the form .
(a) 3.3 (b) 5.6 (c) 3.17 (d) 2.102.
4. Given that p and q are any two rational
numbers, is it true that
(a) p + q, (b) p – q, (c) pq, and
(d) are rational?
Rational and irrational square roots
Some square roots of numbers are rational. For
example, √4 = 2, √1.44 = 1.2 = , etc. are
rational.
Some square roots are irrational, eg √2, √3, √5
etc.
Example 4.1
Find the value of √85 correct to 1 decimal
place.
Solution
85 lies between 81 and 100 (two exact squares).
∴ √81 < √85 < √100
i.e. 9 < √85 < 10.
Now 9.2 × 9.2 = 84.64,
9.3 × 9.3 = 86.49
Thus 9.2 < √85 < 9.3.
9.21 × 9.21 = 84.241
9.22 × 9.22 = 85.008 4
∴ 9.21 < √85 < 9.22
This shows that √85 = 9.2 correct to 1 d.p
Notice that:
To nd the whole number part of a square
root of a non–square number,
1. Find two consecutive exact squares
between which the non-square number
lies, (e.g. 39 lies between 36 and 49).
6
5
–
· · · ·
·
·
‑‑
‑
53
3
3
3
2. The square root of that number lies
between the square roots of the exact
squares,
(e.g. √36 < √39 < √49, i.e. 6 < √39 < 7).
3. The whole number part of the required
square root is given by the lower value,
(e.g. √39 = 6.245).
The decimal part of the square root is found by
trying out values as in Example 4.1.
Exercise 4.2
1. The following numbers are exact squares.
Find their square roots.
(a) 64 (b) 144 (c) 400
(d) 169 (e) 1.96 (f) 0.003 6
2. Find the two closest integers between
which the square root of each of the
following numbers lies.
(a) 134 (b) 430 (c) 1 440
(d) 3 000
3. Which is larger, √7 or 2.6? Test by
squaring the numbers.
4. Show by squaring, that 3.87 < √15 < 3.88
5. Write down the whole number part of the
square root of each of the following.
(a) √5 (b) √11 (c) √44
(d) √125 (e) √999 (f) √9 999
6. Between which two consecutive integers
does √151 lie?
Surds
Many numbers are not exact powers. Their
roots (e.g. square root, cube root, etc.) are,
therefore, irrational. Expressions containing
roots of such numbers are called surds.
√2, √3, √4, 2 + √2, 3 – 2√5 are examples of
surds.
A surd of one term is called quadratic, cubic,
quartic, quintic, etc. depending of whether the
index (or order) of the radical is two, three,
‑
√x
√y
––
4
x
y
–
√
√x
√y
––
x
y
–
√
four, ve, etc., e.g. √7 is a quadratic surd, √18 is
a quartic surd, etc.
A monomial surd is one which contains only
one term; a binomial surd has two terms; a
trinomial surd has three terms, at least two
of which are surds which cannot be combined
without evaluating them, etc. For example, 3√2
is a monomial surd, √2 – √3 is a binomial surd,
2 + √5 – √3 is a trinomial surd.
In this course, we shall deal with only
monomial and binomial surds of order 2.
Simplication of surds
ACTIVITY 4.2
By putting x = 25 and y = 4, determine which
of the following pairs of expression are equal.
1. √xy , √x × √y 2. ,
3. √x + y , √x + √y 4. √x – y , √x – √y
5. 5√x, √5x 6. 5√x, √25x
Note that in Activity 4.2, the pairs of
expressions in 1, 2 and 6, are equal.
Thus in general, given that x and y are positive
values, then
√xy = √x × √y and =
These facts can be used when simplifying surds.
Example 4.2
Simplify (a) √28 (b) √216
(c) √x
2
y (d) √a
4
b
2
c
3
Solution
(a) √28 = √4 × 7 = √4 × √
 ××× 3
= 6 ××
 
2
 
2
× 
54
√12
√6
–––
5
100
–––
√
12
3x
2
–––
√
√60
√10
–––
√12
4
–––
16
√18
–––

4
b
2
c
3
 
4
×
2
×
2
×
= a
2

A surd is said to be in its simplest form when
the number under the radical (√ ) is a prime
number.
The process of simplifying surds can also be
reversed to obtain surds of single numbers.
Example 4.3
Express the following surds as square roots of
single numbers.
 
Solution
××
××
Exercise 4.3
1. Express the following surds in their
simplest forms.
(a) √12 (b) √45 (c) √72 (d) √35
(e) √250 (f) √432 (g) √124 (h) √132
2. Express each of the following surds as a
square root of a single number.
(a) 2√5 (b) 4√2 (c) 3√7
(d) 7√3 (e) 2√3√7 (f) 10√2√5
(g) 10√3√7 (h) 12√3√5
3. Simplify the following.
(a) √10 × √5 (b) √32 × √2
(c) √5 × √6 × √3 (d) √15 × √5
(e) √24 × √6 (f) √5 × √12 × √60
(g) √10 × √18 × √20 (h) √10 × √18 × √6
4. Simplify the following
(a) (b) (c) (d)
(e) (f) (g)
(h) √12√6 ×
5
4
–
√
Operations on surds
The four basic mathematical operations ( +, –,
×, ÷) apply to surds in much the same way as in
algebra, as illustrated in the following sections.
Addition and subtraction of surds
Example 4.4
Evaluate
  

Solution

   

=
–

   
just as there is no common factor in 7x + 2y.
∴
We notice that:
To be able to add or subtract surds, they must
contain roots of the same number. In general,
a√x ± b√x = (a ± b)√x
Multiplication of surds
When two monomial surds have to be
multiplied together
1. rst simplify each surd where possible,
and then
2. multiply whole numbers together and
surds together.
Example 4.5
Work out
(a) √32 × √12 × 2√18 × √20
(c) (3√7)
2
–––
√15
√3
55
3
5
–
1
4
–
5
3
–
2
3
–
(a)
(

)
2
=
(

)(

)
=
(

)
2

(

)
2
    
    
(b)
(

)
2
=
(

)(

)
= 3
2
– 2 ×
(

)
2
   
   
(c)
(

)(

)
= 3
2

(

)
2
= 9 – 2 = 7
(d)
(

)(

)
  
Exercise 4.4
1. Simplify
(a) 2√3 + 7√3 (b) 12√5 + 5√5
(c) 13√2 – 7√2 (d) 12√7 – 17√7
(e) 3√11 + 7√11 – 11√11
(f) a√x – 3a√x + 5a√x
(g) 7√45 – 2√45 (h) 3√128 – 14√128
(i) 8√216 – √216 (j) √54 – √6
2. Work out
(a) √7 × √3 (b) 5√3 × 7√11
(c) 6√5 × √7 (d)
(
√3
)
6
(e)
(
4√3
)
3
(f)
(
5√x
)
2
×
(
2√x
)
3
3. Evaluate, writing your answer in the
simplest form.
(a)
(
√3 + 4
)
2
(b)
(
√2 + √7
)
2
(c)
(
√5 – 3
)
2
(d)
(
√5 – √3
)
2
(e)
(
√2 + √5
)(
√7 – √2
)
(f)
(
√7 – 4
)(
4 – √5
)
(g)
(
√7 + √2
)(
√7 – √2
)
(h)
(
2√5 – 3√3
)(
2√5 + 3√3
)
(i)
(
√48 + √12
)(
√48 + √108
)
(j)
(
√75 + √150
)(
√12 – √24
)
Solution
×× 2 ×× 3
    ×
    
××
   × 3 ×× 2 ×× 5
   × 2 ××
   
 
2
×
   × 7 = 9 × 7 = 63
Example 4.6
×××××
Solution
×××××
 ×××××
= 2 × 3 × 3 × 4 ×× 2 × 2 × 2 × 3 × 3
= 72 × 2 × 2 × 3 = 864

surds to give simpler surds, as follows:
×××××
 × 3 × 8 × 18 × 27 × 32
 × 32) × (3 × 27) × (8 × 18)
 × 81 × 144
= 8× 9 × 12 = 864
Note that the second method is sometimes
simpler and quicker.
Example 4.7
Expand the following and give your answers in
the simplest surd form.
(a)
(

)
2
(b)
(

)
2
(c)
(

)(

)
(d)
(

)(

)
Solution
These products are worked out in the same way
as the algebraic expansions of
(x + y)
2
, (x – y)
2
, (v + w)(x + y)
56
7

–––
7

–––


–––

7
––––
7

–––
×

– –– ––––

25
8
–––
5

––––


–––
5

––––
–––




–––

4
––––
Note that they compare to the algebraic identity
(a + b)(a – b) = a
2
– b
2
, known as a dierence
of two squares. The product of any two such
surds is always a rational number.
Example 4.9
Work out
 

Solution
 
  
  
= 4 – 3 = 1
 
  
  
= 27 – 5 = 22
Pairs of surds such as 2 + √3, 2 – √3 and
3√3 – √5, 3√3 + √5 are called conjugate surds
(or conjugate pairs).
In general:
If the product of two surds is a rational
number, then the surds are said to be con-
jugates of each other or simply conjugate
surds.
Thus, the conjugate of √x + √y is √x – √y.
Note that √x is a conjugate of itself.
Example 4.10
Rationalise the denominator in
(a) (b)
Solution
(a) Note that
(
√2 + √3
)(
√2–√3
)
=
(
√2
)
2
–
(
√3
)
2

squares
= 2 – 3 =
–
1
x
y
–

9

–––
3

––––
7

–––
9

–––

3
––––
9

–––


–––
Division of surds: rationalising the
denominator
If a fraction has a surd in the denominator, it is
usually better to rationalise the denominator.
Rationalising the denominator means making
the denominator a rational number, so
that we divide by a rational number rather
than divide by a surd. When rationalising,
we multiply both the numerator and the
denominator of the fraction by a surd which
makes the denominator rational. It is easier
to divide by a rational number than a surd.
Rationalisation of monomial
denominators
Example 4.8
Rationalise the denominator of
(a) (b) (c) (d)
Solution
×
(a) = × = = 
(b) = =× = =
Note: Multiplying by in (a) and by in
(b) is equivalent to multiplying by 1. This way,
the value of the given fraction is not changed.
(c) = × .
 
(d) = (using the fact that = )
= = ×
=
Rationalisation of binomial denominators
Refer to Exercise 4.4, Questions 3 (g) and (h).
3

––––
3

––––
3

––––


–––

2 × 7
–––– –

14
––––


–––

25
8
–––


–––
7
√2 + √3
–––––––
√2 – √3
3√2 – 2√3
––––––––
57
√45 + √50
√48 – √18
––––––––
√112 – √20
√140 – 5
–––––––––
√252 – √45
√5 – √28
–––––––––
3 × 2 – 1√2√3 – 2 × 3
6
––––––––––––––––––
–
√2√3
6
––––––
5

––––––––
5
2 × 1.414 – 1.732
––––––––––––––
5
1.096
–––––
4
√11
–––
5
√5 – 4
–––––
3√7
3√7 – 2√5
––––––––
3√2 – 5√11
4√18 – 2√2
––––––––––
3√2
2√5
–––
The rst method gives an “exact” answer while
the second one gives an approximation. Thus,
the rst method is more accurate.
Note: You may use a calculator when
evaluating surds, but you still must show all the
steps involved in the process.
Exercise 4.5
1. State the conjugate of
(a) √13 (b) 2 + √17
(c) √3 – 7 (d) √3 – √8
(e) 2√5 + 3√4 (f)
–
3√x + b√y
2. Rationalise the denominator in
(a) (b) (c) (d)
3. Rationalise the denominator in
(a) (b) (c)
(d) (e) (f)
(g) (h) (i)
(j)
4. Given that √2 = 1.414, √3 = 1.732 and
√5 = 2.236, rationalise the denominator and
then evaluate each of the following correct
to 2 d.p.
(a) (b) (c)
(d)
5. Given that x = √5 and y = √7, express the
following in terms of x and y. Use
factorisation to simplify each expression.
(a) √1 260 + 2√5 (b) √140 – √45
(c) (d)
3 + √2
√20
–––––
√2 – √3
3√2 – 2√3
––––––––
√2 – √3
3√2 – 2√3
–––––––––
3√2 + 2√3
3√2 + 2√3
–––––––––
3√2√2 + 2√2√3 – 2√3√3 – 3√2√3
6
––––––––––––––––––––––––––––
(
√2 – √3
)(
3√2 + 2√3
)
6
––––––––––––––––––
5
2√2 – √3
––––––––
5
2.828 – 1.732
–––––––––––
6
√48
–––
13
√13
–––

162
5
–––
7
5 + √11
––––––
2√3 + 3√2
3√3 + 2√2
–––––––––
––––––––
√27
√21 + √33
√7 – 2√3
1 + 3√2
–––––––
2√3
√5 + √2
––––––
3√11 + 4√13
√11 – √13
–––––––––––
√3 – √2
√3 + √2
––––––
5√7
√3 – √2
––––––
7
√2 + √3
–––––––
√2 – √3
√2 – √3
––––––
7
√2 + √3
–––––––
7√2 – 7√3
–
1
––––––––
5
(

)
8 – 3
––––––––––
(Did you get similar results in Questions 3
(g) and (h) of Exercise 4.4?)
∴ = ×
=
= 7√3 – 7√2
(b)
(
3√2 – 2√3
)(
3√2 + 2√3
)
=
(
3√2
)
2
–
(
2√3
)
2

two squares
= 18 – 12 = 6
 ∴ = ×
=
=
=
=
Example 4.11

evaluate
.
Solution
=  ×
=
= =
= 2√2 + √3
= 2 × 1.414 + 1.732 = 4.56
Note that evaluating in this way is much easier
than evaluating as follows:
= =
    
reciprocals
  
5

––––––––
5

––––––––


––––––––
5
(

)
(

)
2
–
(

)
2
–––––––––––
5
(

)
5
––––––––––
58
5
FURTHER LOGARITHMS
1
81
––
1
8
–
x
y
–
Logarithmic notation
In Form 2, we learnt that:
When a number is written in the form b = a
m
,
a is called the base, and
m is called the index or power or
exponent.
The index m is also called the logarithm of b
to the base a, written in short as log
a
b = m.
Now, go through the following example.
Example 5.1
Write down the logarithm of each of the
following numbers to the stated base.
(a) 32, base 2 (b) 125, base 5
(c) 512, base 8 (d) 27, base 9.
Solution

a power of the base.
(a) 32 = 2
5
∴log
2
32 = 5
(b) 125 = 5
3
∴log
5
125 = 3
(c) 512 = 8
3
∴ log
8
512 = 3
(d) 
9
27 = m
Then, 9
m
= 27
 ⇒3
2m
= 3
3
 ⇒2m = 3
 ∴ m = 1
i.e. log
9
27 = 1
Exercise 5.1
1. Rewrite the following expressions in
logarithmic notation.
(a) 3
4
= 81 (b) 3 125 = 5
5
(c) 0.001 = 10
–3
(d) b
0
= 1
2. Express the following in index notation.
(a) log
2
32 = 5 (b) log
10
100 = 2
(c) log
3
x = 8 (d) p = log
q
r
3. Evaluate the following logarithms.
(a) log
2
8 (b) log
10
10 000
(c) log
5
625 (d) log
3
243
(e) log
7
343 (f) log
4
8
(g) log
8
4 (h) log
64
8
(i) log
9
( ) (j) log
4
( )
(k) log
0.2
25 (l) log
8
0.125
Laws of logarithms
In Form 2, we also learnt that the fundamental
laws of indices are:
1. a
m
× a
n
= a
m + n
2. a
m
÷ a
n
= a
m – n
3. (a
m
)
n
= a
mn
Suppose x = a
m
and y = a
n
.
Then, log
a
x = m and log
a
y = n.
From the rst law of indices, xy = a
m + n
.
Thus, log
a
(xy) = m + n
i.e. log
a
(xy) = log
a
x + log
a
y (since m = log
a
x
and n = log
a
y)
Similarly, if x = a
m
and y = a
n
, then
= a
m – n
∴log
a
= m – n
i.e. log
a
= log
a
x – log
a
y.
1
2
–
1
2
–
x
y
–
( )
x
y
–
( )
59
x
y
–
30
2
––
Likewise , if x = a
m
, then
x
n
= (a
m
)
n
= a
mn
∴ log
a
x
n
= mn
i.e. log
a
x
n
= n log
a
x
Thus, the fundamental laws of logarithms are:
1. log
a
(xy) = log
a
x + log
a
y
2. log
a
( ) = log
a
x – log
a
y
3. log
a
x
n
= n log
a
x
Note:
(1) These laws are true for any base.
(2) If the base is 10, the base is often not stated,
e.g. log 0.1 is understood to mean log
10
0.1.
(3) Logarithms to base 10 are known as
common logarithms because 10 is the
most commonly used base.
Simplication and evaluation of
logarithmic expressions
Simplifying a logarithmic expression requires
that we rst write the logarithms as expressions
of logarithms of the same number to the same
base. This means coming up with a single
logarithm in its simplest form.
Examples 5.2
Without using tables or calculators, simplify
(a) (b) log 2 + log 50
(c) log 500 – log 5
Solution
(a) = = = 4.
(b) log 2 + log 50 = log(2 × 50) = log 100
= log 10
2
= 2 log 10
= 2 × 1
= 2.
(c) log 500 – log 5 = log (500 ÷ 5)
= log 100
= 2.
At times, we may be asked to evaluate the
logarithm of a given number, having already
been given the logarithms of its factor. This
requires that we rst express the given number
in terms of the given factors and then apply the
laws of logarithms.
Example 5.3
Given that log 2 = 0.3010 and log 3 = 0.4771,
evaluate
(a) log 6 (b) log 18 (c) log 45
Solution
(a) log 6 = log (2 × 3) = log 2 + log 3
= 0.301 0 + 0.477 1
= 0.778 1.
(b) log 18 = log (2 × 9) = log (2 × 3
2
)
= log 2 + 2 log 3
= 0.301 0 + 2 × 0.477 1
= 1.255 2.
(c) log 45 = log (15 × 3) = log( ×3)
= log 30 – log 2 + log 3
= log 10 + log 3 – log 2 + log 3
= log 10 + 2 log 3 – log 2
= 1 + 2 × 0.477 1 – 0.301 0
= 1.653 2.
Note:
(i) a
1
= a ⇒ log
a
a = 1 [Hence, log 10 = 1 in
(c)]. Also a
0
= 1 ⇒ log
a
1 = 0
(ii) We also need to remember that since
√a = a , then log
(
√a
)
= log a = log a.
Example 5.4
Evaluate to 2 decimal places log
2
5.45
Solution

2
5.45 = x
Then 2
x
= 5.45
log 16
log 2
–––––
log 16
log 2
–––––
log 2
4
log 2
–––––
4 log 2
log 2
–––––
1
n
–
n
n
1
n
–
1
n
–
60
1
2
–
4
3
4
–
1
2
–
1
2
–
∴ x log 2 = log 5.45
 ⇒ x =
=
∴ x = 2.45 (2 d.p.)
Note that you could also obtain this result using
a calculator as follows:
! Caution:
1. is not the same as log
(
)
. We
have to nd the log of each number
separately and then divide the values
obtained.
(2) Another very common error made by
students is as follows:
= = 2.725
Note that ‘log’ is just a word (an adjective)
describing a function and therefore it cannot
be divided.
Exercise 5.2
1. Express each of the following as a single
logarithm
(a) log 2 + log 5
(b) log 2 + log 3 – log 7
(c) 3 log 2 + 2 log 5
(d) 3 log x – 2 – log 4y
(e) log √x + log √x
(f) 3 log x
2
– 8 log
√x.
2. Without using tables or calculators, evaluate
log 6 + log 4 + log 20 – log 3 – log 16.
3. Express log
2
64 – log
2
16 in its simplest form.
4. Given that log y = 3 – log x, express
y in terms of x in a form not involving
logarithms.
5. If x = log (a – by) – log a, nd the value of
y when a = 4, b = 2 and x =
–
2.
6. Without using tables or calculators, nd the
value of
given that log 2 = 0.301 03.
7. Given that log 2 = 0.301 0, log 3 = 0.477 1
and log 7 = 0.845 1, evaluate
(a) log 5 (b) log 8 (c) log 49
(d) log 14 (e) log 42 (f) log 35
8. Use logarithm tables to evaluate the
following, giving your answers to 2 d.p.
(a) log
4
8 (b) log
7
9
(c) log
5
11.5 (d) log
12
2.532
9. Express the following equations in index
form. (Do not solve them).
(a) log
a
y = n
(b) log
m
x + 2 log
m
y = 3
(c) 3 log
b
x + 2 = 1
(d) log x – 2 log y = log z.
10. Write the following formula without using
logarithms.
m log A = 4 – log B.
Equations involving logarithms
When faced with an equation involving indices,
we either
(a) Express both sides of the equal sign as
powers of the same base, equate the
powers and then solve the resulting simple
equation, or
(b) Take logarithms on both sides of the equal
sign and then divide or multiply as may be
appropriate.
At times, the equation may involve sums and/
or dierences of logarithms of numbers. This
simply requires the application of the laws of
logarithms in reverse.
log 5.45
log 2
–––––––
0.736 4
0.301 0
––––––
log 5.45
log 2
–––––––
5.45
2
––––
log 5.45
log 2
–––––––
5.45
2
––––
log 0.8 – log 32 + log 8
log 0.7 + log 7 – log 49
–––––––––––––––––––
3
61
400
4
–––
log 0.04
log 25
–––––––
log 64
log 8
––––––
log √2
log 32
––––––
log 32 – log 128
log 2 – log 8
–––––––––––––
log 81 – log 9
log 27
–––––––––––
1
4
–
3
5
–
log √3
log 81
––––––
3
10
27
––
Example 5.4
Solve for x in:
(a) 3
x
= 81 (b) 5
x
= 4
(c) log
x
3 + log
x
27 = 2.
Solution
(a) 3
x
= 81
⇒ 3
x
= 3
4
⇒ x = 4
(b) 5
x
= 4
⇒ x log 5 = log 4
∴ x = = (from tables)
(c) log
x
3 + log
x
27 = 2
∴ log
x
(3 × 27) = 2
i.e. log
x
81 = 2
∴ 81 = x
2
⇒ 9
2
= x
2
∴ x = 9
Exercise 5.3
1. Solve for x in
(a) log x = 4 (b) log x =
–
3
(c) log
x
81 = 4 (d) 4 log
x
(2 ) = 12
2. Solve the following equations
(a) 2
x
= 128 (b) 3
2x
= 27
(c) 6
x
= 15.36 (d) 16
x
= 0.25
(e) 6.23
x
= 0.618
3. Find the value of y in
2 log
4
y = log
4
3 + log
4
9 – log
4
y.
4. Solve the equation
(a) log (x + 3) = log (x – 6) + log 3.
(b) log (3y – 2) – log (y + 10) + 1 = 0.
5. By taking log 5 ≈ 0.7, obtain an estimate of
y in the equation 10
y – 5
= 5
y + 2
, giving your
answer to the nearest integer.
6. Find the values of x which satisfy the
equation
4 log
3
x = .
7. Solve for x in log
3
x + = 4.
8. Solve for x in the equation
log
2
(x + 4) = 2 – log
2
x.
9. Given that log
2
(x – 5y + 4) = 0 and
log
2
(x + 1) –1 = 2 log
2
y, nd the values of
x and y.
Further computations using
logarithmic laws
This section provides further practice on
simplication and evaluation of logarithmic
expressions without using tables or calculators.
Example 5.5
Simplify (a) (b) log 400 – log 4.
Solution

two numbers as powers of the same base.
(a) = = = 5
(b) log 400 – log 4 = log = log 100
= log 10
2
= 2 log 10
= 2 × 1 = 2.
Exercise 5.4
1. Without using tables or calculators,
evaluate the following.
(a) (b) (c)
(d) (e)
(f)
2. Without using tables or calculators,
evaluate
log 6 + log 1 .
log 4
log 5
––––
0.602 1
0.699 0
–––––––
log 32
log 2
–––––
log 32
log 2
–––––
5 log 2
log 2
––––––
log 2
5
log 2
–––––
3
log
3
x
–––––
9
log
3
x
–––––
62
3. Without using tables or calculators,
evaluate
4. If xy
n
= 475, nd n given that x = 2.5 and
y = √10 giving your answer correct to 2 d.p.
5. Find the least integral value of n for which
0.95
n
is less than 0.1.
6. Without using tables or calculators,
evaluate
4 log 3 + 3 log 2
3 log 6 + log 3
––––––––––––––
√
––––––––––––––––––––
1
2
–
1
3
–
log 409 6
1 + log 0.16 + log 8
63
1-5
REVISION EXERCISES 1
Revision exercise 1.1
1. (a) Use the method of factorisation to
solve
(i) 2x
2
+ 5x + 2 = 0
(ii) 2x
2
– 7x + 6 = 0
(iii) 3x
2
– 4x – 15 = 0
(iv) (x – 2)(2x + 5) = 5
(b) What must be added to each of the
following to make it a perfect square?
Of what expression is each a square?
(i) y
2
+ 5y (ii) x
2
– 13x
2. For what value of k is each of the following
expressions a perfect square?
(a) (i) 4x
2
– 10x + k (ii) 4p
2
+ 8p + k
(iii) 9x
2
– kx + 25
(b) Use the method of completing the
square to solve
(i) 2x
2
– 14x + 9 = 0 (ii) 3x
2
– 4x = 5
3. Use a calculator to evaluate the following
(a) (i) 2.40 + 530.5 + 0.356 89
(ii) 814 069 – 93 740
(iii) 1 – 0.335 9
(iv) 89.67 + 0.563 – 11.34 + 1.000 63
(b) (i) 2.789 1 × 3.1
(ii) 89.18 ÷ 2.4
(iii) 197.33
2
(iv) √1 993
4. (a) (i) Round of 4.358 correct to 2 s.f.
(ii) Truncate 4.358 correct to 2 s.f.
(b) Using your answers to (a), nd
(i) absolute rounding o error
(ii) truncation error
(iii) the respective relative errors.
(c) Find
(i) the absolute percentage rounding
o error
(ii) the percentage truncation error.
5. Three angles of a quadrilateral are stated as
45.5°, 107° and 137.5° to the nearest 0.5°.
Find the range within which the fourth
angle must lie.
6. (a) Find the values of the trigonometric
ratios of 15°. Hence, state the values of
(i) sin 165° (ii) cos 195°
(iii) tan 345°
(b) Find the values of trigonometric ratios
for 70°. Hence state the values of
(i) sin 110° (ii) cos 290°
(iii) tan 250°
(c) Evaluate the basic trigonometric ratios
for the following
(i) 365°
(ii) 550°
(iii) 1 788° (iv) 4 820°
(v)
–
20° (vi)
–
140°
(vii)
–
260° (viii)
–
485°
7. (a) Change the following angles into
radians, correct to 3 s.f.
(i) 60° (ii) 20°
(iii) 276° (iv) 64.5°
(b) Convert the following angles from
radians into degrees, correct to 3 s.f
(i) 0.3
c
(ii) 4.15
c
(iii) 2.4
c
(iv)
c
8. (a) State the conjugate of
(i) √3 (ii)2 – √5 (iii) √3 + 3
(iv) 2√2 + 3√3
(b) By rst rationalising the denominator,
evaluate
to 4 d.p.
9. (a) Express the following in logarithmic
form. (Do not use base 10).
(i) 3
4
= 81 (ii) 5° = 1
(iii) a
3
= (iv) 4
–
= 0.5
π
12
––
2
1 + √2
–––––
1
2
–
1
27
––
64
(b) Evaluate the following
(i) log
5
125 (ii) log
0.1
1 000
(iii) log
10
0.000 1 (iv) log
27
(c) Express each of the following as a
single term and simplify where possible.
(i) log
10
7 + 3 log
10
2 – log
10
28
(ii) log
x
5 – 2 log
x
15 + 3 log
x
3 – log
x
16
10. In a physics lesson, a student measured the
diameters of three objects and recorded
them as a = 20.6 cm, b = 15.2 cm and
c = 0.017 4 cm. Find the percentage error
in
(a) a + b (b) a – c (c) ac (d)
11. (a) Solve for x in the following:
(i) 1 024 × 2
x
= 1
(ii) 4 log x + log 81 = 2 log 6x
(b) Express n in terms of x and y given that
log y = log(10x
n
)
Revision exercise 1.2
1. (a) Use the quadratic formula to solve
(i) 2x
2
+ 7x – 2 = 0
(ii) x
2
– 6x + 9 = 0
(iii) x
2
+ 6x + 13 = 0
(b) Solve the simultaneous equations
(i) y
2
– 8 = xy (ii) x
2
+ y
2
= 25
x + y = 0 x + y = 7
(c) A right angled triangle has its longest
side as 25 cm and the two shorter sides
as x cm and y cm. If one of the shorter
sides exceeds the other by 5 cm, form
two equations in x and y and hence nd
the lengths of the shorter sides of the
triangle.
2. A cyclist travels a distance of 120 km at
an average speed of x km/h. On his return
journey, he reduces his speed by 8 km/h
and takes half an hour more. Write down
algebraic expressions for the time taken by
the cyclist for his “to” and “fro” journeys.
Form an equation in x and solve it to obtain
the average speed for the “to” journey and
the total time taken for the return journey.
3. Use logarithms to nd the value of 3
10
.
Given that the correct value is 59 049, nd
the percentage error, giving your answer
correct to 1 s.f.
4. The measurements of the oor of a
rectangular room are stated as 5.4 m by
3.5 m. Each of these measurements has a
possible error by ± 2 cm. Find
(a) maximum length and width of the
rectangle
(b) (i) perimeter (p cm) of the rectangle
must lie in the form p ± x.
(ii) area (A cm
2
) of the room must lie
in the form A ± x.
5. Solve the following equations for
0° ≤ x ≤ 360°.
(a) cos x = 0.466 7 (b) cos x =
–
0.234 5
(c) cos(2x) = 0.625 (d) sin x = 0.565
(e) sin x =
–
0.485 (f) sin x = 0.85
6. A point P is on a side of a straight river.
Points A and B are 100 m apart on the other
side of the river. If ∠BAP = 38.5°, and 
∠ABP = 46.7°, nd the width of the river.
7. (a) Given that a = √5 + 4 and b = √5 – 4,
evaluate
(i) ab (ii) a
2
– b
2
(b) Simplify each of the following
expressions leaving your answer in the
simplest surd form.
(i)
√5
(ii)
4
√5 + 2 √7 + √2
8. (a) Write each of the following as a sum or
dierence of two logarithms
(i) log (m
4
n ) (ii) log (p√q
–2
)
(b) Without using tables or calculator,
simplify
(i)
3
2
–
1
81
––
b
–
c
log 144
log 12
1
3
–
65
(ii) 2 log 5 + log 15 – log 125 – log 3
9. Use a calculator to evaluate
(a) (i) 10
0.86 95
(ii) 3
4.651 9
(iii) log
10
19.918 (iv) 10
3.666 1
(b) (i) 9.42
2
– 0.942
2
(ii) 0.176 × 0.720 ÷ 0.165
2
(iii) √632 + 3.61 ÷ 2.75
(iv) √8 653 × 128 + 1.93
3
10. (a) Factorise completely 8x
2
– 32y
2
(b) Find the value of m given that
3m
2
– 16 = 13m.
11. Given that sin x =
–
and that
180° < x < 360°, nd cos x and tan x with-
out using tables or calculator.
12. An aircraft ies from airstrip A to airstrip
B, 300 km away on a bearing of 125°.
From airstrip B, the pilot ies on a bearing
of 240° to airstrip C, 200 km away. If the
pilot ies back to A directly, calculate the
distance that he ies and the direction in
which he ies.
Revision exercise 1.3
1. (a) A two digit number is such that twice
the tens digit is three less than the ones
digit. Given also that the product of the
digits is three more than twice the sum
of the digits, nd the number.
(b) If (x + y)
2
is miscopied as x
2
+ y, nd
the percentage error when x = 5 and
y = 3.
2. (a) The sides of a triangle ABC are given
as AB = 4.5 cm, BC = 3.6 cm and
AC = 2.4 cm, measured to the nearest
mm. Find the percentage error in the
perimeter.
(b) A car travels 16 km in 24 minutes.
Find its average speed in km/h. If each
measurement is liable to a 10% error,
nd the minimum and maximum values
of the speed.
3. Draw the graph of y = x
2
– 3x + 2 for values
of x from
–
1 to
+
4. Use your graph to an-
swer the following:
(a) For what range of values of x is
x
2
– 3x + 2 negative?
(b) What is the minimum value of
x
2
– 3x + 2?
(c) What are the roots of the equation
x
2
– 3x + 2 = 0?
(d) What are the solutions of the equation
x
2
– 3x + 1 = 0?
(e) Solve the equation x
2
= 4x
4. Fig. R1.3.1 shows a crane
supporting a load N.
If KL is vertical, nd the
height of the load above
the ground? (State your
answer to 3 s.f.)
Fig. R1.3.1
5. A vehicle is travelling at 90 km/h on a road
running due east. A passenger in the vehi-
cle sights a radio station mast on a bearing
of 010°. Five minutes later, the passenger
notes that the bearing of the same mast is
now 355°. Find the direct distance from
the road to the mast.
6. Simplify
(a)
2
(b)
5
3√5 3√2 – √3
7. (a) Evaluate each of the following leaving
your answer in the surd form where
necessary.
(i)
11
–
5
√7 – √3 √7 – √3
5
13
––
3
–
K
4 m
M
L
N
1.6 m
2.5 m
5.3 m
Ground level
66
(ii)
√5
–
√5
1 – √5 1 + √5
(b) A triangle ABC is right angled at B and
has sides AB = (√5 + 2) cm and
AC = (3√2) cm. Find, in surd form, the
length of side BC, and hence the area
of the triangle.
Hint: 9 can be expressed as (5 + 4)
8. Evaluate log
5
125 without using tables or a
calculator.
9. Solve for x in the equation
(a) 2 log x + log 5 = 1 + 2 log 4
(b) log (2x – 11) – log 2 = log 3 – log x
(c) Solve the simultaneous equations
(i) x – y = 21
log x + log y = 2
(ii) xy = 4
log x + 2 log y = 1.
10. A man stands at point A, a distance of
300 m from a tall building. After walking
towards the building a distance of 205 m, he
observes that the angle of elevation of the top
of the building is 75°. Calculate the angle of
elevation of the top of the building from A.
11. Copy and complete Table R1.3.1 for
y = 5 + 3x – 2x
2
.
x
–
2
–
1.5
–
1
–
0.5 0 0.5 1 1.5 2 2.5 3 3.5
y
–
9 3 3
Table R1.3.1
(a) Use the values from the table to draw
the graph of y = 5 + 3x – 2x
2
.
(b) Use your graph to determine the range
of values which satisfy the inequality
5 + 3x – 2x
2
≥
–
2
(c) In order to solve graphically the
equation 2x
2
– 2x – 3 = 0, a straight line
must be drawn to intersect the curve
y = 5 + 3x – 2x
2
. Obtain the equation
of this straight line, draw the line and
hence obtain the graphical solution of
the equation.
12. ABC is a triangle such that AB = 17.5 cm,
BC = 14.3 cm and ∠ABC = 25.6°.
Calculate
(a) the length of side AC,
(b) the remaining angles of the triangle.
67
6
COMMERCIAL ARITHMETIC
In this chapter, we shall deal with the basic
operations and computations in relation to
commercial activities. These activities include
borrowing and lending money, hire purchase
and income tax. Appreciation and depreciation
are also included.
Principal, rate, time and interest
If you borrow money to buy a car, a house, or
for any other project, you pay the bank or the
nance company for lending you the money.
The amount you pay depends on how much
you borrow, and for how long. This payment is
called interest (I). Interest is a xed percentage
of the amount borrowed usually at xed time
interval, say yearly, half yearly etc. This
percentage is called the rate and the amount
borrowed is called the principal (P). The sum
of the principal and the interest is called the
amount (A), i.e. A = P + I.
If you save money with a bank or with
a building society or any other nancial
institution, your money will be lent to other
people who pay interest and you in turn will be
paid interest for lending the money. The above
terms apply here as well.
Simple interest
If interest is paid at xed periods on a constant
principal, the interest is called simple interest.
Suppose you borrow sh 20 000 and pay interest
at the rate of 9% per annum.
This means that in the rst year, the interest is
9% of sh 20 000, i.e. sh 1 800. If you continue
to pay sh 1 800 each year, then you are paying
simple interest. If you borrow sh 200 000 for
3 years at 15% per annum (p.a.), the interest
you pay each year is
sh 200 000 ×.
For 3 years, the total interest paid is
sh 200 000 × × 3
= 200 000 × 15 × 3
100
Thus, if the principal is sh P, the rate R and time
T years, then
Simple interest = Principal × Rate × Time
100
i.e. I =
P ×R×Τ
=
PRT
100 100
Note: Since interest is the same each year, then
simple interest is proportional to time.
Example 6.1
Find the simple interest on sh 390 000 at the
rate of 12% p.a. for 5 years.
Solution
Interest = where P = sh 390 000
R = 12
T = 5 years
I = 390 000 × 12 × 5
100
= Ksh234 000
∴ simple interest on sh 390 000 at 12% for
5 years is sh 234 000.
Example 6.2
Find the simple interest on sh 196 800 at 10%
p.a. from 1st March to 15th June, excluding 1st
March.
Solution
Time = (30 + 30 + 31 + 15) days (March, April,
May and part of June)
= 106 days
= year (for an ordinary year)
15
100
–––
15
100
–––
PRT
100
––––
106
365
–––
68
I = PRT
= 196 800 × ×
= 5 715.29
∴ Simple interest is sh 5 715.30
(to the nearest 10 cts)
Example 6.3
In what time will sh 14 400 amount to sh 16 920
at 5% p.a.?
Solution
Interest = sh 16 920 – sh 14 400
= sh 2 520
Interest on sh 14 400 for 1 year at 5%
=
= sh 720
Interest on sh 14 400 for n years at 5% is
sh 720.
Thus,720n = sh 2 520.
∴ Time = n = (since simple interest is
proportional to time)
= 3 years.
Alternative method
I = sh (16 920 – 14 400) = Sh 2 520
Since I =
T = (Expressing T in terms of I, P
and R)
=
∴ Time = 3 years
Example 6.4
At what rate will sh 11 000 amount to sh 13 200
in 5 years?
Solution
I = sh (13 200 – 11 000)
= sh 2 200
Since I =
R = (Exprssing R in terms of I. P
and T)
=
= 4
∴ Rate is 4%.
Exercise 6.1
1. Find simple interest to the nearest whole
number on
(a) sh 12 480 at 13% p.a. for 3 years.
(b) sh 91 280 at 9 % p.a. for 5 years.
(c) sh 41 000 at 7 % p.a. for 50 days.
(d) sh 13 600 at 8 % p.a. from June 5th
to 31st October (inclusive).
2. Find the time in which
(a) sh 11 000 earns an interest of
sh 1 100 at 12% p.a. simple interest.
(b) sh 19 600 earns an interest of
sh 1 120 at 11% p.a. simple interest.
(c) sh 7 000 will amount to Sh 7 875 at
5% p.a. simple interest.
3. Find the rate at which
(a) sh 17 600 will amount to sh 19 300
in 3 years.
(b) sh 64 100 amounts to sh 73 060 in
4 years.
(c) sh 48 000 earns an interest of sh 1 680
in 2 years.
4. What sum of money will amount to
sh 10 080 in 3 years at 4% p.a. simple
interest?
5. At the beginning of year 2006, I invested a
certain amount of money (sh x) at 4% p.a.
At the beginning of year 2007, I invested
another amount, three times the earlier one,
at 5% p.a. If at the end of year 2007 the
two investments had amounted to
sh 423 000, nd the value of x.
2 520
720
–––––
106
365
–––
14 400 × 5 × 1
100
––––––––––––
10
100
–––
PRT
100
––––
1
2
–
100 I
PR
––––
100 × 2 520
14 400 × 5
––––––––––
1
2
–
PRT
100
––––
1
2
–
1
2
–
1
2
–
1
4
–
1
4
–
100 I
PT
––––
100 × 2 200
11 000 × 5
––––––––––
69
Compound interest
We have just seen that under simple interest,
principal plus interest gives the amount. If the
amount at the end of one year is taken to be the
principal for the next year, the interest so earned
is called compound interest.
Example 6.5
Find the compound interest on sh 350 000 at a
rate of 4% p.a. for 3 years.
Solution
Since principal changes every year, we will
calculate interest at the end of each year.
sh
Principal for year 1 = 350 000
+
Interest for year 1 is = 14 000
∴ Principal for year 2 = 364 000 +
Interest for year 2 is = 14 560
∴Principal for year 3 = 378 560 +
Interest for year 3 is = 15 142.40
= 393 702.40
Amount at the end of year 3 = 393 702.40–
Initial principal = 350 000
Interest = 43 702.40
∴ Compound interest on sh 350 000 for
3 years at 4% p.a. = sh 43 702.40.
This method of calculating compound interest is
called step-by-step method.
If the principal varies due to fresh borrowing or
repayment at the end of the year, the amount at
the beginning of each year must be calculated
separately and the interest for this amount
worked out. The method is basically the same
as that of Example 6.5.
Note: Just like in simple interest, the period for
compound interest may be shorter than a year.
Example 6.6
May borrows sh 6 000 000 at 3% p.a. She is to
repay sh 1 000 000 at the end of each year. How

repayment?
Solution sh
May borrows = 6 000 000
  
= = 180 000
Amount owing at the end
of 1st year = 6 180 000 –
End of 1st year, May repays 1 000 000
Amount owing at the beginning
of 2nd year = 5 180 000
Interest for the 2nd year +
= = 155 400
Amount owing at the end
of 2nd year = 5 335 400 –
End of 2nd year, May repays 1 000 000
Amount owing at the beginning
of 3rd year = 4 335 400
Interest for the 3rd year +
= = 130 062
Amount owing at the end
of 3rd year = 4 465 462
–
End of 3rd year May repays 1 000 000
3 465 462
Amount still owing after 3rd repayment
= sh 3 465 462.
Note:
If May in Example 6.6 was investing
sh 1 000 000 at the end of each year, the
procedure would have been the same except that
we would be adding instead of subtracting
sh 1 000 000 at each stage.
Example 6.7
Find the amount on sh 4 160 000 invested for
1 years at 3 % p.a., the compound interest
being payable half yearly.
350 000 × 4
100
––––––––––
364 000 × 4
100
––––––––––
378 560 × 4
100
––––––––––
1
2
–
1
2
–
6 000 000 × 3 × 1
100
––––––––––––––
5 180 000 × 3 × 1
100
––––––––––––––
4 335 400 × 3 × 1
100
––––––––––––––
70
( )
Compound interest formula
Suppose a person invests Sh 250 at 4% p.a.
compound interest.
sh
Principal for the rst year = 250
Interest for the rst year = × 250
∴ Amount at the end of rst year
= ×250 + 250 = × 250
= 1.04 × 250
Principal for the 2nd year = 1.04 × 250
Amount at the end of 2nd year
= of (1.04 × 250)
Principal for the 3rd year = 1.04 × 1.04 × 250
Amount at the end of 3rd year
= ×(1.04
2
× 250)
= 1.04× 1.04
2
× 250
= 1.04
3
× 250
Similarly, amount at the end of year 4
= 1.04
4
× 250
and amount at the end of year n
= 1.04
n
× 250
= 250 × 1.04
n
Now, suppose the principal is sh P and the rate
percent per annum is r,
the interest for the rst year = P ×
and amount at the end of the rst year A
1
= P +
= P
(
1 +
)
…………………… (i)
Interest at the end of second year = A
1
×
∴ Amount at the end of second year A
2
= A
1
+ A
1
= A
1
(
1 +
)
……………………(ii)
=
[
P
(
1 +
)]
·
(
1 +
)
[
remember
A
1
= P
(
1 +
)]
A
2
= P
(
1 +
)
2
…………………… (iii)
1
2
–
1
2
–
3
4
–
7
4
–
1
2
–
1
2
–
1
2
–
Solution
3 % per annum = (3 ÷ 2)% per half year
= % = 1 % per half year
Number of half-years in 1 years = 3. Thus,
there are three interest periods.
sh
Principal for 1st half year = 4 160 000
Interest for 1st half-year +
=
= 72 800
Principal for 2nd half-year = 4 232 800 +
Interest for 2nd half-year = 74 074
Principal for 3rd half-year = 4 306 874 +
Interest for 3rd half-year = 75 370.30
= 4 382 244.30
∴ Amount after 1 years (3 half-years)
= sh 4 382 244.30
Exercise 6.2
1. Calculate the amount, in compound
interest, on the following:
(a) sh 10 000 at 10% p.a. for 2 years.
(b) sh 8 000 at 12% p.a. for 3 years.
(c) sh 20 000 at 5% p.a. for 3 years.
(d) sh 60 000 at 3% p.a. for 2 years.
(e) sh 24 000 at 2 % p.a. for 4 years.
2. A man invests sh 150 000 on 1st January of
each year at 4% p.a. Find the total amount
of his investment at the end of 3 years.
3. A woman borrowed sh 400 000 at 4% p.a.
She repaid sh 100 000 at the end of each
year. How much was she still owing after
the 4th repayment?
4. On 1st January 2001, a lady invested
sh 100 000. In January 2002, she invested
another sh 200 000 and on 1st January
2003, she invested another sh 300 000. If
interest is paid at the rate of 4% p.a.,
nd the total amount in her account on
31st December 2003.
4
100
–––
4 160 000 × 1.75
100
––––––––––––––
4
100
–––
104
100
–––
104
100
–––
104
100
–––
r
100
–––
r
100
–––
Pr
100
–––
r
100
–––
r
100
–––
r
100
–––
r
100
–––
r
100
–––
r
100
–––
r
100
–––
71
Interest at the end of 3rd year = A
2
×
∴ Amount at the end of 3rd year A
3
= A
2
+ A
2
×
= A
2
(
1 +
)
=
[
P
(
1 +
)
2
]
·
(
1 +
)
(A
2
from (iii))
A
3
= P(1 + r)
3
.
Similarly, it amounts A
3
, A
4
, A
5
at the ends of
3rd, 4th and 5th years respectively.
By comparing amounts A
1
, A
2
, …, A
5
, what will
be the amount A
n
at the end of the nth year?
You should have noticed that A
4
= P
(
1 +
)
4
A
5
= P
(
1 +
)
5
Hence, at the end of the n
th
year,
amount A
n
= P
(
1 +
)
n
.
Thus:
If A stands for amount, n for time in years,
and r for rate percent p.a. compound interest,
the formula for amount is
A =
P
(
1 +
)
n
= P(1 + R)
n
, where R =
This is known as the compound interest
formula.
We can use this formula to nd any one of the
quantities A, P, r and n if the other three are given.
Suppose we need to nd the time in which
the principal doubles itself at a rate of 4% p.a.
compound interest. Using the formula, we have
the equation
2P =
P
(
1 +
)
n

⇒ 2 =
(
)
n
(dividing both sides by P.)
⇒ 2 = 1.04
n
Let us use logarithms to solve this equation.
Thus, log 2 = log 1.04
n
= n log 1.04
∴n = = 17 yrs
n = = 17.
∴ The time required for any principal to
double itself at 4% p.a. compound interest
is 17.67 years.
Example 6.8
Find the compound interest in 10 years on
sh 35 000 at a rate of 8% p.a.
Solution
A =
P
(
1 +
)
n
=
35 000
(
1 +
)
10
= 35 000 (1.08)
10
A = sh 75 562
Interest = Amount – Principal
∴I = sh (75 562 – 35 000)
= sh 40 562
Example 6.9
Find the principal, which in 12 years at 5% p.a.
Compond interest amounts to sh 450 000.
Solution
A =
P
(
1 +
)
n
450 000
=
P
(
1 +
)
12
= P(1.05)
12
P =
= sh 250 500
∴ The principal is sh 250 500.
Example 6.10
At what rate will sh 55 220 invested at compound
interest amount to Sh 67 200 in 8 years?
Solution
A =
P
(
1 +
)
n
67 200 = 55 220
(
1 +
)
8
(
1 +
)
8
= (dividing both sides by
by 55 220)
r
100
–––
r
100
–––
r
100
–––
r
100
–––
r
100
–––
r
100
–––
r
100
–––
r
100
–––
104
100
–––
4
100
–––
log 2
log 1.04
–––––––
0.301 0
0.017 0
––––––
r
100
–––
8
100
–––
r
100
–––
r
100
–––
67 200
55 220
––––––
450 000
1.05
12
–––––––
r
100
–––
5
100
–––
r
100
–––
r
100
–––
r
100
–––
72
8 log
(
1 +
)
= log
(
)
(taking
logarithms on both sides)
= 0.08527
log
(
1 +
)
= (dividing both
sides by 8)
= 0.010 66
⇒ 1 + = 1025
∴r = 2.5
So, the rate is 2.5% p.a.
Exercise 6.3
1. Find, to the nearest whole number, the
compound interest on sh 350 000 in 10
years at 3% p.a.
2. How long would it take the principal to
double itself at a rate of 5% compound
interest?
3. A man leaves a legacy of sh 4 000 000
to his daughter. After funeral expenses
amounting to 8% of the legacy were paid,
she invested the remainder for 3 years at
the rate of 4% p.a. compound interest.
What was the value of her investment after
the 3 years?
4. An amount of sh 100 000 invested at
compound interest is worth sh 130 000
after 5 years. Find its value after another
5 years.
5. How long will it take a sum of money
invested at 4% p.a. compound interest to
increase in value by 50%?
6. Find the rate percent p.a. at which
sh 40 000 invested at compound interest
amount to sh 50 000 after 5 years.
7. Which is the better investment and by how
much, at the end of 2 years:
(a) sh 100 000 at 10 % compound
interest p.a, or
67 200
55 220
––––––
0.08527
8
––––––
1
2
–
r
100
–––
r
100
–––
r
100
–––
(b) sh 100 000 at 10% p.a. compound
interest, the interest being paid half
yearly?
8. Konyu wishes to save sh 175 000 in three
years time. What sum must she invest now
at 4% p.a. compound interest?
9. If a sum of money invested at compound
interest doubles itself in 10 years, nd how
much longer it will take to be 4 times its
initial value.
Appreciation and depreciation
When the value of an asset increases with time,
we say it appreciates in value. Land is an
example of an asset that appreciates in value.
If an asset decreases in value with time, we
say that its value depreciates. A car is a good
example of an asset whose value depreciates
with time.
Given the rate of appreciation or depreciation,
the value of such assets may be calculated using
a formula similar to the compound interest
formula.
Example 6.11
A car cost sh 600 000 when new. If its value
depreciated at the rate of 10% p.a., calculate its
value after 3 years.
Solution
When new, the value of the car was sh 600 000

would be of sh 600 000.
At the end of the 2nd year, the value of the car
would be 90% of
(
× 600 000
)
= 600 000 × ×
= 600 000
(
1 –
)
2
and so on.
If V
0
stands for the initial value, V for the
new value, r for the rate and t for the time,
then, V = V
0
(
1 –
)
t
.
90
100
–––
90
100
–––
90
100
–––
90
100
–––
10
100
–––
r
100
–––
73
1
2
–
∴ At the end of the 3rd year, the value of the
car was,
V = 600 000
(
1 –
)
3
= 600 000 × 0.9
3
= 600 000 × 0.729
= 437 400
After 3 years the value of the car will be
sh 437 400.
Example 6.12
A residential plot in town appreciates in value
at the rate of 15% p.a. Its current value is
sh 3 500 000. Calculate its value after 3 years
(answer to the nearest sh 1 000).
Solution
Since the asset appreciates in value with time,
in 3 years time it will be worth more than it is
now worth.
V =
V
0
(
1 +
)
t
= 3 500 000
(
1 +
)
3
= 3 500 000(1.15)
3
= 3 500 000 × 1.520 9
= 5 323 063
i.e., after 3 years, the value of the plot will be
sh. 5 323 000.
Note that questions on growth and decay may
be dealt with in a similar way as for appreciation
and depreciation.
Example 6.13
It has been found that the annual rate of
increase in the population of people infected

in a certain country is 3 %. How long will the
infected population take to double itself? (Give
your answer to the nearest year).
Assumption: The rate of increase is not changing.
Solution

Then, double the infected population is A = 2N.
Rate of increase p.a. is r = 3 %.
Using the compounding formula A = P
(
1 +
)
n
gives 2N = N
(
1 +
)
n
 ⇒2 = (1 + 0.035)
n
i.e. 2 = 1.035
n
Taking logs on both sides gives
log 2 = n log 1.035
∴n =
≈20 years. (by calculator)
! Caution:
This problem involves a purely hypothetical
case, but all the same, HIV/AIDS is a killer
disease and has no cure yet. You must,
therefore, abstain from pre-marital sex
in order to avoid contracting this disease.
When married, stick to your partner.
Exercise 6.4
1. A painting increases in value by 12 % each
year. If its value now is sh 30 000, nd its
value after 10 years.
2. The value of a car depreciates by 8%
each year. A man pays sh 900 000 for his
second hand car. Find its value after 4
years.
3. A diamond ring is priced at sh 20 000. It
appreciates by 10% each year. What will it
be worth 3 years after purchase?
4. Wanyingi has added a new bus to his eet
of buses. From experience, he knows that
in 5 years time the bus will be worth
sh 2 658 000, having depreciated at the rate
of 10% p.a. How much did he pay for the
bus? (Answer to the nearest sh 1 000).
10
100
–––
1
2
–
1
2
–
log 2
log 1.035
––––––––
r
100
–––
15
100
–––
3
100
–––
r
100
–––
1
2
–
74
1
2
–
5. The initial cost of a sewing machine was
sh 45 000. Its value depreciated over a
period of 5 years by 8%, 7%, 6 %, 5% and
4% p.a. respectively. Find the value of the
machine after the 5 years.
At the same time the cost of a new machine
increases by 7 % p.a. If the old machine
is sold and a new one bought, how much
money would have to be added to the
proceeds of the sale of the old one in order
to pay for the new one?
6. Mrs. Kazi has a car which is 3 years old.
It is valued at sh 1 700 000 and its rate of
depreciation is 12% p.a. What was the
value of the car when new?
7. A town residential house was sold at
sh 5 000 000. Thereafter, its value
appreciated at the rate of 2% p.a. for the
rst 3 years and 3% p.a. for the next 6
years. Find the value of the property six
years after purchase (answer to the nearest
sh 1 000).
8. An amount of sh 1 500 000 is invested at
4% p.a. compound interest. In how many
years will it amount to sh 2 500 000?
(Give your answer to the nearest year).
9. The present population of a country is
24.65 million. If the population increases
at 3% p.a., obeying the compound interest
law, after how many years will it become
35 million?
10. Everyday, an iceberg, somewhere in
Iceland, loses everyday 5% of its volume at
the beginning of that day by melting. After
how many days will its volume be halved?
11. A certain corrupt person misappropriated
government money to the tune of sh 250
million. When caught, the government
demanded that he pays back the amount in
full. Failure to pay would attract a penalty
of 7.5% every year for any amount owed
that year. If he paid an amount of
sh 50 000 000 at the end of the 2nd year
and had not paid anything more, how much
does he owe the government now, ve
years down the line?
 Note:
This is a hypothetical case but all the same,
1. Corruption is a moral decadence.
2. Corruption kills development.
3. Corruption brings down the nation.
4. Corruption is real and has to be fought.
LET US ALL SAY NO TO CORRUPTION !!
Hire purchase
When goods are bought and paid for by
instalments, we say they are bought on hire
purchase.
Usually, the buyer pays a deposit, which forms
part of the purchase price. The remainder is
paid for by instalments, spread out over a xed
period of time. The goods remain the property
of the seller until all the instalments have been
settled. However, the purchaser is allowed to
use the goods as he continues to pay for them.
In practice, the hire purchase price turns out to
be higher than the marked cash price. Why do
you suppose this is so?
Example 6.14
Mrs. Juma bought a washing machine on
hire purchase. The marked cash price of the
machine was sh 42 500. She paid a deposit
equivalent to 15% of the cash price, followed by
17 equal monthly instalments of sh 2 225.
(a) How much did Mrs. Juma pay for the
machine?
(b) How much would she have saved, had she
bought the machine on cash basis?
(c) Express your answer to (b) as a percentage
of the cash price.
1
4
–
75
Solution
(a) Amount paid by instalments is
2 225 × 17 = sh 37 825
+
Deposit is × 42,500 = sh 6 375
Total paid = sh 44 200
Mrs. Juma paid sh 44,200 for the machine.
(b) Hire purchase price = sh 44 200
–
Cash price = sh 42 500
   
She would have saved sh 1 700.
(c) Saving = sh 1 700
As a percentage of the cash price it is
× 100 = 4%
The hire purchase price was higher by 4%
of the cash price.
Exercise 6.5
1. Atieno bought a deep-freezer on hire
purchase. She paid a deposit of sh 18 000
followed by 15 equal monthly instalments.
The total hire purchase price was sh 243 000.
How much did she pay per month?
2. Mr. Akili bought new sewing machines on
hire purchase. He paid sh 50 000 followed
by ve instalments of sh 6 500 each. If the
seller made a prot of 21% on the machine,
nd the cost price of the machine. Give
your answer to the nearest sh 10.
3. A garden set of easy chairs and a matching
table is advertised at sh 30 000. A discount
of 15% is oered for cash purchase.
(a) Find the cash price for the items.
(b) If the furniture is bought on hire
purchase, an extra 15% of the cash
price is charged. What is the hire
purchase price?
4. The cash price of a carpet 8 m by 6 m is
sh 26 400. Anne purchased the carpet
by making 12 monthly instalments of
sh 2 500 each. Beth bought a similar
carpet by paying half the cash price and the
remaining amount on one year in two equal
instalments that are equally spaced.
(a) Calculate the cash price per square
metre if the cash price is paid.
(b) How much did Anne pay above the
cash price?
What was the rate percent per month?
(c) Beth paid an interest at the rate of
1.5% per month simple interest. How
much did she pay for the carpet?
5. A house was valued at sh 3 000 000 after
an appreciation of 12% in 1 year.
(a) Calculate the value of the house before
appreciation.
(b) Mrs. Koech paid a deposit of 12%
of the appreciated value of the
house, followed by 36 equal monthly
instalments of sh 84 000 each. How
much did he pay for the house in total?
(c) If Mrs. Koech had paid for the house in
cash, how much would he have saved?
(d) Mr. Kirimi bought a similar house at
sh 3 000 000 by taking a loan, payable
in full at the end of two years, at 14%
p.a. compound interest. How much
more did Kirimi pay than Koech?
Give your answers to the nearest thousand.
Taxation
It is the duty of the government of any country
to provide its citizens with such services as
defence, health care, education, roads and so
on. In order to do this, the government needs to
raise funds. It does this through such means as
taxation.
Tax is a charge levied on individuals and
companies for purposes of raising money for
the provision of government services. The
government of Kenya levies two types of taxes,
namely direct and indirect taxes.
15
100
–––
1 700
42 500
––––––
76
Direct taxes include income tax (levied on
the incomes of individual persons), corporate
tax (levied on the incomes or prots made by
companies).
Indirect taxes include Value added tax - V.A.T.
(levied on consumer goods and services), excise
duty (levied on locally manufactured goods),
customs duty (levied on imported goods), and
export duty (levied on exported goods).
The government also raises funds by charging
various fees such as business and trade licences,
road licences, airport tax, parking fee, etc.
For the purposes of this course, we shall only
look at income tax.
Income tax
Every person who earns an income is duty and
legally bound to pay part of it to the government as
income tax. The amount of tax that a person pays
depends on the amount of his income, irrespective
of its source.
Income may be earned in the form of salary,
wages, commission, interest on bonds, dividends
from shares, gain or prot from business activities,
and so on.
The rate at which income tax is charged are
revised, by the government, from time to time
and any new rates take eect on the 1st January
of the following year.
Table 6.1 shows the rates of income tax which
were in eect from 1st January 2017. This table
will be used throughout this book whenever
working out taxation problems.
Monthly Rate of tax in
taxable income % in each year
sh sh
0 – 11 180 10%
11 181 – 21 714 15%
21 715 – 32 248 20%
32 248 – 42 781 25%
42 781 and above 30%
Table 6.1
Taxable income is determined as gross income
(i.e. total that a person earns) less deduction
exempted from tax, called allowable deductions.
i.e.,
Taxable income
= Gross income – Allowable deductions
Such deductions that are exempted from tax
include:
1. Pension / provident fund contribution
2. National social security fund (NSSF)
The deductible contribution is the actual
contribution made to the NSSF
3. Contribution to a home ownership savings
plan.
4. Owner occupied interest
Interest paid on amount borrowed from a
specied nancial institution is deductible
provided the amount was borrowed to
nance either
(a) the purchase of premises, or
(b) improvement of premises, which is
occupied for residential purposes.
Each individual tax payer is entitled to a
personal relief (an amount deducted from
the tax payable to reduce the tax burden on an
individual) of sh 1 280 per month.
An insurance policy holder is entitled to a
further relief, known as insurance relief,
amounting to 15% of the premium paid.
Example 6.15
Mutiso earns a salary of sh 13 000 per month.
(a) How much tax should he pay per month?
(b) If he makes a monthly contribution of
sh 500 to a pension scheme, how much
should he pay in tax?
77
Solution
  
= 10% of sh 10 180 = sh 1 118.00
Tax on remaining sh 1 820 +
= 15% of sh 1 820 = sh 273.00
Total tax payable = sh 1 391.80
Tax due = tax payable – monthly relief
= sh (1 391 – 1 280)
= sh 111
(b) Taxable income
= Income – Allowable deduction
= sh (13 000 – 1 500)
= sh 11 500
    
= 10% of sh 11 118 = sh 1 118.00
Tax on remaining sh 1 320 +
= 15% of sh 320 = sh 198.00
Total tax payable
= sh
1 316.80
Tax due = sh (1 316.80 – 1 280)
= sh 36
Exercise 6.6
1. Find the taxable incomes in each of the
following cases.
(a) A company employee has a monthly
salary of sh 14 800 and is deducted
sh 200 per month towards a pension fund.
(b) An accountant is paid a monthly salary
of sh 24 000 and she contributes sh
200 towards her pension scheme
per month. She pays sh 2 100 life
insurance premium per month.
2. Find the tax due by the following.
(a) A man with an income of sh 25 000
per month
(b) A person who earns sh 36 500 per month
3. The monthly income of a business lady is
sh 336 000. If she contributes sh 1 400 per
month to a Provident Fund and sh 2 500
per month for life insurance, how much tax
should she pay per month?
4. A man pays sh 2 404 as tax after relief. If
he does not have any allowable deductions,
what is his monthly income, in shillings?
Pay as you earn (PAYE)
In Kenya, the law requires that every employer
deducts income tax from the pay of his employees
every month and remit the money to the Income
Tax Department. This system, of collecting tax
at source, is called Pay as you earn (PAYE).
Deductions under PAYE are included in an
individual’s tax assessment, and are therefore
deducted from the tax calculated on his total
income for the year.
Example 6.16
In February 2017 an employees taxable income
in the last bracket was Ksh 12,500. The employee
was entitled to Ksh 1280 personal relief.
(a) Calculate:
(i) The total taxable income for the
employee in the month of february
(ii) The tax payable to the employee in
February
(b) The employees income included a house
allowance of Ksh 11 500 per month. The
employee also paid Sh. 7500 on momthly

pay for the employee.
Solution
(a) Taxable income (use table 6.1 given on
page 77)
Maximum value taxed upto 2nd last bracket
= Ksh 42 781
Value for last bracket = Ksh 12 500
taxable income = Ks 55 281
78
(b)
   
= 11 180 × 10% = sh 1 180.00
Tax on the second +
= 10 534 × 15% = sh 1580.10
Tax on the third +
= 10 534 × 20% = sh 2 106.80
Tax on the fourth +
= 10 533 × 25% = sh 2 633.25
 
= 12 500 × 30% = sh 3 750.00
gross tax = sh 11 188.15
relief = sh 1 280.00
Net tax Sh 9 908.15.
Example 6.17
Mr. Nguru is a teacher employed by private
school for a monthly salary of sh 10 370. He
also gets a house allowance of sh 10 000 and
a medical allowance of sh 990 per month.
Calculate how much he is deducted per month
in P.A.Y.E.
Solution
Monthly salary = sh 10 370
+
Monthly house allowance = sh 10 000
+
Monthly medical allowance = sh 990
∴ Taxable income per month = sh 21 360
   
= 10% of 11 180 = sh 1 118.00
Tax on next sh 10 180 +
= 15% of 10 180 = sh 1 527.00
Tax payable per month = sh 2 645.
–
  
∴ Monthly PAYE deduction = sh 1 365
Notice that any cash benets are added to the
basic salary to give the taxable income.
Example 6.18
Ben is employed as a manager of a Cooperative
Society for a basic salary of sh 44 000 per
month and a house allowance of sh 35 000. In
addition, he is provided with other allowances
totalling to Sh 10 400
Calculate his monthly PAYE deduction.
Solution
Monthly basic salary = sh 44 000
+
Add house allowance = sh 35 000
other allowances = sh 10 400
∴ Taxable pay per month = sh 89 400

= 10% of 11 180 = sh 1 118.00
Tax on next sh 10 534 +
= 15% of 10 534 = sh 1 580.10
Tax on next sh 10 534
= 20% of 10 534 = sh 2 106.80
Tax on next sh 10 533
= 25% of 10 533 = sh 2 633.25
Tax on remaining sh 46 619
= 30% of 46 619 = sh 13 985.70
Tax payable per month = sh 21 432.85
 
∴ Monthly PAYE deduction = sh 20 143.85
Exercise 6.7
In this exercise, use tax rates in Table 6.1 and a
monthly relief of sh 1 180.
1. A man’s monthly pay is sh 29 400.
Calculate his monthly PAYE deduction.
2. Anyang’ is employed by a company for a
monthly basic salary of sh 17 500. He is
given a house allowance of sh 14 000 and a
monthly travel allowance of sh 1 180. Find
his monthly PAYE deduction.
79
3. Mr. Kengo works in a local company
where he earns a salary of sh 90 000 per
month. He also runs a business from which
he earns sh 48 200 per month. Calculate his
(a) PAYE deduction.
(b) income payment per year.
4. The headmaster of a certain private school
is paid a monthly basic salary of sh 64 560.
He lives in a school house for which he
pays a monthly rent of sh 960.
5. Mrs. Chacha is a civil servant. She lives
in a government house for which she pays
a rent of sh 7 250 per month. She enjoys
the benet of a 1 680 cc saloon car, a cook
and a night watchman. If she earns an
annual basic salary of sh 204 000 and gets
a monthly house allowance of sh 20 000,
calculate her monthly PAYE deduction.
6. Miss Nzomo earns a basic salary of
sh 13 200 and a house allowance of
sh 9 000 per month.
(a) Calculate
(i) the tax payable without relief,
(ii) the tax paid after relief
(b) Suppose that apart from income tax,
the following monthly deductions are
also made: A service charge of sh 100,
sh 280 for NSSF, 2% of basic salary as
pension contribution. Calculate
(i) the total monthly deduction made
from Miss Nzomo’s income,
(ii) Miss Nzomo’s net employment
income per month.
7. Mary nds that her employer deducts, from
her monthly salary, sh 2 347 under PAYE
Calculate her monthly salary.
8. Miss Rotich earns a basic salary of
sh 45 000. In addition she gets a house
allowance of sh 25 000 p.m. At the end of
every month she is deducted sh 200
towards NSSF.
(a) How much income tax does she pay
per month?
(b) What is her net salary per month?
9. Mrs. Mwangi, a civil servant, earns sh
38 350 p.m. and lives in a government
house for which she pays sh 14 000 p.m.
She gets a house allowance of sh 40 000,
a medical allowance of sh 4 250 and a
transport allowance of sh 842 every month.
Calculate her PAYE deduction.
80
7
CIRCLES — CHORDS AND TANGENTS
Fig. 7.1
A
B
Minor arc AB
Major arc AB
Arc of circle
In Forms 1 and 2, we learnt about the parts of a
circle. We learnt that:
(1) The circumference is the perimeter
of the circle.
(2) An arc of a circle is a
part of the circumference
of the circle. We saw
that an arc whose length
is longer than half the
circumference is called
a major arc and an arc
whose length is less
than half the circumference is called a
minor arc (Fig. 7.1).
Length of an arc
Fig. 7.2 shows the points of a compass with
N, NE, E, SE, … replaced by A, B, C, D, …
respectively.
What is the size of ∠AOB?
Which arc subtends ∠AOB at the centre of the
circle?
What fraction of 360° is ∠AOB?
What fraction of the circumference is the arc
subtending ∠AOB?
Copy and complete Table 7.1.
G
C
H
B
A
D
F
E
O
Fig. 7.2
Table 7.1
Angle as Arc Length of
Angle
fraction
subtending
arc
of 360°
angle
∠AOB = 45° = AB × 2πr
∠AOC = 90° = ABC × 2πr
∠AOD = 135° = ABCD × 2πr
∠AOE =
Obtuse ∠AOF =
Reex ∠AOG =
Reex ∠AOH =
1
8
–
3
8
–
1
4
–
1
8
–
3
8
–
1
4
–
Fig. 7.3
P
Q
O
θ
A
55°
Fig. 7.4
B′
B
You should note that the length of the arc is
proportional to the size of the angle it subtends
at the centre of the circle, i.e. the
bigger the angle the longer
the arc.
From Table 7.1, derive an
expression for the length
of any arc PQ that subtends
an angle θ at the centre of
a circle (Fig. 7.3)
In general we have the following results;
The length of any arc PQ subtending an
angle
θ
at the centre of a circle of radius r
is × 2πr.
Example 7.1
In Fig. 7.4, AB represents the
position of a door of width
0.8 m when shut. The door
is opened to position AB′
sweeping through an angle
of 55°.
45°
360°
––––
135°
360°
––––
90°
360°
––––
θ
360
–––
81
4 cm
4 cm
100°
A
B
C
Fig. 7.5
7. Find the perimeters of the shapes in Fig. 7.7?
(a)
Fig. 7.7 (b)
Properties of chords
In Form 2, we learnt that a chord is a line
segment joining any two points on the
circumference of a circle. A diameter is a
chord that passes through the centre of the
circle. In this section we shall learn other
properties of chords.
ACTIVITY 7.1
1. Draw a circle of any radius r cm, centre O.
Draw a chord AB (not a diameter). From
O draw a line perpendicular to AB, cutting
AB at N. Measure AN and NB. What do
you notice?
2. Draw a circle of any radius r cm, centre
O. Draw a chord CD (not a diameter).
Construct a perpendicular bisector of CD.
What do you notice?
You should have noticed that:
1. A perpendicular drawn from the centre
of a circle to a chord bisects the chord.
2. A perpendicular bisector of a chord passes
through the centre of the circle.
Example 7.2
Fig. 7. 8 shows a circle,
centre O. M is the midpoint
of the chord AB. Show that
OM is perpendicular to AB.
θ
360°
––––
55°
360°
––––
A
B
M
O
Fig. 7.8

– – – – – – –
6 cm
6 cm
35°
– – – – –
A
B
C
8 cm
Fig. 7.6
N
What is the length of arc BB′? (Take π = 3.142).
Give your answer to 2 d.p.
Solution
× 2πr
θ = 55° r = 0.8 m
∴′
= × 2 × 3.142 × 0.8 m
= 0.768 0 m
= 0.77 m.
Exercise 7.1
In this exercise, take π = 3.142 and state your
answers to 3 s.f.
1. An arc of a circle subtends an angle of 50°
at the centre of a circle. If the circle is of
radius 4 cm, what is the length of the arc?
2. An arc of length 6.4 cm subtends an angle
of 110° at the centre of a circle. What is
the radius of the circle?
3. AB is a chord of a circle of radius 7.7 cm.
If AB subtends an angle of 80°
at the
circumference of the circle, what is the
length of the major arc AB?
4. In Fig. 7.5, ABC is an
equilateral triangle with
vertices on a circle of
radius 4.5 cm. What is
the length of arc ABC?
5. A circle of radius 12 cm
has a chord of length 18 cm. What is the
length of the minor arc?
6. In Fig. 7.6, AB is a chord of a circle.
AN = NB = 18 cm.
What is the radius
of the circle?
What is the
length of arc ACB?
82
3 cm
Fig. 7.11
– – – –
M
O
C
A
B
Solution
Join OA and OB (Fig. 7.9)
   
OA = OB (radii of the
circle)
AM = MB (given that M
is midpoint of AB)
OM (common to both
triangles)
∴ SSS)
∴ ∠OMA = ∠OMB.
But these are adjacent angles on a straight line
∴ ∠OMA is a right angle, i.e. OM is
perpendicular to AB.
Length of a chord
Using the properties noted above, we can
calculate the length of a chord.
Example 7.3
A circle has a radius of 6 cm. A chord of the
circle is 3.9 cm from the centre of the circle.
Find the length of the chord.
Solution
In Fig. 7.10, let PQ be the
chord whose length is
required. O is the centre of
the circle, radius 6 cm.
OM = 3.9 cm (the
perpendicular distance
from O to the chord)

2
= OM
2
+ PM
2
(Pythagoras
theorem)
6
2
= 3.9
2
+ PM
2
⇒PM
2
= 6
2
– 3.9
2
= 36 – 15.21 = 20.79
∴PM = 4.56 cm
∴PQ = 2 × PM = 2 × 4.56 cm
= 9.12 cm.
O
M
P
Q
6 cm
3.9 cm
Fig. 7.10
Exercise 7.2
1. A chord of a circle of radius 13 cm is at a
distance of 5 cm from the centre. Find the
length of the chord.
2. A chord of a circle of radius 9 cm is at a
distance of 4 cm from the centre. What is
the length of the chord?
3. A chord of a circle is 10 cm long and is
12 cm from the centre of the circle. What
is the radius of the circle?
4. A chord of a circle of radius 6 cm is 10 cm
long. How far is the chord from the centre
of the circle?
5. Fig. 7.11 shows a circle,
radius 6 cm, centre
O. Given that M is the
midpoint of AB and is
3 cm from O, what is the
length of AC?
6. A circle has a chord whose length is 9 cm.
The chord is 4 cm from the centre of the
circle. The same circle has a chord which is
3.5 cm from the centre, what is its length?
7. A chord of a circle is 12 cm long and 5 cm
from the centre. What is the length of a
chord which is 3 cm from the centre?
8. A chord of a circle of radius 5.5 cm
subtends an angle of 42° at the centre.
Find the dierence in length between the
chord and the minor arc.
9. P, Q and R are points on the circumference
of a circle. If PQ = 12cm, PR = 12cm and
QR = 8cm, what is the radius of the circle?
Parallel chords and equal chords
ACTIVITY 7.2
1. Draw a circle, centre O, radius 4 cm.
Draw chords PQ and RS such that PQ//RS.
A
B
M
O
Fig. 7.9

– – – – – – –
– – – – – – –
83
1
2
–
Construct a perpendi-
cular bisector of PQ
and let it cut RS at
point T. (Fig. 7.12).
Measure (i) ∠OTR
(ii) RT (iii) TS.
What do you notice?
What can you say about line OT?
2. Draw a circle, centre
O, radius 4 cm. Draw
two chords AB and
CD such that
AB = CD. Construct
perpendicular bisectors
OH and OK of AB and
CD respectively (Fig. 7.13).
Measure (i) OH and OK (ii) ∠AOB and
∠COD. What do you notice?
You should have observed that:
(i) In Fig. 7.12, ∠OTR = 90°, RT = TS, hence
OT is a perpendicular bisector of RS.
(ii) In Fig. 7.13, OH = OK, ∠AOB = ∠COD.
In general:
1. If two chords of a circle are parallel, then
the perpendicular bisector of one is also
perpendicular bisector of the other.
2. The midpoints of parallel chords of a
circle lie on a diameter.
3. If two chords of a circle are equal, then
they are equidistant from the centre.
4. All parts are equidistant
5. If two chords of a circle are equal, then
the angles they subtend at the centre are
equal.
6. If two angles at the centre of a circle are
equal, then they are subtended by equal
chords.
10 cm
P
6 cm
4 cm
S
Q
R
N
M
O
Fig. 7.14
1
2
–
– – – – – – – – –
•
Q
T
O
P
R
S
Fig. 7.12
– – – – – – – – –
•
B
K
O
A
C
D
Fig. 7.13
– – – – – – – – –
– – – – – – – – –
– – – – – – – – –
H
•
X
S
Y
Z
O
T
W
Fig. 7.16
– – – – – – – –
– – – – – – – –
Example 7.4
Two parallel chords of a circle are of lengths
8 cm and 12 cm respectively and are 10 cm
apart. What is the diameter of the circle?
Solution
Consider Fig. 7.14.
     
PO
2
= PM
2
+ MO
2
(Pythagoras theorem).
     
RO
2
= RN
2
+ NO
2
(Pythagoras theorem).
PO = RO = r (radius)
  ⇒MO = 10 – x
Then r
2
= 6
2
+ (10 – x)
2
and r
2
= 4
2
+ x
2
Hence 16 + x
2
= 36 + (100 – 20x + x
2
)
⇒20x = 120
i.e. x = 6
 ∴ r
2
= 4
2
+ x
2
= 4
2
+ 6
2
= 52
⇒ r = 7.211
∴ Diameter = 14.42 cm
Example 7.5
Fig. 7.15 shows a circle
centre O, with two
equal chords WX and
YZ. Given that
∠OSY = ∠OTW = 90°,
show that OS = OT.
Solution
Join OW and OY (Fig. 7.16).
Since the perpendicular
from the centre of a
circle to a chord bisects
the chord,
WT = WX and YS = YZ
But WX = YZ (given)
∴ WT = YS.
•
X
S
Y
Z
O
T
W
Fig. 7.15
84
6. A, B and C are points on the circumference
of a circle. If AB = BC = CA = 6 cm,
(a) what angle is subtended at the centre
of the circle by AB?
(b) what is the radius of the circle?
7. Points P, Q and R are points on the
circumference of a circle. If PQ = PR = 13
cm and QR = 10 cm, what is the radius of
the circle?
8. Two parallel chords of a circle are of
lengths 12 cm and 16 cm. If the radius
of the circle is 10 cm, what are the two
possible perpendiculardistances between
the chords?
9. KL is a chord of length 15 cm and is 10
cm from the centre of a circle. What is the
length of a chord which is 12 cm from the
centre?
Intersecting chords
Chords that intersect inside a circle
ACTIVITY 7.3
Draw a circle of any reasonable radius.
Draw two chords PQ and
RS such that they intersect
at a point X (Fig. 7.18).
Accurately, measure
PX, XQ, RX and XS.
Work out the product
of PX
.
XQ and RX
.
XS.
What do you notice?
You should have noticed that PX
.
XQ = RX
.
XS.
We can show this result
analytically using Fig. 7.19.
In ∆s PRX and SQX:
∠PRX = ∠SQX
(∠s in same segment)
∠RPX = ∠QSX
(∠s in same segment)
N
P
K
L
Q
M
Fig. 7.17

OW = OY (radii)
WT = YS (Fig. 7.16)
∠OTW = ∠OSY (rt ∠s, given)
∴ ≡
∴ OS = OT.
This example shows that
Equal chords of a circle are equidistant from
the centre of a circle.
Conversely, chords which are equidistant
from the centre of a circle are equal.
Note that equal chords of equal circles are also
equidistant from the centres of the respective
circles.
Exercise 7.3
1. Two parallel chords of a circle are of
lengths 3 cm and 5 cm respectively. What
is the radius of the circle if the chords are
(a) 1 cm apart (b) 8 cm apart.
(State your answers to 2 d.p.)
2. Two parallel chords of a circle are each
10 cm long. What is the perpendicular
distance between the chords if the circle
has a radius of 13 cm?
3. A circle of radius 2.5 cm has two parallel
chords of lengths 3 cm and 4 cm. What
is the distance between the chords? (Two
possible answers).
4. PQ and RS are equal chords of a circle
centre O. Show that ∠POQ = ∠ROS.
5. In Fig. 7.17 chords KL
and MN are equal,
and P and Q are
their mid-points
respectively. Show
that ∠KPQ = ∠MQP.
X
S
P
R
Q
Fig. 7.18
X
S
P
R
Q
Fig. 7.19
– – – – – – – – –
– – – – – –
85
PX
XS
–––
RX
XQ
–––
∠PXR = ∠SXQ (vertically opposite ∠s)
∴ ∆PRX is similar to ∆SQX.
It follows that = (ratios of lengths of
corresponding sides). Thus
PX
.
XQ = RX
.
XS
Example 7.6
PQ and RS are two chords of a circle
which intersect at T. What is
the length of RS if PT = 4 cm,
TQ = 7cm and TS = 5cm?
Solution
In Fig.7.20:
PT
.
TQ = RT
.
TS
4 × 7 = RT × 5
⇒RT = = 5.6 cm
Since RS = RT + TS
Then RS = (5.6 + 5)cm
= 10.6cm.
Distance from the centre of a circle to a
point of intersection of chords.
In Fig. 7.21, PQ and RS
are chords which intersect
at X. O is the centre of
the circle and ON is the
perpendicular from the
centre O to PQ.
Hence PN = NQ, since
the perpendicular from the
centre of a circle to a chord bisects the chord.
PX
.
XQ = (PN + NX)
.
(NQ – NX)
= (PN + NX)
.
(PN – NX) (since
PN = NQ)
= PN
2
– NX
2
In ∆PNO: PN
2
= PO
2
– NO
2
(Pythagoras’
theorem)
∴ PX
.
XQ = PO
2
– NO
2
– NX
2
= PO
2
– (NO
2
+ NX
2
)
But in ∆ONX, NO
2
+ NX
2
= OX
2
(Pythagoras
theorem)
∴ PX
.
XQ = PO
2
– OX
2
Since PO = r (radius of the circle), we can write:
PX
.
XQ = r
2
– OX
2
⇒ OX = √r
2
– PX
.
XQ
Similarly RX
.
XS = r
2
– OX
2
OX
= √r
2
– RX
.
XS
Example 7.7
A circle, centre O, radius 6 cm has chords AB
and CD which intersect at X. Find OX if
AX = 3 cm and XB = 2 cm.
Solution
From Fig.7.22,
AX.XB = r
2
– OX
2
.
⇒ OX
2
= r
2
– AX
.
XB
= 6
2
– 3 × 2
= 30
∴OX = √30
= 5.477 cm
  
Chords that intersect outside the circle
In Fig. 7.23, the two
chords AB and CD of a
circle are extended to
meet at X. Lines AD
and CB have been
drawn to complete
triangles ADX and CBX.
We want to show that
AX
.
XB = CX
.
DX as follows.
In ∆s ADX and CBX,
∠DAX = ∠BCX (angles in same segment)
∠AXD = ∠CXB (same angle)
∠ABC = ∠ADC (angles in same segment)
⇒∠ADX = ∠CBX (supplementary angles of
equal angles)
∴ ∆ADX is similar to ∆CBX.
Q
P
R
T
S
4 cm
7 cm
5 cm
Fig. 7.20
4 × 7
5
––––
X
D
A
C
B
Fig. 7.22
O
•
– – – – – – – – –
– – – – – – –
6 cm
2 cm
3 cm
– – – – – – – – – – – – –
– – – – – – – – – – – – – –
Fig. 7.23
C
D
B
X
A
X
S
P
R
Q
Fig. 7.21
O
N
– – – – – – – – –
– – – – – –
•
86
AX
CX
–––
DX
BX
–––
It follows that = (corresponding sides)
Thus
 AX
.
BX = CX · DX
Example 7.8
Chords AB and CD of
a circle meet at X
(Fig. 7.24).
If AB = 8 cm, BX = 5 cm
and DX = 6 cm, what is the
length of chord CD?
Solution
AX · BX = CX
.
DX
AX = AB + BX = (8 + 5) cm
= 13 cm
∴ 13 × 5 = CX × 6
∴ CX = cm
∴CX = 10.83 cm
CD = CX – DX
= (10.83 – 6) cm
= 4.83 cm
In Fig. 7.21, in which the chords PQ and RS
intersect at X inside the circle radius r, centre
O, we saw that
PX · XQ = RX
.
XS = r
2
– OX
2
.
⇒OX = √r
2
– PX
.
XQ
= √r
2
– RX
.
XS
Do you think this relationship holds if the
chords intersect outside the circle?
Let us nd out.
Consider a circle centre
O, radius r.
Let chords PQ, RS
intersect at X (Fig. 7.25).
Let ON be the
perpendicular from O
to PQ. Join OP and OX.
PN = NQ (perpendicular from centre bisects
chord)
PX
.
QX = (PN + NX)(NX – NQ)
= (PN + NX)(NX – PN)
= NX
2
– PN
2
In ∆ONX, NX
2
= OX
2
– ON
2
(Pythagoras
theorem)
∴ PX
.
QX = OX
2
– ON
2
– PN
2
= OX
2
– (ON
2
+ PN
2
)
In ∆ONP, OP
2
= ON
2
+ PN
2
(Pythagoras
theorem)
∴ PX
.
QX = OX
2
– OP
2
= OX
2
– r
2
⇒OX = √r
2
+ PX · QX
It can similarly be shown that
RX
.
SX = OX
2
– r
2
i.e. OX = √r
2
+ RX
.
XS
Exercise 7.4
For Questions 1–8 refer to
Fig. 7.26 in which O is the
centre of the circle and chords
KL and MN intersect at P
inside the circle.
1. Find MP if KP = 8 cm,
PL = 9 cm and PN = 12 cm.
2. Find KL if MP = 2 cm, PN = 18 cm and
PL = 4 cm.
3. Find KP if MP = 3 cm, PN = 12 cm and
KL = 13 cm.
4. Find the radius of the circle, to 2 d.p., if
MP = 3 cm, PN = 8 cm and OP = 2 cm.
5. If the diameter of the circle is 16 cm,
KP = 10 cm and PL = 4 cm, what is OP?
6. If the diameter of the circle is 24 cm,
OP = 8.5 cm and PL = 5.5 cm, what is KL?
7. If the diameter of the circle is 10 cm and
OP = 4 cm, and if P is 2 cm from the
midpoint of MN, what is the length of MN?
– – – – – – – – – – – –
– – – – – – –
Fig. 7.25
R
S
Q
X
P
– – – –
N
O
13 × 5
6
–––––
Fig. 7.24
C
D
B
X
A
Fig. 7.26
L
P
O
N
K
M
•
87
8. If the radius of the circle is 18 cm,
KL = 24 cm and P is 4 cm from the
midpoint of KL, what is OP?
For Questions 9–13, refer to
Fig. 7.27, in which O is the
centre of the circle and
chords PQ and RS intersect
at X outside the circle.
9. If PQ = 8.5 cm,
RX = 13.5 cm and
QX = 2.5 cm, what is the
length of SX?
10. If PX = 15.8 cm, QX = 2.4 cm and
SX = 3.6 cm, what is the length of RS?
11. What is the length of PX if PQ = 9.5 cm,
RS = 8.5 cm and SX = 5.6 cm?
12. What is the radius of the circle if
PQ = 7.6 cm, QX = 3.6 cm and
OX = 6.4 cm?
13. What is the length of OX if RS = 10.9 cm,
SX = 4.4 cm and the radius of the circle is
13.5 cm?
14. In Fig. 7.28, AB is a
diameter of the circle and
PQ is perpendicular to AB.
If AQ = 5 cm and
QB = 8.9 cm, what is the
length of PQ?
15. A circle, centre O, has a radius of 4.5 cm.
C is a point outside the circle such that
OC = 5.5 cm. A chord AB of length 4 cm
passes through C when produced. Find the
length of BC.
Tangent to a circle
Consider Fig. 7.29.
In Fig. 7.29(a), lines KL and PQ have only one
point common with the circle. A line with at
Fig. 7.29
least one point common with the circle is said
to meet the circle at that point.
In Fig. 7.29(b), line RS has two distinct points
common with the circle. Such a line is said to
meet and cut the circle at the two points.
In Fig. 7.29(c), the line TV has one point of
contact with the circle. Line TV is said to meet
and touch the circle at that point of contact.
Point U is called the point of contact.
1. A line which cuts a circle at two distinct
points (as in Fig. 7.29(b)) is called a secant
of the circle.
2. A line which has one, and only one point
in contact with a circle (as in Fig. 7.29(c)),
however far it is produced either way, is
called a tangent of the circle.
Consider Fig. 7.30.
(a) (b)
(c)
(d)
Fig. 7.30
Fig. 7.28
P
Q
B
A
S
R
(b)
L
P
Q
K
(a)
(c)
U
V
T
Fig. 7.27
R
P
O
X
Q
S
•
S
R
P
Q
O
•
S
1
R
1
P
1
Q
1
O
•
S
2
R
2
P
2
Q
2
O
•
S
3
R
3
P
3
Q
3
O
•
88
Fig. 7.30(a) to (d) shows what happens when
the secant PQRS moves away from the centre
of the circle. As the secant moves further away,
the points Q and R get closer to each other and
the chord QR gets shorter each time. Eventually
Q and R coincide at one point [Fig. 7.30(d)]. On
the other hand, angles OQP and ORS become
smaller and smaller. Eventually when Q and R
coincide, angles OQR and ORS each becomes
90°.
Note that in ∆OQR, since OQ = OR, 
∠OQR = ∠ORQ.
It follows that ∠PQO = ∠SRO.
Therefore, when Q and R coincide [Fig.
7.30(d)], ∠PQO =∠SRO = 90°.
Hence the radius is perpendicular to the tangent
PS.
Note that:
1. A tangent to a circle is perpendicular to
the radius drawn through the point of
contact.
2. At any point on a circle, one, and only
one, tangent can be drawn to the circle.
3. The perpendicular to a tangent at its point
of contact passes through the centre of the
circle.
Example 7.9
In Fig. 7.31, AC is a tangent to the circle, centre
O. If ∠ABD = 120°,
(a) what is the size
of ∠ODB?
(b) what is the
length of OA
if OB = 6 cm
and AB = 7.5 cm?
Solution
(a) Since AC is a tangent
and OB is a radius,
AC is perpendicular to OB.
∴∠ABO = 90°.
∠OBD = ∠ABD – ∠ABO
= 120° – 90°
= 30°
Since OB = OD (radii)
∠ODB = ∠OBD (base angles of an


∴ ∠ ODB = 30°
(b) OA
2
= AB
2
+ BO
2

Pythagoras theorem)
= 7.5
2
+ 6
2
= 92.25
∴
Exercise 7.5
1. Fig. 7.32 shows a circle,
centre O. PR is a tangent
to the circle, at P and PQ
is a chord. Calculate
(a) ∠RPQ given that  
∠POQ = 85°.
(b) ∠RPQ given that
∠PQO = 26°.
(c) ∠POQ given that ∠RPQ = 54°.
(d) ∠POQ given that ∠QPO = 17°.
2. In Fig. 7.33, ABC is a
tangent and BE is a
diameter to the circle.
Calculate
(a) ∠EBD if 
∠CBD = 33°.
(b) ∠BED if 
∠ABD =150°.
(c) ∠DBC if ∠DEB = 65°.
(d) ∠ABD if ∠BED = 38°.
3. PQ is a diameter of a circle. Let R be a
point on the circumference of the circle.
Show that PR is a tangent to a circle,
centre O, radius OR.
4. AB is a chord of a circle, centre O. If BC
is a perpendicular to the tangent at A, show
that ∠OBA = ∠ABC.
D
A
C
B
– – – – – – – – – – – – – –
Fig. 7.31
O
D
E
O
Fig. 7.33
•
A
B
C
Q
R
P
O
Fig. 7.32
89
P
O
Fig. 7.35
A
B
T
5. Two circles have the same centre O, but
dierent radii. PQ is a chord of the bigger
circle but touches the smaller circle at A.
Show that PA = AQ.
6. Two circles have the same centre O and
radii of 13 cm and 10 cm. AB is a chord of
the bigger circle, but a tangent to the small
circle. What is the length of AB?
7. A tangent is drawn from a point 17 cm
away from the centre of a circle of radius
8 cm. What is the length of the tangent?
Construction of tangents to a circle
Constructing a tangent at any given point
on the circle
To construct a tangent to a circle, we use the
fact that a tangent is perpendicular to the circle
at the point of contact.
Procedure
1. Draw a circle, centre
O, using any radius.
2. Draw a line OB
through any point A
on the circumference,
with B outside the circle.
3. At A, construct a line
PQ perpendicular to OB.
The line PQ (Fig. 7.34) is a tangent to the
circle at A.
Constructing tangents to a circle from
a common point
Procedure
1. Draw a circle of any radius, centre O.
2. Mark a point T outside the circle.
3. Join OT. Construct the perpendicular
bisector of TO to meet TO at P.
4. With centre P, radius PO, construct arcs to
cut the circle at A and B.
5. Join AT and BT. These are the required
tangents from the external point T (Fig. 7.35).
B
O
Fig. 7.34
A
Q
P
ACTIVITY 7.4
1. Draw a circle of any radius, centre O.
2. Choose any points A and B on the circle.
Construct tangents at A and B
3. Produce the tangents till they meet at a
point T.
4. Join OA, OB and OT.
5. Measure (a) AT, BT.
(b) ∠ATO, ∠BTO,
(c) ∠AOT, ∠BOT
What do you notice?
Which points on a circle would have
tangents that do not meet?
You should have observed that:
If two tangents are drawn to a circle from a
common point,
(a) the tangents are equal;
(b) the tangents subtend equal angles at
the centre;
(c) the line joining the centre to the
common point bisects the angles
between the tangents.
Example 7.10
In Fig. 7.36, TA and TB
are tangents to the
circle, centre O.
If ∠ABO = 28°,
what is the size
of ∠ATO?
O
T
P
A
B
Fig. 7.36
90
O
P
Q
R
C
A
B
– – – – – – – –
– – – – – – – –
– – – – – – – –
Fig. 7.38
Solution

∠ABO = ∠BAO = 28°

∴∠AOB = 180° – 56° = 124°
∠AOT = ∠AOB (tangents subtend equal
angles at the centre of circle)
= 62°
∠OAT = 90° (tangent is
perpendicular to radius)
∴ ∠ATO = 90° – ∠AOT
= 90° – 62°
= 28°.
Exercise 7.6
1. In Fig. 7.37, O is the centre of the circle
and PT, RT are tangents to the circle.
Calculate
(a) ∠POT if 
∠OTR = 34°.
(b) ∠PRO
if ∠PTR = 58°.
(c) ∠TPR 
 if ∠PRO = 15°.
(d) ∠RTO if ∠POR = 148°.
2. Draw a circle, centre O, and radius 2.5 cm.
Mark points A and B on the circle such that
∠AOB = 130°. Construct tangents at A
and B. Measure
(a) the lengths of the tangents
(b) the angle formed where the tangents
meet.
3. In Fig. 7.38, O is the centre of the circle. If
BO = 19.5 cm, BQ = 18 cm, QC = 8.8 cm
and AO = 9.9 cm, what are the lengths of
(a) AB (b) BC (c) AC?
4. A tangent is drawn to a circle of radius
5.8 cm from a point 14.6 cm from the
centre of the circle. What is the length of
the tangent?
1
2
–
O
T
Q
P
R
Fig. 7.37
5. Tangents are drawn from a point 10 cm
away from the centre of a circle of radius
4 cm. What is the length of the chord
joining the two points of contact?
6. Tangents TA and TB each of length 8 cm,
are drawn to a circle of radius 6 cm. What
is the length of the minor arc AB?
7. Construct two tangents from a point A
which is 6 cm from the centre of a circle of
radius 4 cm.
(a) What is the length of the tangent?
(b) Measure the angle subtended at the
centre of the circle.
8. Draw a line KL = 6 cm long. Construct a
circle centre K radius 3.9 cm such that the
tangent LM from L to the circle is 4.5 cm.
Measure ∠KLM.
Common tangents to two circles
Fig. 7.39 shows non-intersecting pairs of circles
which have common tangents.
(a)
(b)
Fig. 7.39
P
1
Q
P
A
•
B
•
Q
1
R
1
S
R
C
•
D
•
S
1
91
1
1
11
1
1
1
1
1
1
The tangents in Fig. 7.39(a) are called exterior
or direct common tangents, while those in
Fig. 7.39(b) are called transverse or interior
common tangents.
Constructing direct common tangents to
two circles
Procedure
1. Draw two circles with centres A and B, and
radii r
1
and r
2
respectively where r
1
> r
2
.
2. With centre A, radius r
1
– r
2
, construct a
circle.
3. Join AB.
4. Bisect AB to obtain the point N.
5. With centre N, radius AN, draw arcs to cut
the circle of radius r
1
– r
2
at P and Q.
6. Join AP and AQ and produce them to cut
the circle of radius r
1
at P
1
and Q
1
.
7. Draw BR parallel to AP
1
and BS parallel
to AQ
1
.
8. Join RP
1
and SQ
1
.
RP
1
and SQ
1
are the required direct (or
exterior) common tangents (Fig. 7.40).
Fig. 7.40
Example 7.11
Using Fig. 7.40, show analytically that RP
1
is
actually a tangent to both circles.
Solution
Join PB.
AP = r
1
– r
2
and AP
= r
1
∴ PP
= AP
– AP = r
1
– (r
1
– r
2
) = r
2
= BR.
Since PP
//BR (by construction), then PBRP
is
a parallelogram (opp. sides equal and parallel).
Since, by construction, PB is a tangent and AP
is a radius, then   
∠APB = 90°
(radius is perpendicular to tangent).
∴∠P
PB = 90° (adj. angles on straight line)
∴∠PBR = 90°.
Thus PBRP
must be a rectangle
 ∴∠BRP
= ∠PP
R = 90°.
Since this is true only when radius and tangent
meet, then RP
must be a tangent to both
circles.
Similarly it can be shown that SQ
1
is a direct
common tangent.
Give an example on direct common tangents.
Constructing interior common tangents
Let the circles have centres A and B and radii r
1
and r
2
.
Procedure
1. Draw two circles with centres A and B, radii
r
1
and r
2
respectively.
2. With centre A, radius r
1
+ r
2
, draw another
circle.
3. Join AB and bisect it to get N
4. With centre N and radius AN, draw arcs to
cut the circle of radius r
1
+ r
2
at P and Q.
5. Join AP and AQ to cut the circle of radius r
1
at P
1
and Q
1
respectively.
6. Through B draw BR//AP and BS//AQ
to meet the circle, centre B, at R and S
respectively.
7. Join P
1
R and Q
1
S
P
1
R and Q
1
S are the required transverse or
interior common tangents (Fig. 7.41)
Fig. 7.41
– – – – – – – – – – – – – – – – – – – – –
B
A
Q
1
P
1
S
R
N
Q
P
– – – – – – – – – – – – – – – – – – – – –
B
A
Q
1
P
1
S
R
N
Q
P
92
(a) Since MNRP is a rectangle, MN = PR.
 
PR
2
= PQ
2
– QR
2

= 21
2
— 5
2
= 416
∴ 
 
cos ∠PQR =
5
= 0.238 1
21
Hence, ∠PQN =

∠PQR = 76.23° (4 s.f.).
Example 7.14
Find the length of a transverse common tangent
to two circles of radii 7 cm and 3 cm given that
the centres are 12 cm a part.
Solution
Refer to Fig. 7.44.
Fig. 7.44


Draw AR ⊥ BQ produced
Since ∠ APQ = ∠ PQR = 90°,
APQR is a rectangle.
BR = BQ + QR = BQ + PA
= (3 + 7) cm = 10 cm
AR
2
+ BR
2
= AB
2
(Pythagoras theorem)
 ⇒ AR
2
= AB
2
– BR
2
= 12
2
– 10
2
= 144 – 100
= 44
 ⇒ AR = √44 = 6.633cm.
But PQ = AR (opposite sides of rectangle)
∴PQ = 6.633 cm.
Example 7.12
Using Fig. 7.41 show analytically that P
1
R is
actually a tangent to both circles.
Solution
Join BP and BQ.
AP = r
1
+ r
2
(by construction) and AP
1
= r
1
.
∴ PP
1
= AP – AP
1
= (r
1
+ r
2
) – r
1
= r
2
= BR.
But PP
1
//BR (by construction). Since PP
1
and
BR are equal and parallel, then PBRP
1
is a
parallelogram.
By construction, PB is the tangent at P.
∴ ∠APB = 90°
(radius perpendicular to
tangent)
∴ ∠PBR = 90°.
∴ PBR P
1
is a rectangle.
∴ P
1
R is perpendicular to the radii through
P
1
and R.
∴ P
1
R is a common tangent.
Similarly it can be shown that Q
1
S is a common
tangent
Example 7.13
Two circles, of radii 4 cm and 9 cm, are
positioned in such a way that the distance
between their centres is 21 cm, as shown in
Fig. 7.42. Calculate
(a) the length of the
tangent MN.
(b ∠PQN.
Solution
In Fig 7.43, line PR has been drawn such that
PR⊥ QN, hence forming rectangle MNRP.
Fig 7.43
– – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – –
•
•
– – – – – – – –
– – – –
Q
N
P
M
Fig 7.42
R
– – – – – – – – – – – – – – – – –
•
•
– – – – – – – –
– – – –
Q
N
P
M
– – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – –
5 cm
4 cm
4 cm
R
7 cm
Q
P
A
B
3 cm
93
Exercise 7.7
1. The centres of two circles of radii 10 cm and
6 cm are 20 cm a part. Find the length of
(a) a direct common tangent to the circles.
(b) a transverse common tangent of the
circles.
2. Draw two circles, radii 5 cm and 2 cm,
such that their centres are 8.5 cm apart.
Construct a common tangent direct to the
circles. Measure its length.
3. Draw two circles, radii 3.5 cm and 2.5
cm, such that their centres are 9 cm apart.
Construct a transverse common tangent.
Measure its length.
4. Two circles of radii 6.5 cm and 1.5 cm
have their centres 10 cm apart. What angle
does
(a) the direct common tangent make with
the line joining the centres?
(b) the transverse common tangent make
with the line joining the centres?
5. The centres of two circles of radii R and r,
are d units apart. What is the length of
(a) a direct common tangent to the two
circles?
(b) a transverse common tangent to the
two circles?
6. Two circles with radii 3 cm and 8 cm are
positioned in such a way that their
centres are 13 cm apart. What is the length
of their common direct tangent?
7. Two circles, with radii 12 cm and 4 cm, are
placed such that the length of their direct
common tangent is 15 cm. What is the
distance between their centres?
Contact of circles
Consider Fig. 7.45.
(a) (b)
Fig. 7.45
Fig. 7.45 shows pairs of circles which have only
one point in common. They are said to be in
contact or to touch each other at that point T.
In Fig. 7.45(a), the circles touch each other
externally or have external contact.
In Fig. 7.45(b), the circles touch each other
internally or have internal contact.
Let P and Q be the centres of the smaller and
the bigger circles respectively, and let XTY be
the common tangent.
Since a radius is perpendicular to a tangent at
the point of contact,
∠PTX = 90° and ∠QTX = 90°.
But ∠PTX + ∠QTX = ∠PTQ
∴ ∠PTQ = 180° ⇒ PTQ is a straight line.
We note that:
1. If two circles touch each other, the
line joining their centres (produced if
necessary) passes through the point of
contact.
2. If two circles touch each other externally,
the distance between the centres is equal
to the sum of the radii.
In Fig. 7.45(a), PQ = PT + TQ.
3. If two circles touch each other internally,
the distance between the centres is equal
to the dierence of the radii.
In Fig. 7.45(b), QP = QT – PT.
•
P
•
Q
T
– – – – – – – – – – –
X
Y
•
P
•
Q
T
– – – – – – –
X
Y
94
O
r
4 cm
Fig. 7.51
•
C
B
A
•
•
Fig. 7.49
•
O
– – –
•
M
– – – – – – – – – – – –

P
K
Fig. 7.47
•
O
•
M
K
N
Fig. 7.46

Example 7.15
In Fig. 7.46, KN is
the diameter of a
circle centre M,
radius 8 cm.
Find the radius
of the circle, centre O,


Solution
   
the circle O be
r cm. Join MO
and produce it
to P.

MP = 8 cm (given)
MO = MP – OP = (8 – r) cm
 
∠ ° (radius ⊥ tangent)
∴ OM
2

2

2

i.e. (8 – r)
2
= r
2
+ 6
2
⇒64 – 16r + r
2
= r
2
+ 36
 ⇒64 – 16r = 36
 ⇒28 = 16r
 ⇒
Exercise 7.8
1. In Fig. 7.48, P, Q and R
are the centres of the
three circles. If
PQ = 4 cm, QR = 6 cm
and PR = 7 cm, what
are the radii of the circles?
2. Draw a line PQ = 6 cm. Construct a circle
centre Q such that the tangent from P to
the circle is 4 cm. What is the radius of the
circle?
3. Two circles of radii 5 cm and 12 cm touch
each other externally. What is the length of
their direct common tangent?
4. In Fig. 7.49, two circles
centres A and B touch
externally. They each
touch a circle, centre
C internally.
If AB = 6.8 cm,
BC = 9.8 cm and
AC = 10.2 cm,
calculate the radii of the three circles.
5. Two circles of radii R and r touch
externally. Find an expression in terms
of R and r for the length of the common
tangent to the circles.
6. In Fig. 7.50, the
circles touch at D,
line AC touches
the circles at A
and C and meets
the tangent BD at B.
Show that (a) AB = BC (b) ∠ADC = 90°.
7. Draw two circles, each of radii 2 cm with
their centres 5 cm apart. Draw a circle
of radius 6 cm such that it touches one
circle internally and the other externally.
Measure the largest angle of the triangle
formed by joining the three centres.
8. In Fig. 7.51, O is the centre
of the two circles radii
r cm and 4 cm. A circle
of radius 3 cm is drawn
such that it touches both
circles internally. Find
the value of r.
9. The centres of two circles of radii 5 cm and
9 cm are 20 cm apart. Find the radius of
the smallest circle that would touch
(a) both circles internally.
(b) the smaller circle internally and the
bigger circle externally.
R
•
Q
•
P
•
Fig. 7.48
D
C
B
A
Fig. 7.50
95
of the circle. This is called the alternate
segment theorem.
2. If a straight line is drawn at the end of a
chord of a circle making with chord an
angle equal to an angle in the alternate
segment, the straight line touches the
circle (i.e. it is a tangent to the circle).
Analytic proof of the alternate segment
theorem
(a) We use Fig. 7.54(a) to show that 
∠RQT = ∠QST.
(a) (b)
Fig. 7.54
Draw diameter QU. Join UT.
Since QU is a diameter and PR is a tangent,
∠RQT + ∠TQU = 90° (tangent ⊥ radius)
∠QTU = 90° (∠ in semi-circle)
 ∴∠QUT + ∠TQU = 90°
(∠ sum of ∆)
 ∴∠RQT + ∠TQU = ∠QUT + ∠TQU
⇒∠RQT = ∠QUT.
But ∠QUT = ∠QST (∠s in same segment)
∴∠RQT = ∠QST.
(b) We use Fig. 7.54(b) to show that 
∠PQT = ∠QMT
∠PQT + ∠RQT = 180°
(adj. ∠s on
straight line).
∠QMT + ∠QST = 180° (opp. ∠s cyclic
quadrilateral).
∴∠PQT + ∠RQT = ∠QMT + ∠QST.
But ∠RQT = ∠QST (shown in (a) above)
∴∠PQT = ∠QMT
Angles in alternate segment
In Fig. 7.52, ABC is a tangent
to the circle at B. The
chord BD divides the
circle into two segments
BED and BFD.
We say that BFD is the
alternate segment
to ∠ABD.
Similarly, BED is the alternate segment to ∠CBD.
ACTIVITY 7.5
1. Draw a circle of any radius.
2. Draw a tangent at any point B.
3. Draw a chord BD.
4. Mark points P, Q, R on the circumference
in the same segment as in Fig. 7.53.
Join BP, BQ, BR, DP, DQ and DR.
5. Measure angles ABD, BPD, BQD and
BRD.
What do you notice?
You should have observed that
∠ABD = ∠BPD = ∠BQD = ∠BRD.
In general:
1. If a straight line touches a circle, and
from the point of contact a chord is
drawn, the angle which the chord makes
with the tangent is equal to the angle the
chord subtends in the alternate segment
A
Q
C
P
B
D
Fig. 7.53
R
A
F
C
D
B
E
Fig. 7.52
T
U
S
Q R
P
– – – – – – – – – – – – –
– – – – – –
T
U
S
Q R
P
– – – – – – –
– – – – – – –
M
96
4. In Fig. 7.59, KLM is a
tangent to the circle.
If ∠LPN = 38° and 
∠KLP = 85°, nd
∠PQN.
5. In Fig. 7.60, DC is a
tangent to the circle.
Show that 
∠CBD = ∠ADC.
6. In Fig. 7.61, AB and
DE are tangents to the
circle. ∠ABC = 40°
and ∠BCD = 38°.
Find ∠CDE.
7. In Fig. 7.62, ABC
is a tangent to the
circle at B and ADE
is a straight line.
If ∠BAD = ∠DBE,
show that BE is a
diameter.
8. In Fig. 7.63, AD is a
tangent to the circle.
BC is a diameter
of the circle and 
∠BCD = 30°.
Find ∠DAB.
9. In Fig. 7.64, AD is a tangent to the circle at
D, ∠DAB = 28° and ∠ADC = 112°. Find
Example 7.16
In Fig. 7.55, PQR is a tangent
to the circle at Q. QT is a
chord and PST is a straight
line. Given that 
°, ∠TPQ = 25°,
∠SQP.
Solution
In ∠PQT, ∠PQT + ∠QTP + ∠TPQ = 180°
(∠
But ∠PQT = 110°, ∠TPQ = 25°
∴ ∠QTP = 180° – 135° = 45°
But ∠SQP = ∠QTP (∠s in alternate segment)
∴ ∠SQP = 45°.
Exercise 7.9
1. In Fig. 7.56, AC
is a tangent to
the circle and
BE//CD.
(a) If ∠ABE = 42°,  
∠BDC = 59°,
nd ∠BED
(b) If ∠DBE = 62°, ∠BCD = 56°, nd 
∠BED.
2. In Fig. 7.57, PR is a tangent to the circle.
(a) If ∠PQT = 66°, nd ∠QST.
(b) If ∠QTS = 38° and ∠QRS = 30°, nd 
∠QST.
(c) If ∠QTS = 35°
and ∠TQS = 58°,
nd ∠QRS.
(d) If ∠PQT = 50°
and ∠PRS = 30°,
nd ∠SQT.
3. In Fig. 7.58, AB, BC and AC are tangents
to the circle. If ∠BAC = 75° and ∠ABC =
44°, nd ∠EDF, ∠DEF and ∠EFD.
D
30°
B
C
A
Fig. 7.63
A
E
D
B
C
Fig. 7.62
E
40°
38°
C
A
B
D
Fig. 7.61
C
B
A
Fig. 7.60
D
N
P
K
L
Fig. 7.59
Q
M
Fig. 7.57
Q
P
S
T
R
A
E
D
C
B
Fig. 7.56
R
T
Q
P
S
Fig. 7.55
C
A
D
E
F
B
44°
75°
Fig. 7.58
97
A
C
Fig. 7.66
B
O
•
D
28°
B
C
A
Fig. 7.64
centre of the required circle.
2. From O, construct a perpendicular to BC,
meeting BC at N. (The perpendicular from
O could be drawn to any of the sides).
3. With centre O, radius ON, draw a circle.
This is the required circle (Fig. 7.67).
Note that to locate the centre of the circle we
use the fact that:
The internal bisectors of the three angles of
a triangle are concurrent, i.e. they meet at a
common point which is equidistant from the
three sides of the triangle.
This is referred to as the in-centre theorem.
Circumscribed circles
A circle which passes through all the three
vertices of a triangle, as in Fig. 7.68, is called
the circumscribed circle or the circumcircle of
the triangle. The centre of the circle
is called the circumcentre.
The radius of this circle is
called the circumradius.
For any ∆ABC, if O is
the circumcentre
(Fig. 7.68), then the
circumradius
= OA = OB = OC.
Constructing the circumcircle of a given
triangle
Procedure
1. Draw any ∆ABC.
2. Construct perpendicular bisectors of any
the angle subtended at
the centre of the circle
by the chord BC.
10. Points A, B and C are on a circle such that
∠ABC = 108°. Find the angle between the
tangents at A and C.
11. In Fig. 7.65, O is
the centre of the
circle. AB and
CD are chords
that meet at X.
XT is a tangent to
the circle.
Show that (a) XT
2
= XA
.
XB
(b) XT
2
= XC
.
XD.
Inscribed circles
An inscribed circle is a circle drawn inside a
triangle such that the circle touches the three
sides of the triangle (Fig. 7.66).
The centre of an inscribed circle is called the
in-centre and the circle is also called the
in-circle.
The radius of the circle is called the in-radius.
Constructing the inscribed circle of a
given triangle
Draw any triangle ABC.
Procedure
1. Bisect any two angles, say ∠ABC and
∠ACB. The point of intersection O is the
A
C
Fig. 7.67
B
N
O
B
O
•
A
C
Fig. 7.68
A
C
T
X
Fig. 7.65
O
•
B
D
98
Constructing an escribed circle of a given
triangle
1. Draw any ∆ABC and produce two sides,
say AB and AC to D and E respectively.
2. Construct the bisectors of ∠DBC and
∠BCE to meet at O.
3. Construct ON, perpendicular to AB
produced or AC produced.
4. With centre O, radius ON, draw a circle
(Fig. 7.71).
This is the required circle.
The centre O is called the e-centre.
The bisectors BO and CO of ∠DBC and ∠BCE
respectively are called the external bisectors of
∠ABC and ∠ACB respectively.
There are three possible escribed circles to any
triangle. In Fig. 7.71, the circle is said to be
escribed to side BC. Similarly, circles can be
constructed, escribed to AB and to AC.
Centroid and orthocentre of a triangle
ACTIVITY 7.6
1. Draw any ∆PQR. From P construct PM
perpendicular to QR meeting QR at M.
Similarly, construct QN and RS perpendi-
cular to PR and PQ respectively.
What do you notice about the perpendi-
culars PM, QN and RS?
These are the altitude of the triangle.
2. Draw any ∆ABC. Locate the point P, Q
and R, the midpoints of AB, BC and AC
respectively.
B
A
C
D
E
Fig. 7.70
two sides, say BC and AC, to meet at O.
3. With O as centre, radius OA or OB or OC,
draw a circle.
The circle passes through all the three vertices
A, B and C.
This is the circumcircle of ∆ABC (Fig. 7.69).
Note that if the perpendicular bisector of AB
was drawn, it would also pass through the
circumcentre O.
The perpendicular bisectors of the three sides
of a triangle are concurrent.
This is called the circumcentre theorem. It
leads as to the following theorem.
There is one, and only one circle, which
passes through three given points which are
not on the same straight line.
Escribed circles
We have seen that an in-circle touches the
sides of a triangle internally. In Fig. 7.70, we
have a circle that touches the sides of ∆ABC
externally, two of the sides having been
produced. Such a circle is called
an escribed circle or
an e-circle of ∆ABC.
B
A
D
E
Fig. 7.71
C
•
O
N
B
O
•
A
C
Fig. 7.69
99
QG
AQ
–––
RG
BR
–––
PG
CP
–––
1
3
–
1
3
–
1
3
–
Join AQ, BR and CP. Label the point where
they meet as G.
Measure AQ, QG, BR, RG, CP and PG.
Find the ratios , and .
What do you notice?
Note that the lines AQ, BR and CP are medians
of the triangle.
We summarise the results as follows:
1. The altitudes of a triangle are concurrent
and the point where they meet is the
ortho-centre [Fig. 7.72-(a)].
This is referred to as the ortho-centre
theorem.
2. (a) The three medians of a triangle are
concurrent.
(b) The point at which the medians
intersect is one-third of the way along
each median, measured towards the
vertex i.e. QG = QA, RG = RB
and PG = PC [Fig. 7.72(b)].
(c) The point at which the medians
intersect is called the centroid of the
triangle [Fig. 7 72(b)].
This property is referred to as the
centroid theorem.
(a) (b)
Fig. 7.72
Example 7.17
Find the radius of the circle which
circumscribes an isosceles triangle of sides
15 cm, 15 cm and 9 cm.
Solution
In Fig. 7.73, O is the circumcentre and AD is
the perpendicular from A to BC.
∴ AD bisects BC ⇒ BD = 4.5 cm


2
= AB
2
– BD
2
(Pythagoras theorem)
= 15
2
– 4.5
2
= 225 – 20.25
= 204.75 or 204 .8
∴AD = 14.31 cm
OD = AD – r = 14.31 – r

2
= OD
2
+ BD
2
i.e. r
2
= (14.31 – r)
2
+ 4.5
2
r
2
= 204.8 – 28.62 r + r
2
+ 20.25
 ⇒ 28.62 r = 225.05
r =
225.05
28.62
= 7.863 cm
  
Exercise 7.10
1. An equilateral triangle of sides 3 cm is
inscribed in a circle. Find the radius of the
circle.
2. An equilateral triangle is inscribed in a
circle of radius 4 cm. Find the length of
the side of the triangle
3. An isosceles triangle has sides 5 cm, 5 cm
and 6 cm. Find the radius of its circumcircle.
4. Draw a triangle ABC such that AB = 6 cm,
BC = 8 cm and ∠ABC = 60°. Construct the
inscribed circle. Measure its radius.
5. Draw a triangle PQR such that PQ = 6.9 cm,
QR = 5.4 cm and PR = 7.8 cm. Construct
the inscribed circle. Measure its radius.
6. Draw a triangle with sides 8 cm, 8 cm and
5 cm and construct its inscribed circle.
Measure the radius of this circle.
7. Draw a circle of radius 5 cm. Inscribe an
equilateral triangle in the circle. Measure
its side.
A
C
B
Q
G
P
R
– – – – – – – – – – – –
– – – – – – –
Fig. 7.73
A
B C
15 cm
D
9 cm
15 cm
O
S N
– – – – – – – – – – – – –
– – – – – – – – – – –
– – – – – – – – – – – – –
P
R
Q
M
O
100
8. Draw a circle of radius 2.5 cm and
circumscribe an equilateral triangle about
the circle. Measure its sides.
9. Draw ∠ABC = 55°. Construct a circle of
radius 2.5 cm to touch AB and BC. Measure
the length of the tangent to the circle.
(Hint: First make sketch and calculate the
distance of the centre of the circle from
point B.).
10. Draw ∆ABC such that AB = 6 cm,
BC = 5 cm and ∠B = 90°. Construct the
circle escribed to BC and measure its radius.
11. Triangle PQR is such that ∠P = 60°, 
∠Q = 40° and ∠R = 80°. A circle is
escribed to QR to touch it at A, and touch
PQ and PR produced at C and B
respectively. Find the angles of ∆ABC.
12. Construct ∆PQR such that PQ = 6 cm,
QR = 7 cm and PR = 8 cm. By
construction, determine the distance from
P to the ortho-centre of the triangle.
13. In ∆ABC, AB = 7 cm, BC = 9 cm and 
∠B = 33.7°. AD is a median of the
triangle, and the area of ∆ABD = 8.7 cm
2
.
Find the distance of the centroid of the
triangle from the point A.
14. Given that XP, YQ and ZR are medians of
∆XYZ, construct ∆XYZ where XY = 5 cm,
YQ = 4.2 cm and XP = 3.6 cm. Measure YZ.
(Hint: First make a sketch and use Pythagoras
Theorem to nd the lengths XA and YP).
Application of arcs, chords and
tangents to real life situations
In real life, we often encounter problems
involving objects, structures or situations that
incorporate circles, parts of circles, chords and
tangents. Such problems require that we use
the knowledge learnt in this chapter to solve
them.
In this section, we give some examples and an
exercise on such problems.
Example 7.18
The main doorway of a church is to be made
up of a circular arc, which subtends an angle
of 120° at the centre, two upright frames each
of length 2.8 m and a horizontal base frame of
width 3 m. Find the total length of frame timber
to be used.
Solution
Fig. 7.74 shows the doorway.
OM is a perpendicular to chord AC from the
centre of the circle. Thus, OM bisects AC.
∴ AM = MC = 1.5 m
∠OAC = ∠OCA = 30° (base angles


cos 30° =
∴ Radius OA = =
= = 1.73 m (2 d.p.)
Hence, length of arc ABC
= × 2 × 3.142 × 1.73
= 3.62 m (2 d.p.)
Total length of frame timber to be used
= perimeter of the doorway
= 3.62 + 2.8 + 3 + 2.8
= 12.22
= 12.2 m (1 d.p.).
– – – – – – – – – – – – – – – –
C
A
M
B
2.8 m
3 m
120°
E
D
– – – – – – – – – –
– – – – – – – – – –
– – – – –
O
Fig. 7.74
AM
OA
–––
AM
cos 30°
––––––
1.5
–––

2
––
1.5 × 2

––––––
120°
360°
––––
101
4.4 m
Fig. 7.77
4.4 m
Since CD and BE are opposite sides of a
rectangle, then
CD = 1.296 m
FG and CD are a pair of common tangents to
the same circles.
 ∴ FG = 1.296 m

cos ∠EAB = = = 0.076 9
∴ ∠EAB = 85.59°
 ∴ ∠DAF = 171.18°
∠DAF = 188.82°
∴ 
=
× 3.142 × 1.4
= 2.307 m
∠CBG = ∠DAF = 171.18°
∴
= × 3.142 × 1.2
= 1.793 m
Thus, the total length of strap is
1.296 + 2.307 + 1.296 + 1.793 + 0.45
= 7.142 m
= 7.14 m (2 d.p.)
Exercise 7.11
1. A section of a ring of radius 8 cm subtends
an angle of 150° at the centre. The section
is cut and bent to make another ring. What
is the radius of the new ring?
2. What angle does the hand of a clock sweep
through when the time changes from 11.55
p.m to 12.13 a.m?
3. Fig. 7.77 shows a
cross-section of a
tunnel, in the form
of a major segment
of a circle. The
perpendicular
height and the
chord are each
4.4 m.
Calculate the diameter of the tunnel.
EA
AB
–––
0.1
1.3
–––
171.18°
360°
–––––––
188.82°
360°
–––––––
•
– – – – – – – – – – – – –
strap
knot
– – – – – – – – – – – – –
•
Example 7.19
Jemima wants to transport two drums, of
diameters 1.4 m and 1.2 m, using a pickup
truck. To avoid rolling, she ties them together,
using a strap, as shown in Fig. 7.75. Given that
she needs 45 cm of strap for knotting, what is
the total length of strap that she uses?
Fig. 7.75
Solution

a direct common tangent to two circles. The
solution refers to Fig. 7.76.
Fig. 7.76

CD be a common tangent (part of the strap).
BC and AD are perpendicular to CD.
Draw BE ⊥ AD.

right-angled at E.

AE
2
+ BE
2
= AB
2

Now, AB = 0.7 + 0.6 + = 1.3 m (sum of radii)
∴ BE
2
= 1.3
2
– 0.1
2
= 1.69 – 0.01
= 1.68
∴ 
F
•
– – – – – – – – – – – – –
– – – – – – – – – – – – – – –
– – – – – – – – –
– – – – – – – –
– – – – – – – – – – – – –
A
C
B
E
D
0.6 m
0.6 m
0.1 m
– – – – – – – – –
– – – – – – – –
– – – – – – – – – – – –
G
•
102
is the length of the bicycle’s chain?
10. In order to have the two pulleys in Fig.
7.81 rotate in opposite directions, a belt
is tted so that it is crossed over between
the pulleys. The straight parts of the belt
form transverse common tangents. If the
diameters of the pulleys are 22 cm and 8 cm
and the distance between their centres is
25 cm, calculate the length of the belt.
Fig. 7.81
11. A crime was committed at a certain area.
To mark o the area, the scene of crime
police used three drums and a rope. By
coincidence, they positioned the drums
in such a way that their centres formed a
triangular area with sides 13 m, 12 m and
5 m. To keep o intruders, they put a rope
round the three drums as shown in Fig. 7.82.
If the rope was taut and given that the
diameter of each drum was 1 m, nd the
length of rope used.
(Ignore the length
used in knotting)
12. A manufacturer
of shoe polish has
a special package
comprising of
three cylindrical
tins of radii
8.1 cm, 3.6 cm and
3.6 cm, packed in a rectangular box as
shown in Fig. 7.83. Find the interior
dimensions of the box.
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – –
•
•
•
12 m
13 m
5 m
Fig. 7.82
•
•
Fig. 7.83
•
•
•
10 cm
Fig. 7.78
30 cm
4. Fig. 7.78 shows a
hemispherical
container of
internal diameter of
30 cm, partly lled
with water. If the water
surface is 10 cm below the
top of the container, what is the diameter of
the water surface?
5. A ball of radius 10 cm rests in a hole whose
radius is 5 cm. What is the height of the
centre of the ball above the ground level?
6. The rim of a hemispherical bowl, placed as
the one in Question 4 above, forms a circle
of radius 37 cm. Water is poured in until
its surface forms a circle of radius 35 cm.
Calculate the depth of the water.
7. Fig. 7.79 shows two
gears of radii
3 cm and 5cm.
A chain which
is taut, passes
round them. What
is the length of the chain?
8. Two pulleys of radii
5 cm and
12 cm have
their centres
30 cm apart.
A belt which is
taut passes round them
(Fig. 7.80). What is the length of the belt?
9. The propulsion mechanism of a bicycle
comprises of two gears. The one rotated by
the peddling action of the cyclist is known
as the chain wheel while the one that is
rotated by the chain (i.e. the one that rotates
the rear wheel) is known as the rear sprocket.
Now, the radius of the chain wheel of
a bicycle is 8.2 cm and that of the rear
sprocket is 3.7 cm. If the distance between
the centres of the two gears is 46 cm, what
Fig. 7.79
•
12 cm
Q
P
•
5 cm
30 cm
– – – – – – – – – – – – – – –
– – – – – – – – – – – – – – –
Fig. 7.80
– – – – – – – – – – –
– – – – – – – – – – –
•
3 cm
•
5 cm
B
A
103
8
MATRICES
Matrix
There are many instances when large quantities
of numeric information have to be stored. Very
often this information is arranged in the form
of tables, which is one of the most convenient
ways of arranging information.
Table 8.1 shows how dierent types of
packets of biscuits are packed in a certain
manufacturing factory.
Name of Type of biscuit
packet
Britania Shortcake Marie Glucose
Economy 14 14 10 10
Family 5 8 9 14
Standard 8 3 7 6
Table 8.1
This means, for example, that the “Economy”
packet contains 14 Britania, 14 Shortcake, 10
Marie and 10 Glucose biscuits.
With time, the packers get to know what each
row and each column refers to, and they need to
remember only the patterns:
14 14 10 10
5 8 9 14
8 3 7 6
Such an arrangement is called a matrix (plural:
matrices).
Thus:
A matrix is a rectangular array of numbers
whose value and position in the arrangement
is signicant.
A matrix is usually shown in brackets either as:
1 8
or
1 8
3 7 3 7
(
)
[
]
(
)
(
)
(
)
(
)
Normally, a capital letter in bold type e.g. A,
B, is used to denote a matrix. However, it is
dicult to bold in our normal hand writing, we
use a wavy underline on the letter, e.g. A, B.
Each number in a matrix is called an element of
the matrix.
Order of a matrix
A convenient way of describing the shape or
size of a matrix is by using rows and columns.
For example, the matrix from Table 8.1 has 3
rows and 4 columns. It is said to be a matrix of
order 3 × 4 (read “three by four”) or a “three by
four matrix”.
Thus:
The order of a matrix denotes the number of
rows and columns in the matrix. A matrix of
order m × n has m rows and n columns.
How many rows and columns does a matrix of
order 4 × 3 have?
State the order of each of the following
matrices.
7 0 0
7 0 3
1. 2. 1
–
4 4
2
–
1 5
–
5 2 4
4
3. (5 5 9) 5.
–
8
9
A matrix of order 1 × n is called a row matrix,
a matrix of order m × 1 is called a column
matrix, and a matrix of order n × n is called a
square matrix.
The position of an element inside a matrix is
described using suces (plural of sux).
˜
˜
104
( )
1
2
–
1
2
–
1
2
–
( )
Thus, if A is a matrix and a
m,n
is an element
in it, then a
m,n
is the element in the m
th
row
and n
th
column.
A sux is also known as a subscript.
Exercise 8.1
1. Table 8.2 shows the number of times that
three couples attended various types of
entertainment in one year.
Type of Couple
The The The
entertainment
Kinyuas Ronos Jumas
Cinema 7 2 5
Dance 1 2 9
Play 5 8 1
Circus 0 3 2
Table 8.2
(a) Write down the information in the
table in the form of a matrix and state
the order of the matrix.
(b) Write The Ronos’ attendance as a
column matrix. What is the order of
this matrix?
(c) Write, as a row matrix, the number of
times that plays have been attended,
and state the order of the matrix.
2. Make up a matrix of order
(a) 3 × 5 (b) 5 × 4 (c) 3 × 1
(d) 1 × 2 (e) 2 × 2 (f) 3 × 3
3. How many elements are there in a matrix
of order
(a) 2 × 4 (b) 4 × 2 (c) 4 × 4
(d) 1 × 3 (e) 1 × 1 (f) m × n?
3 4 5 6
4. Given A = 2 1
–
1
–
2 , what is the
–
4
–
3 0 7
element (a) a
1,3
(b) a
2,1
(c) a
3,4
(d) a
3,3
?
5. Write down, as a
m,n
, the following elements
of A in Question 4.
(a)
–
1 (b)
–
4 (c) 3 (d) 1
6. Three salesgirls sold the following numbers
of bottles of perfume on a certain day:
Ivy sold 9 bottles of She, 13 of Rosy and
6 of Shield.
Liz sold 8 bottles of Yu, 7 of Rosy and
10 of Shield.
Meg sold 15 bottles of Yu, 1 of She and
18 of Rosy.
Show this information in a 3 × 4 matrix.
Addition and subtraction of matrices
Over a period of two weeks, two families used
the amounts of bread, milk and sugar shown in
Table 8.3.
Week 1 Week 2
Item Owour’s Pandit’s Owour’s Pandit’s
family family family family
Bread 17 12 7 16
(loaves)
Milk
10 14 23 16
(litres)
Sugar
2 3 12 4
(kg)
Table 8.3
Regard the table as two 3 × 2 matrices and
use these to complete the 3 × 2 matrix shown
below, giving the total amounts of each item
used by each family in the two weeks.
14 28
* *
* *
To do this you must add together the elements
in corresponding positions in the rst two
matrices. This is how matrices are added.
The method of subtraction follows the same
pattern as that of addition. For, example, to nd
out how much more of each food item that the
families in Table 8.3 used in the second week
1
2
–
105
( )
( )
( )
( )
( )
( )
( )
( )
( ) ( )
( )
( )
( )
( )
( ) ( )
( ) ( )
( )
( )
( )
(d) B – A = 3 7
–
2 – 2
–
5 7
3
–
2
–
4
–
3 2 4
= 1 12
–
9
6
–
4
–
8
From Example 8.1, we notice that:
Matrix addition is commutative, i.e
A + B = B + A, but
Matrix subtraction is non-commutative, i.e.
A – B ≠ B – A.
Exercise 8.2
1. Add the following pairs of matrices
where possible.
(a)
–
3 2 and 4
–
3
4 0
–
2
–
1
(b) 3 5 and
–
5 4 1
5 4 0 5 2
3 1 1
(c) 2 3 and
–
3 2
1 2
–
1
(d) (3
–
2 5) and 4
–
5
Explain why in some cases it is not
possible to add.
2. Work out the following, where possible.
(a) 3 2 5 6 – 5 4 2 1
4 6 2 4 4 3 5 5
4 5
(b) 3 –
–
2
–
2
–
4
4
5
(c) – 2
5
3
(d) 4 8 + 1 3 – 6 4
9 4 2 7 5 3
( ) ( )
1
2
–
1
2
–
( )
( )
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
( )
( )
(
) (
) (
)
than in the rst week, each quantity in the rst
matrix is subtracted from the corresponding
quantity in the second matrix.
7 16 17 12 7 – 17 16 – 12
i.e. 23 16 – 10 14 = 23 – 10 16 – 14
12 4
2 3 12 – 2 4 – 3
10 4
= 13 2
Compatibility in addition or subtraction
Matrices can be added or subtracted only if
they are of the same order. Such matrices
are said to be compatible for addition or for
subtraction.
The resulting matrix is of the same order.
The operation is done as follows:
a b c
±
g h i
=
a ± g b ± h c ± i
d e f j k l d ± j e ± k f ± l
Example 8.1
If A = 2
–
5 7 and B = 3 7
–

–
3 2 4 3
–
2
–
4
(a) A + B (b) B + A
(c) A – B (d) B – A
Solution
(a) A + B = 2
–
5 7 + 3 7
–
2
–
3 2 4
3
–
2
–
4
= 5 2 5
0 0 0
(b) B + A = 3 7
–
2 + 2
–
5 7
3
–
2
–
4
–
3 2 4
= 5 2 5
0 0 0
(c) A – B = 2
–
5 7 – 3 7
–
2
–
3 2 4 3
–
2
–
4
=
–
1
–
12 9
–
6 4 8
(
)
(
)
(
)
(
)
2
3
–
5
6
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
( )( )
( )
10
106
( )
( )
( )
( )
( ) ( )
( )
( ) ( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( )
To multiply a matrix by a scalar, we multiply
each element of the matrix by the scalar.
e.g. k
a b c
=
ka kb kc
d e f kd ke kf
Example 8.2
Find the values of m, n, p and q if
3
3 4
– 4
1 2
=
m n
2 5
3 4
p q
Solution
3
3 4
– 4
1 2
=
9 12
–
4 8
2 5 3 4 6 15
12 16
= 5 4
–
6
–
1
Thus, m n = 5 4
p q
–
6
–
1
Hence, m = 5, n = 4, p =
–
6 and q =
–
1.
Two matrices are equal if they are of the
same order and their corresponding elements
are equal.
Exercise 8.3
1. Given that A = 2 0 , B =
1 3 and
5 1
5
2
C =
–
1
–
2
, nd
1 0
(a) 3A (b)
–
4C (c) A + 2B (d) B – 2C
2. Find k if k
3 0
+ 3
–
1 2
=
6 6
1 2 3
–
4 12
–
6
3. Find the matrix M for which
(a)
3 4
+ 2
–
4 2
= 4M
–
2 0 7
–
1
(b)
1 9
– 2M =
7 3
+ M
6 2 8 6
3
(e) (3 5
–
4) + (2 5 5) – 2
–
1
(f) (5 4 5) – (3
–
2
–
4) + (3
–
5
–
2)
3. Write down any three matrices A, B and C
which have the same order. Work out
(a) A + (B + C) (b) (A + B) + C
(c) Now complete the following statement
about the implied property of matrix
addition. Matrix addition is _______
4. Copy the following, replacing the stars
with appropriate values.
(a)
2 5
+
*
*
=
3 1
0 1
*
*
2 5
3 1
*
*
1 2
(b)
–
2 0 –
*
*
= 3 1
4
–
7
*
*
2
–
3
5. If
2 6
+
a b
=
5 2 ,
7
–
7 b a a
–
4
nd the values of a and b.
6. In relation to matrix addition or subtraction,
what does it mean to say “the matrices are
incompatible”?
Scalar multiplication of matrices
If A =
3 4
, A + A =
3 4
+
3 4
2 5 2 5 2 5
=
6 8
4 10
But A + A = 2A.
Thus, we see that 2A = 2
3 4
=
6 8
2 5 4 10
The number, for example 2, multiplying the
matrix is called a scalar.
( )
( )
( )
( )
( )
( )
( )
( )
( )( )
( )
( )
107
( )
( )
( )
( )
(
)
( )
( )
( )
( )
( )
( )
( )
Note: 1. The rst matrix is written as a row
matrix and the second as a column
matrix but in the same order.
2. No multiplication symbol is placed
between the two matrices.
Notice that:
To combine a row matrix and a column
matrix with the same number of elements, we
multiply their corresponding elements and
x
add, i.e. (a b c) y = (ax + by + cz).
z
Example 8.4
In a national soccer league, the result of two
soccer clubs, Breweries and Rivatex, were as
shown in Table 8.4
    
Breweries 8 2 2
Rivatex 7 4 1
Table 8.4
If three points are awarded when a match is
won, 1 point when it is draw, and no point if it

points obtained by each club.
Solution
3
(8 2 2) 1 = (8 × 3 + 2 × 1 + 2 × 0)
0
= (26)
3
(7 4 1) 1 = (7 × 3 + 4 × 1 + 1 × 0)
0
= (25)
Thus, Breweries has 26 points and Rivatex has
25 points.

3
?
1
4. Find x and y if
5
x 3
– 2
x 4
=
18 7
7 3 y 2 29 11
p + q + r 8
5. Find the unknowns in p + 2q = 6
3p 6
Multiplication of matrices
Multiplication of row and column
matrices
We have seen how matrices are added or
subtracted. Sometimes, it is necessary to
combine matrices in a dierent way. Now
consider the following example.
Example 8.3
Mrs. Kamau bought 2 kg of meat at sh 140 per
kilogram, 3 packets of Unga maize meal at
sh 52 per packet and 4 loaves of bread at sh 19
per loaf. How much did she spend?
Solution
Two
kinds of information are given: the
quantities of food bought and the costs. This
information can be shown in a matrix form as:
M U B M U B
Quantity
2 3 4
Price
140 52 19

Cost = sh (2 × 140 + 3 × 52 + 4 × 19)
= sh (280 + 156 + 76)
= sh 512
When using matrices, this calculation is written
in the form
140
(2 3 4) 52 = (2 × 140 + 3 × 52 + 4 × 19)
19
= (5 1 2)
The matrix (5 1 2) is known as the product of
the two matrices on the left.
108
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
Exercise 8.4
1. Where possible, work out the following
12 2
(a) (0 2 1) 5 (b) (
–
5 1
–
2) 0
2
–
5
(c) (5 5)
6
(d) (3
–
6 5)
–
3
5 0
–
2
–
3
(f) (6 2 3 4)
6
(e) (0 3)
0
–
4
3
2. Find the value (or values) of the unknown
in each of the following cases.
(a) (x 5)
2
= (
–
24)
–
4
(b) (x
–
5)
2
= (9)
x
x
(c) (2 x 5) 3 = (28)
4
5
(d) (3 0 x) 1 = (40)
x
3. A wholesaler sells salt in packets of two
sizes: small and large. The amounts of salt
contained in these packets are 500 g and
1 kg. A retailer bought them at sh 7 and
sh 15 respectively.
(a) Write down two column matrices, one
for amount and one for cost.
(b) The retailer ordered 2 dozen large and
3 dozen small packets. Write this
information as a row matrix.
(c) By multiplying matrices, calculate the
total amount of salt ordered.
(d) How much did the order cost?
4. A farmer took the following to the market:
5 boxes of tomatoes, 5 sacks of cabbages
and 15 packets of onions. He sold the
vegetable at the rates of sh 700 per box,
sh 450 per sack and sh 600 per packet
respectively. Use matrix to nd how much
the farmer got.
5. It costs an average sh 12 to feed a goat
per day and sh 30 to feed a cow per day.
One farmer has 5 cows and 15 goats and
another has 10 cows and 10 goats. Find the
dierence in their expenditure per day on
feeding their animals.
General matrix multiplication
Consider the following example.
Example 8.5
Mrs. Kamau bought 2 kg of meat at sh 140
per kilogram, 3 packets of Unga maize meal
at sh 52 per packet and 4 loaves of bread
at sh 19 per loaf. At the same time and at the
same store, Mrs. Kones bought 3 kg of meat,
2 packets of Unga maize meal and 5 loaves of

the same quantities of food items at a store
where the prices were sh 150 per kilogram of
meat, sh 50 per packet of Unga maize meal and
sh 20 per loaf of bread. Use matrix method to

Solution
Mrs. Kamau, 1st store:
140
(2 3 4) 52 = 2 × 140 + 3 × 52 + 4 × 19
19
= (512) i.e. she spent sh 512.
Mrs. Kamau, 2nd store:
150
(2 3 4) 50 = (2 × 150 + 3 × 50 + 4 × 20)
20
= (530) i.e. she spent sh 530.
Mrs. Kones, 1st store:
140
(3 2 5) 52 = (3 × 140 + 2 × 52 + 5 × 19)
19
= (619) i.e. she spent sh 619.
( )
( )
( )
1
2
–
109
( )
( )
( )
( )
( )
( )
( )
( )
( )
Mrs. Kones, 2nd place:
150
(3 2 5) 50 = (3 × 150 + 2 × 50 + 5 × 20)
20
= (650) i.e. she spent sh 650.
Note that in calculating Mrs. Kamau’s
expenditure, the rst matrix is the same in each
case. We can combine these multiplications and
write for Mrs. Kamau.
140 150
(2 3 4) 52 50 = (512 530)
19 20
Likewise, we can write for Mrs. Kones:
140 150
(3 2 5) 52 50 = (619 650)
19 20
Finally, since the matrix giving the prices is the
same in each case, we can now represent the
whole calculation by a single product as:
2 3 4
140 150
512 530
3 2 5
52 50 =
619 650
19 20
We see that:
In working out the product of two matrices,
the rows of the left-hand matrix are
combined, in turns, with the columns of the
right-hand matrix.
Example 8.6
If A =
3 2
and B =
2
–
1
, work out the
0
–
2
–
2 4
product AB.
Solution
AB =
3 2 2
–
1
.
0
–
2
–
2 4

row of A with the columns of B in turn to give
3 2 2
–
1
=
2 5
.
*
*
–
2 4
* *
(
)
( )
( )
(
)
(
)
(
)
(
)
(
)
( )(
)
We then combine the second row of A with the
columns of B to give
* * 2
–
1
=
* *
.
0
–
2
–
2
4
4
–
8
Thus,
3 2 2
–
1
=
2 5
.
0
–
2
–
2 4 4
–
8
In the product AB, we say that B has been
pre-multiplied by A.
If the product matrix AB, which is in Example
8.6, is
a
11
a
12
, each of its elements is
a
21
a
22
obtained as shown in Table 8.5.
Elements Got by combining
of product Row of Column of
matrix 1st matrix 2nd matrix
a
11
1st row 1st column
a
12
1st row 2nd column
a
21
2nd row 1st column
a
22
2nd row 2nd column
Table 8.5
This enables us to write down any required
element in the product immediately when
dealing with more complicated products.
Example 8.7
Write down the values of c and e in the product
3 1 2
1 3
–
2 0 7
–
2 1
=
a b c
2 7
–
4 1 1
–
2 1 d e f
4 0
–
3
Solution
Element c is obtained by combining the 1st row
of the 1st matrix with the 3rd column of the 2nd
matrix.
∴ c = 1 × 2 + 3 × 1 +
–
2 × 1 + 0 ×
–
3 = 3.
Element e is obtained by combining the 2nd row
of the 1st matrix with the 2nd column of the 2nd
matrix.
∴ e = 2 × 1 + 7 ×
–
2 +
–
4 ×
–
2 + 1 × 0 =
–
4.
(
)
(
)
(
)
( )
110
( )( )
( )
( )
( )
( ) ( )
( )
( )( )
Compatibility in multiplication
0 5
1 3
Given that P = and Q = 4 2 , which
2 1
5 0
of the products PQ and QP is possible?
Work out the one which is possible.
0 5
1 3
Try PQ: PQ = 4 2 = ?
2 1
5 0
1 3
0 5
Try QP: QP = 4 2 = ?
5 0
2 1
How many rows and columns are in the
possible product?
You should have now noticed that:
Two matrices can be multiplied if, and only if,
the number of columns in the matrix to the
left is the same as the number of rows in the
matrix to the right. When this is the case,
the two matrices are said to be compatible
for multiplication. For example, a (p × q)
matrix and a (q × r) matrix are compatible
in that order but a (q × r) matrix and a (p ×
q) matrix are not compatible.
You should also have noticed that:
The product of two matrices has the same
number of rows as the rst matrix and the
same number of columns as the second
matrix. Thus, a (p × q) matrix will pre-
multiply a (q × r) matrix to give a (p × r)
matrix,
i.e (p × q)(q × r) ⇒ (p × r)
Example 8.8
If P =
1 3
and Q =
2 0
PQ and
–
2
–
1
–
1 2
QP. What do you notice?
( )
( )
( )
( )
( )( )
( )
( )
(
)
( )
( )
( )
( )
Solution
PQ = 1 3 2 0
–
2
–
1
–
1 2
= 1 × 2 + 3 ×
–
1 1 × 0 + 3 × 2
–
2 × 2 +
–
1 ×
–
1
–
2 × 0 +
–
1 × 2
=
–
1 6
–
3
–
2
QP = 2 0 1 3
–
1 2
–
2
–
1
= 2 × 1 + 0 ×
–
2 2 × 3 + 0 ×
–
1
–
1 × 1 + 2 ×
–
2
–
1 × 3 + 2 ×
–
1
= 2 6
–
5
–
5
We notice that:
PQ ≠ QP, i.e matrix multiplication is not
commutative
Exercise 8.5
1. In which of the following pairs of matrices
is it possible to pre-multiply the second
matrix by the rst? Work out the product
where possible.
(a) 3 2 and 3 (b) 2 and 1 2
1 4 2 3
4 2
(c) 3 and (3 7) (d) (3 7) and 3
1 1
4 8
2 4
(e) and 1 2
1 5
3 5
2 1 3
2 5 0
(f) and 1 1 4
4 2 1
3 0 1
2 1 3 6 1 0
2. Given that A = 0 4 2 and B = 2 3
5
7 1 8 4 4 1
nd AB and BA.
Do your results lead you to the same
conclusion as that of Example 8.8?
(
)
(
)
– – – – –
– – –
– – – – –
– – –
( )
(
)
( )
111
3. Given that
4 3 1
= k
2.5
, nd
2 1 2 1
the value of k.
4. Find the values of x and y if
(a) x 2 3 = 8
–
1 y 1 0
(b) (x y)
3 1
= (9 6)
0 3
5. (a) A matrix B is such that
(2 3
–
1)B = (9 2 10).
What is the order of the matrix B?
(b) A 3 × 5 matrix is multiplied by a 5 × 8
matrix. What is the order of the
product matrix?
6. Work out the following products.
(a) 4 3 1 0 (b) 1 0 1 2
5 6 0 1 0 1 3 4
(c) 0 2 5 2 (d) 2 0 0 3
0 5 0 0 0 2 3 0
1 0 0 1 2 3
(e) 0 1 0 3 2 1
0 0 1 2 1 3
(f) p q w x
r s y z
7. Given that A = 3 0 , B =
–
1 0 and
0 3
1 2
C = 1 3 2 ,
2 4 1
(a) Work out BC and A(BC)
(b) Work out AB and (AB)C
(c) Complete the following statement
about the implied property of matrix
multiplication. Matrix multiplication
is ___________.
8. If P = 4 0 , Q = 0
–
1 and R = 1 3
0 4 2 1 2 4
calculate
(a) PR (b) QR (c) PR + QR and
(d) (P + Q)R
(e) Complete the following statement
about the implied property of matrix
multiplication. Matrix multiplication
is ___________ over addition.
(f) Show that (3P + 5Q)R = 3PR + 5QR.
9. When shopping for Christmas, Penina
bought 2 skirts, 3 blouses and 1 pair of
shoes at her local urban centre where the
prices were sh 360 per skirt, sh 300 per
blouse and sh 850 per pair of shoes. In the
main town, a bargain shop was oering the
same commodities at sh 350, sh 270 and
sh 700 respectively.
(a) How much would she have saved by
going to buy the items at the bargain
shop if the fare was sh 50 return?
(b) Tessie’s purchases were 3 skirts, 2
blouses and 2 pairs of shoes.
(i) Express Penina’s and Tessies’s
purchases as a 2 × 3 matrix (A)
and
(ii) the prices in the urban centre and
bargain shop as a 3 × 2 matrix (B).
(c) Find the matrix product P = AB. What
does P tell you?
Find also the matrix product BA. What
does this tell you?
10. In Form 4N, there are 5 candidates for
Physics, 12 for Chemistry, 20 for Biology
and 16 for Commerce. The numbers in
Form 4S are 6 for Physics, 10 for Chemistry,
25 for Biology and 12 for Commerce.
Each Physics pupil is required to buy 4
textbooks, 3 exercise books and 2 graph
books. Each Chemistry pupil buys 3
textbooks, 3 exercise books and 1 graph
book. A Biology pupil buys 2 textbooks,
3 exercise books and 1 graph book. A
Commerce pupil buys 2 textbooks,
2 exercise books and no graph book.
Find, by matrix multiplication, the total
number of each kind of book bought by
each class.
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( ) ( )
( )
( )
( )
( )
( )
( )
( )
(
)
( )
( )
( )
112
Identity and zero matrices
Look at the results that you obtained in Exercise
8.5, Question 6, parts (a), (b) and (e). What do
you notice?
Multiply any other 2 × 2 matrix by
1 0
, and
1 0 0
0 1
any other 3 × 3 matrix by 0 1 0 .
0 0 1
Does it matter which matrix is the pre-multiplier?
The matrix
1 0
is called the identity (or unit)
0 1
matrix of order 2. It is denoted by I
2
.
1 0 0
The matrix
0 1 0 is the identity (or unit)
0 0 1
matrix of order 3, denoted by I
3
.
Note that an identity matrix is a square matrix
which has only 1’s in the diagonal from the top
left corner to the bottom right corner and 0’s
elsewhere. The 1’s are said to be in the leading
or main diagonal.
Note: The matrices I
2
and I
3
behave like the
number 1 in the multiplication of numbers, e.g.
3 × 1 = 1 × 3 = 3, a × 1 = 1 × a = a.
Thus:
If M is a square matrix, and I is the identity
matrix of the same order as M, then
IM = MI = M
I is, more specifically, a multiplicative
identity of the same order as M.
In Question 6(c) of Exercise 8.5, did you obtain
a matrix with all its elements zeros? Such a
matrix is called a zero matrix.
A zero matrix is a matrix of any order, not
necessarily square, with all its elements zero.
It is denoted by O, and has the property that
OM = MO = O, provided that O and M are
compatible both ways.
( )
( )
( )
O is also an additive identity, i.e.
M + O = O + M = M, provided that O and
M are of the same order. Thus, O behaves
like the number zero.
Exercise 8.6
1. Write down the additive identity of each of
the following matrices.
2 3 3
5 4 3
4 5
(a) 4 1 7 (b) (c) 5 4
2 7
–
2
2
–
9 8
3 8
2. For each of the following matrices, write
down a multiplicative identity, making
sure to specify the order in which it is an
identity.
5 1 7
3 2 1
(a) M = 2
–
1 4 (b) N =
3 6
–
8
5 4 3
1 3
(c) P = 3 4
5 5
6 1 2 3
3. If P =
2 3 1 6
,
2 2 3 7
8 4 3 1
what is (a) I
4
P (b) PI
4
4. Work out the product
3 1 1 2
.
4 3
–
2
–
4
What do you notice? Is it always true that if
XY = Y, then X = I?
5. If A =
1 2
and B =
–
2
4
, work out
2 4
1
–
2
the product AB.
Is it always true that if AB = O, then
A = O or B = O?
Is it true that if X = O and Y = O, then
XY = O?
( )
( )
( )
( )
(
)
( )
( )
( )
( )
(
)
( )
(
)
113
( )
( )
( )( )
( )
|
|
( )
( )
2
3
–
3
2
–
m
n
––
n
m
––
( )
( )
( )
( )
( )
(
)
6. Given that
–
1 2 a b
= O, nd
–
2 4 c d
possible values of a, b, c and d which are
non-zero.
Determinant and inverse of a 2 × 2 matrix
Determinant
a b
c d
is a 2 × 2 matrix. The elemments a and d
form the leading (or main or principal)
diagonal. The elements b and c form the other
(or secondary) diagonal.
The number obtained by subtracting the product
of the elements of the other diagonal from the
products of the elements of the leading diagonal
is known as the determinant of the matrix.
If A =
1 2
, the determinant of A is
3 4
1 × 4 – 2 × 3 =
–
2. This is written in short as
det A =
–
2
An alternative notation is
1 2
=
–
2 where the
3 4
straight lines down along the sides indicate
“determinant”. This is not to be confused with
the matrix
1 2
itself.
3 4
Thus:
If M =
a b
, the determinant of M is
c d
det M ≡ |M| ≡
a b
= ad – bc
c d
Inverse of a 2 × 2 matrix
Consider the matrices A =
2 1
and B =
3
–
1
5 3
–
5 2
Find the product AB. Find also the product BA.
What do you notice?
The matrices A and B are said to be inverses of
each other. They behave like numbers which are
reciprocals of each other e.g.
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
|
|
( )
( )
( )
(
)
( )
× = 1, × = 1
Thus:
If A is a square matrix, then another matrix B,
of the same order, is called the inverse of A if
AB = BA = I, where I is the identity matrix
of the same order as A and B. The inverse of
A is denoted as A
–
1
.
Note that if B is the inverse of A, then A is the
inverse of B.
Example 8.9
Verify the following pairs of matrices are
inverses of each other.
(a)
2 1
and
1
–
1
(b)
4 7
and
2
–
7
1 1
–
1 2 1 2
–
1 4
(c)
5 3
and
–
1
–
3
(d)
–
5 3
and
–
2
–
3
–
2
–
1 2 5 3
–
2
–
3
–
5
Solution
(a)
2 1 1
–
1
=
2 – 1
–
2 + 2
1 1
–
1 2 1 – 1
–
1 + 2
=
1
0
= I
0 1
1
–
1 2 1
=
2 – 1 1 – 1
–
1 2
1 1
–
2 + 2
–
1 + 2
=
1 0
= I
0 1
∴ the two matrices are inverses of each
other.
(b)
4 7 2
–
7
=
8 – 7
–
28 + 28
1 2
–
1 4 2 – 2
–
7 + 8
=
1 0
= I
0 1
2
–
7 47
=
8 – 7 14 – 14
–
1 4 12
–
4 + 4
–
7 + 8
=
1 0
= I .
0 1
∴ the two matrices are inverse of each
other.
(
)
( )
114
( )
Compare, for each pair of matrices in Example
8.9, the elements in the leading diagonals. What
do you notice? Compare also the elements in
the other diagonals.
You should notice that:
1. The elements in the main diagonals are
interchanged.
2. The signs of the elements in the other
diagonals are reversed.
Example 8.10
Find the inverse of each of the following matrix.
Check your answer by multiplication.
(a)
6 11 2 4
1 2 3 8
Solution
Interchange the elements in the leading
diagonal:
2
6
Reverse the signs of the elements in the other
diagonal:
2
–
11
is the required inverse matrix.
–
1 6
Check:
6 11 2
–
11
=
12 – 11
–
66 + 66
1 2
–
1 6 2 – 2
–
11 + 12
=
1 0
= I
0 1
2
–
11 6 11
=
12 – 11 22 – 22
–
1 6 1 2
–
6 + 6
–
11 + 12
=
1 0
= I
0 1
∴ the two matrices are inverses of each other.
Notice that the determinant of each of the
matrices in Example 8.9 and 8.10 is 1.
(b) Now consider the matrix A =
2 4
.
3 8
Find det A. Can we nd A
–
1
in the same
manner as in Example 8.10?
Let us try it.
Interchange the elements in the leading
diagonal.
8
2
Reverse the signs of the elements in the other
diagonal.
8
–
4
–
3
2
Check:
2 4 8
–
4
=
4 0
= 4
1 0
= 4I
3 8
–
3 2 0 4
0 1
and
8
–
4 2 4
=
4 0
=
4
1 0
= 4I
–
3 2 3 8 0 4 0 1
∴
8
–
4
is not the inverse of
2 4
but it
–
3 2 3 8
almost is!
We have obtained 4I, and det A = 4 (conrm!).
Therefore, to get A
–1
we divide
8
–
4
by 4.
3 2
∴ A
–1
=
8
–
4
=
2
–
1
–
3 2
Check this by working out AA
–
1
and A
–
1
A.
Now consider B =
4 6
.
2 3
Let us nd the inverse in the same way as in
Example 8.10.
det = 0
Its inverse should be
3
–
6
–
2 4
Check:
4 6
3
–
6
=
12 – 12
–
24 + 24
2 3
–
2 4
6 – 6
–
12 + 12
=
0 0
= 0
0 0
This matrix does not have an inverse.
( ) ( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( ) ( )
–
3
4
––
1
2
–
( )
( )
( )( )
( )
( )
(
)(
) ( )
( )
(
)( ) ( )
(
)
( )
(
)
(
)
(b)
115
Exercise 8.7
1. Find the determinants of each of the
following matrices
(a)
3
–
2
(b)
4
–
3
(c)
4 5
5 0
–
2 7 4
–
1
(d)
2 5
(e)
1 2
(f)
0 1
–
2
–
4
2. Which of the following matrices are
singular and which are non-singular?
(a)
3
–
2
(b)
2 3
(c)
–
1 2
7
–
5 2 3 4
–
6
(d)
–
6 2
(e)
1 1
(f)
–
1
–
1
3
–
1 1 1 1
–
1
3. Find, where possible, the inverse of each
of the following matrices. Verify in each
case that if the matrix is M, then
MM
–
1
= M
–
1
M = I.
(a)
4 5
(b)
1 4
(c)
12 4
3 4 0 2 6 2
(d)
2 2
(e)
–
3 6
(f)
20 20
4
–
1
–
2 4 30 50
4. Solve for the unknowns in
(a)
2
–
2 a b
=
1 0
3 1 c d 0 1
(b)
p q 3 4
=
1 0
r s 5 7 0 1
5. Solve for the unknowns given that the
following are singular matrices.
(a)
p 3
(b)
x 4
p p 1 x – 3
Simultaneous linear equations in two
unknowns
Consider the simultaneous equations
3x + y = 4
4x + 3y = 7
Thus, the procedure for getting the inverse of a
2 × 2 matrix is:
Find the determinant of the matrix. If it
is zero, then there is no inverse. If the
determinant is non-zero,
1. interchange the elements in the leading
diagonal,
2. reverse the signs of the elements in the
other diagonal, and
3. divide the matrix obtained by the
determinant of the initial matrix.
Thus,
if A =
a b
, A
–1
=
1 d
–
b ,
c d ad – bc
–
c a
provided that ad – bc ≠ 0.
Note:
If a matrix does not have an inverse, it is said to
be a singular matrix. A matrix with an inverse
is said to be non-singular.
Example 8.11
Do the following matrices have inverses? If so,

(a) A =
2 4
(b) B =
3
–
2
3 5
–
9 6
Solution
(a) det A = 2 × 5 – 4 × 3 =
–
2
∴ the matrix has an inverse.
A
–1
=
–
5
–
4
=
–
3 2
=
–
2.5 2
1.5
–
1
(b) det B = 3 × 6 – (
–
2) × (
–
9) = 0
∴ the matrix does not have an inverse,
i.e. B is a singular matrix.
( )
( )
( )
( )
( )
–
4
–
2
––
–
2
2
––
–
3
–
2
––
–
5
2
––
1
2
–
(
) (
)(
)
(
) ( )
(
)
(
) ( )(
)
(
) (
)(
)
( )
( )
( )
–
1
4
––
1
8
–
1
4
–
1
2
–
(
) (
)(
)
(
) ( )(
)
( )
(
)
(
)
( ) (
)
(
)
(
)
116
( ) ( )
( )
( )
( ) ( )
( )
( )
( )
( )
( )
( )( )
( )
( )
( )
( )
( ) ( )( )
The solutions are given by
x
=
1 d
–
b p
,
y ad – bc
–
c a q
provided that ad – bc ≠ 0.
Example 8.12
Use matrix method to solve the simultaneous
equations
m + 4n = 70
3m + 2n = 60.
Solution
The matrix equation is
1 4 m
=
70
3 2 n 60

–
2
–
4
–
3 1
Premultiplying both sides of the matrix
equation by the inverse gives
m
=
–
2
–
4 70
n
–
3 1 60
=
–
–
100
–
150
i.e.
m
=
10 .
n 15
∴ m = 10 and n = 15.
Note: Division by the determinant is left until
the last step. This often eases the calculation,
especially where the inverse matrix contains
fractions.
Example 8.13
Use matrix method to solve the simultaneous
equations
x + 3y = 6
2x + 6y = 3.
Solution
The matrix equation is
1 3 x
=
6 .
2 6 y 3
( )
( )
( )
1
10
––
1
10
––
1
10
––
( )
Using elimination, substitution or graphical
method, we obtain the solution x = 1, y = 1.
Matrices can also be used to solve the given
simultaneous equations. The equation can be
written in matrix form as:
3 1 x
=
4
4 3 y 7
Conrm this by working out the product on
the LHS and equating its elements to the
corresponding elements of the matrix on the
RHS.
The matrix
3 1
is called the coecients
4 7
matrix of the simultaneous equations. The
matrix 4 is called the constants matrix.
7
The inverse of the coecients matrix is
Pre-multiply both sides of the matrix equation
by the inverse. Thus,
3 1 x
=
4
4 3 y 7
⇒
1 0 x
=
× 4 + × 7
0 1
y
× 4 + × 7
i.e.
x
=
1
y 1
∴ x = 1 and y = 1.
Thus,
Given the two simultaneous equations
ax + by = p
cx + dy = q
Write down the matrix equation as
a b x
=
p
c d y q
( )
–
4
5
––
( )
( )
( )
–
1
5
––
3
5
–
3
5
–
–
4
5
––
( )
–
1
5
––
3
5
–
3
5
–
( )
–
1
5
––
3
5
–
3
5
–
–
4
5
––
( )
(
)
( )
( )
( )
3
5
–
–
1
5
––
3
5
–
–
4
5
––
117
1
2
–
Fig 8.1
Since the lines do not intersect, there is no
solution.
Example 8.14
Use matrix method to solve the simultaneous
equations 2x – y = 2
4x – 2y = 4.
Solution
The matrix equation is
2
–
1 x = 2
4
–
2 y 4

2 ×
–
2 – 1 × 4 = 0.
Thus, the matrix has no inverse.
We re-examine the equations and notice that one
equation is simply a multiple of the other. Hence,
these are equations of the same line.
Every point on this line gives a solution to the

1 × 6 – 3 × 2 = 0
∴ the matrix has no inverse.

written in the gradient-intercept form as:
y = x + 2
y = x +
We see that the gradients are the same, but the

are parallel and are as shown in Fig 8.1.
two equations.
solutions.
Examples 8.12 to 8.14 lead us to the geometric
interpretation of the solution of two simultaneous
linear equations. Thus
1. If the determinant of the coefficients
matrix is non-zero, there exists a unique
solution. The two lines represented by
the given equations intersect at a point.
2. If the determinant of the coefficients
matrix is zero, either
(a) the simultaneous equations have no
solution, and represent parallel lines,
or
(b) they have innitely many solutions,
and they represent the same line (i.e
the lines are coincident).
Exercise 8.8
1. Use matrix method to solve the following
pairs of simultaneous equations
(a) x – y = 2 (b) x – 2y = 4
2x + y = 7 2x – y =
–
1
(c) 3x + 2y = 12 (d) 3m + n = 10
5x – 3y = 1 2m + 4n = 0
(e) x + y = 24 (f) 3a – 4b = 1
x – 3y =
–
12 6a – 6b = 5
(g) 4x + 6y = 136 (h) 3p = 4
–
4q
6x + 5y = 164 2q = 5p + 15
(i) 2x + 5y + 1 = 0 (j) 5v = 11 + 3u
3x + 7y = 1 2u + 7v = 3
2. Examine each of the following pairs of
simultaneous equations, give a geometric
interpretation, and solve them where the
solutions exist.
(a) x + y = 5 (b) 6m – 3n = 9
x + y = 10
–
4m + 2n =
–
6
(c) 2x + 3y = 4 (d) 3x – 2y = 4
–
3y = 2x – 2 3y + 2x =
–
6
(e) 3p – q = 6
p – q = 2
–
1
3
––
–
1
3
––
1
2
–
( )
( )
( )
3
2
1
0
–
1
–
2
1 2 3 4- 5 6
–
1
x
y
y = x + 2
–
1
3
––
y = x +
1
2
–
–
1
3
––
118
3. For each of the following pairs of
simultaneous equations, state whether the
solution exists, without necessarily solving
the equations.
(a) 10x – 2y = 50 (b) 3x – 2y = 12
x + 4y = 10 6x – 4y = 24
(c) 7x + 7y = 48 (d) 100x + 7y = 54
2y + 2x = 12 10y +
x = 24
(e) 8x + 4y = 54 (f) 9x + 18y = 108
2y + 4x – 27 = 0 3y + 6x = 36
(g) 3x + 4y = 7 (h) 5x – 33y = 90
2x – 6y = 3 x – 1
y = 3
7
10
––
1
5
–
3
5
–
8
25
––
4. Solve the following pairs of simultaneous
equations using matrix method and,
conrm their solutions using an alternative
method.
(a) x – y – 1 = 0 (b) 2x + 7y = 3
x + y – 2 = 0 x – 3y = 2
(c) 3x + 4y = 7 (d) 2m + 7n = 12
2x + 6y = 3 m – 2n = 5
(e) 3p + 9q = 18 (f) 3a – 7b = 5
p + 2q = 10 a – 4b = 6
119
9
FORMULAE AND VARIATIONS
Formulae
Suppose a rectangle is l cm long and b cm wide.
We know that its area is l × b cm
2
. If we call the
area A cm
2
, then A = l × b, i.e. A = lb.
This relation is called a formula. We use it
whenever we wish to nd the area of a
rectangle.
The formula is said to express A in terms of l
and b.
Example 9.1
If a motorist covers a distance, D km in T hours
at an average speed of S km/h, write down a
relation between D, T and S.
Solution
We know that Total distance travelled (km)
Total time (h) taken
gives average speed in km/h.
∴ S =
D .
T
This formula expresses S in terms of D and
T. In this case, S is called the subject of the
formula.
Using this formula, we can nd the average
speed when the distance travelled in a given
time is known.
Most formulae are given or stated in a
standard form. However, we can construct
our own formulae given specic situations and
conditions.
Example 9.2
The length of a room is n times its width. Its

2
and its breadth is x m long.
Obtain a formula that relates A, n and x.
Solution
Fig. 9.1
 
= nx
Since area is equal to length × breadth,
A = (x) × (nx)
i.e. A = nx
2
∴ the relation between A, n and x is
A = nx
2
.
Exercise 9.1
1. Julia bought x oranges at sh y each. If she
spent sh C, nd a formula relating x, y and C.
She also bought P pawpaws at sh 4y each.
If altogether she spent a total of sh T, nd a
formula for T.
2. A cylindrical metal drum is rolled along
level ground in a straight line. If its
diameter is d cm, nd the distance covered
in one revolution. After n revolutions, the
drum has covered a distance of S metres.
Obtain a formula for this distance.
3. Two pieces of string have lengths S
1
and
S
2
cm. A length of x cm is cut o from
each piece. The rst piece of string is
now half the length of the second. Find a
formula connecting S
1
, S
2
and x, and give
your answer in the simplest form.
4. A car does m km to a litre of petrol, which
costs s shillings per litre. Find the cost of
petrol for a journey of x km.
Breadth = x m

A m
2
120
√
5. A man is paid x shillings per hour for a
normal week of p hours and double this
rate for overtime. Find his wages for a week
of n hours given that n is greater than p.
The subject of a formula
For the purpose of solving a particular problem,
it may be necessary or more convenient to
express a given formula dierently.
For example, the circumference (C) of a circle
of radius r is given by the formula
C = 2πr.
Here, C is the subject of the formula.
Changing the subject of the formula
Suppose we are given the circumference of a
circle and are required to nd the radius. Using
the same notation as before,
2πr = C.
Dividing each side by 2π:
r =
C
2π
This is a formula whose subject is r. It has
been obtained from the rst formula by using
the same method as is employed in solving an
equation. This process is called changing the
subject of the formula.
Example 9.3
Make P the subject of the formula
S =
PR .
l – P
Solution
Write down the formula: S =
PR
l – P
Multiply both sides by (l – P):
(l – P)S =
PR
(l – P)
l – P
Expand the expressions:
lS – PS = PR
Collect the terms containing P to one side of the
equation: lS = PR + PS
Factor out P: lS = P(R + S)
Divide both sides by (R + S):
lS
=
P(R + S)
R + S R + S
   
∴ P =
lS
R + S
Example 9.4
The volume V cm
3
of a right cone of radius r cm
and height h cm is given by the formula
V = πr
2
h.
Make r the subject of the formula.
Solution
Write down the formula: πr
2
h = V
Divide both sides by πh:
r
2
=
V
πh
Multiply both sides by 3: r
2
= 3V
πh
Take the square root of both sides: r =
3V
πh
Note: We only take the positive square root of
both sides because r is a length. In this case, the
negative square root has no meaning.
Example 9.5
Make t the subject of the formula
s – a =
ab
+ a
2
t
Solution
Remove the square root sign by squaring both
sides: (s – a)
2
=
ab
+ a
2
t
Subtract a
2
from both sides:
(s – a)
2
– a
2
=
ab
t
Multiply both sides by t: t{(s – a)
2
– a
2
)}
= ab
Divide both sides by (s – a)
2
– a:
t =
ab
(s – a)
2
– a
2
1
3
–
1
3
–
1
3
–
√
121
√
√
Note: It is not necessary to expand (s – a)
2
. But
if the required subject was within that bracket,
then it would have been necessary to expand
and simplify as far as possible.
Exercise 9.2
1. (a) Use Fig. 9.2 to write a relation that
connects x and y. Hence make x the
subject of the formula.
Fig. 9.2
(b) If the perimeter of the triangle in
Fig. 9.2 is P cm, express P in terms of
b and z. Hence make b the subject of
the formula.
2. The area of a trapezium is given by the
formula A = h(x + y), where A represents
area, x and y the lengths in centimetres
of the parallel sides, and h is the distance
between them. Make h the subject of the
formula.
3. The radius of a sphere is r cm. Its surface
area S is given by the formula S = 4πr
2
,
and its volume V is given by the formula
V = πr
3
.
Find r in terms of (a) S (b) V.
In question 4 to 10, make the letter in the
bracket the subject.
4. P =
wv
2
(a) (r) (b) (v)
32r
5. PT =
w
(v – u) (a) (w) (b) (u)
g
6. d = t
(u – 1)
(u)
u
7. t =
2pr
(p)
p + 2s
8. t =
2s
(s)
g
9.
1
= 1 –
r
(i) (r) (ii) (n)
n 100
10. A = P +
PRT
(i) (P) (ii) (T)
100
11. From the formulae P =
w
+ 1.5 and E =
w
7.5 5P
express E in terms of P only.
12. Given that S =
3t(t – x)
make x the
8
subject of the formula.
Find the value of x when S = 3 and t = 67 .
13. Rearrange the formula
ml = w(T – t) – m(100 – T) to give m
in terms of the other symbols. Find the
value of m to the nearest whole number if
l = 540, w = 1 000, T = 20 and t = 0.
14. If the perimeter of a semi-circular plate of
radius r cm is P cm, write down a formula
connecting P and r. Hence make r the
subject of the formula.
Variation
When used in mathematics, variation is
concerned with the ways in which one variable
depends on one or more other variables. It is
very much related to the ratio and proportion
that we did in Form 1. It may, therefore, be
useful to revise these ideas rst.
If a car travels at a uniform speed of 60 km/h,
in 10 minutes, it travels 10 km;
in 20 minutes, it travels 20 km;
in 30 minutes, it travels 30 km; and so on.
In this case, we see that distance varies as
the time varies, i.e. if the time is doubled, the
distance travelled is also doubled.
4
3
–
1
2
–
b cm
z cm
y
x
x
1
2
–
122
2
3
–
1
2
–
Thus, the ratio of any two distances is the same
as that of the corresponding times taken,
i.e.
10 km
=
and
10 minutes
=
20 km 20 minutes
20 km
=
and
20 minutes
=
30 km 30 minutes
Direct variation
We say that distance is directly proportional
to the time taken, or distance varies directly as
time.
The symbol used for ‘varies as’ is ∝.
Thus, if distance is D km and time is T hours,
then D varies directly as T, i.e. D ∝ T.
This means that
D
is a constant.
T
In general, if M ∝ L, then L ∝ M and
M
= k and
L
= c, where k and c are
L
M
constants of variation or proportionality.
Example 9.6
Given that A ∝ B and that A = 1 when B = ,

A when B = 0.4 and B when A = 7.5.
Solution
A ∝B. Therefore,
A
is a constant
B

Thus,
A
= k Therefore, A = Bk.
B
1 = k (substituting given values of A and B)
∴k =
×
k = (This is the constant of proportionality)
∴ A = B. (This is the required law connecting
A and B)
When B = 0.4,
A = × 0.4 (since A = k × B)
= 0.9
1
2
–
2
3
–
5
6
–
7
8
–
5
6
–
7
8
–
6
5
–
15
8
––
9
4
–
9
4
–
9
4
–
7.5 × 4
9
––––––
1
3
–
5
6
–
7
8
–
1
3
–
1
2
–
9
4
–
y – y
1
= k
x – x
2
or
y
=
x
= k
y
1
x
1
When A = 7.5
7.5 =
B
B =
= 3 .
Plot the points (1 , ), (0.9 , 0.4) and (7 , 3 ),
and join them using a straight line. What
do you notice about the gradient of the line?
Which special point does the line pass through?
You should have noticed that:
If the values of A are plotted against the
corresponding values of B, the result will be a
straight line through the origin. The gradient
of the line will be equal to the consant of
variation.
In general, if two variables x and y are directly
proportional, then y = kx where k is a constant.
Since y = kx is a linear relation, a graph of y
against x would be a straight line through the
origin, gradient k (Fig. 9.3).
Conversely, if the graph of y against x is linear
through the origin, then y varies directly as
x, and the gradient of the line represents the
constant of proportionality.
Fig. 9.3
Exercise 9.3
1. If A ∝ B and A = 140 when B = 35, nd the
law of variation and the value of B when
A = 176.
2. If x ∝ y and x = 15 when y = 18, nd x
when y = 15 and y when x = 21.
•
•
(x , y)
(x
1
y
1
)
x
y
0
123
3. Table 9.1 shows corresponding values of
two variables x and y. Show graphically
that y ∝ x. Find the law of variation in the
form y = kx.
x 3 4 6 9 12
y 2.6 3.5 5.2 7.8 10.4
Table 9.1
Use your graph to read o x when y = 7,
and y when x = 3.8.
4. The quantity, Q, of a certain commodity,
in kilograms, and the corresponding cost, P,
in shillings are given in Table 9.2.
Q 2 4 5 7
P 80 160 200 280
Table 9.2
Show graphically that Q ∝ P. Find the law
connecting Q and P. Read o the value of
Q when P = 450.
Non-linear direct variation
Direct variation does not always give rise to a
linear relation.
Consider Table 9.3.
A 20 45 80 125 180 245 320
h 2 3 4 5 6 7 8
Table 9.3
If A ∝ h, then the ratio should be a constant.
But in this case, it is not!
∴ A and h are not directly proportional.
Now copy and complete Table 9.4.
A 20 45 80 125 180 245 320
h 2 3 4 5 6 7 8
h
2
4 9
Table 9.4
From Table 9.4, you should notice that A varies
directly as the square of h, i.e. = 5.
∴ A = 5h
2
We can also say that h
2
varies directly as A.
Thus h
2
∝ A
i.e. h
2
= kA
Thus 4 = 20k (from 1st pair of values in
Table 9.4)
∴k = = .
In this case, the constant of proportionality is .
Note: If h
2
varies directly as A, then h varies
directly as the square root of A.
i.e h
2
∝ A
⇒ h
2
= kA
⇒ h = ± √kA
= ± √k
.
√A
Since k is a constant, ± √k must also be a
constant, say c.
∴h = c√A
This means h ∝ √A
By graphing A against h
2
, verify that A and h
2
are directly proprtional.
Note also that: If A ∝ h
2
, the result of
multiplying h by 2 means multiplying A by 2
2
.
In general, if y ∝ x
n
, = k ⇒ y = kx
n
where
k is a constant. Plotting y against x
n
gives a
straight line through the origin, gradient k.
Exercise 9.4
1. If y = 3 when x = 5, nd the relationship
between x and y if
(a) y ∝ x (b) y ∝ x
3
(c) y ∝ √x
In each case, nd the value of y when x = 7.
2. Copy and complete the following:
(a) If A ∝ r
2
, r ∝ __ (b) If V ∝ r
3
, r ∝__
(c) If x ∝ √y, y ∝ __
3. If y ∝√x and y = 0.4, when x = 4,
(a) nd y in terms of x.
(b) nd y when x = 100 and x when y = 1.4.
A
h
2
––
A
h
––
4
20
––
1
5
–
1
5
–
y
x
n
––
124
1
4
–
4. If y varies directly as the cube of x, and
y = 50 when x = 5,
(a) nd y in terms of x.
(b) nd y when x = 1 and x when y = 6 .
Inverse variation
Suppose a motorist is to cover a xed distance,
driving non-stop. We know that the faster he
drives, the shorter the time he takes.
Suppose the distance is 240 km.
If his average speed is 40 km/h, he takes
= 6 hours.
If his average speed is 60 km/h, he takes
= 4 hours.
If his average speed is 80 km/h, he takes
= 3 hours.
If his average speed is 120 km/h, he takes
= 2 hours, and so on.
In this case, the higher the speed, the shorter
the time taken to cover the same distance. The
average speed is said to vary inversely (or
indirectly) as time.
If S represents average speed and T represents
time in hours, then ‘S varies inversely as T’ is
written as
S ∝.
Just as in direct variation, S ∝ means that
ST is a constant.
i.e. S × = constant
or ST = constant.
If S ∝ , then ST = k where k is a
constant.
Note: If the values of S are plotted against the
corresponding values of T, the resulting graph
would be a curve as in Fig. 9.4.
This graph shows that as the values of T increase,
the values of S decrease (approaching zero).
Similarly, as the values of S increase (towards
240 km), the values of T decrease.
Now construct a table of values giving
corresponding values of S and for the
motorist travelling a distance of 240km. Draw
the graph of S against . Describe fully the
resulting graph.
In general, if y varies inversely as the nth
power of x, written
y ∝
then y ÷ is a constant
i.e.
= c
⇒ yx
n
= c
or y =
.
The graph of y against is a straight line.
Example 9.7
The number of beats per minute of a pendulum
varies inversely as the square root of its length.
If a pendulum which is 49 cm long makes 36
beats per minute, how many beats per minute
would a pendulum which is 144 cm long make?
Solution

number of beats per minute be n.
Thus, n ∝.
120
T
S
2 4 6 8 10 12 14
100
80
60
40
20
0
Time in hours
Speed in km/h
240
40
–––
240
120
–––
240
80
–––
240
60
–––
1
x
n
––
1
x
n
––
1
x
n
––
y
––
1
x
n
––
1
T
––
T
1
––
1
T
––
1
T
––
1

––
c
x
n
––
1
T
––
1
T
––
125
Solution

Since x ∝ y
3
z,
then x = ky
3
z
When y = 2 and z = 3, x = 4 ,
x = ky
3
z becomes
4 = k × 2
3
× 3
k = ×
=
∴x = y
3
z
(a) When y = 5 and z = 4,
x = × 5
3
× 4
= 93 .
When x = 21 and z = 3 , 21 = × y
3
×
⇒ y
3
=
 = 32
∴ y = √32 = 3.175
= 3.18 (2 d.p.)
x = y
3
z
(b) y is doubled: let the new y be y
1
= 2y.
z is halved: let the new z be z
1
= z.
 
1
.
Since x = y
3
z
x
1
= y
1
3
z
1
= (2y)
3
( z) (since y
1
= 2y and
z
1
= z.)
= × 2
3
y
3
× z
= 2
3
× ( y
3
z)
= 4( y
3
z)
But y
3
z = x
∴ x
1
= 4x
∴when y is doubled and z is halved, x is
multiplied by 4.
1
3
–
21 × 16 × 2
3 × 7
–––––––––
252
12
–––
1
2
–
1
2
–
3
16
––
3
16
––
1
2
–
1
2
–
3
16
––
3
4
–
1
2
–
7
2
–
1
2
–
3
16
––
3
16
––
3
16
––
9
2
–
1
8 × 3
––––
3
3
16
––
1
3
–
i.e, n = , where c is a constant
∴ n√l = c
When l = 49, and n = 36,
36√49 = c
⇒ c = 252 (only the positive √49 is
meaningful)
∴ n√l = 252
When l = 144
n√144 = 252
∴ n =
=
= 21
Thus, a pendulum 144 cm long makes 21 beats
per minute.
Joint variation
It is common for one variable to depend on
variables. For example, the volume of a cone
is given by the formula V = πr
2
h. In this
formula, π is the constant.
We say that V varies directly as h and as the
square of r. This is called joint variation.
Note:
In joint variation, the subject of the formula
is on one side of the equal sign while all the
other quantities are on the other side and
make only one term, no matter how many
quantities there are.
Example 9.8
The variable x varies jointly as z and as the
cube of y. When y = 2 and z = 3, then x = 4 .
(a) Find x when y = 5 and z = 4 and y when
x = 21 and z = 3 .
(b) What happens to x when y is doubled and
z is halved?
252
√144
––––
1
√ l
––
1
2
–
3
16
––
1
2
–
1
2
–
1
2
–
3
16
––
3
16
––
3
16
––
3
16
––
126
Exercise 9.5
1. Write down the relationship between the
variables given below.
(a) a varies directly as b.
(b) m is proportional to n.
(c) x is inversely proportional to y.
(d) P varies inversely as y.
(e) r varies jointly as s and as t.
(f) R is proportional to m and p.
(g) w varies directly as the square of x and
inversely as y.
(h) c varies directly as d and inversely as g
and the square of f.
2. If y varies inversely as x and y = 12 when
x = 6, nd y in terms of x. Find y when
x = 8 and x when y = .
3. Suppose y varies inversely as the square of
x. If y = 9 when x = 10, nd y in terms of x.
Also nd y when x = 6 and x when y = 4.
4. P varies directly as the cube of t and
inversely as the square root of n. When
P = 16, t = 2 and n = 9. Find n in terms of
P and t.
5. Suppose z varies directly as the square of x
and inversely as the square root of y.
(a) Find z in terms of x and y.
(b) If x increases by 20% and y decreases
by 19%, nd the percentage change in z.
(c) If z = 3 when x = 6 and y = 18, nd z
when x = 12 and y = 25.
6. A quantity w varies jointly as h and the
square of r.
(a) Given that w = 15 when h = 3, r = 2 ,
nd
(i) w in terms of r and h.
(ii) w when h = 1 and r = 10.
(b) If h increases by 20% and r by 50%,
nd the percentage change in w.
7. If z varies directly as x and inversely as y,
(a) nd the percentage change in z if x
increases by 20% and y increases by 25%.
(b) if z = 9 when x = 12 and y = 2, nd
(i) z in terms of x and y.
(ii) z when x = 20 and y = 15.
8. If P ∝ , and x ∝ vt, how does P vary
(a) with v and t? (b) with x and t?
9. In the formula d = k
.
, k is a constant.
If w is increased by 50% and B by 20%,
nd the percentage change in l if d remains
unaltered.
Partial variation
If a relation consists of three or more terms, it
represents neither direct nor inverse variation.
It does not represent joint variation either,
although the separate terms may satisfy one
or the other type of variation. For example, a
motorist calculated that the total weekly cost of
running his car was given by the formula
C = 500 + 45d
where d represents the distance travelled, in
kilometres, in the week and C represents the
cost in shillings. In this formula, C is the sum
of the two terms, one of which is a constant,
and the other varies as d. We say that the cost is
partly constant and partly varies as d.
Similarly, in the formula
y = ax +
where a and b are constants and x and y are
variables, we say that y partly varies as x and
partly varies inversely as x.
Relations like C = 500 + 45d and y = ax + are
examples of partial variation.
In these examples, the variation has two terms
and each term must have a constant. Therefore,
we have to nd two constants of proportionality.
1
2
–
1
2
–
v
2
x
––
wl
2
B
–––
b
x
–
b
x
–
127
Example 9.9
The variable x is partly constant and partly
varies inversely as y. When y = 3, x = 1 and
when y = 6, x = 3. Find the equation connecting
x and y. Find x when y = 4.
Solution
Since this variation is in two parts, there must
be two constants, one for each part.
x ∝a + (where a is a constant)
∴x = a + (where b is another constant)
When y = 3, x = 1
∴x = a + becomes 1 = a +
When y = 6, x = 3
∴x = a + becomes 3 = a +
By substituting the given conditions, we have
two simultaneous equations to solve for a and b:
 
 
Subtracting (ii) from (i) gives
=
–
2
∴ b =
–
12
Substituting
–
12 in (i) gives
a – = 1
∴a = 5
∴x = 5 – is the required equation
When y = 4
x = 5 –
= 2.
Example 9.10
The variable w varies partly as x and partly as
the square of x. When x = 2, w = 14 and when

when w = 2.
Solution
Since this is partial variation, there must be two
constants.
w ∝ x + x
2
i.e. w= ax + bx
2
(where a and b are constants).
When x = 2, w = 14
∴w = ax + bx
2
becomes 14 = 2a + 4b ...(i)
When x = 4, w = 44
∴w = ax + bx
2
becomes 44 = 4a + 16b

Now solve equations (i) and (ii) simultaneously:
 

8b = 16 [subtracting 2 × (i) from (ii)]
b = 2
2a + 8 = 14 [substituting 2 for b in (i)]
2a = 6
a = 3
∴w = 3x + 2x
2
is the required relationship.
When w = 2, 2x
2
+ 3x = 2
2x
2
+ 3x – 2 = 0
(Now solve the resulting quadratic equation)
(x + 2)(2x – 1)
= 0
x = or
–
2.
Exercise 9.6
1. The variable y is partly constant and partly
varies as x. When x = 2, y = 16 and when
x = 7, y = 31. Find y in terms of x.
2. The variable x is partly constant and partly
varies as y. When y = 2, x = 0 and when
y = 6, x = 20. Find the law of variation, and
also x when y = 3.
3. P varies partly as Q and partly as the square
root of Q. When Q = 4, P = 22 and when
Q = 9, P = 42. Find P when Q = 25.
4. The cost of mass production of cars is
partly constant and partly varies as the
number of cars produced. The total cost of
making 400 cars is sh 3 740 000 and that of
making 1000 cars is sh 7 100 000. Find the
cost of making 2 000 cars.
1
y
–
b
3
–
b
y
–
b
y
–
b
y
–
b
6
–
b
3
–
b
6
–
b
6
–
12
y
––
12
4
––
1
2
–
12
3
––
128
5. The total cost of making a solid metal
sphere is partly constant and partly varies
as the cube of its radius. When the radius
of the sphere is 8 cm, its total cost is
sh 960 and when the radius is 12 cm, the
cost if sh 2 480. Find the total cost of
making a sphere whose radius is 10 cm.
6. The cost of silver-plating articles of a given
shape is the sum of two parts; one is a
constant and the other varies as the square
of the height of the article. It costs sh 1 850
to plate one of these articles, which is 10 cm
high and sh 5 600 to plate one which is
20 cm high. What will it cost to plate one
which is 14 cm high? How high will the
article be if it costs sh 2 400 to plate?
7. The fee per student at a certain college is
partly proportional to the number of lessons
given and partly inversely proportional to
the number of students in the class. The
fee for 3 lessons with a class of 45 is sh
240, and that for 8 lessons with a class of
65 is sh 520. Find the fee for 5 lessons
with a class of 8.
129
10
SEQUENCES AND SERIES
• • • • • •
• • • • • •
• • •
• • •
• • •
• • •
• • •
• •
•
(a)
(e)(d)
(c)
(b)
• •
•
• • •
• • •
•
•
• • •
• • •
•
•
• •
• •
• • •
•
•
•
•
1
16
9
4
Number patterns have been used since ancient
times. For example, the early Arabs, Hindus,
Greeks and Babylonians studied certain number
patterns which they believed had mystic
powers.
In this unit, we shall look at some numbers
which have some interesting observations and
usefulness in real life.
Number patterns
The number 6 can be represented using patterns
of dots as shown in Fig. 10.1.
Fig. 10.1
Fig. 10.1(a) and (b) are said to be linear
patterns. Fig. 10.1(c) and (d) are said to be
rectangular patterns. Fig. 10.1(e) is said to be
a triangular pattern.
Numbers which can be represented by a
rectangular pattern of dots are called rectangle
numbers. Each rectangle number has two or
more factors other than itself.
Numbers which can be represented by a square
pattern of dots are called square numbers.
Fig 10.2 shows some square numbers and their
patterns of dots.
Fig 10.2
Since 1 = 1
2
, 4 = 2
2
, 9 = 3
2
, 16 = 4
2
and so on,
we can write the square numbers 1, 4, 9, 16, …
as 1
2
, 2
2
, 3
2
, 4
2
, …
We can generate other number patterns from the
number pattern in Fig. 10.2.
Note that 4 = 2
2
is obtained by adding 3 to 1
2
,
9 = 3
2
is obtained by adding 5 to 2
2
, and so on.
So we have 2
2
= 1
2
+ 3 or 2
2
– 1
2
= 3
3
2
= 2
2
+ 5 or 3
2
– 2
2
= 5
4
2
= 3
2
+ 7 or 4
2
– 3
2
= 7.
Now copy and ll in the gaps in the following:
5
2
= 4
2
+ … or 5
2
– 4
2
= …
6
2
+ 5
2
+ … or 6
2
– 5
2
= …
7
2
= 6
2
+ … or 7
2
– 6
2
= …
:
:
50
2
= 49
2
+ … or 50
2
– 49
2
= …
:
.
99
2
= 98
2
+ … or 99
2
– 98
2
= …
Thus, from this pattern, we see that
a
2
– b
2
= a + b, provided that a and b are
consecutive positive integers and a > b.
Using the pattern, evaluate
(i) 111
2
– 110
2
(ii) 888
2
– 887
2
Copy and complete the following pattern
1 = 1
2
1 + 3 = 2
2
1 + 3 + 5 =
1 + 3 + 5 + 7 =
1 + 3 + 5 + 7 + 9 =
1 + 3 + 5 + 7 + 9 + 11 =
Numbers like 6 which form a triangular
pattern [as in Fig. 10.1(e)] are called triangle
numbers.
130
Copy and complete the following pattern
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 =
1 + 2 + 3 + 4 + 5 =
1 + 2 + 3 + 4 + 5 + 6 =
1 + 2 + 3 + 4 + 5 + 6 + 7 =
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 =
What type of numbers do you get?
Exercise 10.1
1. Copy and complete the following number
pattern.
3
2
– 1
2
= 8
4
2
– 2
2
= 12
5
2
– 3
2
= 16
6
2
– 4
2
= …
7
2
– 5
2
= …
8
2
– 6
2
= …
What relationship can you deduce from the
set of numbers 3, 1, 8; 4, 2,12; 5, 3, 16; …
2. When 11 is squared, we get 121 which is
said to be symmetrical. Similarly,
11
3
= 1 331 is symmetrical
(a) Which of the following is symmetrical
11
4
, 11
5
, 11
6
?
(b) Which integer between 1 and 11 has a
cube which is a symmetrical number?
Sequences
General sequences
Examine the following lists of numbers
(a) 1, 2, 3, 4, 5, …
(b) 2, 4, 6, 8, 10, …
(c) 1, 3, 5, 7, 9, …
(d) 1, 3, 9, 27, 81, …
(e) 21, 19, 17, 15, 13, …
(f) 15, 14, 12, 9, 5, …
(g) 1, 3, 10, 4, 19, …
(h) 0, 5, 1, 10,
–
47, …
Five numbers are given in each list. We refer to
each number in a list as a term. Thus, in (a), 1
is the rst term, 2 is the second term, etc.
What could be the next three terms in each list?
How do you obtain them?
The terms could be obtained as shown below.
(a) 1, 2, 3, 4, 5, 6, 7, 8: (any term) + 1 = next
term, or since 1st term = 1, 2nd term = 2,
3rd term = 3 and so on, then 6th term = 6,
7th term = 7, 8th term = 8.
(b) 2, 4, 6, 8, 10, 12, 14, 16: (any term) × 2 =
next term; or since 1st term = 1 × 2, 2nd
term = 2 × 2, 3rd term = 3 × 2 and so on,
then 6th term = 6 × 2, 7th term = 7 × 2,
8th term = 8 × 2.
(c) 1, 3, 5, 7, 9, 11, 13, 15: (any term) + 2 =
next term; or 1st term = 1 × 2 – 1,
2nd term = 2 × 2 – 1, 3rd term = 3 × 2 – 1,
so 6th term = 6 × 2 – 1, 7th term = 7 × 2 –
1, 8th term = 8 × 2 –1.
(d) 1, 3, 9, 27, 81, 243, 729, 2 187:
(any term) × 3 = next term.
(e) 21, 19, 17, 15, 13, 11, 9, 7:
(any term) – 2 = next term.
(f) 15, 14, 12, 9, 5, 0,
–
6,
–
13: 1st term – 1 = 2nd
term, 2nd term – 2 = 3rd term, 3rd term – 3
= 4th term.
(g) and (h) have no pattern that would help us
obtain the next term
A list of numbers, arranged in order and
having a rule which gives the terms in the list
[as in (a) – (f)] is called a sequence.
Thus, the lists of numbers in (a) – (f) are
sequences while (g) and (h) are not sequences.
Since the terms in the sequences 1, 2, 3, 4, 5
are increasing in magnitude, the sequence is
an increasing sequence. It follows that the
sequence 21, 19, 17, 13, ... is a decreasing
sequence.
131
Exercise 10.2
1. For each of the following sequences, nd
the next three terms and the rule which
gives the terms.
(a) 0, 3, 6, 9, 12, …
(b) 0, 4, 8, 12, 16, …
(c) 1, 4, 7, 10, 13, …
(d) 90, 85, 80, 75, 70, …
(e) 88, 84, 80, 76, …
(f) 30, 20, 10, 0,
–
10, …
2. For each of the following sequences, nd
the next three terms and the rule which
gives the terms
(a) 1, 4, 16, 64, 256, …
(b) 2, 4, 8, 16, 32, …
(c) 3, 6, 9, 12, 15, …
(d) 1, , , , , …
(e) 1,
–
, ,
–
, , …
(f) 1, , , , , …
3. For each of the following sequences, nd
the indicated terms.
(a) 5, 8, 11, 14, 17, …; 8th and 12th terms
(b) 11, 9, 7, 5, 3, …; 9th and 13th terms
(c)
–
1, 3,
–
5, 7,
–
9, …; 10th and 12th terms
(d) 39, 33, 27, 21, 15, …; 8th and 10th
terms
(e)
–
2,
–
5,
–
8,
–
11, …; 12th and 15th terms
(f) 2,
–
6,
–
14,
–
22,
–
30, …; 10th and 12th
terms
4. For each of the following sequences, nd
the indicated terms and the general term of
the sequence.
(a) 1, 4, 9, 16, 25, …; 8th and 11th terms
(b) 2,
–
6, 18,
–
54, 162, …; 7th and 10th
terms.
(c) 2, 4, 8, 16, 32, …; 7th and 11th terms.
(d) 4, 2, 1, , , …; 9th and 14th terms.
(e) 1, 8, 27, 64, 125, …; 10th and 15th
terms.
Note that the sequence 1, 3, 5, 7, 9, 11 has 6
terms only while the sequence 1, 3, 5, 7, 9, 11, …
has many terms that cannot be exhausted. The
rst sequence is called a nite sequence while
the second one is an innite sequence.
Examine the following sequences again.
(a) 1, 2, 3, 4, 5, …
(b) 2, 4, 6, 8, 10, …
(c) 1, 3, 5, 7, 9, …
(d) 1, 3, 9, 27, 81, …
How would you obtain the 10th, 20th, 100th
and nth terms?
(a) 1, 2, 3, 4, 5, …
We saw that 1st term = 1, 2nd term = 2,
3rd term = 3, 4th term = 4, etc. It follows
that 10th term = 10, 20th term= 20, 100th
term = 100 and nth term = n.
(b) 2, 4, 6, 8, 10, …
We saw that 1st term = 1 × 2, 2nd term =
2 × 2, 3rd term = 3 × 2, 4th term = 4 × 2,
etc. Hence 10th term = 10 × 2, 20th term
= 20 × 2, 100th term = 100 × 2 and nth
term = n × 2 = 2n.
(c) 1, 3, 5, 7, 9, …
We saw that 1st term = 1 × 2 – 1, 2nd term
= 2 × 2 – 1, 3rd term = 3 × 2 – 1, 4th term
= 4 × 2 – 1, etc.
∴10th term = 10 × 2 – 1,
20th term = 20 × 2 – 1,
100th term = 100 × 2 – 1 and hence
nth term =
n × 2 – 1 = 2n – 1.
(d) 1, 3, 9, 27, 81, …
1st term = 1, 2nd term = 3, 3rd term = 9 = 3
2
,
4th term = 27 = 3
3
, 5th term = 81 = 3
4
, etc.
10th term = 3
9
, 20th term = 3
19
,
100th term = 3
99
and nth term = 3
n–1
.
We see that the nth term in (a) – (d) are n, 2n,
2n – 1 and 3
n — 1
respectively.
We say that these are the rules or laws which
give the terms of the sequences or the general
terms of the sequences.
1
2
–
1
16
––
1
4
–
1
8
–
1
3
–
1
5
–
1
7
–
1
9
–
1
2
–
1
16
––
1
4
–
1
8
–
1
2
–
1
4
–
132
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
4
–
Solution
First term a =
–
 
d = 2 – (
–
2) = 4, nth term = a + (n – 1)d
∴ 16th term =
–
2 + (16 – 1) × 4 = 58.
Example 10.2
The seventh term of an arithmetic sequence is
80. If the eleventh term is 68, what is the 4th
term?
Solution
In this example there are two unknowns, the


The nth term = a + (n – 1)d


Equation (ii) – equation (i) gives 4d =
–
12
 ∴ d =
–
3
Substituting in (i) a + 6d = 80
a = 80 – 6d = 80 – 6(
–
3)
= 98
Thus, 4th term is a + 3d = 98 + 3(–3)
= 98 – 9
= 89.
Exercise 10.3
1. For each of the following sequences, nd
the indicated terms.
(a) 1, 3 , 6, 8 , …; 20th and 30th terms.
(b) 4, 3 , 3, 2 , …; 15th and 25th terms.
(c) 1, 2 , 4, 5 , …; 10th and 20th terms.
(d) 1, 0.8, 0.6, …; 10th and 15th terms.
2. How many terms are in each of the
following sequences?
(a) 1, 4, 7, …, 61
(b) 4, 4 , …, 7
(c) 20, 17 , …,
–
15
(d) 5, 5.9, 6.8, …, 23
(f) 1, , , , , …; 8th and 11th terms.
5. Write down the rst four terms of the
sequence for which the general term is
(a) 2n – 3 (b) 6n + (c) 3 × 4
n
(d) n
2
+ 1 (e) (n – 1) 2
n
(f) 2 × 3
n – 1
(g) (h) 6 +
Arithmetic sequences
Examine the following sequences:
(a) 1, 3, 5, 7, 9, … (b) 2, 4, 8, 16, 32, …
(c) 1, 4, 9, 16, 25, …
In the sequence 1, 3, 5, 7, 9, …, any two
consecutive terms dier by 2, i.e.
3 – 1 = 5 – 3 = 7 – 5 = 9 – 7 = 2, etc.
Note that the dierence in consecutive terms of
sequences (b) and (c) is not constant.
A sequence in which any two consecutive
terms differ by the same number, i.e. a
constant, is called an arithmetic sequence.
The constant number by which the consecutive
terms dier is called the common dierence.
An arithmetic sequence is generally expressed
in the form a, a + d, a + 2d, a + 3d, …, where a
is the rst term and d is the common dierence.
Noting that 1st term = a, 2nd term = a + d, 3rd
term = a + 2d, 4th term = a + 3d. What is the
5th term, 10th term, 15th term, nth term?
In general
The nth term (denoted as U
n
or T
n
) of an
arithmetic sequence is a + (n – 1)d, i.e.
T
n
= a + (n – 1)d and d = T
n
– T
n – 1
.
Example 10.1
In the sequence
–

term?
1
2
–
n
n + 1
––––
–
1
n
2
3
–
4
9
–
8
27
––
16
81
––
133
3. The 10th term of an arithmetic sequence is
–
15, and the 20th term is
–
35. What is
(a) the rst term,
(b) the common dierence?
4. The 9th term of an arithmetic sequence is
37 and the 16th is 65. What is the 20th
term?
5. An arithmetic sequence consists of all
integers between 1 and 100 which are
divisible by 3. How many terms are there?
6. Which term of the arithmetic sequence 4,
13, 22, … is 139?
7. What are the next three terms of the
sequence 1, 5, 13, 25, 31, …?
[Hint: Find the dierence between
consecutive terms of the sequence. What
type of sequence do you get?]
Geometric sequence
Examine the sequences
(i) 2, 4, 8, 16, 32, … and
(ii) 4, 9, 16, 25, …
Neither of them is an arithmetic sequence. In
sequence (i) it can be seen that each term is
obtained by multiplying the preceding one by 2
and so, if we divide each term by the preceding
one, we get a constant value. Thus, we have
4 ÷ 2 = 8 ÷ 4 = 16 ÷ 8 = 32 ÷ 16 = 2.
A sequence in which the ratio between any
two consecutive terms is a constant value is
called a geometric sequence. This constant
value is called the constant ratio or common
ratio.
Note that sequence (ii) above is neither an
arithmetic nor a geometric sequence.
In general, the rst term is represented by a and
the common ratio is represented by r. Since in
3
2
–
a geometric sequence each term is obtained by
multiplying the preceding one by the common
ratio, then we have:
1st term = a, 2nd term = ar, 3rd term = ar
2
,
4th term = ar
3
.
What is the 5th term, 12th term, 15th term, nth
term?
We note that:
The nth term (U
n
or T
n
) of a geometric
sequence is ar
n – 1
, i.e. T
n
= ar
n – 1
.
In general, a geometric sequence is in the form
a, ar, ar
2
, ar
3
, ar
4
, …, ar
n – 1
Example 10.3
What are the 8th, 10th and nth terms in the
sequence
–
8, 24,
–

Solution

a =
–
8 and common ratio r = =
–
3.
∴ 8th term =
–
8 × (
–
3)
8 – 1
=
–
8(
–
3)
7
= 17 496
10th term = 8 × (
–
3)
10 – 1
=
–
8(
–
3)
9
= 157 464
nth term =
–
8 × (
–
3)
n – 1
Example 10.4
The 3rd term of a geometric sequence is 3 and

of the sequence.
Solution
3rd term = ar
3 – 1
= ar
2

5th term = ar
5 – 1
= ar
4

(ii) ÷ (i) gives r
2
= ⇒ r = or
–
∴ar
2
= 3 ⇒ a = 12.

12,
–
6, 3,
–

1
4
–
1
2
–
1
2
–
3
4
–
3
2
–
3
4
–
24
–
8
––
134
9. Which is the rst term of the geometrical
sequence 2 187, 729, 243, … which is less
than 1?
Series
A series is the sum of the terms of a sequence.
Thus, from the sequences (a) 1, 3, 6, 7, …
(b) 2, 4, 8, 16, … (c) 1, 4, 9, 16, … we obtain
the series
(a) 1 + 3 + 5 + 7 + …
(b) 2 + 4 + 8 + 16 + …
(c) 1 + 4 + 9 + 16 + … respectively.
Note that the series 1 + 4 + 9 + 16 + 25 + … +
100, has a nite number of terms. A series with
a nite number of terms is called a nite series.
Such a series as 1 + 3 + 5 + 7 + … has many
terms that cannot be counted exhaustively.
Such a series is called an innite series.
Arithmetic progression (A.P.)
[Arithmetic series]
When the terms of an arithmetic sequence
are added, the series obtained is called
an arithmetic progression (A.P.) or an
arithmetic series.
Thus the arithmetic sequence a, a + d, a + 2d,
a + 3d, … becomes the Arithmetic progression
a + (a + d) + (a + 2d) + (a + 3d) + …
Just as for an arithmetic sequence, for an A.P,
the rst term is a, the common dierence is d
and the nth term is a + (n – 1)d.
We use the symbol S
n
to denote the sum of the
rst n terms in a series. Thus for the sum of
the arithmetic series 1 + 3 + 5 + 7 + …, S
1
is
equivalent to the rst term, S
2
is the sum of the
rst 2 terms, S
3
is the sum of the rst 3 terms,
etc. Thus:
S
1
= 1, S
2
= 1 + 3 = 4, S
3
= 1 + 3 + 5 = 9
S
4
= 1 + 3 + 5 + 7 = 16, S
5
= 1 + 3 + 5 + 7 + 9
= 25, and so on.
Exercise 10.4
1. For each of the following sequences nd
the common ratio and the indicated terms.
(a) 1, 3, 9, …; 5th and 9th terms
(b) 1, , , …; 7th and 10th terms
(c) 1, 0.8, 0.64, …; 4th and 6th terms
(d) 2,
–
4, 8, …; 8th and 12th terms
(e) , , , …; 8th term
(f) 6, –0.6, 0.06, …; 6th term
2. Find how many terms each of the following
sequences has.
(a) 2, , , …,
(b) 20, 4, 0.8, …, 0.006 4
(c) 60, 20, 6 , …,
(d) 80, 32, 12.8, …, 0.819 2
3. 12, b, 75 are consecutive terms of a
geometric sequence. What is the common
ratio for the sequence?
4. The second term in a geometric sequence
is 64. The fth term is . Find the rst 4
terms of the sequence.
5. The rst term of a geometric sequence is
16 and the fth term is 9. What is the value
of the 9th term?
6. The rst term of a geometric sequence is
1.1 and the fourth term is 1.464 1. What is
the common ratio for the sequence?
7. The second term of a geometric sequence is
–
1 and the fth term is . What is the tenth
term?
8. Ounga saves sh 100 on his son’s rst
birthday. He saves sh 200 on the second
birth day and sh 400 on the third birthday
and so on doubling the amount on every
birthday. How much will he be saving on
the boy’s 10th birthday?
2
3
–
4
9
–
2 401
16
––––
343
8
–––
49
4
––
1
512
–––
1
2
–
1
8
–
20
729
–––
2
3
–
1
8
–
1
8
–
135
∴ S
10
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +
17 + 19
= 100
We get the sum S
10
in two ways:
S
10
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19
and
S
10
= 19 + 17 + 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1
Adding the two vertically gives:
2S
10
= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 +
20 + 20 = 200.
∴ S
10
= = 100.
We use the same method to nd the sum S
n
of n
terms of an A.P.:
S
n
= a + (a + d) + (a + 2d) + (a + 3d) + … +
[a + (n – 1)d].
For simplicity, we write the last term
[a + (n – 1)d] as l.
So we have S
n
= a + (a + d) + (a + 2d) + … +
(1 – 2d) + ( l – d) + l ……………………… (i)
Writing the same equation for S
n
but starting
with the last term, we have
S
n
= l + (l – d) + (l – 2d) + …+ (a + 2d) +
(a + d) + a …………………………… (ii)
Adding the terms on the left of equations (i) and
(ii) together and also adding the terms on the
right hand side of the two equations, gives:
2S
n
= (a + l) + ( a + l) + (a + l) + … +
(a + l) + (a + l) + (a + l)
i.e. 2S
n
= n(a + l)
∴ S
n
= (a + l) where a is the rst term and l is
the last term of the series.
Substituting a + (n – 1)d for l gives
S
n
= [a + a + (n – 1)d]: Thus
S
n
= [2a + (n – 1)d
]
Example 10.5

(a) the 100th term
 
200
2
–––
n
2
–
n
2
–
n
2
–
Solution
(a) nth term l = [a + (n – 1)d], a = 5, d = 4,
n = 100
∴ 100th term = [5 + (100 – 1) × 4]
= 5 + 99 × 4
= 401
 S
90
S
n
= [2a + (n – 1)d] where n = 90, q = 5
and d = 4.
Thus, S
90
= [2 × 5 + (90 – 1) × 4]
= 90(5 + 89 × 2)
= 16 470.
Exercise 10.5
1. State which of the following are A.P.s, and
state their common dierence.
(a) 1 + 2 + 4 + 8 + …
(b)
–
9 – 7 – 5 – 3 + …
(c) 1
2
+ 2
2
+ 3
2
+ 5
2
+ …
(d) 19 + 20 + 21 + …
(e) 2 + 5 + 8 + …
(f) 1
4
+ 2
4
+ 3
4
+ …
(g) 8 + 14 + 20 + …
(h) 7 + 14 + 28 + …
2. Given the A.P. 7 + 13 + 19 + …
(a) write down the 15th term
(b) nd the sum to the 15 terms and to n
terms
3. Given the A.P. 9 + 13 + 17 + … + 85, nd
(a) the number of terms in the series
(b) the sum of the series
4. The rst term of an A.P. is 10 and the
sum of the rst 15 terms is 465. Find an
expression for the sum of the rst n terms.
5. The second term of an A.P. is 18 and the
10th term is 106. What is the sum of the
rst 10 terms?
n
2
–
90
2
––
136
Rewriting the two sums and subtracting as
shown, we have
2S
10
= 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1 024 + 2 048–
S
10
= 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1 024 +
2S
10
– S
10
=
–
2 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 2 048
Thus 2S
10
– S
10
= 2 048 – 2
(2 – 1)S
10
= 2046
 ∴ S
10
= 2046
We use the same method to nd the sum S
n
of n
terms of a G.P.
S
n
= a + ar + ar
2
+ ar
3
+ … + ar
n – 2
+ ar
n – 1
…(i)
Multiplying (i) by r, the common ratio, we have
rS
n
= ar + ar
2
+ ar
3
+ … + ar
n – 1
+ ar
n
………(ii)
Thus, (ii) – (i) gives rS
n
– S
n
= ar
n
– a
S
n
(r – 1) = a(r
n
– 1)

∴ S
n
=
a(r
n
– 1)
r – 1
Alternatively, (i) – (ii) gives
S
n
– rS
n
= a – ar
n
S
n
(1 – r) = a(1 – r
n
)

∴ S
n
=
a(1 – r
n
)
1 – r
Note:
When the common ratio |r| > 1, we use the
formula S
n
=
a(r
n
– 1)
r – 1
and when the common ratio |r| < 1, we use the
formula S
n
=
a(1 – r
n
) .
1 – r
This helps avoid having to divide by a negative
denominator.
Example 10.6

following G.P.s.


Solution
 
4
3
–
6. The sum of the A.P. 10 + 7 + 4 + … is
–
4.
How many terms does the A.P. have?
7. How many terms of the A.P. 19 + 16 + 13
+ … must be taken, at the most, before the
sum becomes less than zero?
8. Find the sum of the even numbers between
100 and 300 that can be divided by 7?
9. Which term of the A.P. 15 + 12 + 10 + …
is the rst to be less than zero?
Geometric progression (G.P.)
(Geometric series)
When the terms of a geometric sequence
are added, the series obtained is called a
Geometric progression (G.P.) or Geometric
series.
Thus from the geometric sequence 2, 4, 8, 16, …,
we obtain the G.P. 2 + 4 + 8 + 16 + …
In general, a geometric sequence a, ar, ar
2
, ar
3
,
…, ar
n – 1
gives the geometric progression
a + ar + ar
2
+ ar
3
+ … + ar
n – 1
where a is
the rst term and r is the common ratio.
Taking S
n
as the sum of the rst n terms of a
G.P. we have
S
n
= a + ar + ar
2
+ … + ar
n – 1
Thus, for the series 2 + 4 + 8 + 16 + …
S
1
= 2, S
2
= 2 + 4 = 6, S
3
= 2 + 4 + 8 = 14,
S
4
= 2 + 4 + 8 + 16 = 30, etc.
Consider the sum
S
10
= 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 +
512 + 1 024 = 2 046.
We can also obtain S
10
in the following way.
S
10
= 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 +
512 + 1024 + ………………………… (i)
Multiplying each term of equation (i) by 2, the
common ratio we have
2S
10
= 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 +
1 024 + 2 048 ………………………(ii)
1
2
–
137
First term a = 3, common ratio r = 2.
Since r = 2 is greater than 1, we use the
formula S
n
=
a(r
n
– 1)
r – 1

∴ S
10
=
3(2
10
– 1)
= 3 × 1 023 = 3 069
2 – 1
(b) 12 + 4 + + ...
a = 12, r =
Since r = is less than 1, we use the
formula S
n
=
a(1 – r
n
)
1 – r

∴ S
10
=
12[1 – ( )
10
] = 12[1 – ( )
10
]
1 –
= 18[1 – ( )
10
]
× 1
Thus, S
10

Exercise 10.6
1. Find the last term and the sum of the
following G.P.s.
(a) + 2 + 6 + …; (6 terms)
(b) 9 + 3 + 1 + …; (8 terms)
(c) – 1 + 2 – …; (10 terms)
(d) 1 + + + …; (8 terms)
(e) 1 + 0.9 + 0.81 + …; (5 terms)
(f) 9 – 6 + 4 – …; (9 terms)
(g) + 10 + + …; (10 terms)
2. Find the sums of the following G.P.s.
(a) 1 – 2 + 4 – … + 1 024
(b) 100 – 10 + 1 – … + 0.000 001
(c) 50 + 25 + 12.5 + … + 0.781 25
(d) 20 – 4 + … – 0.000 256
(e) 0.5 – 2.5 + 12.5 – … – 1 562.5
(f) 9 + 3 + … +
3. The second term of a G.P. is 8 and the 4th
term is 128. Write down the rst 4 terms of
the GP.
2
3
–
1
2
–
2
3
–
4
9
–
40
3
––
15
2
––
1
243
–––
4
3
–
1
3
–
1
3
–
1
3
–
2
3
–
1
3
–
1
3
–
1
3
–
4. The 3rd term of a GP is
–
and the 6th term
is . What is the rst term of the G.P.?
5. The 3rd term of a G.P. is and the 5th term
is . Find the sum of the rst six terms.
6. The rst three terms of a G.P. are the rst,
fourth and tenth terms of an A.P. Given
that the rst term is six, and that all the
terms of the G.P. are dierent, nd the
common ratio.
7. What is the smallest number of terms of
the G.P. 1 + 4 + 16 + 64 + …, for which the
sum is more than 25,000?
Application of A.P. and G.P. to real
life situations
There are many real life problems which can be
solved using the knowledge of A.P.s and G.P.s.
Example 10.7 illustrates this.
Example 10.7
It is said that the population of some of the
developed countries are declining. Suppose
that the population of one such country was 75
million eight years ago and has been declining

the current population of that country.
Solution
Each year the population declines by 3% to
97% of its size at the end of the previous year.
Thus, the common ratio, r, is 0.97.

number of the terms, n, is 8 (equal to the
number of years).
∴ The current population is
T
n
= ar
n – 1
= 75 million × 0.97
7
= 60.6 million
1
2
–
1
32
––
1
3
–
1
81
––
138
Exercise 10.7
In this exercise, do not use the compound
interest formula.
1. Mwaura was employed in January 1994 at
a basic salary of sh 79 200 per annum.
If he was given an annual increment of
sh 6,600 each year, in which year would his
salary be double his starting salary?
2. Kiprono starts a savings account by
depositing sh 250 in the rst month. Each
subsequent month, his deposit will be sh
75 more that the preceding one. After how
many months will his savings be sh 1 825
a month?
3. Chebet joined a savings and co-operative
society and deposited sh 3 000 in the rst
month. She intended to save a total of
sh 27,000 in 16 months. How much did
she have to contribute per month?
4. In what time will sh 5 500 amount to
sh 19 500 if invested at 3 % per annum
compound interest?
5. The value of a machine depreciated each
year by 10% of its value at the beginning
of that year. If its value when new was
sh 15 000, what was its value after 8 years?
6. A machine depreciated from sh 36 000 to
sh 12 000 in 12 years. What was the yearly
rate of depreciation, assuming it to be
constant?
1
2
–
7. In what time will sh 8 000 amount to
sh 9 440 at 6% p.a. simple interest?
8. A man opens a savings account and
deposits sh 2 000 each year at 5% compound
interest p.a. Find an expression for the
amount in his account at the end of
(a) the rst year,
(b) the second year, and
(c) the third year.
Hence, nd the amount that he will have at
the end of the tenth year.
9. A father puts sh 10 000 in a bank account
for his child at the beginning of the year
at 4% compound interest p.a. How much
money will be in the account at the end of
the tenth year?
10. A researcher nds that the population of
bacteria in a culture that he is studying
doubles itself every hour. At one point,
he estimated that the population was 80
million. Estimate the number of bacteria in
the culture four hours earlier.
11. For some unexplained reason, the
population of the country in Example
10.7 has started growing at a steady rate
of 1.5%. Assuming that it will continue
to grow at that rate, after how many years
(to the nearest whole number) would you
expect it to hit the 75 million mark again?
139
REVISION EXERCISES 2
Z
O
Y
Q
R
P
X
•
Fig. R2.1.1
1 2 3
4 3 2
( )
( )
2 4
1
–
3
( )
1 6
2
3
( )
3 1
–
3 5
6 1
1 4
4 0
2
–
8
( )
3. In Fig. R2.1.1,
PY is a diameter
of the circle and
XYZ a tangent.
Find
(a) (i) ∠QYP
(ii) ∠QPY if 
 ∠QYX = 42°
(b) ∠QYX if ∠QRY = 38°
4. A, B and C are points on a circle centre O.
The tangent at C meets AB produced at T,
∠AOB = 88°, ∠CBT = 68°.
Find ∠BCT.
5. (a) Given that A = and B =
show that
(i) A + B = B + A
(ii) A – B ≠ B – A.
(b) If P = and Q = , show that
PQ ≠ QP.
6. (a) Find the matrix M such that
5M = + 2
(b) Given that = , nd
the values of x, y and z.
7. A rectangular lawn measures l m by b m.
It is surrounded by a path of uniform width
x m and has an area A m
2
. Find an
expression for A in terms of l, b and x.
8. (a) Make y the subject of the formula
2 – 3y
= B
y
x 2 1
3 y 1
4 z 2
( )
( )
3
1
2
16
11
15
( )
3
√
6-10
Revision exercise 2.1
1. (a) Find the simple interest on
(i) sh 25 000 at 3% p.a for 5 years
(ii) sh 20 000 at 5 % p.a for 4 years
(iii) sh 240 000 at 3 % p.a for
33 months
(iv) sh 750 000 at 6 % p.a for 4 years
3 months.
(b) Find the rate percent per annum, simple
interest at which
(i) sh 34 500 will amount to
sh 37 375 in 2 years
(ii) sh 276 000 will amount to
sh 368 000 in 5 years
(iii) sh 570 000 will amount to
sh 903 450 in 7 years
(iv) sh 30 million will amount to sh 36
million in 4 years
2. (a) Find the time in which
(i) sh 250 000 at 4% p.a produces
sh 20 000 simple interest
(ii) sh 414 000 at 2 % p.a produces
sh 41 400 simple interest
(iii) sh 386 400 at 5% p.a produces
sh 67 620 simple interest
(iv) sh 23 200 at 2 % p.a produces
sh 3 857 simple interest
(b) Find the principal which, at simple
interest, amounts to
(i) sh 750 000 after 10 years at 2 %
(ii) sh 524 898 after 6 years at 3 %
(iii) sh 69.125 million after 4 years
at 8 %
(iv) sh 385.2 million after 5 years
at 11%
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
3
–
2
3
–
( )
–
2 3
–
4
–
1 2 4
140
√
( )
–
20 30
40
–
10
( )
1 2
4 3
(b) Make B the subject of the formula
M = 2 +
3
.
N – B
9. Write down the next three terms in each of
the following sequences.
(a) 3, 7, 11, 15, … (b) 42, 35, 28, 21, …
(c) 6, 12, 24, 48, … (d) 3, 4, 6, 9, 13, …
(e) 2, 2, 4, 6, 10, … (f) 0, 3, 8, 15, …
10. Find a formula for the nth term of each of
the following sequences
(a) , , , , … (b) , , , , …
(c) , , , , … (d) , , , , …
11. The cash price of a bicycle is sh 32 000.
It can be bought on hire purchase with a
deposit of 25% of the cash price followed
by 18 monthly instalments of sh 1 550.
price and hire purchase price?
12. Mr. Dede earned a basic salary of sh 26 240
and a house allowance of sh 40 000 per
month. He claimed a tax relief of sh 1 280
per month.
(a) Using Table 6.1, calculate
(i) the tax payable without relief
(ii) the tax paid after the relief
(b) Mr. Dede has the following deductions
made on his salary as well. A life
insurance policy of sh 1 500, a
university loan of sh 650 and 2% of
his basic salary as widow and children
pension scheme. Calculate
(i) the total monthly deductions made
from his salary.
(ii) the net income from his
employment.
Revision exercise 2.2
1. Find the compound interest payable on the
following loans for the stated period of
time and rates of interest:
(a) sh 30 000; 2 years; 5% p.a
(b) sh 800 000; 4 years; 3% p.a
(c) sh 70 000; 8 years; 12% p.a
(d) sh 50 million; 2 years; 4% p.a
(e) sh 2 million; 3 years; 25% p.a
(f) sh 105 000; 3 years; 40% p.a
2. Angweny borrowed sh 400 million payable
at 5% p.a compound interest, yearly. He
paid sh 200 million at the end of the rst
year. If he was required to pay o all his
debt at the end of the second year, how
much did he pay?
3. Shikanda borrowed sh 48 000 from his
Savings and Co-operative Society for an
emergency. He was to repay the loan in
equal instalments for 12 months, at the
beginning of each month. The loan
attracted an interest of 1% per month on
the balance owed by the end of that month.
How much total interest did he pay on the
loan if he started paying one month after
taking the loan?
4. In Fig. R2.2.1, ∠QPR,  
∠PQR and ∠PRQ are
in the ratio 3 : 4 : 5.
Find ∠XYZ, ∠YXZ
and ∠XZY.
Fig. R2.2.1
5. Two circles of radii 4 cm and 8 cm have
their centres 20 cm apart. Find the length
of a transverse common tangent.
6. (a) A and B are two matrices. If A =
nd B given that A
2
= A + B.
(b) Given that A = and B =
(i) nd A
–1
B,
–
2 3
4
–
1
( )
P
Q
R
X
Y
Z
1
81
––
1
27
––
1
9
–
1
3
–
7
10
––
5
8
–
3
4
–
1
2
–
4
5
–
3
4
–
2
3
–
1
2
–
1
5
–
1
4
–
1
3
–
1
2
–
141
2x x
2
2 1
( )
(ii) hence or otherwise, write down
the inverse of B.
7. (a) Determine the values of x for which the
matrix has no inverse.
(b) Use matrix method to solve the
simultaneous equations
(i) 2x + 5y = 9 (ii) 3s + 4t = 6
3x + 4y = 6 5s + 3t =
–
7
8. (a) The variable y varies directly as x.
When y = 6, x = 4. Find
(i) x when y = 8 (ii) y when x = 12
(b) If w ∝r
2
h, and w = 5 when r = 3 and
h = 2, express w in terms of r and h.
Hence nd w when r = 2 and h = 3
9. The variable y varies directly as the square
of z and inversely as the square root of x.
When z = 2, y = 8 and x = 9. Find the value
of x when y = 27 and z = 3.
10. The variable A varies partly as B and partly
as the square root of B. When B = 9,
A = 30 and when B = 4, A = 16. Find A
when B = 36
11. Write down the rst four terms and the
tenth term of the sequences whose nth term
is dened as
(a) 3n – 1 (b) n
2
+ 1
(c) 2
n
+ 1 (d) 2
n
– 5n + 1
12. A ball falls from a height of 10 m and re-
bounds each time to a height equal to four
fths of its previous height. What distance
will the ball have travelled when it touches
the ground for the
(a) 3rd time (b) 5th time
(c) 20th time?
Revision exercise 2.3
1. Langat got a loan of sh 351 000 from a
building society to buy a property. The
loan was to be repaid in equal monthly
instalments for ten years. The loan was
charged an interest of 1% per month of the
balance by the end of the month.
(a) How much did he repay at the end of
the (i) rst month (ii) rst year?
(b) At the end of the fth year, the interest
rate was increased from 1% to 2%. In
total how much had Langat paid by the
end of the tenth year?
2. Nyangaya was employed in a company that
was paying him a basic monthly salary of
sh 21 225, house allowance of 15 000 and
medical allowance of sh 3 500. How much
does he pay in PAYE if he gets a monthly
insurance relief of sh 600?
3. (a) Two circles of radii R and r touch
externally. Find an expression for the
length of the direct common tangent to
the circles.
(b) Two circles of radii R and r have their
centres d cm apart. A direct com-
mon tangent meets the line of centres,
produced, at A. If R > r what is the
distance from the centre of the larger
circle to A?
4. (a) Two circles of radii 8 cm and r cm
touch externally. The common tan-
gent XY to the two circles at X and Y
respectively is 12 cm long . Find r.
(b) ∆ABC is inscribed in a circle.
AB = 6 cm, ∠BAC = 40° and 
∠ABC = 60°. A tangent to the circle
at C meets AB produced at D.
Calculate CD.
5. Use matrix method to solve the simultaneous
equations
6x + 4y = 22
9x + 6y = 33
What can you say about the two lines?
142
6. Three ladies A, B and C went shopping. A
bought 2 kg of sugar, kg of tea leaves
and 1 loaf of bread; B bought 5 kg of sugar,
1 kg of tea leaves and 2 loaves of bread;
C bought 5 kg of sugar, kg of tea leaves
and 3 loaves of bread.
(a) Write this information in the form of a
matrix.
(b) If the prices were; sugar – sh 45 per kg;
tea – sh 196 per kg and bread – sh 20
per loaf, use matrix multiplication to
nd the amount of money spent by
each lady.
7. (a) The sum of two quantities A and B is
y. A varies inversely as x and B varies
directly as x. When x = 4, y = 17 and
when x = 6, y = 13. Find y when x = 10
and x when y = 11.5.
(b) Z ∝ . If x is increased by 20% and y
is decreased by 20%, nd the percentage
change in Z.
8. F is directly proportional to the square root
of T and inversely proportional to L and D.
If T is increased by 50% and L is halved,
nd the percentage change in F.
9. The population of a certain country is 5
million and it increases by 5% each year.
After how many years will the population
be 10 million?
10. The series 1 + 3 + 5 + 7 + … of odd
numbers yields some interesting results
1 = 1 = 1
3
3 + 5 = 8 = 2
3
7 + 9 + 11 = 27 = 3
3
13 + 15 + 17 + 19 = 64 = 4
3
(a) Using the same pattern, obtain 5
3
, 6
3
, 7
3
.
(b) Examine the rst and last odd numbers
in each case and obtain a relationship
between the odd numbers and the num-
ber cubed, i.e.
First odd Last odd
number number
1
2 … … 3 5
3 … … 7 11
4 … … 13 19
etc.
(c) Using the relationship, obtain the rst
and last numbers to obtain
(i) 9
3
(ii) 12
3
(iii) 13
3
Check your results by working the long
way.
(d) What would be the rst and last odd
numbers added when obtaining 20
3
?
11. On his birthday celebration, Chanzu blew
out 1 candle. On his second birth day, he
blew out 2 candles. On his third birth day
celebration, he blew out 3 candles, and so it
went on each year. His father would, each
time, wish him to live long enough to have
blown 1 001 candles. If each year a new
set of candles were used, by which birthday
will he have blown 1 001 candles in total?
12. A motorist passes a street clock on his way
to work every morning at 7.00 am. On 2nd
February, he noticed that the clock had lost
2 minutes compared to the time shown on
1st. On the 3rd, it had lost 4 minutes. On
4th, it had lost 8 minutes and on 5th it had
lost 16 minutes. If the clock continued
losing time in that way, what time was the
clock showing when the motorist passed by
on the 8th February.
1
4
–
1
2
–
x
2
y
––
143
11
VECTORS
1
2
–
1
2
–
1
2
–
1
2
–
( )
3
5
Coordinates in two and three
dimensions
In Form 2, we saw that the position of a point
in a plane can be specied by an ordered
pair of numbers (x, y). These numbers are
called cartesian coordinates. They specify,
respectively, the number of units in the
direction of the x-axis and in the direction of the
y-axis, that one has to move from the origin to
get to the point. Thus, they describe the position
of a point using two dimensions.
Now consider Fig. 11.1, which shows parts of
the oor and two adjacent walls of a hall.
Fig. 11.1
Taking two adjacent edges of the oor of the
hall as the x- and y-axes, we can describe, using
two coordinates, the position of any point on
the oor. For example, the position of point A
is (2 , 4).
Suppose we wish to specify the position of an
electric light bulb B, hanging from the ceiling.
We need a third measurement. This should be
the vertical height of the bulb above the oor.
The vertical height is designated the
z-coordinate and in Fig. 11.1, the intersection of
the two walls is the z-axis (Oz).
In terms of the three directions in Fig 11.1,
the coordinates of point A are (2 , 4 , 0). The
coordinates of the bulb B are (2 , 4 , 4 ). These
are the coordinates of A and B given in three
dimensions.
Fig. 11.2 represents a cuboid. With respect to
the x-, y- and z-axis, the coordinates of points A,
B and C are as shown.
What are the coordinates of
(a) D?
(b) E?
(c) F?
(d) G?
(e) O?
Fig 11.2
Write down the position vectors of points A, B
and C in Fig. 11.2.
Column vectors and position vectors
in three dimensions
In two dimensions, the column vector
represents a change in position (i.e. a
displacement) of 3 units in the direction of
the x-axis and 5 units in the direction of the
y-axis.
Likewise, in three dimensions the column
vector represents the displacement of
3 units in the x-direction, 5 units in the
y-direction and 4 units in the z-direction. It
could, for example, describe the displacement
from (1 ,
–
2 , 2) to (4 , 3 , 6)
What is the column vector which represents the
displacement from (
–
1 , 1 , 4) to (3 , 5 ,
–
1)?
– – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – –
O
O
x
A
B
4 m
1
2
–
2 m
1
2
–
4 m
z
y
– – – – – – – – – – – – – – – – – – –
– – – – – – –
– – – – – – – –
O
(2 , 0 , 0)
A
(2 , 4 , 3)
C
(2 , 4 , 0)
B
F
G
D
E
x
z
y
– – – – – – – – – – – – – –
– – – – – – – – – – – – – –
( )
3
5
4
144
Example 11.2
Given points A(6 , 5 , 7) and B(5 , 8 , 9), state
AB
and |AB|.
Solution
Position vector of A is OA =
Position vector B is OB =
∴ AB = – =
|AB| = √(
–
1)
2
+ 3
2
+ 2
2
= √14
     
Note that addition and subtraction of vectors in
three dimensions is carried out in the same way
as in two dimensions, i.e. we add corresponding
components together.
Exercise 11.1
1. State the coordinates of points A, B , C
and D in Fig. 11.3.
Fig. 11.3
2. Using Fig. 11.3, nd
(a) DF (b) BD (c) CG
(d) AF (e) GA
3. In Fig. 11.3, calculate the length of
(a) OE (b) OF (c) OG
(d) DF (e) CG
( )
6
5
7
( )
5
8
9
( )
6
5
7
( )
5
8
9
( )
–
1
3
2
( )
a
b
c
– – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – –
– – – – – – –
– – – – – – –
– – – – – – –
– – – – – – – – – – – – –
F
B
G
D
A
C
z
5
3
y
x
2
If the displacement of a point is from the
origin (0 , 0 , 0), the column vector for the
displacement is called a position vector.
For example, the position
vector of point P(3 , 2 , 6) is OP =
Length or magnitude of a vector in three
dimensions
Consider the following example.
Example 11.1
Find the length of OC in Fig. 11.2.
Solution

Hence, by P
OB
2
= OA
2
+ AB
2
= 2
2
+ 4
2
= 20

Hence, by P
OC
2
= OB
2
+ BC
2
= 20 + 3
2
= 29
∴ OC
Notice that in Example 11.1, C is (2, 4, 3), its
position vector is and OC = √2
2
+ 4
2
+ 3
2
.
OC is the length of vector OC.
The length of a vector AB is called the
modulus or magnitude of the vector AB and
it is often written as |AB|.
In general, if a point P has coordinates (x , y
, z), then
OP = √x
2
+ y
2
+ z
2
and if PQ = is a vector joining any two
points P and Q in three dimensions, then
|PQ| = √a
2
+ b
2
+ c
2
.
( )
3
2
6
( )
2
4
3
145
( )
1
0
( )
0
1
( )
1
0
( )
0
1
10. A tree 20 m high is located 120 m from a
point P and it is on a bearing of W45°N
from P. Taking eastwards as the positive x
direction and northwards as the positive y
direction,
(a) state the coordinates of an eagle
resting at the top of the tree.
(b) nd the speed of a bullet, red from a
rie at point P, that kills the eagle
0.4 seconds after it is released.
The unit vectors i, j and k
In two dimensions, the vectors and
are position vectors of the points I and J on the
x- and y-axis, respectively. Each of the points I
and J are one unit from the origin, i.e. each of
the vectors and has magnitude of 1
unit.
In three dimensions, , and are
the unit vectors in the x-, y- and z- directions,
respectively. The three vectors are denoted
as:
i = , j = and k = .
The unit vectors i, j and k are illustrated in
Fig. 11.4.
Fig. 11.4
– – – – – – – –
– – – –
– – – – – – – –
x
y
z
K (0 , 0 , 1)
I (1 , 0 , 0)
J (0 , 1 , 0)
•
i
•
•
k
j
O
4. Given that P is (3,
–
2, 4) and Q is (
–
3, 5, 3),
nd
(a) the position vector of P,
(b) the position vector of Q,
(c) PQ, (d) QP.
(e) What is the relationship between PQ
and QP?
5. Given that BC = and B is (2, 3,
–
5),
nd the coordinates of C.
6. How far from the origin is point
(a) (3, 2, 7) (b) (
–
4, 2,
–
3)
(c) (
–
x,
–
y,
–
z)?
7. Given that a = and b = , nd
(a) |a| (b) |b| (c) |a + b| (d) |a| + |b|
What do you notice?
8. A mosquito is sitting on the tip of my
nose. When I brush it o, it ies away in a
motion described by the vectors:
, , .
What is the nal position of the mosquito,
assuming that I remain in the same position?
9. The dimensions of a rectangular room
are 4 m × 3 m × 3 m. The corners of the
oor of the room are A, B, C and D, and
AB and AD are taken as the x- and y-axis
respectively. If 1 m represents 1 unit,
(a) What is the position vector of
(i) a cockroach resting on the ceiling
of the room at a point vertically
above the centre of the oor.
(ii) the corner of the room furthest
from A?
(b) Find the distances of the points in
(a)(i) and (ii) from point A.
( )
1
0
0
( )
0
1
0
( )
1
0
0
( )
0
0
1
( )
0
1
0
( )
0
0
1
–
5
2
0
( )
5
–
2
1
( )
–
2
3
7
( )
2
2
3
( )
1
–
1
–
2
( )
( )
–
3
–
1
–
1
146
Vectors expressed in terms of i, j and k
Consider point C in Fig, 11.2.
OC =
But OC = OA + AB + BC
= + +
= 2 + 4 + 3
∴ OC = 2i + 4j + 3k.
Thus, OC has been expressed in terms of the
unit vectors i, j, and k.
Any column vector is expressed in
terms of the unit vectors i, j and k as
ai + bj + ck.
Example 11.3
Given that a = 3i – 2j + k and b =
–
i + 3j + 4k,
 a + b (b) 2a – b.
Solution
(a) In column vector form, a = , b = .
Thus, a + b = + =
i.e. a + b = 2i + j + 5k.
This could equally well be worked as:
a + b = (3i – 2j + k) + (
–
i + 3j + 4k)
= (3i – i) + (
–
2j + 3j) + (k + 4k)
= (3 – 1)i + (
–
2 + 3)j + (1 + 4)k
= 2i + j + 5k
(b) 2a = 2 =

∴ 2a – b = – =
= 7i – 7j – 2k.
The same result is obtained as follows:
2a – b = 2(3i – 2j + k) – (
–
i + 3j + 4k)
= (6i – 4j + 2k) – (
–
i + 3j + 4k)
= (6 –
–
1)i + (
–
4 – 3)j + (2 – 4)k
= 7i +
–
7j +
–
2k
= 7i– 7j – 2k
From Example 11.3, we see that:
Given a = xi + yj + zk,
ma = mxi + myj + mzk, where m is a
scalar.
Also, given that a = pi + qj + rk and
b = xi + yj + zk,
a + b = (p + x)i + (q + y)j + (r + z)k.
Example 11.4
A and B are the points (1 , 2 , 3) and (9 , 8 , 10),
respectively.
(a) Find in terms of i, j and k the displacement
vector AB.
(b) Calculate the distance from A to B.
Solution
(a) OA = and OB =
AB = – =
= 8i + 6j + 7k.
(b) The distance from A to B = |AB|
= √8
2
+ 6
2
+ 7
2
= √149
= 12.21 units (2 d.p.)
( )
2
4
3
( )
2
0
0
( )
0
0
3
( )
0
4
0
( )
1
0
0
( )
0
0
1
( )
0
1
0
( )
3
–
2
1
( )
–
1
3
4
( )
3
–
2
1
( )
–
1
3
4
( )
2
1
5
( )
a
b
c
( )
3
–
2
1
( )
6
–
4
2
( )
1
2
3
( )
9
8
10
( )
9
8
10
( )
1
2
3
( )
8
6
7
( )
–
1
3
4
( )
7
–
7
–
2
( )
6
–
4
2
147
(i) AB (ii) BA
What is the relationship between the
two vectors?
(b) Find |AB| and |BA|.
What do you notice?
7. Given that p = i + j, q =
–
4i + 2k and
r = 2i – j + 3k, and that O is the origin,
determine the coordinates of A if OA is
(a) p + q + r (b) p – q – r
(c) 2p – 3q + r (d) p – 2q – 2r
8. Find the distance of A from B(2 ,
–
5 , 2) in
each of the cases in Question 7.
Combination of vectors
Consider two vectors a and b as shown in
Fig. 11.5.
In the gure, a parallelogram OCED has been
drawn with the sides being parallel to the vectors
a and b, and with a diagonal vector r = OE.
Given that AD = OA and BC = 2OB, then
OD = 2 OA = 2a, and
OC = 3 OB = 3b
Since DE = OC, then OE = OD + DE becomes
OE = OD + OC
i.e. r = 2a + 3b
The vector r is said to have been expressed as
a linear combination of vectors a and b, and
since r is the result of the combination, it is
called a resultant vector.
Thus:
If r = ma + nb, where m and n are scalars,
then r is said to be a linear combination of
a and b.
( )
x
2
– x
1
y
2
– y
1
z
2
– z
1
From Example 11.4, we notice that:
Given two points A(x
1
, y
1
, z
1
) and
B(x
2
, y
2
, z
2
), the displacement vector AB
is given by
AB =
= (x
2
– x
1
)i + (y
2
– y
1
)j + (z
2
– z
1
)k,
and the distance between points A and B,
i.e. the magnitude of vector AB, is given by
|AB| = √(x
2
– x
1
)
2
+ (y
2
– y
1
)
2
+ (z
2
– z
1
)
2
Exercise 11.2
1. Express in terms of i, j and k the position
vector of the following points
(a) (3 ,
–
5 ,
–
7) (b) (
–
2 , 0 , 5)
(c) (0 , 0 , 7) (d) (
–
5 , 0 , 0)
2. State in terms of i, j and k the vector
representing a displacement from:
(a) (0 , 0 , 0) to (3 , 5 ,
–
2)
(b) (3 , 5 , 4) to (2 , 7 , 8)
(c) (
–
5 , 3 , 5) to (0 , 0 , 0)
(d) (
–
3 ,
–
3 ,
–
3) to (4 , 4 , 4)
3. Express in terms of i, j and k
(a) + (b)
3 – 4
4. Find the magnitude of
(a) a =
–
3i + 3j – 2k
(b) b = 4i + 3j – 6k
5. Given that OP = 3i + 8j – 3k and
OQ = 3i + 2j + 2k, nd
(a) PQ in terms of i, j and k (b) |PQ|
6. A and B are given by (4 , 2 , 1) and
(
–
3 , 2 ,
–
2) respectively.
(a) Find in terms of i, j and k the vectors
( )
5
2
–
5
( )
3
–
3
7
( )
1
–
2
4
( )
–
2
1
–
3
Fig. 11.5
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – –
– – – – – – – – – –
– – – – – – – – – – – – – –
a
r
B
b
C
O
A
D
E
•
•
148
Example 11.5
Given that p = , q = and r = ,
express
(a) r as a linear combination of p and q.
(b) p as a linear combination of q and r.
(c) q as a linear combination of p and r.
Solution
(a) Required is an expression of the form
r = mp + nq, where m and n are scalars.
Thus, = m + n
=
Equating the components gives
5m + n = 5
3m – 2n = 16
Solving simultaneously gives
m = 2 and n =
–
5
Thus, r = 2p – 5q.
(b) Since r = 2p – 5q
p = r + q.
(c) Similarly,
q = p – r.
Example 11.6
a = , b = and c = .
Express each of the vectors as a linear
combination of the other two.
Solution
(a) a = mb + nc
 ⇒  = m + n
⇒15m + 7n = 5, and
3m + 2n = 1
Solving simultaneously gives
m = and n = 0.
Thus, a = b + oc.
(b) Since a = b + oc,then
b = 3a + oc
(c) It is not possible to express c as a linear
combination of a and b.
Parallel vectors and collinear points
If two vectors are such that one is a scalar
multiple of the other, then they are parallels. If
they are not multiples of each other, they are not
parallel (Fig. 11.6).
Fig 11.6
This can be summarised as follows:
If a and b are two vectors and a = kb, then
a is k times as long as b, and a and b are
parallel.
The idea of parallel vectors may be used to test
if any three given points are collinear (i.e. if
they lie on the same straight line).
How?
1. Determine any two vectors using the
three points.
2. Show that one vector is a scalar multiple
of the other. This indicates that the
vectors are parallel.
3. Since the vectors share a common point,
and are parallel, then conclude that the
vectors lie in the same straight line (Fig.
11.7).
Fig. 11.7
Say, |AB| = 1 unit long and |BC| = 2 units long.
Then, BC = 2AB and B is a common point.
Vectors not multiples
of each other
One vector a multiple
of the other
( )
5
3
( )
1
–
2
( )
5
3
( )
1
–
2
( )
5
16
( )
5m + n
3m – 2n
5
2
–
1
2
–
1
5
–
2
5
–
( )
5
1
( )
7
2
( )
15
3
( )
5
1
( )
7
2
( )
15
3
1
3
–
1
3
–
( )
5
16
1
3
–
C
A
B
•
149
( )
5
3
( )
2
–
2
( )
3
4
A C B
Example 11.7
Show that the points A(
–
2 ,
–
2), B(2 , 1) and
C(10 , 7) are collinear.
Solution
AB = (2i + j) – (
–
2i – 2j)
= 4i + 3j
BC = (10i + 7j ) – (2i + j)
= 8i + 6j
= 2(4i + 3j)
i.e BC = 2AB
Thus, AB and BC are in the same direction and
since B is a common point, then A, B and C are
collinear, i.e. they are on the same line.
Note: 1. The above working using column
vectors.
2. Collinearity in three dimensions is
proved in the same way.
Exercise 11.3
1. Given a = , b = and c = ,
express
(a) a as a linear combination of b and c.
(b) b as a linear combination of a and c.
2. Express c as a linear combination of a and
b in each of the following cases.
(a) a = 3i + 2j, b = i + j, c = i + 2j
(b) a = 3i + 4j, b = 2i + 2j, c = j
3. (a) Given that 3p + 2q – 3r = 0, express
each of the following as a linear
combination of the other two
(i) p (ii) q (iii) r
(b) If kp + mq + nr = 0, express q as a
linear combination of the other two
vectors
4. Given that p = 4i + 5j + 6k and
q = i + 2j + k, nd values of m and n such
that mp + nq = 15i + 24j + 19k
5. Given that p = 8a + 6b, q = 10a – 2b and
2
6
–
r = 2ma + 2(m + n)b, where m and n are
scalars, nd values of m and n such that
r = 6p – 8q.
6. If OA = 6p – 4q, OB = 2p + 14q and
AB = 4mp + (2m – n)q, nd the scalars m
and n.
7. Show that the following points are collinear.
(a) A(3 , 1), B(6 ,
–
1) and C(
–
3 , 5).
(b) A(7 , 20 ,
–
3), B(1 , 2 , 3) and
C(3 , 8 , 1).
8. Determine if the following points are
collinear.
A(5 , 3), B(
–
3 , 2), C(9 , 5).
9. In Fig. 11.8, DE = 2 EB, AB = p, DC = 2p
and DA = q.
(a) Find in terms of
p and q
(i) DB
(ii) EB
(iii) CB
(iv) AE
(v) EC
(b) What do your answer to (iv) and (v)
tell you about the points A, E and C?
Proportional division of a line
Internal division of a line
Consider Fig. 11.9.
Fig. 11.9
Line AB is divided into six equal parts.
Points C lies of the way along AB.
∴the ratio AC : CB = 2 : 4.
Taking the direction from A to B to be positive,
we say that C divides AB internally in the
ratio 2 : 4.
In what ratio does C divide BA?
B
p
C
E
D
A
2p
•
Fig. 11.8
q
150
A CB
External division of a line
Fig. 11.10 shows a line AB produced to a
point C.
Fig. 11.10
Point C is outside the interval AB, i.e. it is
external to line AB.
Line AB is divided into 6 equal parts. BC is
equal to 3 of these parts.
Taking the direction from A to B as positive,
then the direction from C to B is negative.
Thus, AC : CB = 9 :
–
3
In this case, we say that C divides AB externally
in the ratio 9 : 3 or simply, C divides AB in the
ratio 9 :
–
3.
In what ratio does C divide BA?
In what ratio does A divide BC?
If a point X lies on a line segment PQ such
that X is between P and Q and PX : XQ =
m : n, we say that X divides PQ internally in
the ratio m : n (Fig. 11.11(a)), and X divides
QP internally in the ratio n : m. Thus,
direction matters.
If X lies on PQ produced and PX : QX = m
:
–
n, we say that X divides PQ in the ratio m
:
–
n or that, X divides PQ externally in the
ratio m : n (Fig. 11.11(b)). X divides QP in
the ratio
–
n : m. The negative sign indicates
that XQ is in the opposite direction to PQ.
(a)
(b)
Fig. 11.11
The ratio theorem
Consider Fig. 11.12 which shows a point R
which divides line PQ in the ratio 4 : 3. O is the
origin, chosen arbitrarily.
With reference to the origin O, the position
Fig. 11.12
vectors of P, Q and R are, respectively,
OP = p, OQ = q and OR = r.
We know that OR = OP + PQ.
But PQ = PO + OQ =
–
OP + OQ
= OQ – OP
∴OR = OP + (OQ – OP)
= OP + OQ
i.e. r = p + q.
In the same way, we can nd the position vector
of a point that divides a line externally. For
instance, in Fig. 11.13, R divides PQ externally
in the ratio 5 : 2.
Fig. 11.13
Taking any point O as the origin,
OP = p, OQ = q and OR = r.
Now OR = OP + PR
= OP + PQ
= OP + (OQ – OP), since
PQ = OQ – OP
=
–
OP + OQ
i.e. r =
–
p + q
We now consider the general case, in which a
point R divides a line PQ in the ratio m : n
(Fig 11.14).
With any point O as the origin,
OP = p, OQ = q and OR = r.
P
m
X
Q
n
P
Q
X
m
O
– – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
q
P
R
Q
p
r
4 units
3 units
4
7
–
4
7
–
3
7
–
4
7
–
4
7
–
3
7
–
2
3
–
5
3
–
5
3
–
2
3
–
5
3
–
5
3
–
q
P
O
R
Q
p
r
– – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
3 units
2 units
151
L P Q MR
Example 11.8
Use ratio theorem to work out the example in
Figs. 11.12 and 11.13.
Solution
r = p + q, by ratio theorem.
In Fig. 11.12, m : n = 4 : 3, i.e. m = 4 and n = 3.
∴r = p + q
i.e. r = p + q
In Fig 11.13, m:n = 5 :
–
2, i.e. m = 5 and n =
–
2
∴r = p + q
i.e. r =
–
p + q
Exercise 11.4
1. In Fig. 11.15, line LM is divided into 10
equal parts.
Fig. 11.15
State the ratio in which
(a) P divides LM. (b) Q divides LR.
(c) Q divides MP. (d) M divides LR.
(e) R divides LP.
2. Draw a line segment AB and show the
position of a point P which divides
(a) AB internally in the ratio 2 : 7.
(b) AB externally in the ratio 7 :
–
2.
(c) BA in the ratio 5 : 4.
(d) AB in the ratio
–
6 : 7.
(e) BA in the ratio 6 :
–
1.
3. A point P divides a line AB in the ratio 4 :
3 and a point Q divides the same line AB in
ratio 9 :
–
2.
(a) Draw a sketch showing points A, B, P
and Q.
(b) In what ratio does
(i) B divide QP?
(ii) A divide PQ?
m
m + n
–––––
m
m + n
–––––
n
m + n
–––––
m
m + n
–––––
n
m + n
–––––
m
m + n
–––––
1
2
–
1
2
–
m
m + n
–––––
m
m + n
–––––
m
m + n
–––––
m
m + n
–––––
m
m + n
–––––
m + n – m
m + n
––––––––
m
m + n
–––––
n
m + n
–––––
m
m + n
–––––
n
m + n
–––––
m
m + n
–––––
n
m + n
–––––
m
m + n
–––––
Then, OR = OP + PR
But PR = PQ
∴ OR = OP + PQ
= OP + (OQ – OP)
= OP – OP + OQ
= (1 – )OP + OQ
= OP + OQ
= OP + OQ
i.e r = p + q
This is called the ratio theorem.
The ratio theorem states that if a point R
divides a line PQ in the ratio m : n, then the
position vector of R is
OR = r = p + q,
where p and q are the position vectors of P
and Q respectively.
Note:
1. The result is independent of the position
chosen for the origin.
2. + = 1. If + = 1,
P, Q and R are not collinear.
3. If R is the mipoint of PQ, then m = n and
r = p + q.
This is a special case of the ratio theorem,
called the midpoint theorem.
Specifying the position of a point using the ratio
theorem is quicker, especially when working
in three dimensions, in which case, calculation
and drawing diagrams may otherwise become
cumbersome or impossible.
n
m + n
–––––
m
m + n
–––––
4
4 + 3
––––
3
4 + 3
––––
3
7
–
4
7
–
–
2
5 +
–
2
–––––
2
3
–
5
3
–
5
5 +
–
2
–––––
– – – – – – – – – – – – – – –
– – – – – – – – –
O
P
R
m
n
p
r
– – – – – – – – – – – – – – –
q
Q
Fig. 11.14
152
4. In Fig. 11.15, nd
(a) the position vector of P in terms of the
position vectors L and M.
(b) the position vector of R in terms of the
position vectors Q and M.
(c) the position vector of Q in terms of the
position vectors of L and P.
(d) the position vector of P in terms of the
position vectors of Q and R.
5. The coordinates of A and B are (1 , 5) and
(5 , 2) respectively. A point X divides
AB in the ratio 5 : 2. Find, using the ratio
theorem, the position vector of X.
6. P and Q are the respective points (5 , 3 ,
–
2)
and (
–
1 , 3 , 4). If R is a point that divides
PQ externally in the ratio 8 : 5, use the ratio
theorem to nd the position vector of R.
7. Given that P is (
–
3,
–
5,
–
4) and Q is (5, 2,
5). nd the coordinaes of point M, the
midpoint of PQ.
8. The coordinates of points A and B are
(2 ,
–
3 , 4) and (
–
4 , 7 ,
–
2), respectively. A
point P is the midpoint of AB and a pointQ
divides PB in the ratio 9 :
–
5. Find the
coordinates of P and Q.
9. The position vectors of A and B are
respectvely, a = 6i + 8j and b = 4i – 3j.
Use the ratio theorem to determine
(a) the position vector of a point C which
divides AB in the ratio 3 :
–
2.
(b) the position vector of a point D which
divides BA in the ratio 3 : 2.
Further problems on ratio theorem
In some problems, it may be necessary to use
the ratio theorem more than once. In such
cases, simultaneous equations may be formed in
the process of solving the problem.
Example 11.9
A triangle OPQ is such that OP = p and
OQ = q. A point R divides OP in the ratio 1 : 2
and a point S divides PQ in the ratio 5 : 2.
OS and RQ meet at T. Determine
(a) the ratio OT : TS,
(b) the ratio QT : TR,
(c) t in terms of p and q.
Solution
Figure 11.16 shows the triangle and all the
points described.
OT = mOS, and
QT: TR = n : 1 – n, i.e. QT = nQR.
Using the ratio theorem,
OS = p + q
∴ OT = mOS
= m( p + q)
OR = p and OQ = q
∴ by ratio theorem,
OT = nOR + (1 – n)OQ
= np + (1 – n)q.
Equating the two expressions for OT gives
m( p + q) = np + (1 – n)q
i.e. p + q = p + (1 – n)q.

= and = 1 – n.
Solving simultaneously for m and n, we get
m = and n = .
Thus,
(a) the ratio OT : TS = : = 7 : 4,
Q
R
S
P
O
T
q
p
Fig. 11.16
7
11
––
4
11
––
6
11
––
7
11
––
5m
7
––
2m
7
––
n
3
–
2m
7
––
5m
7
––
n
3
–
1
3
–
2
7
–
5
7
–
1
3
–
1
3
–
5
7
–
2
7
–
5
7
–
2
7
–
153
(b) the ratio QT : TR = : = 6 : 5,
(c) t = OT = ( p + q)
= p + q.
Note that it was necessary to nd the same
vector OT in two dierent ways.
Exercise 11.5
1. In a triangle OAB, OA = a and OB = b.
C divides AB in the ratio 4 : 3 and D
divides OB in the ratio 1 : 5. OC and AD
intersect at E. Use the method of Example
11.9 to nd OE in terms of a and b.
2. In triangle ABC, AB = a and BC = b.
M and N are points on BC and AC such
that BM = MC and AN = NC.
(a) Express AM and AN in terms of a
and b.
(b) Find the ratio AE : EM, where E is the
point at which AM and BN intersect.
3. ABC is a triangle with AB = a and AC = b.
D is a point on BC such that BD : DC = 5 : 2.
ABED is a trapezium with BE parallel to
AD and 7AD = 4BE.
(a) Express BE and AE in terms of a
and b.
(b) If AE and BC meet at F, nd the ratio
in which
(i) E divides AF,
(ii) C divides BF.
Application of vector methods in
geometry
Vector methods can be used to establish some
well known geometric results as illustrated in
the following examples.
Example 11.10
Using vector methods, show that the diagonals
of a parallelogram bisect each other.
Solution
Fig 11.17 shows
a parallelogram
OACB on which
OA = a and OB = b.
If its diagonals
bisect each other, then the
midpoint of the two diagonals should coincide.
By midpoint theorem, the position vector of the
midpoint of diagonal AB is
a + b.
AC = OB = b (opposite sides of a
parallelogram)
∴ OC = a + b.
Hence, the position vector of the midpoint of
diagonal OC is
(a + b) = a + b
The two midpoints coincide. Hence the
diagonals bisect each other.
Example 11.11
AB = b and AD = d.
If P and Q are the midpoints of BC and AD
respectively, show that APCQ is a
parallelogram.
Solution
Fig. 11.18 shows the
required square.
PC = BC = AD = d
AQ = AD = d
Thus, vectors PC and AQ
are equal and in the same
direction, therefore parallel.
AP = AB + BP = AB + BC
= b + d
QC = QD + DC = AD + DC
= d + b = b + d
Thus, vectors AP and QC are equal and in the
same direction. Since AP and QC are opposite
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
Fig. 11.17
B
A
C
O
a
b
– – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – –
D
b
P
Q
B
C
A
d
Fig. 11.18
•
•
7
11
––
2
7
–
5
7
–
5
11
––
6
11
––
5
11
––
2
11
––
1
2
–
1
2
–
1
2
–
154
sides of the quadrilateral APCQ, then the
quadrilateral is a parallelogram.
Exercise 11.6
In this exercise, use vector methods only.
1. Show that the diagonals of a rhombus
bisect each other.
2. In Fig. 11.19, OA = a, OB = b, OA = OC
and OB =
OD.
Show that AB is
parallel to CD.
If A and B are the
midpoints of OC
and OD respectively,
show that even in this
case, AB//CD.
3. The midpoints of the sides AB, BC, CD
and DA of a quadrilaterial are P, Q, R and
S, respectively. Taking AB = b, AC = c
and AD = d, show that PQRS is a
parallelogram.
4. A(3 , 8), B(11 , 14), C(11 , 4) and D(5 , 2)
are the vertices of a quadrilateral.
(a) By considering the lengths of the sides
– – – – – – –
B
Fig.11.19
C
A
O
D
of the quadrilateral, show that the
quadrilateral is a kite.
(b) If P,Q, R and S are the midpoints of
AB, BC, CD and DA respectively
show that PQ//AC and QR//
BD.
(c) Hence, show that PQRS is a rectangle.
5. Draw a kite ABCD with AB = AD and
BC = CD. P, Q, R and S are the midpoints
of AB, BC, CD and DA, respectively. If
AB = a, BC = b, CD = c and DA = d,
show that PQRS is a rectangle.
6. Fig. 11.20 shows the edges of a box.
AB = x, AD = y and AE = z.
If P and Q are the midpoints of BF and
DH respectively, show that EPCQ is a
parallelogram.
1
3
–
Fig. 11.20
– – – – – – – –
– – – – – – – –
– – – – – – – –
– – – – – – – –
– – – – – – – –
– – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – –
H
F
B
C
Q
D
E
y
P
G
A
x
z
•
•
– – – – – – – –
1
3
–
155
12
BINOMIAL EXPANSION
Binomial expansion
An expression such as (a + b), where a and
b are distinct terms, is called a binomial
expression. The expansion of an expression
such as (a + b)
n
where n = 1, 2, 3,…, n is called
a binomial expansion.
In Form 2, we learnt how to expand binomial
expressions, such as
(i) (a + b)
2
= a
2
+ 2ab + b
2
(ii) (a – b)
2
= a
2
– 2ab + b
2
(iii) (a + b)(x + y ) = ax + ay + bx + by, and so on.
In this chapter we shall develop a formula, or
method, of expanding binomial products of the
form (a + b)
n
when n is a natural number.
We begin by looking for patterns in the
expansion of (a + b)
n
for n = 1, 2, 3, 4.
1. (a + b)
1
= a + b (since any number to
power 1 equals the
number itself.)
2. (a + b)
2
= a
2
+ 2ab + b
2
(as in (i) above).
3. By writing (a + b)
3
as (a + b) (a + b)
2
, we
have
(a + b)
3
= (a + b) (a
2
+ 2ab + b
2
)
= a(a
2
+ 2ab + b
2
) + b(a
2
+ 2ab + b
2
)
= a
3
+ 2a
2
b + ab
2
+ a
2
b + 2ab
2
+ b
3
= a
3
+ 3a
2
b + 3ab
2
+ b
3
(4) Similarly, expand (a + b)
4
and verify that
(a + b)
4
= a
4
+ 4a
3
b + 6a
2
b
2
+ 4ab
3
+ b
4
Looking at these results we should note the
following patterns:
1. In each expansion, there are (n + 1) terms
where n is the power.
2. The expansion begins with a
n
and ends
with b
n
.
3. The coefficients of the terms in the
expansion read exactly the same forwards
and backwards, e.g. in (a + b)
3
= 1a
3
+
3a
2
b + 3ab
2
+ 1b
3
, the coecients 1, 3,
3, 1 form the same sequence from left to
right and from right to left.
4 The sum of the exponents (powers) in
each term is equal to n.
5 The exponent of a decreases by 1 from
one term to the next.
6 The exponent of b increases by 1 from
one term to the next.
7 The coecient of the second term and
that of the second last term are always
equal to n.
8 When n is even, the coefficients are
symmetrical about the middle term.
9 When n is odd, the coefficients are
symmetrical about the two middle terms.
The patterns we have observed apply for all
natural numbers, n.
We could continue this process to nd
expansions for (a + b)
5
, (a + b)
6
…, (a + b)
n
.
However, as n becomes large the multiplication
process becomes cumbersome and the need for
an alternative method becomes evident.
Pascal’s triangle
Let us use the pattern that we have observed
from the previous section to expand (a + b)
6
.
(a + b)
6
= a
6
+ (?)a
5
b + (?)a
4
b
2
+ (?)a
3
b
3
+
(?)a
3
b
4
+ (?)ab
5
+ b
6
To be able to complete the expansion, we need
to discover additional patterns in the expansion
of (a + b)
n
.
156
Using the expansion for (a + b)
n
for n = 1, 2, 3,
4, 5, let us write down only the coecients, as
shown in Pattern 12.1.
(a + b)
0
............................................... 1
(a + b)
1
........................................... 1 1
(a + b)
2
.......................................1 2 1
(a + b)
3
.................................. 1 3 3 1
(a + b)
4
..............................1 4 6 4 1
(a + b)
5
.......................... 1 5 10 10 5 1
(a + b)
6
...................... 1 6 15 20 15 6 1
(a + b)
7
..................1 7 21 35 35 21 7 1
(a + b)
8
.......... 1 8 28 56 70 56 28 8 1
(a + b)
9
........ 1 9 36 84 126 126 84 36 9 1
(a + b)
10
.... 1 10 45 120 210 252 210 120 45 10 1
Pattern 12.1
The resulting triangle of coecents is called
Pascal’s triangle. For reasons of symmetry, we
begin with (a + b)
0
rather than (a + b)
1
. In the
Pascal’s triangle,
1. The numbers along the sides are 1s.
2. Each entry in the pattern is the sum of
the two numbers diagonally above it. For
example, the 6 in the 5th row is the sum
of the two 3
’
s diagonally above it.
Using these observations, we can form as many
additional rows as we wish.
Pattern 12.1 can be extended indenitely, each
row being derived from the one immediately
before it.
The coecients for (a + b)
n
will appear in the
row (n + 1). For example
Row 6: (a + b)
5
1 5 10 10 5 1
Row 7: (a + b)
6
1 6 15 20 15 6 1
∴ the coecients for (a + b)
6
are 1, 6, 15, 20,
15, 6, 1
Thus
(a + b)
6
= a
6
+ (6)a
5
b
1
+ (15)a
4
b
2
+ (20)a
3
b
3
+
(15)a
2
b
4
+ (6)ab
5
+ b
6
= a
6
+ 6a
5
b + 15a
4
b
2
+ 20a
3
b
3
+
15a
2
b
4
+ 6ab
5
+ b
6
Now, copy Pascal’s triangle (in Pattern 12.1)
and continue the pattern up to the 12th row.
Point of interest
The process of generating binomial expansions without
the tedious multiplication was devised by Blaise Pascal
(1623–1662), hence the name. Pascal was a French
mathematician, physicist and a theologian. Among his
other discoveries, he invented the calculating machine.
Also to his tribute, a computer programming language
has been named after him.
Example 12.1

7
.
Solution
First, we write the expansion of (a + b)
7
using
Triangle.
Thus (a + b)
7
= a
7
+ 7a
6
b + 21a
5
b
2
+ 35a
4
b
3
+
35a
3
b
4
+ 21a
2
b
5
+ 7ab
6
+ b
7
.
Now, we make the substitutions a = 2x and
b =
–
y.
∴ (2x +
–
y)
7
= (2x)
7
+ 7(2x)
6
(
–
y) + 21(2x)
5
(
–
y)
2
+
35(2x)
4
(
–
y)
3
+ 35(2x)
3
(
–
y)
4
+
21(2x)
2
(
–
y)
5
+ 7(2x)(
–
y)
6
+ (
–
y)
7
= 128 x
7
+ 448x
6
(
–
y) + 672x
5
(
–
y)
2
+
560x
4
(
–
y)
3
+ 280x
3
(
–
y)
4
+ 84x
2
(
–
y)
5
+
14x(
–
y)
6
+ (
–
y)
7
= 128x
7
– 448x
6
y + 672x
5
y
2
– 560x
4
y
3
+
280x
3
y
4
– 84x
2
y
5
+ 14xy
6
– y
7
.
This is the required expansion. Notice how

terms in even positions are negative, otherwise
positive.
Exercise 12.1
Without using Pascal’s triangle expand the
expressions in Questions 1–4.
1. (1 + 2x)
2
2. (√2 – a)
3
3. (b + √5)
4
4. (√2 + √3)
4
Use Pascal’s triangle to expand the expressions
in Questions 5–14.
157
Solution
In this example, the exponents of x and (
–
3y)
must add up to 10.
∴ The required term is ( )(x)
6
(
–
3y)
4
.


term is 210.
∴ 5th term = ( )(x)
6
(
–
3y)
4
= (210)(x
6
)(
–
3y)
4
= 210 x
6
× (
–
3)
4
y
4
= 210 × 81x
6
y
4
= 17 010x
6
y
4
∴ 
Exercise 12.2
In this exercise, do not write out the whole
expansions.
1. Find
(a) the 6th term in the expansion of
(x + y)
9
.
(b) the 9th term in the expansion of
(n – 2m)
10
.
(c) the 4th term in the expansion of
(3x + y)
8
.
2. Find the coecient of the term in
(a) b
8
in (a – 2b)
12
.
(b) the coecient of the term in a
4
in
(
√a – √x
)
10
and (3a – 5x)
12
.
3. Find
(a) the coecient of the term in x in the
expasion
(
1 – √x
)
8
(b) the 5th
term in
(
x – y
)
7
4. Write down (i) the third term (ii) the term
in x
3
for each of the following binomial
powers.
(a) (2 – x)
8
(b)
(
2 +
)
6
x
3
–
x
3
–
5. (1 + x)
10
6. (a – b)
9
7. (x – 2)
7
8. (1 + 2x)
6
9. (2x + y)
6
10. (3x – y)
5
11.
(
1 –
)
6
12.
(
1 + x
)
8
13.
(
2 +
)
4
14.
(
y
2
+
)
8
Finding coecients of individual
terms
Any single term of a binomial expansion can
be determined without writing out the whole
expansion.
For example, in the expansion
(a + b)
3
= a
3
+ 3a
2
b + 3ab
2
+b
3
the second term has b raised to power 1, the
third term has b raised to power 2, and the
fourth term has b raised to power 3.
Thus:
In the expansion of (a + b)
n
, the rth term
has b raised to power (r – 1), and a raised to
power {(n – (r – 1)}.
Example 12.2
Find the 7th term in the expansion of (a + 2b)
10
.
Solution
In the 7th term, the second term in the binomial,
i.e. (2b), is raised to power 6.
∴ a is raised to power 10 – 6 = 4.
The 7th term, therefore, is ( )(a)
4
(2b)
6
.

7th term is 210.
∴ 7th term = ( )(a)
4
(2b)
6
= 210a
4
(64)b
6
= 13 440a
4
b
6
.
Example 12.3

6
in the
expansion of (x – 3y)
10
.
1
x
–
1
2
–
1
y
–
1
5
–
158
5. Write down the constant term in
(a) (x – )
8
(b) (2x
2
+ )
10
Applying binomial expansions
Binomial expansion can be used in computation
to nd approximate values of powers of
numbers which are close to integers. The
following example illustrates this.
Example 12.4
Find the value of (1.01)
8
correct to 4 d.p.
Solution
Since 1.01 = 1 + 0.01, then (1.01)
8
= (1 +
0.01)
8
.
To work out (1.01)
8

of (1 + x)
8
.
Thus (1 + x)
8
= 1 + 8x + 28x
2
+ 56x
3
+ 70x
4
+
56x
5
+ 28x
6
+ 8x
7
+ x
8
Now, we make the substitution x = 0.01 in the
expansion of (1 + x)
8
, so that
(1 + 0.01)
8
= 1 + 8(0.01) + 28(0.01)
2
+
56(0.01)
3
+ 70(0.01)
4
+
56(0.01)
5
+ 28(0.01)
6
+
8(0.01)
7
+ (0.01)
8
= 1 + 8 × 0.01 + 28 × 10
–4
+
56 × 10
–6
+ 70 × 10
–8
+ 56 × 10
–10
+ 28 × 10
–12
+ 8 × 10
–14
+ 10
–16
= 1 + 0.08 + 0.002 8 + 0.000 056 +
   
Note that the 5th and further terms will not


∴ (1.01)
8
= 1.082 9 correct to 4 d.p.
Exercise 12.3
1. Find the value of
(a) (1.1)
10
(b) (1.02)
8
(c) (1.01)
9
(d) (3.02)
6
giving your answer correct to 4 d.p.
2. Use the expansion of (x – 0.2)
5
to nd the
exact value of 9.8
5
.
3. By using an appropriate expansion, nd
the value of each of the following correct
to 4 s.f.
(a) 1.99
8
(b) 0.99
12
(c) 2.99
4
(d) 0.98
6
(e) 0.9
10
4. Write down the expansion of (x + 1)
7
.
Use the substitution x = 10, to nd the
exact value of 11
7
. Find 101
7
by a similar
method and use logarithms to check your
answers correct to 4 s.f.
5. Use logarithms to calculate (0.9)
8
correct to
4 s.f.
Now, expand (1 – x)
8
to evaluate (0.9)
8
giving your answer to 4 s.f. What is the
dierence between your two answers?
6. (a) Use the expansion of (1 + x)
4
to
evaluate (1.028)
4
correct to 3 d.p.
(b) Evaluate (10.01)
6
correct to 2 d.p.
7. Use appropriate expansions to evaluate
(a)
(
)
6
(b)
(
)
5
giving your answers correct to 3 d.p.
8. Write down the expansion of (1 + x)
8
to
the term of x
4
. Hence evaluate (1.0125)
8
giving your answer correct to 5 s.f.
9. Without using tables evaluate
(a)
(
1 + √2
)
3
–
(
1 – √2
)
3
(b)
(
2 + √3
)
4
+
(
2 – √3
)
4
(c)
(
√5 + √3
)
5
–
(
√5 – √3
)
5
leaving your answers in surd form.
1
x
–
1
x
2
––
201
100
–––
98
100
–––
1
8
–
1
4
–
159
13
PROBABILITY
What is probability?
Consider the following statements:
1. The sun will rise tomorrow morning.
2. If you put a shirt in water, it will get wet.
3. If you put your exercise book in re, it will
burn.
Under normal circumstances would you
imagine the sun not rising in the morning or
putting a shirt in water and it remains dry, or
putting an exercise book in re and it does not
burn? Since these events will denitely take
place, they are said to be sure or certain
events.
State any three other events that are certain.
Now think about the following statements:
1. It will rain tomorrow.
2. The national football team will win their
next match.
3. Mwangi will win the elections.
These statements are about events which may
or may not take place. We cannot be very sure
or certain that they will or will not take place.
Such events are said to be uncertain events.
We use words like probable, likely, unlikely,
chance etc, when talking about such events.
For example, we could say, “It is likely to rain
tomorrow” or “It will probably rain tomorrow”
or “There is a chance of it raining tomorrow.”
What can you say about the following?
1. In the last two weeks it has been raining
everyday. It will rain tomorrow.
2. The national football team has won 5 out
their last 6 matches. The team will win the
next match.
3. This medicine has helped 8 out of 10
patients who had a problem similar to
yours. Try it, you will be helped.
Note that in these statements, there are numerical
values that help us to predict whether the events
are more or less likely to occur.
For example, from the fact that the national
team has won 5 out of 6 matches, we could
predict that their chance of winning the next
one is very high. If 8 out of 10 patients have
been helped by a certain medicine we are able
to predict that another such patient could be
helped by the same medicine.
We usually assign numerical values to our
predictions. Thus in the case of the national
football team, we would say that the chance of
the team winning is . Similarly, in the case of
the medicine, we would say that the chance of it
helping is .
The numerical values and are called
probabilities.
Thus:
Probability is the branch of mathematics
in which appropriate numerical values are
assigned as measures of the chances of
uncertain events occuring or not occuring.
The numerical values assigned are called
probabilities (singular: probability), derived
from the word “probable”.
Thus we can say: “The probability of the
national football team winning is ”. “The
probability of the medicine helping a patient
is ”.
The knowledge of probability has many
practical uses. The following are a few
examples.
1. An agricultural oce may need to know
whether a dierence in crop yield is due to
dierent conditions or just to chance.
2. A manufacturer may need to know what
percentage of his/her products needs to be
8
10
––
5
6
–
8
10
––
5
6
–
8
10
––
5
6
–
160
tested to check whether the machines are
still working eciently.
3. A medical researcher may need to know
what percentage of tests are needed to de-
cide that a newly discovered drug produces
the desired eect.
State more such examples.
Experimental probability
For us to be able to dene experimental prob-
ability, let us consider the Kenyan coin. It has
the President’s portrait on one side and the coat
of arms on the other side. The side with the
Presidential portrait will be referred to as heads
(H) and the other side as tails (T). The tossing
of a coin and recording the observation as either
H or T is referred to as an experiment. Each
toss of a coin is called a trial. Each observation
or result of a trial is called an outcome. Thus,
H is one outcome and T is another outcome. A
fair coin is one which does not favour any side
when tossed.
Suppose a fair coin is tossed 100 times and the
number of heads and tails showing are recorded
as 48 and 52 respectively.
Then, the number of trials is 100.
The number of times “heads” appears is 48 and
the number of times “tails” appears is 52.
The fraction is referred to as the experimen-
tal probability of obtaining heads. In short, we
write P(Heads) or P(H) = .
Similarly, P(Tails) = P(T) = .
Probabilities which are determined from ex-
periments or from practical events are called
experimental probabilities.
In general, the experimental probability of
an outcome
=
the number of outcomes
the total number of trials
This ratio is also called relative frequency.
Thus, P(H) =
Number of heads
Total number of tosses
P(T) =
Number of tails
Total numbers of tosses
Table 13.1 shows part of the results of a football
competition.
P = Number of matches played
W = Number of matches won
D = Number of matches drawn
L = Number of matches lost
Teams P W D L
Utala 18 9 7 2
Mumiad 18 9 4 5
Brewers 18 5 9 4
Shabina 18 2 5 11
Table 13.1
In this case, each match is a trial. The outcome
of each trial is either a win (W), a draw (D) or a
loss (L). Thus the experimental probability of
1. Utala winning =
2. Utala losing =
3. Shabina winning =
What is the probability of:
(a) Mumiad drawing?
(b) Brewers winning?
(c) Shabina losing?
Example 13.1

arrives at work on time is 0.2. How many times
would you expect her to be on time in the next
20 days?
Solution
 20
days be x. Then, the experimental probability of
her being on time
=
Number of days she is on time
=
.
Total number of days
52
100
–––
48
100
–––
48
100
–––
9
18
––
2
18
––
2
18
––
x
20
––
161
Since the probability that she is on time is 0.2,
then = 0.2
⇒ x = 0.2 × 20 days = 4 days.
Exercise 13.1
1. Open a textbook. Write down the last digit
of the right hand page. Do this 40 times.
For example, if the book opens at pages
120 and 121, write down 1.
(a) What is the probability of getting
(i) 0 (ii) 1 (iii) 2 (iv) 4?
(b) What is the probability of getting a
number less than 5?
(c) What is the probability of getting an
odd number?
2. Toss a bottle top 50 times. Each time
record whether the bottle top faces up or
down.
(a) What is the probability of the bottle
top facing up?
(b) What is the probability of the bottle
top facing down?
3. Toss two bottle tops together. Record
whether both face down, both face up or one
faces down while the other faces up. Do
this 40 times. What is the probability that
(a) both face down?
(b) both face up?
(c) one faces down and one faces up?
4. Toss two coins together. Record whether
they show 2 heads (HH), 2 tails (TT) or
1 head and 1 tail (HT). Do this 40 times.
What is the probability of obtaining
(a) HH (b) TT (c) HT?
5. Open a text book. Record the number of
letters of the rst and last words on a page.
Do this 30 times. What is the probability
that
(a) a word has more than 5 letters?
(b) a word has 4 letters?
(c) a word has less than 4 letters?
6. In the past twenty days that Rukwaro has
been going to work, he has had a lift seven
times, used a bus nine times and walked
four times. What is the probability that
(a) he gets a lift?
(b) he does not pay to get to work?
7. In the last season, it was found that the
probability of Ranscom Football Club
winning a match was . How many
matches would you expect the team to win
if they play 24 matches in the next season?
8. In 30 ghts, a boxer has won 24, drawn 2
and lost 4. What is the probability that
(a) he wins?
(b) he does not win the next ght?
9. The probability that it rains on Christmas
day in town X is 0.3. What is the
probability that it will not rain on Christmas
day in that town?
10. If the probability of the school’s volleyball
team winning is 0.4, how many games has
the team possibly played if it has won 16
games?
11. (a) In Question 2, what do you get when
you add the answers to (a) and (b)?
(b) In Question 3, what do you get when
you add the answers to (a), (b) and (c)?
(c) In Question 4, what do you get when
you add your answers to (a), (b) and
(c)?
(d) In Question 8, what do you get when
you add your answers to (a) and (b)?
12. Toss a coin 20 times. Each time, record
whether heads or tails show up. Find the
experimental probability
(a) P(H) (b) P(T) (c) P(H) + P(T)
What do you notice in each case?
x
20
––
1
8
–
162
Repeat the process when the coin is tossed
(i) 40 times (ii) 60 times
(iii) 80 times and (iv) 100 times
Copy and complete Table 13.2.
No. of tosses 20 40 60 80 100
P(H)
P(T)
P(H) + P(T)
Table 13.2
What do you notice about the values of
P(H) and P(T) as the number of tosses
increases?
What do you notice about P(H) + P(T) in
each case?
13. (a) Place 10 bottle tops (3 of drink A, 5 of
drink B, 2 of drink C) in a bag. Mix
them, then pick one at random and
note which brand it is and then return
it. Do this 20 times. Find the experi-
mental probability
(i) P(A) (ii) P(B)
(iii) P(C)
(b) Repeat part (a) using 10 bottle tops, all
of drink A.
Range of probability measure
In Question 13(b) of Exercise 13.1, you should
have obtained P(B) = 0.
Since the bag did not have any drink B bottle
tops, it is impossible to pick a drink B bottle
top from it. Hence we say that “Picking a drink
B bottle top from the bag is an impossible
event. Thus:
The probability of an impossible event
is 0 (zero).
In Question 13(b) of exercise 13.1, you should
have obtained P(A) = 1.
Since the bag contains only drink A bottle tops,
it is certain that whatever bottle top you pick it
is of drink A. We say that “picking a drink A
bottle top” is a certain (or sure) event.
Thus:
If an event is certain (or sure), its
probability is 1.
1. What is the probability that if you put a
shirt in water it will get wet?
2. What is the probability that the sun will
rise tomorrow?
3. What is the probability that if you kick a
ball up, it will not come down?
Look at the answers you obtained in Exercise
13.1. Did you obtain any probability less than
0 or greater than 1?
You should have noticed that for any event, the
probability lies between 0 and 1 (inclusive).
In general:
For any event A, 0 ≤ P(A) ≤ 1
In Questions 11 and 12(c), you should have
obtained the value 1 in each case.
In general:
If the only possible events are A and B and
when A takes place B does not, and vice
versa, then P(A) + P(B) = 1.
A′ or ~A is used to denote that event A has
not taken place (has not occured). Thus
P(A) + P(A′) = 1.
Probabilities can be expressed either as fractions
or as decimals.
Theoretical probability
In this section, we see how certain probabilities
can be found without experimenting.
163
In Question 12 of Exercise 13.1, you should
have noticed that as the number of tosses
increased, the values of P(H) and P(T) were
each closer and closer to 0.5 (or ). If the
number of tosses is very large, for all practical
purposes we have P(H) = P(T) = (or 0.5).
When a fair coin is tossed, there are only two
possible outcomes H or T. The symmetry of
the coin tells us that “heads” and “tails” have
equal chances of occuring. We say that they are
equally likely. It follows that P(H) = P(T).
Since P(H) + P(T) = 1 (from the earlier section),
then P(H) = P (T) = .
Alternatively
All the outcomes are H, T.
Number of all outcomes = 2.
Number of times H appears = 1.
Number of times T appears = 1.
∴ P(H) =
Number of times H appears
=
Total number of outcomes
and P(T) =
Number of times T appears
=
Total number of outcomes
Since these values can be obtained without
tossing a coin, they are called theoretical
probabilities.
In general, if N = number of all possible
outcomes of A and n = Number of times a
particular outcome appears, then P(A) = .
Note that equally likely events (or outcomes)
have equal probabilities.
Example 13.2
A coin is tossed twice. Find the probability of
obtaining
(a) 2 heads (b) 1 head (c) 2 tails.
Solution

in the second toss, we indicate the outcome as
HH.

in the second toss, we indicate the outcome as
HT, etc.
So, all the possible outcomes are HH, HT, TH, TT.
Number of all possible outcomes = 4.
Number of outcomes with two heads = 1.
Number of outcomes with 1 head and 1 tail = 2.
Number of outcomes with two tails = 1.
Hence
(a) P(2 heads) = P(HH) = .
(b) P(1 head, 1 tail) = P(HT) = = .
(c) P(2 tails) = P(TT) = .
Note: The same results would be obtained if
two coins were tossed together.
Example 13.3
A bag contains 10 bottle tops of which 3 are of
drink A, 5 are of drink B and 2 are of drink C.
If a bottle top is picked from the bag at random,
what is the probability that it is
(a) a drink A (b)
a drink B (c)
a drink C?
Solution
A represent drink A, B represent drink B and
C represent drink C.
All the possible outcomes are
A, A, A, B, B, B, B, B, C, C.
Thus, the number of possible outcomes is 10.
Each bottle top is equally likely to be picked.
Number of possible outcomes for drink A is 3.
Number of possible outcomes for drink B is 5.
Number of possible outcomes for drink C is 2.
(a) P(A)
=
No. of possible outcome for drink A
No. of all possible outcomes
i.e. P(A) =
No. of drink A bottle tops
=
No. of all bottle tops
(b) P(B) =
No. of drink B bottle tops
=
=
No. of all bottle tops
(c) P(C) =
No. of drink C bottle tops
=
=
No. of all bottle tops
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
n
N
––
2
10
––
5
10
––
3
10
––
1
2
–
1
5
–
1
2
–
2
4
–
1
4
–
1
4
–
164
Second toss
(Compare these results with your answers for
Question 13(a) of Exercise 13.1).
The possibility or sample space
When a coin is tossed, the only possible out-
comes are H,T. When a coin is tossed twice or
when two coins are tossed at the same time, the
only possible outcomes are HH, HT, TH, TT.
The list of all possible outcomes of an
experiment is called the possibility space or
the sample space. Each outcome is called a
sample or a sample point.
Thus, we have the following as examples:
Experiment Possibility
space
(a) Tossing a coin once H,T
(b) Tossing a coin twice (or two HH, HT, TH,
coins at the same time) TT
(c) Tossing a die once 1, 2, 3, 4, 5, 6
What would be the possibility space when a
coin is tossed three times?
Table 13.2 shows another way of displaying the
possibility space when a coin is tossed twice.
First toss
H T
H HH HT
T TH TT
Table 13.2
Example 13.4
Two dice are tossed at the same time. What is
the possibility space?
Solution
Note that each die has the faces marked 1, 2,


for the second die, the possibility space could
be written as:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2),
.
The same possibility space could be presented
as shown in Table 13.3. Copy and complete the
table.
2
nd
die
1
st
die
1 2 3 4 5 6
1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
3 (3,1)
4 (4,1)
5 (5,1)
6 (6,1)
Table 13.3
Exercise 13.2
Dene a suitable possibility space for each of
the following experiments.
State how many outcomes there are in each
case. Discuss:
1. A coin and a die are tossed at the same time
and the faces showing recorded.
2. Two dice are tossed at the same time. The
sum of the faces showing up is recorded.
3. Three coins are tossed and the faces show-
ing recorded.
4. A natural number is chosen at random.
5. Two coins are tossed and the number of
heads showing recorded.
6. Two coins are tossed onto the oor and the
distance between the coins recorded.
7. Three coins are tossed and the number of
heads observed is recorded.
8. The height of each pupil in your class is
recorded.
9. Four coins are tossed and the number of
heads observed is recorded.
10. The life time of an electric bulb is measured.
165
Example 13.5
A die is tossed and the number showing on top
is observed. What is the probability that the
number is greater than 4?
Solution
The possibility space is 1, 2, 3, 4, 5, 6.

than 4. Then, E has outcomes 5 and 6.
P(E) =
No. of outcomes in favour of E
=
=
.
Total number of outcomes
Example 13.6
A card is selected from an ordinary pack of


Jack (J), Queen (Q) or King (K)”.
Find (a) P (A) (b) P (B).
Solution
Total number of outcomes
= total number of cards = 52 cards.
(a) No. of spades = 13

∴
P(A) =
No. of spades
=
=
Total no. of cards
(b) Number of (Jack + Queen + King) = 12
P(B) =
No. of (J + Q + K)
=
=
Total No. of cards
Note:
1. The probability that an event occurs may
also be expressed as
“The odds that an event occurs are
p to q”.
This means that the probability that the
event occurs is .
For example, in Example 13.6 the odds that
the event A occurs are 1 to 3 and the odds
that the event B occurs are 3 to 10.
2. There are certain terms that are commonly
used in probability.
p
p + q
–––––
13
52
––
1
4
–
12
52
––
3
13
––
Finite and innite sample space
From Exercise 13.2, you should have noticed
that some possibility spaces have a nite num-
ber of sample points while others do not. A
possibility space that contains a nite number
of sample points is said to be nite. Otherwise
it is innite.
Examine the experiments in Exercise 13.2 and
state which possibility spaces are nite and
which are innite.
Events
An event is a set of outcomes which are a part
of the possibility space of an experiment.
For example, when two coins are tossed, the
possibility space is HH, HT, TH, TT. We may
describe event A as getting one head. Then the
outcomes of event A are HT, TH.
We may describe event B as getting at least one
head. Then the outcomes of event B are HH,
HT, TH.
Equally likely outcome
We have seen that because of the symmetry of
a coin, “ heads” and “tails” have equal chances
of occuring when the coin is tossed. We say that
the two outcomes are equally likely. Hence
P(H) = P(T) = .
In general
If an experiment has N equally likely
outcomes a
1
, a
2
, …, a
n
, then the probability
of each outcome is
P(a
1
) = P(a
2
) … = P(a
n
) = .
If an event E has f outcomes, then the
probability of event E is
P(E) = + + … + =
f times
= No. of outcomes in favour of E
Total No. of outcomes in the possibility space
1
2
–
1
N
––
1
N
––
1
N
––
{
f
N
––
1
N
––
2
6
–
1
3
–
166
These include:
Term Meaning
(a) Selection at Each item has the same
random chance of being selected.
(b) At least Starting with the given
value onwards.
(c) At most Up to and including the
given value.
(d) Not more than Same as “at most”.
(e) Not less than Same as “at least”.
Exercise 13.3
1. A die is tossed and the number showing
up on top is observed. E
1
is the event “the
score is 5”, E
2
is the event “the score is less
than 5”. Find
(a) P(E
1
) (b) P(E
2
).
2. Two dice are rolled and the sum of the
numbers showing on top is recorded. You
are given the following events:
E
1
: the score is a prime number;
E
2
: the score is a multiple of 4;
E
3
: the score is greater than 8;
E
4
: the score is a factor of 9.
Find:
(a) P(E
1
), (b) P(E
2
),
(c) P(E
3
), (d) P(E
4
).
3. Two dice are rolled and the numbers show-
ing on top observed. You are given the
following events.
E
1
: the sum is 5;
E
2
: both dice show the same score;
E
3
: the total score is greater than 7.
Find:
(a) P(E
1
) (b) P(E
2
) (c) P(E
3
).
4. In a room there are 4 men and 6 women.
One person is picked at random. What is
the probability that the person is a woman?
5. Mutua bought shirts from a wholesaler.
He found that 10 were good, 4 had minor
defects and 2 had major defects. One shirt
was chosen at random. Find the probabil-
ity that:
(a) it had no defects
(b) it had no major defects
(c) it was either good or had major defects.
6. An integer is chosen at random from the
numbers 1, 2, 3, …, 50. What is the
probability that the chosen number is
divisible by 6 or 8?
7. Kebiyet had 3 black biro pens and 5 blue
ones. He took one at random and gave it to
Mbue. What is the probability that Mbue
got a blue biro pen?
8. Two coins are tossed. What is the
probability of at least one tail appearing?
9. Two sets of cards X and Y are numbered
from 1 to 5. A card is drawn at random
from each set. The two cards are placed
side by side to form a two-digit number.
What is the probability that the number
formed
(a) is divisible by 5?
(b) contains at least one 4?
(c) is prime?
10. In a school of 700 pupils, 150 are in Form
4. There are 25 prefects in Form 4. What
is the probability that a pupil chosen at
random is in Form 4 but is not a prefect?
11. What is the probability that your teacher
was born on a Monday?
12. A boy made savings by keeping
one-shilling and ve-shilling coins in
a bag. A coin is picked from the bag at
random. The probability that the coin is a
ve-shilling coin is .
(a) What is the probability that the coin is
a one-shiling coin?
(b) If the boy’s savings were sh. 500, how
much of it was in ve-shilling coins?
3
8
–
167
outcomes are HH,HT,TH,TT. Since all the
outcomes are equally likely, each has a prob-
ability of .
Thus, P(HH) = P(HT) = P(TH) = P(TT) = .
Note that we are interested in the number of
heads appearing. Thus, the possibility space
is 0, 1, 2, since TT means 0 heads, HT or TH
means 1 head and HH means 2 heads. We get
1 head by getting HT or TH.
Thus P(0) = P(TT) = , P(1) = P(HT or TH) = ,
P(2) = P(HH) = .
∴ Probability space is , , .
Exercise 13.4
1. Suppose the possibility space of an
experiment is a
1
, a
2
, a
3
, a
4
. Which of
the following will dene a probability
space? Explain why others do not dene
probability spaces.
(a) P(a
1
) = , P(a
2
) = , P(a
3
) = , P(a
4
) =
(b) P(a
1
) = , P(a
2
) = , P(a
3
) = , P(a
4
) =
(c) P(a
1
) = , P(a
2
) = , P(a
3
) = , P(a
4
) =
(d) P(a
1
) = 0, P(a
2
) = , P(a
3
) = , P(a
4
) =
2. A coin is made in such a way that heads
is three times as likely to appear as tails.
Find (a) P(H), (b) P(T)
3. Let the possibility space of an experiment
be x
1
, x
2
, x
3
, x
4
.
(a) Find P(x
4
) if P(x
1
) = , P(x
2
) = ,
P(x
3
) = .
(b) If P(x
1
) = 3 P(x
2
), P(x
3
) = and
P(x
4
) = , nd P(x
1
) and P(x
2
)
4. Two boys and three girls are in a table
tennis tournament. Those of the same sex
have equal chances of winning, but each
boy is three times as likely to win as any
girl. Find the probability of a girl winning
the tournament.
13. A class consists of 5 Kenyans, 4 Ugandans,
8 Tanzanians and 3 Zambians. A student
is chosen at random to represent the class.
What is the probability that the student is
(a) Ugandan? (b) Zambian?
(c) Tanzanian or Zambian?
14. Find the probability of an event if the odds
that it will occur are
(a) 3 to 1 (b) 6 to 7.
Discrete probability space
Earlier, we said that the list of all possible
outcomes of an experiment is called a possibility
space or a sample space.
Let the possibility space of an experiment be
a
1
, a
2
, …, a
n
.
Let the corresponding probabilities be P(a
1
),
P(a
2
), …,P(a
n
) respectively.
Then the list P(a
1
), P(a
2
), …, P(a
n
) is called the
probability space of the experiment. Since
the sample points are countable, we call such a
sample space a discrete sample space.
Note:
1. For each P(a
i
), 0 ≤ P(a
i
) ≤ 1.
2. P(a
1
) + P(a
2
) + … + P(a
n
) = 1.
Example 13.7
A die is tossed. What is the probability space?
Solution
The possibility space is 1, 2, 3, 4, 5, 6.
All the outcomes are equally likely.
So the probability of each is .
The probability space is , , , , , .
Example 13.8
Two coins are tossed and the number of heads
observed. What is the probability space?
Solution
When two coins are tossed, all the possible
1
6
–
1
6
–
1
6
–
1
6
–
1
6
–
1
6
–
1
6
–
1
4
–
1
4
–
1
4
–
1
2
–
1
4
–
1
4
–
1
4
–
1
2
–
1
2
–
1
4
–
1
2
–
1
–
4
––
1
15
––
1
4
–
1
3
–
1
2
–
1
5
–
1
8
–
1
4
–
1
4
–
1
2
–
1
4
–
1
10
––
1
5
–
1
6
–
1
9
–
1
8
–
1
4
–
168
12
7
8
9
10
11
4
5
6
1
2
3
– – – – – – –
– – – – – – –
•
Fig 13.1
R
Q
P
Fig 13.3
Continuous possibility space
At the end of Exercise 13.2, you were asked
how many outcomes there were in each
possibility space in Questions 1–10. What
were the results?
You should have noticed that in some
possibility spaces,
(a) the outcomes are specic numbers. Such
a possibility space is said to be a nite
possibility space.
(b) the outcomes are countable but innite.
Such a possibility space is said to be a
countably innite possibility space.
(c) the outcomes cannot be counted. For
example, height, area, volume, etc.
would not, under normal circumstances,
be specic values. A height, an area
or a volume could be a value within a
certain range. Such outcomes make
up a continuous possibility space or
uncountable possibility space.
Note: A nite or countably innite space is
said to be discrete and a continuous (uncount-
able) space is said to be non-discrete.
If S is a continuous possibility space and A is
a given range within S, then the probability
of the event A (i.e. the probability that a point
selected at random belongs to A) is given by
P(A) =
Length of A
or P(A) =
Area of A
or
Length of S Area of S
P(A) =
Volume of A
.
Volume of S
Example 13.9
What is the probability that at a given
moment the minute hand of a
clock is between 2 and 3?
Solution
Fig 13.1 shows the face of a
clock. The area shaded is
the region the minute hand
would be in.

and 3”
P(A) =
Area of A
.
Area of the face of the clock
Since area of A is of the area of the face of
the clock, then P(A) = .
Exercise 13.5
1. Two concentric circles have radii of 4 cm
and 6 cm respectively. A point is selected
at random inside the bigger circle. What is
the probability that the point is outside the
smaller circle?
2. A point is selected at random inside an
equilateral triangle ABC whose side is 5
cm. What is the probability that the point
is more than 2 cm away from any vertex?
3. Fig 13.2 shows a hollow cylinder of radius
1 m and length 6 m, having lids on both
ends. The circular openings A and B,
which are the only openings into the
hollow of the cylinder, each has a radius of
0.3 m. A buttery enters through A and
gets out through B. Given that a buttery
does not keep on a straight path as it ies,
nd the probability that at any given time
the buttery is in the shaded space A and B?
Fig 13.2
4. Fig 13.3 shows an
inscribed circle radius
1.2 cm, of a ∆PQR in
which PQ = 3.6 cm,
QR = 4.6 cm and
PR = 5.8 cm.
A point is selected at random in the
triangle. What is the probability that the
point lies outside the circle?
1
12
––
1
12
––
– – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – –
6 m
2 m
A
B
169
O
A
B
10 cm
70°
Fig. 13.4
8
18
––
4
9
–
4
18
––
2
9
–
6
18
––
1
3
–
12
18
––
2
3
–
5. Fig 13.4 shows a circle,
centre O, radius 10 cm
and ∠AOB = 70°.
An insect is observed
to be moving around
within the circle.
What is the probability
that the insect is in the
shaded region?
6. The school clock is faulty. It works and
stops at random. What is the probability
that it stops when the minute hand is be-
tween 10.43 a.m and 11.02 a.m?
7. Fig 13.5 shows a darts
board of radius 24 cm.
The inner ring A has
inner radius 7 cm and
is 1cm thick. The next
ring B has inner radius
15 cm and is 1cm thick.
Angweny throws darts which land on the
board in a random manner. What is the
probability that the dart lands
(a) in the sector labelled 20?
(b) either in the sector labelled 20 or in the
one labelled 12?
(c) on ring A?
(d) on ring B?
(e) on the part of ring A which is in sector 1?
(f) on ring A or in sector 11?
Combined events
We have so far been dealing with cases
involving single events. Now we consider cases
in which two or more events are involved.
Mutually exclusive events
We have seen that when a coin is tossed, the
outcomes are H,T. It is not possible to have
heads and tails appear at the same time.
When a die is tossed, the outcomes are 1, 2, 3,
4, 5, 6. If the outcome is 1, then none of the
others can occur at the same time. Similarly,
when any other number appears, none of the
others can occur.
Two or more events in which the occurence
of one eliminates the possibility of the other
one occuring are called mutually exclusive
events.
Thus, when a coin is tossed, the events “getting
heads” and “getting tails” are mutually exclusive.
Similarly, when a die is tossed, the events
“getting a one”, “getting a two”, “getting a
three”, “getting a four”, “getting a ve” or
“getting a six” are mutually exclusive events.
Example 13.10
A bag contains 8 oranges, 4 mangoes and 6
lemons. A fruit is taken from the bag at ran-
dom. What is the probability that it is
(a) an orange, (b) a mango,
(c) a lemon , (d) an orange or a mango,
(e) an orange or a lemon,
(f) a mango or a lemon?
What relationship is there between the answers
you got in (a)–(c) and those you got in (d)–(f)?
Solution
(a) P(an orange) =
Number of oranges
Total number of fruits
= = .
(b) P(a mango) =
Number of mangoes
Total number of fruits
= = .
Similarly
(c) P(a lemon) = = .
(d) P(an orange or a mango)
=
Number of oranges and mangoes
Total number of fruits
= = .
Fig 13.5
1
20
5
17
12
2
9
15
14
10
11
6
8
13
16
4
7
18
3
19
A
B
170
14
18
––
7
9
–
(e) P(an orange or a lemon)
=
Number of oranges and lemons
Total number of fruits
= = .
Similarly
(f) P(a mango or a lemon) = = .
The relationship between the answers is as follows:
+ = i.e. P(an orange) + P(a mango)
= P(an orange or a mango)
+ = i.e. P(an orange) + P(a lemon)
= P(an orange or a lemon)
+ = i.e. P(a mango) + P(lemon)
= P(a mango or a lemon)
We note that:
If A and B are mutually exclusive outcomes
of a random experiment, the probability
that A or B will occur is the sum of their
probabilities. We write
P(A or B) = P(A) + P(B)
or P(A + B) = P(A) + P(B)
This property is also called the addition
rule for probabilities of mutually exclusive
events. It follows that if events A, B, C, …,
are mutually exclusive, P(A or B or C or …)
= P(A) + P(B) + P(C) + …
Exercise 13.6
1. A group of tourists arrived at Jomo Kenyatta
International Airport. 5 were English,
4 were French, 8 were American and
3 were German. One was chosen at
random to be their leader. What is the
probability that the one chosen was
(a) English (b) American
(c) German (d) French or German
(e) English or French
(f) not English?
2. In a bag, there are some blue pens, some
red pens and some of other colours. The
probability of taking a blue pen at random
is . If the probability of taking a blue pen
or a red pen at random is , what is the
probability of taking
(a) a red pen?
(b) a pen which is neither red nor blue?
3. In a factory, machines A, B and C produce
identical balls. The probability that a ball
was produced by machine A or B is . The
probability that a ball was produced by
machine B or C is . If the probability that
a ball was produced by machine A is ,
what is the probability that a ball was
produced by machine C?
4. A card is chosen at random from an
ordinary pack of playing cards. What is
the probability that it is
(a) either hearts or spades?
(b) either a club or a jack of spades?
5. When playing netball, the probability that
only Ann scores is , the probability that
only Betty scores is and the probability
that only Carol scores is . What is the
probability that none of them scores?
6. Two dice are tossed. Find the probability
that
(a) an odd number shows on the second die,
(b) a two or a ve shows on the rst die,
(c) a two or a ve shows on the rst die
and an odd number on the second die.
What connection is there between the
answers to parts (a), (b) and (c)?
7. In a certain school of 1 000 pupils, 20 are
colour blind and one hundred are
overweight. A pupil is chosen at random.
What is the probability that the pupil is
(a) colour blind, (b) overweight?
10
18
––
5
9
–
7
9
–
1
3
–
4
9
–
5
9
–
1
3
–
2
9
–
6
9
–
2
9
–
4
9
–
1
8
–
1
12
––
1
4
–
2
3
–
1
3
–
8
21
––
1
7
–
11
15
––
171
8. In a certain race, the odds that Cheruiyot
wins are 2 to 3 and the odds that Rono
wins are 1 to 4. What is the probability that
Cheruiyot or Rono wins the race?
Independent events
Examine Question 6 of Exercise 13.6 again.
The question shows that the events A “a two or
ve on the rst die” and B “an odd number on
the second die” are not mutually exclusive.
Note that in this question P(A) × P(B) equals
the probability of the events occuring together
[answer to part (c)].
Two events A and B are said to be indepe-
ndent if the probability of them occuring
together is the product of their individual
probabilities. We write:
P(A and B) = P(A) × P(B) = P(A) P(B) or
P(A.B) = P(A) × P(B) = P(A) P(B).
This property is also called the product law
for probabilities of independent events.
Given a number of events, we conrm for their
independence by
1. checking that any two of them are
independent, and
2. checking that the probability of all of them
occuring together is equal to the product of
their individual probabilities
Example 13.11
A box contains 10 bolts. It is found that 4 of
them are substandard. If two bolts are taken
from the box at random, what is the chance that
both are substandard?
Solution
 
substandard is .
Since 3 substandard bolts are remaining in the
box, the probability that the second bolt taken is
substandard is .
∴ Probability that both bolts are substandard
is × = .
Exercise 13.7
1. A bag contains 7 black and 3 white balls.
If two balls are drawn from the bag, what is
the probability that
(a) one is black and one is white?
(b) they are of the same colour?
2. Three pupils were asked to solve a
problem. Their chances of solving the
problem independently were , , and .
(a) What is the chance that all of them
solved the problem independently?
(b) What is the probability that only two
solved the problem independently?
3. Two dice are tossed giving the events:
A: the rst die shows a six, B: the second
die shows a three, C: the sum of the
numbers on the two dice is 7. Check these
events for independence.
4. A class has 18 boys and 12 girls. Three
pupils are chosen at random from the class.
What is the probability that
(a) they are all boys?
(b) one is a boy and the others are girls?
5. In an oce there are 3 men and 7 women.
Three people are chosen at random. What
is the probability that two are women and
one is a man?
6. Events A and B are such that P(A) = and
P(A and B) = . What is P(B) if A and B
are independent?
7. A bag contains 7 lemons and 3 oranges. If
they are drawn one at a time from the bag,
what is the probability of drawing a lemon
then an orange, and so on, alternately until
only lemons remain?
1
2
–
1
3
–
1
4
–
3
9
–
2
15
––
3
9
–
4
10
––
4
10
––
1
5
–
2
15
––
172
8. The probability that machine A will be
working at the end of the year is . The
probability that machine B will be working
at the end of the year is . Find the
probability that
(a) both will be working
(b) only machine B will be working at the
end of the year.
9. A school has two old typewriters. One is
under repair 10% of the time and the other
for 15% of the time. What is the probability
of both being out of action at the same
time?
Tree diagram
Consider the case where we are given three
bags as follows:
Bag A contains 10 watches of which 4 are
defective.
Bag B contains 12 watches of which 2 are
defective.
Bag C contains 15 watches of which 3 are
defective.
If a watch is chosen at random from a bag, what
is the probability that the watch is defective?
Note that a bag must be selected rst before a
watch is chosen. Hence there are two
experiments:
(i) Selecting a bag
(ii) Selecting a watch from the bag. It is either
defective (D) or not defective (N).
We represent this information
on a tree diagram as
shown in Fig. 13.6
Since the bags have
equal chances of
being selected,
the probability of
selecting a bag is
P(A) = P(B) = P(C) = .
Let P(D) be the probability of a watch being
Since choosing a bag and selecting a watch
from the bag are independent events, we
multiply the probabilities along the branches.
Since the branches are mutually exclusive
alternatives, we add the products on the branches.
Thus, the probability of having a defective
watch is,
P(D) = P(A) × P(defective watch in A) + P(B) ×
P(defective watch in B) + P(C) × P(defective
watch in C).
i.e. P(D) = × + × + ×
= + +
= .
A tree diagram is very useful when working out
problems involving successive experiments.
Exercise 13.8
1. A class has 15 boys and 10 girls. If three
pupils are selected at random from the
class, what is the probability that
(a) they are all boys?
(b) two are boys and one is a girl?
(c) they are all of the same sex?
A
N
D
C
B
N
D
N
D
Fig. 13.6
1
3
–
3
5
–
2
3
–
Fig. 13.7
defective and P(N) be the probability of a watch
not being defective.
From bag A, P(D) = and P(N) =
From bag B, P(D) = and P(N) =
From bag C, P(D) = and P(N) =
Fig. 13.7 shows the same tree diagram as in Fig.
13.6, but with probabilities shown on it.
4
10
––
3
15
––
6
10
––
10
12
––
12
15
––
2
12
––
A
N
D
C
B
N
D
N
D
— P(A and D) = ×
— P(A and N) = ×
— P(B and D) = ×
— P(C and N) = ×
— P(B and N) = ×
4
10
––
6
10
––
2
12
––
10
12
––
3
15
––
12
15
––
1
3
–
1
3
–
1
3
–
— P(C and D) = ×
4
10
––
1
3
–
6
10
––
1
3
–
2
12
––
1
3
–
10
12
––
1
3
–
3
15
––
1
3
–
12
15
––
1
3
–
1
3
–
4
10
––
1
3
–
1
6
–
1
3
–
1
5
–
1
15
––
1
18
––
2
15
––
23
90
––
173
3
7
–
2. A container has 7 red balls and 3 white
balls. Three balls are taken from the
container one after the other. What is the
probability that the rst two are red and the
third is white?
3. Box A contains 8 oranges of which 3 are
unripe. Box B contains 5 oranges of which
2 are unripe. An orange is taken from each
box. What is the probability that
(a) both oranges are unripe?
(b) one orange is unripe and the other is
ripe?
4. The probabilities that three soccer players
score penalties are , , respectively. If
each shoots once, nd the probability that
only one of them scores.
5. The probability that the school team wins a
match is 0.6. The probability that the team
loses is 0.3 and the probability that the
team ties is 0.1. The team plays 3 games.
What is the probability that the team
(a) wins two matches?
(b) either wins all the matches or loses all
the matches?
(c) wins one match, loses one and ties in
one?
6. On average, Langa Estate misses
water once a fortnight and experiences
power failure once a week. What is the
probability that there is either lack of water
or a power failure on a given day if the
events are independent.
7. The chance that it will rain tomorrow is .
The chance that Okongo football team wins
is . What is the chance that
(a) it rains and the team loses?
(b) it does not rain and the team wins?
8. There are 7 men and 3 women waiting for
an interview. They are called for the inter-
view one by one at random. What is the
probability that
(a) the rst one is a woman?
(b) the second one is woman?
(c) of the rst two, only one is a woman?
9. Some cards numbered 1 to 9 are put in a
bag. Three cards are taken from the bag
one after the other without replacement.
What is the probability that
(a) only the rst two are odd?
(b) the last two are odd?
10. The chance that the school’s volleyball
team wins a game is . If the team plays
three games, what is the chance that the
team
(a) wins only two games?
(b) loses at least one game?
3
5
–
2
3
–
1
6
–
1
3
–
1
4
–
174
14
COMPOUND PROPORTION, MIXTURES
AND RATES OF WORK
Proportion
In Book 1, we learnt about ratios and
proportions. We learnt that a proportion is a
statement expressing the equality of two ratios.
For example, = is a proportion.
Recall that the ratio of two quantities means
the fraction obtained by expressing one as a
fraction of the other, simplifying the result, and
writing it in the form a : b or .
Example 14.1
Solve the proportion = .
Solution
Solving a proportion means looking for the

Thus, =
⇒x = × 24 = 15.
Hence, = is a true proportion.
Proportional parts
In Book 1, we learnt that:
If a quantity is divided or shared into parts
related by a given ratio, the parts thus
obtained are called proportional parts.
To divide a quantity in the ratio a : b, we rst
split it into (a + b) equal shares. The required
parts then are
and of the given quantity.
Similarly, if a quantity n is to be divided
proportionally in the ratio a : b : c, the
proportional parts are
n,
n and n.
Example 14.2
Divide 450 in the ratio 4 : 5 : 6.
Solution
Number of shares = 4 + 5 + 6 = 15.
∴ the proportional parts are:
First part:  ×450 = 120
Second part:
× 450 = 150, and
Third part:  ×450 = 180
Exercise 14.1
1. Solve the proportion = .
2. If 5 : 9 = 20 : x, nd x.
3. Find y given that 33 : y = 22 : 48.
4. Find n if n : 1 = 12 : 5.
5. Express the ratio 9 : 4 in the form 1 : n.
6. Express the scale 5 cm represents 1 km in
the form 1 : n.
7. If x : 3 : 5 = 8 : y : 9, nd the values of x
and y and express the ratios as simply as
possible.
8. (a) Divide sh 5 200 in the ratio 9 : 4.
(b) Find the proportional parts when a
mass of 560 kg is shared in the ratio
1 : 2 : 3 .
9. Four people share sh 91 500 as follows:
The rst gets three times as much as the
second, the third gets twice as much as the
second and the fourth gets twice as much
as the rst.
What is the size of the largest share?
a
a + b + c
–––––––
c
a + b + c
–––––––
b
a + b + c
–––––––
24
x
––
8
5
–
a
b
–
24
x
––
8
5
–
24
x
––
8
5
–
8
5
–
24
15
––
5
8
–
a
a + b
––––
b
a + b
––––
1
2
–
1
2
–
x
12
––
3
2
–
5
15
––
4
15
––
6
15
––
175
1
2
–
8
23
––
5
23
––
10. A chemical compound is made of 185
parts copper, 14 parts tin and one part
phosphorus. Find the mass of tin in 555 kg
of the compound.
Ratios and weighting
When sharing quantities in proportional parts,
there are cases in which it is necessary to
scale up, i.e. weigh the proportion according
to specic conditions. The ratio in which the
quantity is divided thus becomes a weighted
ratio. Example 14.3 illustrates this.
Example 14.3
Stella and Cate entered into a partnership into
which they contributed sh 100 000 and
sh 80 000 respectively. After six months, they
were joined by Joe who contributed sh 60
000. After 3 years of business, their books of


Solution
There are two things to consider: the sum
invested and the duration of investment.
Stella invested sh. 100 000 for 3 years,
Cate invested sh. 80 000 for 3 years, and
Joe invested sh 60 000 for 2 years.
 
100 000 × 3 = 300 000 shilling-years
  
80 000 × 3
= 240 000 shilling-years
   
60 000 × 2 = 150 000 shilling-years
Note that it is necessary to invent the compound

working.
Their investments are in the ratio
300 000 : 240 000 : 150 000 = 10 : 8 : 5.
∴ × 193 200
= sh 84 000.
× 193 200 = sh 67 200.
× 193 200 = sh 42 000.
Contrast the working of Example 14.3 with
the working in the case where all the partners
joined at the same time, in which case the prot
would be shared in the ratio 5 : 4 : 3.
In Example 14.3, 10 : 8 : 5 is the weighted
ratio, weighted according to the duration of
investment of each partner’s contribution.
Exercise 14.2
1. Atieno invests sh 245 000 in a business for
6 months and Bii invests sh 441 000 in
the same business for 2 months. If the
business makes a prot of sh 101 920 in
the rst six months, how much should each
person get?
2. Adhu invests sh 126 000 in a business
venture. Six months later, Boiyo joins him
and invests sh 216 000. After two more
months, Chuma joins in and contributes
sh 162 000. If Adhu, Boiyo and Chuma
share the year-end prot of sh 648 000 in
a ratio which reects the amount of money
and investment duration, calculate how
much each person receives.
3. Kamau, Maina and Thuo started a business.
Kamau contributed sh 60 000 right away.
Three months later, Maina brought in his
contribution of sh 75 000 while Thuo made
his contribution of sh 105 000 after the fth
month of business. If the rst nine months’
prot was sh 141 000, nd how much each
person received.
4. Songok, Otieno and Welime are employed
to harvest maize on a farm. Songok works
for one hour, 33 minutes each day. Otieno
works for four hours, eight minutes each
day. Welime works for two hours,
35 minutes each day. Due to shortage of
10
23
––
1
2
–
176
13
26
––
1
2
–
2
4
–
cash, the owner of the farm proposes to pay
them using 16 bags of maize, to which they
agree.
If they are to share, the maize according
to the proportion of work input, nd how
many bags each man gets.
Continued proportion
We have seen that given the numbers 2, 4, 13
and 26, we can write them in proportion form as
= = or 2 : 4 = 13 : 26 = 1 : 2.
Generally, any four numbers a, b, c and d form
a proportion if =
.
Notice also that = .
We now look at another type of proportion.
If the numbers a, b, c, d, … are such that
= = = …,
they are said to form a type of proportion
called continued proportion.
Numbers that are in continued proportion
form a geometric progression. For example,
1, 2, 4, 8 and 16 are in continued proportion
and they form a geometric progression.
If the numbers a, b and c are in continued
proportion, then
=
∴b
2
= ac
⇒ b = ±√ac
The number b is called the mean proportional
to a and c while c is called the third
proportional to a and b.
Example 14.4
Find (a)
the mean proportional,
(b) the third proportional,
(c) the fourth proportional,
to 3 and 9.
Solution
 
Then =
⇒ b
2
= 3 × 9
∴ b = +√27 (since 3 and 9 are positive)
= 3√3.
 
Then =
 ⇒3c = 81
  ∴ c = = 27.
 
Already, we know the third proportional is
27.
∴ =
⇒9d = 27 × 27
∴d = = 81
Check: 3, 9, 27, 81 are in geometric progression.
 ∴ 27 and 81 are the third and fourth
proportionals to 3 and 9.
Derived proportion
Consider the following example.
Example 14.5
 
(2a – b) : (a + b)
Solution
a : b = 3 : 5
∴ = = k (where k is a constant)
⇒ a = 3k and b = 5k
Substituting into gives
=
=
= =
∴ (2a – b) : (a + b) = 1 : 8.
a
b
–
b
c
–
a
b
–
b
c
–
c
d
–
a
c
–
b
d
–
a
b
–
c
d
–
1
8
–
b
5
–
3
b
–
b
9
–
a
3
–
3
9
–
9
c
–
2a – b
a + c
–––––
2a – b
a + b
–––––
6k – 5k
8k
–––––
k
8k
––
9
27
––
27
d
––
27 ×27
9
––––––
81
3
––
2 × 3k – 5k
3k + 5k
––––––––
177
w
x
–
y
z
–
The proportion (2a – b) : (a + b) = 1 : 8 in
Example 14.5 is said to be derived from the
proportion a : b = 3 : 5.
Exercise 14.3
1. Find the mean proportional to
(a) 3 and 27 (b) 81 and 9
(c) 8 and 16 (d) pq
2
and p
5
2. Find the third proportional to
(a) 8 and 16 (b) 27 and 3
(c) √6 and √3 (d) a and b
3. Find the fourth proportional to
(a) 3, 6, 12 (b) 10, 8, 6.4
(c)
–
, ,
–
(d)
–
, ,
–
4. Given that p : q = 2 : 3, nd the ratio
(5p – 4q) : (p + q).
5. If x : y = 5 : 1, nd (4x + 5y) : 3x.
6. Find the ratio (7x + 2y) : 5x given that
x : y = 8 : 5.
7. If = , nd the ratio a : b.
8. Given that (5m – 2n) : (2m – n) = 7:5, nd
the ratio n : m.
9. If (x + 3) : x = (x – 3) : (x – 4), nd the
value of x.
10. Given that = , show that
(w – 3x) : (x – 3w) = (y – 3z) : (z – 3y).
Unitary method
Problems involving proportions can be solved
using either ratio method or unitary method, as
illustrated in Examples 14.6 and 14.7.
Example 14.6
A car travels 130 km in 2 hours. How far does
it go in 3 hours?
Solution
Ratio method
Write the problem as a proportion;
i.e. 130 : x = 2 : 3, where x is the distance
travelled in 3 hours.
Thus =
⇒ x = × 130 = 195 km.
Unitary method
In 2 hours, the car goes 130 km.
∴ in 1 hour, the car goes (130 ÷ 2) km = 65 km
∴ in 3 hours, the car goes (65 × 3) km = 195 km
Example 14.7
Five labourers till a piece of land in 12 days.
How long would eight labourers take to till the
same piece of land?
Solution
Ratio method
5 labourers take 12 days.
8 labourers take x days.
Ratio of numbers of labourers is 5 : 8.
Ratio of numbers of days is 12 : x.
∴ =
∴ x = × 12 = 7
i.e. 8 labourers take 7 days.
Note that the ratio of numbers of labourers ( )
is inverted to in the calculation since the
number of labourers and the number of days
taken are in inverse proportion.
Unitary method
5 labourers take 12 days.
∴ 1 labourer takes (12 × 5) days = 60 days.
∴ 8 labourers take (60 ÷ 8) days = 7 days.
In the unitary method , the rst line of the
working should be a statement of what
is given. The working is arranged so that
whatever is to be found comes last.
a + b
a – b
––––
11
4
––
1
6
–
1
12
––
1
3
–
8
9
–
16
27
––
4
3
–
3
2
–
3
2
–
x
130
–––
12
x
––
8
5
–
5
8
–
1
2
–
1
2
–
5
8
–
8
5
–
1
2
–
178
Exercise 14.4
In this exercise, use only the unitary method.
1. If 10 ballpoint pens cost sh 75, how much
would 7 pens cost?
2. If 13 electric bulbs cost sh 585, how many
can be bought for sh 360?
3. The food supply in a besieged camp will
last a garrison of 189 men 33 days. How
long would the supply last 231 men?
4. A man, working 7 hours a day, completes
a job in 24 days. How long would it take
him if he worked for 6 hours a day?
5. If 7 hockey sticks cost sh 875, what is the
cost of 9 sticks?
6. If 5 men can dig a ditch in 14 days, how
long would 7 men take?
7. 2 kg of a certain type of sweets cost
sh 234. What quantity of sweets can
be bought for sh 390?
8. A shelf holds 21 books, each 5.2 cm thick.
How many books 7.8 cm thick will it hold?
9. What is the rent for 196 hectares of land if
the rent for 126 hectares is sh 15 300?
10. Moving at an average speed of 55 km/h,
a motorist reaches his destination in 2.4
hours. How long would he have taken if he
moved at 60 km/h? Give your answer in
hours and minutes.
Compound proportions and rates
of work
Sometimes, a quantity may be proportional to
two or more other quantities. In such a case,
the quantities are said to be in compound
proportion.
Problems involving rates of work, and other
similar problems, often contain quantities that
are in compound proportion. Such problems
are solved using either the ratio or unitary
method.
Example 14.8
Eighteen labourers dig a ditch 80 metres long
in 5 days. How long will it take 24 labourers
to dig 64 metres long? What assumptions have
you made?
Solution
Notice that the number of days depends on the
number of labourers as well as on the length of
the ditch. Thus, this is a problem in compound
proportion.
Ratio method

in the ratio 64 : 80.
A shorter ditch takes a shorter time.
∴ multiply days by .
Number of labourers increases from 18 to 24,
i.e. in the ratio 24 : 18.
More labourers take a shorter time.
∴ multiply days by .
80 metres of ditch are dug by 18 labourers
in 5 days.
64 metres of ditch are dug by 24 labourers in
5 × × days = 3 days.
Unitary method
18 labourers dig 80 metres of ditch in 5 days.
∴ 1 labourer digs 80 metres of ditch in
(5 × 18) days = 90 days.
∴ 1 labourer digs 1 metre of ditch in
(90 ÷ 80) days = 1 days.
∴ 1 labourer digs 64 metres of ditch in
(1 × 64) days = 72 days.
∴ 24 labourers dig 64 metres of ditch in
(72 ÷ 24) days = 3 days.
64
80
––
18
24
––
1
4
–
64
80
––
18
24
––
1
8
–
1
8
–
179
Exercise 14.5
1. 16 men dig a trench 92 m long in 9 days.
What length of trench can 12 men dig in
15 days?
2. To unclog a silted drain 85 m long, 15
workers take 10 days. Find how many
workers are required to unclog a similarly
silted drain 51 m long drain in 5 days.
3. Six people pay sh 40 320 for a 7 day stay at
a hotel. How much would eight people pay
for a 3 day stay?
4. A car hire company with 24 cars uses 2 940
litres of petrol in 5 days. How long would
4 116 litres of petrol last if the company
had 28 cars and the consumption rate does
not change?
5. Rift Valley Railways charges sh 5 480 to
move a load of 2.8 tonnes for 350 km. For
what load will the corporation charge
sh 4 704 for 400 km?
6. A man, standing next to a railway line,
nds that it takes 6 seconds for a train,
105 m long, travelling at 63 km/h, to pass
him. If another train, 100 m long, takes
5 seconds to pass him, at what speed is it
moving?
7. It takes 15 days for 24 lorries, each of
which carries 8 tonnes, to move 1 384
tonnes of gravel to a construction site.
How long will it take 18 lorries, each of
which carries 10 tonnes, to move 1 903
tonnes of the gravel?
8. A car moving at 65 km/h takes 2 h 24 min
to travel 156 km. What distance does the
car travel in 48 min moving at 55 km/h?
9. Twelve men, working 8 hours a day, can
do a piece of work in 15 days. How many
hours a day must 20 men work in order to
do it in 8 days?
1
5
–
LET US
GO GREEN
1
7
–
The assumptions made in this calculation are:
1. The width and depth of the ditch is uniform.
2. The hardness of the ground is the same
throughout.
3. The labourers work at the same rate.
Example 14.9
To plant a certain number of tree seedlings
Otieno takes 5 hours. Wanjiku takes 7 hours
to plant the same number of seedlings. If
Otieno and Wanjiku worked together, how long
would they take to plant the same number of
seedlings?
Solution
In 1 hour, Otieno plants of the seedlings.
In 1 hour, Wanjiku plants of the seedlings.
∴ in 1 hour, Otieno and Wanjiku together
plant + = of the seedlings.
of the seedlings are planted in 1 hour.
∴ of the seedlings are planted in
( ÷ ) hours = hours.
∴ Otieno and Wanjiku working together
take hours
= 2 hours 55 minutes.
Note: We rst found out the fraction of the
number of seedlings that each person plants
in 1 hour, then the fraction of the number of
seedlings that they plant together in 1 hour.
Invert this fraction, and the result is the number
of hours they take working together.
All problems of the types in Examples 14.8 and
14.9 are solved in similar ways.
Trees are life.
If our forests go, we also go
with them.

conserve our environment
and to preserve ourselves.
1
5
–
12
35
––
12
35
––
35
35
––
12
35
––
35
35
––
35
12
––
35
12
––
1
7
–
180
10. If x men earn sh P in m months, in how
many months will y men earn sh Q if they
are paid at the same rate?
11. One pipe can ll a bath in 5 minutes and
another can empty the same bath in 10
minutes. Both pipes are opened at the
same time and after 5 minutes, the second
pipe is turned o. What fraction of the
bath is then full? How long will it take
for the rst pipe to ll the bath completely
from then?
12. A man can do a job in 4 days. Another
man can do the same job in 9 days. How
long will the two men take on the job if
they work together?
13. Workmen A and B, working together, do
a certain job in 1 hour. Workman A alone
does the job in 3 hours. How long does it
take workman B alone to do the job?
14. Working alone, Kimeu can do some job in
6 days. Mutua, also working alone, can
do the same job in 9 days. Kimeu starts
alone, but is joined by Mutua after 1 day.
How long do they take to nish the work
together?
15. Three masons, Amayo, Maduli and Owiti,
build a certain length of wall in 2 days.
Amayo and Maduli together could build
the same length of wall in 4 days, and
Maduli and Owiti together would take
3 days.
(a) Find the fraction of the length of the
wall that Maduli and Owiti build in
2 days and hence nd how long
Amayo would take by himself.
(b) Find the fraction of the length of the
wall that Amayo and Maduli build in
2 days and hence nd how long Owiti
would take by himself.
(c) Similarly, nd how long Maduli would
take by himself.
Mixtures
It is common practice, in business, to mix
dierent quantities or grades of a commodity,
valued at dierent prices, in order to come
up with a mixture or blend valued at an
intermediate price. This is usually done in order
to be able to meet the interests of the middle
class of consumers. For example, two grades of
coee, one intended for the lower income group
and the other intended for the higher income
group, may be blended to produce a grade that
caters for the middle income group.
The proportion in which blending is done are
determined using ratio methods.
Example 14.10
Two grades of seed, one costing sh 42 per
kilogram and the other costing sh 47 per
kilogram are to be mixed in order to produce
a blend worth sh 45.50 per kilogram. In what
proportion should they be mixed?
Solution

lower priced grade and y kg of the higher
priced grade.
x kg of the lower priced grade cost sh 42x.
y kg of the higher priced grade cost sh 47y.
Total cost of (x + y) kg of mixture
= sh (42x + 47y).
∴ cost of 1 kg of mixture = sh .
This cost is the same as sh 45.50
i.e. = 45.50
 ∴42x + 47y = 45.50x + 45.50y
⇒45.50x – 42x = 47y – 45.50y
⇒(45.50 – 42)x = (47 – 45.50)y
i.e. 3.50x = 1.50y
 ∴ = =
 ⇒x : y = 3 : 7
i.e. mix the two grades in the ratio 3 : 7.
1
2
–
1.50
3.50
––––
x
y
–
3
7
–
42x + 47y
x + y
––––––––
42x + 47y
x + y
––––––––
181
!
In the working of Example 14.10 there were
two dierences that were calculated, i.e.
45.50 – 42 and 47 – 45.50.
The dierence 45.50 – 42 = 3.50 is the amount
that needs to be added to the cost per kilogram
of the lower priced grade to raise it to the cost
per kilogram of the blend. It is called gain.
The dierence 47 – 45.50 = 1.5 is the amount
that needs to be subtracted from the cost per
kilogram of the higher priced grade to lower
it to the cost per kilogram of the blend. It is
called a loss.
If we mix 1.50 kilogram of the lower priced
grade seed, at a gain of sh 3.50 per kilogram,
with 3.50 kilogram of higher priced grade at a
loss of sh 1.50 per kilogram, the total gain is
sh 1.50 × 3.50 = sh 5.25, and the total loss is
sh 3.50 × 1.50 = sh 5.25.
Thus, total gain and total loss balance. We,
therefore, need to mix the two grades in the
ratio 1.50 : 3.50 = 3 : 7.
Example 14.11
Two grades of a commodity are priced at sh
p per unit and sh q per unit where p < q. In
what proportion should they be mixed in order
to produce a blend priced at sh n per unit such
that p < n < q?
Solution

grade priced at sh p and y units of the grade
priced at sh q.
∴ total cost of (x + y) units of mixture
= sh (px + qy).
∴ cost per unit of mixture = sh .
∴

(n – p) x = (q – n)y
∴ =
 ⇒ x : y = (q – n ) : (n – p)
i.e. mix the two grades in the ratio
(q – n) : (n – p).
Notice that (q – n) is the loss and (n – p) is the
gain involved.
Thus:
Given that the cost of the lower grade is
sh p and the cost of the higher grade is sh
q, to produce a blend costing sh n, such that
p < n < q, mix the two grades in the ratio
loss : gain = (q – n) : (n – p)
The use of gain and loss is a short-cut method
for solving problems on mixtures.
Caution:
The “gain and loss” method should be used
only after the underlying concept (i.e. the long
method) is well understood.
Example 14.12

ratio in which two types of juice, one containing
5.2% sweetener and the other 7.5%, are to be
mixed so that the mixture contains 6% sweetener.
Solution
Gain = (6 – 5.2)% = 0.8%
 
∴ mix in the ratio 1.5 : 0.8 = 15 : 8.
Exercise 14.6
1. Two brands of coee costing sh 180 and
sh 210 respectively are mixed to produce a
brand costing sh 190 per kilogram. In what
ratio are they mixed?
2. In what proportion should tea costing
sh 78 and sh 84 per kilogram be mixed to
obtain a blend worth sh 82 per kilogram?
3. In what proportion should coee priced at
sh 176 per kilogram be mixed with milk
powder priced at sh 44 per kilogram in order
to obtain a blend worth sh 165 per kilogram.
px + qy
x + y
––––––
x
y
–
px + qy
x + y
––––––
q – n
n – p
–––––
182
4. Liquid X contains 80% laboratory spirit
and liquid Y contains 55% laboratory spirit.
In what ratio must X and Y be mixed so that
the mixture contains 70% spirit?
5. Two brands of tea costing sh 160 and
sh 140 per kilogram, respectively,
are mixed in the ratio 2 : 3 by mass. The
mixture is sold at sh 240 per kilogram.
Find the prot percent.
6. Three types of our, maize, millet and
sorghum, costing sh 30, sh 38 and sh 35
per kilogram respectively, are mixed in the
ratio 3 : 4 : 3. Find the price per kilogram
of the mixture.
7. A merchant mixes 50 kilogram of beans
costing sh 38 per kilogram with 150
kilogram of beans costing sh 34 per
kilogram. At what price per kilogram must
he sell the mixture to make a prot of 30%?
8. A grocer mixes two brands of rice costing
sh 40 and sh 85 per kilogram in the ratio
3 : 2 by mass, respectively. Find the prot
percent if he sells the blend at sh 90 per
kilogram.
9. Three grades of rice, costing sh 61, sh 71
and sh 75 per kilogram are mixed together,
taking equal amounts of the rst two, to
produce a blend worth sh 72 per kilogram.
What is the ratio in which the three grades
are mixed?
10. A paintmaster makes a prot of 20% by
selling his “supermix” at sh 540 for a
300 ml can. He makes “supermix” by
blending two paints which cost him
sh 1 200 and sh 1 600 per litre respectively.
In which proportion does the paintmaster
mix the two paints?
183
15
GRAPHICAL METHODS
In Book 1, we learnt that a relation of the form
ax + b where a and b are constants is called a
linear expression. Thus, an equation of the
form y = ax + b is called a linear equation and
the graph of y against x is a straight line.
In the rst chapter of this book, we dealt with
quadratic expressions, quadratic equations and
graphs of quadratic relations.
In this chapter, we shall further our knowledge
of graphs and graphical methods.
Tables of values and graphs of given
relations
In this section, we shall review how to make
and use tables of values of given relations. We
shall also draw and use graphs to solve given
equations. This is best done using examples.
Example 15.1
Table 15.1 gives the corresponding values of x
and y which are related by an equation of the
form y = mx + c, where m and c are constants.
x 2 4 6 8 10 12 14
y 10 16 22 28 34 40 46
Table 15.1
(a) Using an appropriate scale, plot the points
and draw the graph of y = mx + c.
 
(i) m and c,
(ii) x when y = 13,
(iii) y when x = 5.
Solution
(a) By considering the range of the given
values of x and y, choose a suitable scale,
e.g. vertical scale: 1 cm to represent 5 units,
horizontal scale : 1 cm to represent 2 units.

Fig. 15.1
the chosen scales.
Plot the points.
Since the given relation is linear, the points
must lie on a straight line (Fig. 15.1).
(b) (i) Since the equation y = mx + c is linear,
the value of c equals the y-intercept, i.e.
y value when x = 0.
 ∴from the graph, when x = 0, y = 4
⇒c = 4.
  
of x represents the gradient of the line.
We use any two points, e.g (2 , 10) and
(4 , 16), on the line and calculate the
gradient as
change in y
.
change in x
30
20
10
0
2 4 6 8 10 12 14
40
50
y
x
45
35
25
15
5
×
×
×
×
×
×
×
184
x
y
8
4
0
–
4
–
2
–
1
1 2 3 4
y
– 2
x = 2
y =
–
2
x
2
+ 3
x
+ 9
Thus, the gradient = = = 3.
i.e m = 3.
(ii) Using the graph, when y = 13
x = 3.
(iii) When x = 5
y = 19.
Example 15.2
On the same axes, draw the graphs of
y = 9 + 3x – 2x
2
and y = 2x + 2 for values of x
from
–
2 to
+
3.
Use your graphs to solve the equations
(a)
–
2x
2
+ 3x + 9 = 0,
(b)
–
2x
2
+ 3x + 9 = 2x + 2.
Solution
In order to draw the graph of y =
–
2x
2
+ 3x + 9,
we draw up a table of values of x and y for
–
2 < x < 3 (Table 15.2).
x
–
2
–
1 0 1 2 3
–
2x
2
–
8
–
2 0
–
2
–
8
–
18
3x
–
6
–
3 0 3 6 9
9 9 9 9 9 9 9
y
–
5 4 9 10 7 0
Table 15.2
As in Example 15.1, represent the values in
Table 15.2 graphically. Refer to Chapter 1 of
this book for the properties of this type of curve.
To draw the graph of y = 2x + 2, either
(i) use the gradient and y-intercept form
method, or
(ii) make a table of values using at least three
points (Table 15.3).
x
–
1 0 1
y 0 2 4
Table 15.3
Both graphs are shown in Fig. 15.2.
Fig. 15.2
(a) Solutions of
–
2x
2
+ 3x + 9 = 0 are found
at the points where the graph of
y =
–
2x
2
+ 3x + 9 meets the x–axis, i.e. they
are values of x when y = 0.
Thus, x =
–
1.5 or x = 3.
(b) Solutions of
–
2x
2
+ 3x + 9 = 2x + 2 are
found at the intersection of the two graphs.
Thus, x = 2.1 or
–
1.6 (1 d.p.).
Exercise 15.1
1. (a) Copy and complete Table 15.4 for the
relation y = 5 – x.
x
–
3
–
2 0 1 2 3 5
y 5 2 0
Table 15.4
Use your table to draw the graph of
y = 5 – x.
(b) Make a table of values for the relation
y = x
2
–1 for
–
3 ≤ x ≤ 3. Hence draw
the graph of y = x
2
–1.
16 – 10
4 – 2
×
×
×
×
×
×
6
2
–
185
2. Two variables x and y are related by the
equation (i) y = x
2
(ii) y = 25 + .
(a) Make a table of values for each
relation for 1 ≤ x ≤ 6.
(b) On the same axes draw their graphs.
(c) Read the values of x and y at the point
where the graphs intersect.
(d) State the equation whose solution the
value of x represents.
3. Water is poured into a container at a steady
rate. The height h cm of water after t
seconds is as shown in Table 15.5.
t 0 0.5 1.0 1.5 2.0 2.5 3.0
h 0 0.1 0.3 0.6 1.0 1.7 3.0
3.5 4.0 4.5 5.0 5.5 6.0
4.3 5.0 5.4 5.7 5.9 6.0
Table 15.5
Draw the graph of h against t, taking 1 cm
for 1 unit on each axis. From your graph
state the height of water after
(a) 1.7 seconds (b) 4.3 seconds.
Graphs of cubic relations
This section aims at nding graphical solutions
to cubic equations.
A cubic expression is one in which the
highest power of x is 3. Thus, x
3
, 2x
3
+ 2,
2x
3
+ 5x
2
– 3x + 4 are cubic expressions. The
equation y = 2x
3
+ 5x
2
– 3x + 4 is a cubic
relation.
The graph of y = x
3
The method of drawing a curve such as y = x
3
is
exactly the same as that used to draw the graph
of a quadratic relation.
The relation y = x
3
is the simplest cubic relation
and its graph is equally simple to draw.
Example 15.3
Draw the graph of y = x
3
for values of x from
–
3
to
+
3.
Solution
1. First, take integral values of x in the given
range and tabulate your results for x and y
(Table 15.6).
x
–
3
–
2
–
1 0 1 2 3
–
2
–
1 1 2
y
–
27
–
8
–
1 0 1 8 27
–
15.63
–
3.38 3.38 15.63
Table 15.6
2. To make the drawing of the graph easier

corresponding to x =
–
2 ,
–
1 , 1 and 2 .
These extra values of y will reduce the
large gaps between some consecutive
values of y.
3. Choose a suitable scale, large enough to
accommodate all the values on both x- and
y- axes, and plot the points.
4. Join the points using a smooth continuous
curve (Fig. 15.3).
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
x
y
30
20
10
0
–
10
–
20
–
30
1 2 3
–
3
–
2
–
1
y = x
3
36
x
––
Fig. 15.3
As the values of x increase, the corresponding
values of y increase very rapidly. Similarly as
the values of x become larger in magnitude,
though negative, the values of y also become
×
×
×
×
×
×
186
larger in magnitude and negative.
Use the graph in Fig. 15.3 to answer the
following questions.
1. Find the value of y when x = 2.8.
2. Find the value of x when y = 25.
3. Find the number whose cube root is 2.3.
4. Find the number whose cube root is 24.
The graph of y = ax
3
+ bx
2
+ cx + d.
The relation y = ax
3
+ bx
2
+ cx + d is the general
cubic relation. We now draw the graph of such
a relation and observe its properties. Using the
relation y = x
3
– 3x
2
– 9x + 2, as an example, we
work in the same way as in Example 15.3.
1. Make a table of values for
–
3 ≤ x ≤ + 5
(Table 15.7).
x
–
3
–
2
–
1 0 1 2 3 4 5
x
3
–
27
–
8
–
1 0 1 8 27 64 125
–
3x
2
–
27
–
12
–
3 0
–
3
–
12
–
27
–
48
–
75
–
9x 27 18 9 0
–
9
–
18
–
27
–
36
–
45
+
2 2 2 2 2 2 2 2 2 2
y
–
25 0 7 2
–
9
–
20
–
25
–
18 7
Table 15.7
If the gap between any two consecutive y
values is too large, it is advisable to use
extra values of x within the range, for a
better curve.
2. Now, plot the corresponding values of x and
y, and join the points with a smooth curve
(Fig 15.4).
Note that the peak and the trough in the graph
in Fig. 15.4 is a common property of cubic
curves. The curve has rotational symmetry
of order 2, about the point (1 ,
–
9), which is
halfway point between the peak and the trough
along the curve.
The graph of the general cubic relation is of
the form shown in Fig. 15.5(a) if a is positive
and is of the form shown in Fig 15.5(b) if a is
negative.
x
x
y y
Fig. 15.4
(a) (b)
Fig. 15.5
Note:
Some cubic curves do not exhibit the peak and
trough property. A good example of such a
curve is the curve of y = x
3
. In general, any
cubic relation which does not have a quadratic
term (i.e. b = 0) results in a curve without a
peak and a trough.
–
3
–
4
–
8
–
12
–
16
–
20
–
24
–
28
12
8
4
0
–
2
–
1
1 2 3 4 5
x
y
×
×
×
×
×
×
×
×
×
y
=
x
3
– 3
x
2
+ 9
x
+ 2
187
35
x
y
30
25
20
15
10
5
0
–
5
–
10
–
15
–
20
–
25
–
30
–
35
–
5
–
2
–
1
1 2 3
•
•
•
•
•
•
y
=
x
3
y
= 10 – 5
x
Fig. 15.6
Exercise 15.2
For each cubic relation given in this exercise,
make a table of values for the given range, and
hence draw the graphs accurately. Save your
graphs for later use.
1. y = x
3
for 0 ≤ x ≤ 6
2. y = x
3
– x for
–
3 ≤ x ≤ 3
3. y = x
3
+ 3x for
–
4 ≤ x ≤ 4
4. y = x(x – 1)(x – 2) for
–
2 ≤ x ≤ 3
5. y = x
2
(4 – x) for
–
1 ≤ x ≤ 4
6. y = x
3
– 3x + 1 for
–
3 ≤ x ≤ 3
7. y = x
3
– 2x
2
+ 3x – 4 for
–
2 ≤ x ≤ 3
8. y =
–
2x
3
+ x
2
– 5x + 2 for
–
2 ≤ x ≤ 2
9. y = x
3
– 4x
2
+ 7x – 2 for
–
1 ≤ x
Solving cubic equations
The graph of y = x
3
may be used to solve an
equation such as x
3
= 10 – 5x. This involves
drawing an appropriate line on the same axes as
y = x
3
.
Using the same scale and on the same axes,
draw the graphs of y = 10 – 5x and y = x
3
as in
Fig. 15.6.
188
1
4
–
1
2
–
1
2
–
1
2
–
1
2
–
1
8
–
1
8
–
1
2
–
27
8
––
27
8
––
15
2
––
15
2
––
1
2
–
At the point of intersection of the curve and the
line, the value of y on the curve is equal to the
value of y on the line,
i.e. y = x
3
= 10 – 5x.
At the point of intersection, x = 1.4. Thus,
1.4 is an approximate root of the equation
x
3
= 10 – 5x.
Note
If such an equation has more than one real root
in the given range, then the line would meet the
curve at more than one point.
Example 15.4
(a) Draw the graph of y = x
3
+ 5x – 10.
(b) Use your graph to solve the equation
x
3
+ 5x = 15.
Solution
(a) To draw the graph of y = x
3
+ 5x – 10,
(i) Make a table of values for
–

+
3.
x
–
2
–
1 0 1 2 3
–
1
–
1
x
3
–
8
–
1 0 1 8 27
– –
5x
–
10
–
5 0 5 10 15
–
2 2
–
–
10
–
10
–
10
–
10
–
10
–
10
–
10
–
10
–
10
–
10
–
10
y
–
28
–
16
–
10
–
4 8 32
–
12.6
–
7.4 0.88
–
20.9
Table 15.8
If the gap between any two consecutive
y values is too large, it is advisable to
use extra values, of x within the given
range, for a better curve.
(ii) Plot the corresponding values of x and
y and join the points with a smooth
curve (Fig 15.7).
(b) The equation x
3
+ 5x = 15 can be written as
x
3
+ 5x – 15 = 0.
Adding 5 to both sides:
x
3
+ 5x – 15 + 5 = 0 + 5
⇒x
3
+ 5x – 10 = 5.
But, we have already drawn the graph of
y = x
3
+ 5x – 10.
Fig. 15.7
So, on the same axes, draw the graph of
y = 5 (see Fig. 15.7).
 
the line meets the curve. The value of x at
this point is 1.8.
∴The root of the equation x
3
+ 5x – 15 = 0
is 1.8 (approximately).
Note:
If an equation has more than one root, the line
meets the curve as many times as the number of
roots.
Exercise 15.3
In Questions 1–6, use the graphs you drew in
Exercise 15.2.
1. Use the graph of y = x
3
– x to solve the
equation
(a) x
3
– x = (b) x
3
– x = 1
35
30
25
20
15
10
0
–
5
–
10
–
15
–
20
–
25
–
30
5
y
=
x
3
= 5
x
–10
y = 5
–
2
1
–
1
2 3
x
y
×
×
×
+
+
+
189
1
4
–
3
3
1
2
–
1
4
–
Average rate of change
Consider the graph of motion in Fig. 15.8. It
is the travel graph of motion by a boy cycling
from his home to visit a friend.
Fig. 15.8
What can you say about the boy’s speed over
the various sections of his journey represented
by the portions AB, BC, CD, DE and EF?
(i) What distance did the boy cover in the rst
one hour?
In the rst one hour he covered 10 km.
(ii) What is the rate of change between B and C?
At B, the distance is 10 km; at C, the
distance is 10 km.
∴ change in distance = 0 while
change in time = 1 h
∴ average rate of change = = 0.
This rate is equivalent to the gradient of
BC, which is equal to zero.
(iii) What is the average rate of change between
C and D?
At C, the distance is 10 km. At D, the
distance is 15 km.
∴ the change in distance is (15 – 10) km
while the change in time is 2.00 p.m to
4.00 p.m, i.e. 2 h.
The average rate of change of distance is
= = 2.5 km/h.
(iv) What is the rate of change between D
and E?
2. Using the graph of y = x
3
+ 3x, solve the
equation
(a) x
3
+ 3x = 4 (b) x
3
+ 3x =
–
4
3. Using the graph of y = x(x – 1)(x – 2), solve
the equation
x
3
– 3x
2
+ 2x – 4 = 0
4. Use the graph of y = x
2
(4 – x) to nd the
(a) maximum value of x
2
(4 – x) in the
given range.
(b) values of x for which x
2
(4 – x) is
equal to 2.
(c) roots of the equation x
2
(4 – x) = 1.
5. (a) Use the graph of y = x
3
– 3x + 1 to
solve the equation x
3
– 3x + 1 = 0.
(b) Solve the equation in (a) using a
dierent cubic curve and a line.
6. (a) Given that y = x
3
, use graphical
method to nd approximate values of
√100 and √200.
(b) On the same axes as the graph of
y = x
3
, draw the graph of y = 30 – 5x,
and nd the value of x at the
intersection of the two graphs. State
the equation which has this value as
one of its roots.
7. (a) Draw the graph of y = x
2
(4 – x) for
–
1 ≤ x ≤ 4, plotting points at half unit
intervals. Use your graph to state the
roots of x
3
– 4x
2
+ 4 = 0.
(b) On the same axes draw the graph of
2y = x + 6 and state the values of x at
its intersections with the graph of
y = x
2
(4 – x). What equation has these
values as its roots?
(c) What linear graph would enable you to
nd the roots of 2x
3
– 8x
2
– x + 8 = 0
using the graph in (a)?
1
2
–
y
noon 1.00 2.00 3.00 4.00 5.00 6.00
15
10
5
C
F
A
B
E
D
Distance in kilometres
Time in hours
x
7.00
0
1
–
15 – 10
2
––––––
5
2
–
190
•
y
2
4
0
6 8 10
80
60
40
20
x
•
•
•
•
•
•
•
•
•
100
Temperature (°C)
Time (min)
0 – 15
7.00 – 5.00
–––––––––
–
15
2
–––
1
2
–
1
2
–
(v) What is the gradient of EF?
Between E and F, the boy has changed his
direction to return home. He is reducing
the distance and therefore the change is
negative.
At E, the distance is 15 km and time is
5.00 p.m. At F, the distance is 0 km and
time is 7.00 p.m.
∴ Gradient of EF =
change in distance
change in time
= =
=
–
7
Thus, between E and F, the rate of change
in distance is 7 km/h in the opposite
direction.
In the above discussion we are only concerned
with average rates (i.e. average speeds).
Rate of change at an instant
So far we are familiar with the gradient of a
straight line. But given a travel graph that is
not a straight line, how do we nd the rate of
change? The rate of change in a non-linear
graph varies at every instant. Thus, we talk
about the rate of change at an instant.
If a straight line touches a curve at a point, the
line is said to be a tangent to the curve at that
point. The gradient of the tangent represents
the gradient of the curve at that point.
To draw a tangent to a curve at a given point,
place your ruler against the curve so that you
can draw a line which touches the curve at that
point, as in Fig. 15.9.
In this gure, AB is the tangent to the curve and
P is the point of contact.
The gradient of a straight line is the same at all
points on the line, while the gradient of a curve
changes from point to point. So, the gradient
of the graph at any point is the same as the
gradient of the tangent at that point.
Using points A and P, the gradient of tangent
AB is
Fig. 15.9
vertical change
=
4 – 0
=
5.7
horizontal change 1.8 – 1.1
∴ The gradient of the curve at x = 1.8 is 5.7.
The gradient at point P gives the rate of change
of y with respect to x at x = 1.8.
In general:
The rate of change at an instant is given by
the gradient of the tangent to the curve at
that instant.
Exercise 15.4
1. The graph in Fig. 15.10 shows the change
of temperature (°C) with time (min).
Fig 15.10
P
B
A
•
y
x
–
2
–
1
0
1 2 3
10
8
6
4
2
191
(a) What was the average rate of change
of temperature during
(i) the rst 3 minutes
(ii) the full 9 minutes?
(b) Over what interval was the
temperature changing most rapidly?
2. Two lorries start simultaneously from two
towns A and B, 96 km apart, and move
towards each other at 48 km/h and 64 km/h
respectively. Use, an appropriate graph to
nd the interval of time after which they
meet and their distance from A when they
meet.
3. A man cycles at the rate of 22 km/h taking
a rest of 12 minutes at the end of each hour.
A motorist, starting from the same place
1 h later, goes by the same road as the
cyclist and catches up with him after 64
km. Use graphical method to nd the rate
at which the motorist travel?
4. A party of school boys set out for a camp,
5 km away, and walked at the rate of 5 km
per hour. After going 0.8 km, one of them
had to return to the starting point, moving
at the same rate, before starting again for
the camp. Use graphical method to nd the
rate at which he walked to the camp if he
reached the camp at the same time with the
others.
5. Make a table of values of y = x
2
for
–
4 ≤ x ≤ 4 and draw the graph of y against
x. Use your graph to nd the gradient of
the curve at
(a) x = 3, (b) x =
–
2.5.
6. Draw the graph of y = x
2
+ x – 5 for
–
5 ≤ x ≤ 5. Use your graph to nd the
gradient of the curve at
(a) x =
–
3, (b) x =
–
, (c) x = 2.
7. Draw the graph of y = x
3
for
–
4 ≤ x ≤ 5.
Use your graph to nd the gradient of the
curve at
(a) x =
–
2, (b) x = 3.
8. Table 15.9 shows how a radioactive
substance decays with time.
Time(s) 0 1 2 3 4
Mass (g) 1 0.37 0.14 0.05 0.02
Table 15.9
With time on the horizontal axis, draw the
graph of this data. Use your graph to nd
the instantaneous rate of decay at 2.5 s.
9. An object is thrown vertically upwards and
a series of photographs taken at various
times, i.e. at t = 0, 1, 2, …, 8 s, showing the
positions of the particle as in Table 15.10.
Times, t (s) 0 1 2 3 4 5 6 7 8
Height, h (m) 0 7 12 15 16 15 12 7 0
Table 15.10
With t on the horizontal axis, plot the
points and join them with a smooth curve.
Use your graph to estimate the instantaneous
rates of change at
(a) t = 2.5, (b) t = 6.7.
What do your answers represent?
Empirical data and their graphs
Empirical data is data obtained through
experimentation, observation and practical
experience.
Readings observed and recorded during
experiments are rarely exact. When these values
are plotted, they can only approximately lie on
a straight line or a curve.
Graphs from experimental data are called
empirical graphs.
If points approximately lie along a line or a
curve, it is possible to draw a line or curve that
best ts (represents) the points. Such a line is
known as a line of best t. If a point is too far
from the rest of the points, it is advisable to
1
4
–
1
2
–
192

2.35
2.30
2.25
2.20
2.15
t
80 100 120 140
+
× (10
6
)
+
+
+
+
+
check the readings again, or ignore it. It may
suggest that the error in the reading was too big.
Linear relations: line of best t
In this section we will only deal with linear
relations.
Table 15.11 shows a set of data suspected to
satisfy the relation L = at + b, where a and b are
constants. Example 15.5 illustrates how to use
a line of best t to evaluate the constants a and
b and hence conrm that the given data satisfy
the given relation.
Example 15.5

are given in Table 15.11.
t 70 80 90 110 130 140
×10
6
) 2.34 2.31 2.29 2.23 2.17 2.14
Table 15.11
It is suspected that the two quantities are


Solution
The points are shown in Fig. 15.11 and a line of

Fig 15.11
Since the points lie approximately on a straight


suitable distinct points on the line. For
example,
× 10
6
and when t = 120
× 10
6
. Form two equations and solve
them simultaneously.
Thus, 80a + b = 2.31 × 10
6
and 120a + b = 2.20 × 10
6
–
40a = 0.11 × 10
6
∴a =
–
2 750
80(
–
2 750) + b = 2.31 ×10
6
.
b = 2.53 × 10
6
(solving the equations
simultaneously)
The law can now be stated as
 
–
2 750t + 2.51 × 10
6
.
Note:

and b is as follows:
1. Choose two points on the line that you have
 
gradient of the line. This represents the
value of a.
2. Extend the line until it cuts the vertical axis
and read the value at which it does so. This
is the value of b. Note, however, that this
cannot be done in this example because the
horizontal axis does not begin at zero.
Other linear relations
If two variables x and y are such that y varies as
the nth power of x, then y = kx
n
where k and n
are constants. Taking logarithms of both sides,
y = kx
n
becomes
log
10
y = log kx
n
= log k + log x
n
= nlog x + log k
The graph of log
10
y against log
10
x gives a
straight line whose gradient is n and whose
log y-axis intercept is log k.
Example 15.6
Table 15.12 shows corresponding values of two
variables s and v. Given that S = kV
n
, where k
193
–
–
In the relation log y = xlog b + log a, log y and
x are the variables. The graph of log y against x
is a line whose gradient is log b, and log y-axis
intercept is log a.
Example 15.7
Table 15.14 shows two variables x and y such
that y = ab
x
. Find a linear relation to represent
y = ab
x
. By means of an appropriate graph,


when x = 90.
x 10 20 30 40 50 60 70 80
y 20.6 13.3 8.6 5.5 3.6 2.3 1.5 1
Table 15.14
Solution
Taking logarithms of both sides, y = ab
x
becomes
log y = xlog b + log a.
Table 15.15 shows the two variables, log y and x.
and n are constants, express the relation in
logarithmic form and hence draw an appropriate

of k and n and hence state the relation that
connects S and V.
V 10 20 30 50 70 80 100
S 0.2 0.5 0.9 1.9 3.2 3.9 5.5
Table 15.12
Solution
If S = kV
n
,
log s = log k + nlog V
This is a linear relation where log s and log V
are the variables. The graph of log s against
log V is a line of gradient n, and log s-axis
intercept is log k. Table 15.13 shows the
corresponding values of log V and log S.
V 10 20 30 50 70 80 100
log v 1 1.3 1.48 1.7 1.85 1.9 2
S 0.2 0.5 0.9 1.9 3.2 3.9 5.5
log s –0.7 –0.3 –0.05 0.28 0.51 0.59 0.74
Table 15.13
Fig. 15.12 shows the graph of log s agaist log v.

P(0.2 ,
–
1.96) and Q(1 ,
–
0.76), gradient of the
line is =
–
3 –
–
1
1.3 – 0.8
=
0.7
=
1.4
0.5
∴ n = 1.4
log s-axis intercept is
–
2.2
Thus, log k
∴ k = 0.006 3
The law connecting S and V is S = 0.006 5x
1.4
A non linear relation of the form y = ab
x
where
a and b are constant, can be transformed into a
linear relation using logarithms.
Given that y = ab
x
,
log y = log (ab
x
)
= log a + log b
x
= xlog b + log a.
Fig. 15.12
log s
2
0
–
1
–
2
1
2
×
×
×
×
×
×
×
– – – –
– – – – – –
Q
P
log v
1
194
x 10 20 30 40 50 60 70 80
log y 1.31 1.12 0.93 0.74 0.56 0.36 0.18 0
Table 15.15

from the data in Table 15.15.
log y
x
1.5
0
20 40 60 80
1.0
0.5
– – – – – – – –
B
– – – – – –
A
×
×
×
×
×
×
×
The graph meets the log y axis at 1.5
∴ log a = 1.5
a = 10
1.5

Using two points on the graph, say, A(0, 1.5)
and B(20 , 1.12),
gradient =
1.12 – 1.5
20
=
–
0.38
20
=
–
0.019
Thus, log b =
–
0.019
∴ b = 0.957 2
The relation connecting y and x is
y = 31.62 × 0.957 2
x
When x = 90,
y = 31.62 × 0.957 2
90
= 0.616 9
Exercise 15.5
1. The results in Table 15.16 were obtained
in an experiment to test the connection
between the diameter (D cm) of a driving
pulley and the power (H watts) transmitted
by a belt passing round it.
D 6 7 8 9 10 11 12
H 900 1 140 1 320 1 480 1 640 1 820 1 980
Table 15.16
Draw a graph showing the relation between
D and H and determine the linear relation
in the form H = aD + b.
2. The electrical resistance (R ohms) of a wire
at a temperature (T
°C) is given in Table
15.17.
T 10 20 40 60 80 100
R 11.3 11.7 12.5 13.3 14.1 14.9
Table 15.17
Draw a graph showing the relation between
T and R. Hence determine the equation
connecting R and T. Use this equation to
calculate R when T = 140.
3. Corresponding measurements of two
observed quantities are shown in Table
15.18.
x 1 2 3 4 5
y 1.5 8.0 19.5 36.0 57.5
Table 15.18
Draw the graph of against x.
If x and y satisfy an equation of the form
y = ax + bx
2
, nd a and b and state the law.
4. Plot the points given in Table 15.19. Draw
the best straight line to t these points and
determine its equation.
x 2 5 8 12 16 19 22
y 15.5 24 33 42.5 52 62 70
Table 15.19
Fig. 15.13
y
x
–
195
O
P(x , y)
•
x
y
r
Fig. 15.15
N
5. Two variables x and y are related by the
formula y = ax
n
. Log y and log x are
plotted to give a line of best t, whose
gradient is 2 and log y-axis intercept is 1.4.
Find the values of the constants a and n.
6. Two variables x and y are such that y = ab
n
.
Given that the graph of log y against x is a
straight line, nd the constant a and b and
state the law connecting x and y if
(a) the gradient of the line is 0.36 and the
log y-axis intercept is 1.23.
(b) the gradient is
–
0.42 and log y intercept
is
–
2.
7. Table 15.20 shows the corresponding
values of two variables H and Q
H 0.5 0.9 1.5 2.2 3.1 4.1
Q 0.283 1.229 4.409 11.49 27.07 54.46
Table 15.20
Use graphical method to show that the
relation between H and Q is of the form
Q = kH
n
. Hence nd the values of k and n.
8. Table 15.21 shows the corresponding
values of x and y.
x 0 1 2 3 4 5 6
y 8.9 15.3 26.3 45.3 77.9 134 230
Table 15.21
(a) Given that y = ab
x
, use an appropriate
graph to nd
(i) the values of a and b.
(ii) the value of y when x = 3.5.
(b) Use the relation between x and y to
nd
(i) the value of y when x = 7.
(ii) the value of x when y = 3.
The equation of a circle
In this section, our objective is to be able
to identify the equation of a circle, nd the
equation of a circle and determine the centre
and radius of a circle.
Consider a path, or a set of points, traced out
on a Cartesian plane by a variable point P(x, y).
Let the point P be r units from a xed point O
(Fig 15.14).
The path traced out by
point P represents a
circle, centre O, radius
r units.
Using Pythagoras’
theorem, we can nd
a relation connecting
x, y and r, i.e. the
Cartesian equation of a circle.
Now, consider Fig. 15.15.
From point P, drop a perpendicular to meet
the x-axis at a point N, so that triangle OPN is
right-angled at N.
∴ ON = x units,
NP = y units,
and OP = r units.
∴ ON
2
+ NP
2
= OP
2
(Pythagoras’ relation)
⇒ x
2
+ y
2
= r
2
Thus, in general:
The equation of a circle whose centre is at
the origin and whose radius is r units is
x
2
+ y
2
= r
2
Example 15.8
Find the equation of a circle centre (0 , 0) and
radius 3 units.
Solution
Fig. 15.16 is a sketch
of the circle.
   
on the circle.
The distance from the
point P(x , y) to the
centre (0 , 0) is 3 units.
O
P(x , y)
•
x
y
3
x
y
Fig. 15.16
O
P(x , y)
•
x
y
r
Fig. 15.14
196
A(4 ,
–
3)
P(x , y)
•
x
y
Fig. 15.18
6
y + 3
O
x – 4
By the Pythagorean relationship, the distance, d,
between any two points A(x
1
, y
1
) and B(x
2
, y
2
)
is given by the formula
√(x
2
– x
1
)
2
+ (y
2
– y
1
)
2
= d
Thus, the distance between O(0 , 0) and P(x , y)
is
√(x – 0)
2
+ (y – 0)
2
= 3
⇒ √(x
2
+ y
2
= 3
⇒ x
2
+ y
2
= 9 (squaring both sides)
∴ The required equation of the circle is
x
2
+ y
2
= 9.
Now consider a circle whose centre, C, is at a
point other than the origin and passing through
the variable point P, and of radius r units.
With reference to
Fig 15.17,
PS = y
OS = x
and triangle CNP is
right angled at N.
∴ CN = x – a
NP = y – b and
CP = r is the hypotenuse.
∴ CN
2
+ NP
2
= CP
2
⇒ (x – a)
2
+ (y – b)
2
= r
2
∴ The equation of the circle is
(x – a)
2
+ (y – b)
2
= r
2
.
In general
The equation of a circle centre (a , b),
radius r is (x – a)
2
+ (y – b)
2
= r
2
.
Example 15.9
Find the equation of the circle centre A(4 ,
–
3)
and radius 6 units.
Solution


√(x – 4)
2
+ (y –
–
3)
2
= 6.
Thus, (x – 4)
2
+ (y + 3)
2
= 36 (squaring
both sides)
∴ (x – 4)
2
+ (y + 3)
2
= 36 is the required
equation.
In general:
If (x, y) is a point on the circle centre (h , k)
and radius r units, then
√(x – h)
2
+ (y – k)
2
= r.
i.e. (x – h)
2
+ (y – k)
2
= r
2
.
If the centre is (0 , 0), the equation becomes
x
2
+ y
2
= r
2
.
Note that (x – 4)
2
+ (y + 3)
2
= 36 in Example
15.9 may be written in the expanded form as
x
2
– 8x + y
2
+ 6y + 25 = 36
⇒ x
2
+ y
2
– 8x + 6y = 11.
This result suggests that an equation containing
x
2
and y
2
may represent a circle.
Note:
To identify the equation of a circle, note the
following:
1. The presence of x
2
+ y
2
.
2. The absence of a term in xy.
3. Equality of the coecients of x
2
and y
2
.
4. Being reducible to either of the following
forms:
A(x
2
+ y
2
) + Bx + Cy + D = 0 and
(x – a)
2
+ (y – b)
2
= r
2
.
P(x , y)
•
x
y
r
N
– – – – – – – –
C(a , b)
S
O
– – – –
R
Fig. 15.17
197
Example 15.10
Investigate whether x
2
+ y
2
+ 2x + 6y – 15 = 0
represents a circle.
Solution
(a) x
2
and y
2
suggest the possibility that the
equation represents a circle.
(b) Complete the square on x and the square

Thus, x
2
+ y
2
+ 2x + 6y – 15 = 0 can be
written as
x
2
+ 2x + ( )
2
+ y
2
+ 6y + ( )
2
= 15 + ( )
2
+ ( )
2
x
2
+ 2x + 1 + y
2
+ 6y + 9 = 15 + 1 + 9
(completing squares on x and y)
 ∴ (x + 1)
2
+ (y + 3)
2
= 25 = 5
2
.
This equation represents a circle, centre (
–
1 ,
–
3),
radius 5 units (Fig. 15.19).
Fig. 15.19
Example 15.11
Determine the centre and the radius of a circle,
whose equation is given as
x
2
+ y
2
– 6x + 10y = 2.
Solution

x
2
+ y
2
– 6x + 10y = 2 should be expressed in
the form (x – a)
2
+ (y – b)
2
= r
2
where a, b and
c are constants, (a , b) is the centre of the circle,
and r is the radius of the circle.
x
2
+ y
2
– 6x + 10y = 2
⇒ x
2
– 6x + y
2
+ 10y = 2
⇒ x
2
– 6x +(
–
3)
2
+ y
2
+ 10y + (5)
2
 = 2 + (
–
3)
2
+ (5)
2
(completing squares on
x and y)
⇒
(x – 3)
2
+ (y + 5)
2
= 2 + 9 + 25 = 36
 ∴ x
2
– 6x + y
2
+ 10y = 2
 ⇔(x – 3)
2
+ (y + 5)
2
= 6
2
∴ The centre of the circle is (3 ,
–
5) and the
radius of the circle is 6 units.
If the end points of a diameter of a circle are
known, the equation of the circle can be found.
Example 15.12
Find the equation of a circle whose diameter
has the end points (
–
2 , 5) and (4 , 1).
Solution
If A is (x
1
, y
1
) and B is (x
2
, y
2
), the midpoint of
AB is given by
, .
Thus, the mid point of the line segment joining
(
–
2 , 5) and (4 , 1) is
,
= (1 , 3).
Since the points (
–
2 , 5) and (4 , 1) are the end
points of a diameter, the midpoint (1 , 3) must
be the centre of the circle.
∴ using C(1 , 3) and (4 , 1),
(1 – 4)
2
+ (3 – 1)
2
= r
2
⇒ 9 + 4 = r
2
∴ r = √13
Using the point C(1 , 3) and any point P(x , y)
on the circle, the equation is
(x – 1)
2
+ (y – 3)
2
=
(
√13
)
2
∴ (x – 1)
2
+ (y – 3)
2
= 13.
Note:
–
2 , 5) could
also have been used with (1 , 3).
Exercise 15.6
1. Find the equation of each of the following
circles.
(a) Centre (0 , 0), radius 4 units.
(b) Centre (2 , 4), radius 5 units.
(c) Centre (0 , 3), radius √2 units.
(d) Centre (
–
2 ,
–
1), radius 3 units.
6
2
–
2
2
–
6
2
–
2
2
–
(
)
–
2 + 4
2
–––––
5 + 1
2
––––
O
x
y
(
–
1 ,
–
3)
•
( )
y
1
+ y
2
2
–––––
x
1
+ x
2
2
–––––
198
(e) End points of a diameter are A(4 , 5)
and B(6 , 9).
2. Find the centre and radius of each of the
following circles.
(a) x
2
+ y
2
= 25.
(b) x
2
+ y
2
+ 6x – 2y + 6 = 0.
(c) x
2
+ y
2
+ 8x + 4y – 29 = 0.
(d) x
2
+ y
2
– 2x + 6y + 1 = 0.
3. Sketch the following circles.
(a) x
2
+ y
2
= 16
(b) 2x
2
+ 2y
2
= 8
(c) y
2
= 144 – x
2
(d) (x – 1)
2
+ (y + 3)
2
= 4.
(e) (x + 4)
2
+ (y + 5)
2
= 36.
4. AB is the diameter of a circle centre C.
Given that A(3 ,
–
2) and B(5 , 1), draw
the circle on a Cartesian plane. State the
coordinates of the centre, and nd the
equation of the circle.
5. Given that 4x
2
+ (a – 1)
2
y
2
+ 2ax + 6y =
0, for what values of a does the equation
represent a circle? Find the centre and
radius of the circle.
6. Given that the circle whose equation is
x
2
+ y
2
– 7x + 2y + c = 0 passes through
(7 , 1), nd the value of c. State the
coordinates of the centre of the circle.
If (7 , 1) is one end point of the diameter of
the circle, nd the other end point.
199
11-15
REVISION EXERCISES 3
( )
2
–
1
( )
–
1
2
( )
–
1
5
Revision exercise 3.1
1. Given that points P and Q are (
–
2,
–
4, 6) and
(6, 0, 4) respectively, determine
(a) PQ (b) |2PQ|.
2. Given that a = , b = and c = ,
nd the value of m such that a + mb = c.
3. Vectors a, b and c are expressed in terms of
vectors p and q as follows;
a = 3p + 2q, b = 5p – q and
c = hp + (h – k) q,
where h and k are constants. Given that
c = 2a – 3b, nd the values h and k.
4. (a) Expand, and simplify, the following
binomial expressions.
(i) (a – 2b)
3
(ii) (1 + )
4
(b) Use Pascal’s triangle to expand
(i) (2x – 3y)
5
(ii) (2x – )
6
(c) Write down the expansion of (1 + )
4
.
Hence evaluate (1.05)
4
correct to 4 d.p.
5. Two drums contained ten identical balls each.
The balls in each drum were numbered
0 to 9. A ball was picked at random from
each drum, the number noted and the ball
replaced. This was done 50 times and the
results were recorded as in Table R3.1.1.
(2, 2) (6, 9) (8, 6) (6, 1) (9, 8)
(9, 6) (8, 1) (7, 8) (7, 3) (6, 2)
(1, 1) (9, 4) (0, 9) (2, 0) (6, 9)
(7, 0) (1, 7) (6, 7) (0, 3) (9, 8)
(0, 2) (2, 2) (1, 5) (9, 2) (7, 6)
(8, 6) (3, 5) (0, 0) (6, 4) (1, 6)
(9, 0) (2, 7) (1, 2) (9, 8) (8, 0)
(3, 4) (2, 2) (0, 3) (2, 2) (7, 9)
(3, 9) (2, 8) (0, 5) (0, 1) (9, 0)
(4, 3) (4, 4) (3, 4) (1, 7) (7, 5)
Table R3.1.1
By adding the numbers, nd the probability
that the sum of the two numbers is
(a) 3 (b) 13 (c) less than 10
(d) even (e) greater than 15
(f) divisible by 3.
6. (a) A whole number from 1 to 40, inclusive,
is selected. If each number has the same
chance of being selected, what is the
probability that the number is prime?
(b) A fair die is thrown 60 times. How
many times is
(i) a 1 expected to show up?
(ii) a number divisible by 3 expected
to show up?
7. Three people X, Y and Z share sh 25 300 in
such a way that Y gets one and a half times
as much as X and Z gets twice as much as
Y. Find the amount that each person gets.
8. (a) Solve the proportion
(i) 22 : x = 198 : 54
(ii) 285 : 95 = y : 1
(b) A man takes 2 hours to arrive home
travelling at 70 km/h. How long would
he take if he travelled at 90 km/h?
9. (a) Find the proportional parts when
sh 87 400 is divided in the ratio 3 : 5 : 11.
(b) Share sh 2 990 among three girls in the
ratio : : .
10. It has been found, in a certain country, that
the probability of a child getting measles
is 0.13. Out of 1 200 children, how many
would be expected to catch measles?
Caution: Measles is an immunisable disease.
Anybody who gives birth at home should be
advised to take the infant to the nearest hospital
or dispensary for immunisation.
x
2
–
1
x
–
1
x
–
!
1
3
–
3
4
–
5
6
–
200
Time (sec)
25
20
15
10
5
0
5 10 15 20
y
x
Speed (m/s)
11. Two variables, x and y, are related by the
equation
y = 4x(4 – x)(3 – x).
(a) Copy and complete Table R3.1.2 for
this relation.
x 0 0.25 0.5 1.0 1.5 2 2.5 3
y 10.3 17.5 7.5
Table R3.1.2
(b) Draw the graph of y against x.
(c) Use your graph to nd the
(i) values of x when y = 15.
(ii) range of x for which y ≤ 17.5.
(iii) gradient of the curve at x = 2.
12. Fig. R3.1.1 is a graph of motion of a
particle over a period of 20 seconds.
Calculate the distance travelled by the
particle and its average speed.
2. In ∆ABC, E lies on BC such that = ,
F lies on CA such that = , and G
lies on AB produced such that = .
The position vectors of A, B and C, relative
to the origin O, are a, b and c respectively.
(a) Determine the position vectors of E, F
and G in terms of a, b and c.
(b) Hence, deduce that E, F and G lie on a
straight line.
3. (a) Find the coecient of x
3
in the
expansion of (x + )
8
.
(b) Find the term independent of x in the
expansion of (x + )
10
.
(c) Write down the 6th term of the expansion
of (3x + 2y)
10
, in ascending powers of x
and evaluate the term x = and y = .
4. I have ve dierent pairs of socks mixed
up in a suitcase. I pick one sock. Without
looking, I pick a second sock.
(a) What is the probability that I pick a
matching pair of socks?
(b) If the second sock does not match the
rst one, what is the probability of
picking a third sock, at random, which
matches either of the rst two (which I
have kept aside)?
5. A newspaper vendor had three papers;
Daily Nation, the East African Standard
and Kenya Times, to deliver to three people
A, B and C, respectively. He forgot the list
of who was to receive which paper and so
he delivered the papers at random. What is
the probability that
(a) everyone received the correct paper?
(b) only one person got the correct paper?
(c) no one got the correct paper?
6. (a) Find the mean proportional between 16
and 9.
(b) Find the third and fourth proportionals
to 36 and 48.
Fig, R3.1.1
Revision exercise 3.2
1. Given that A and B are the points whose
position vectors, are i + 2j + k and 3i + 4j
+ 2k respectively, determine AB and |AB|.
GB
GA
–—
CF
FA
–—
3
4
–
1
2
–
BE
EC
–—
2
3
–
1
4
–
1
3
–
1
x
–
1
2
–
201
1
3
–
3
4
–
S
Q
X
P
V
O
7. Ken started a business, investing sh 61
250. After four months, he was joined by
Jim, who put in sh 110 250. If the business
made a prot of sh 25 480 in the rst six
months, how much should each partner
get?
8. Two farmers, Surum and Tipis, share the
grazing of a eld. Surum puts in 120 cows
for 18 days and Tipis puts in 150 cows for
16 days. How should they share the pay-
ment of sh 4 275 rent?
9. (a) Find the equation of a circle with
(i) centre (0, 0), radius 2 units
(ii) centre (3,
–
2), radius 3 units
(b) Find the radius and the centre of the
circle whose equation is
x
2
+ y
2
+ 4x – 6y + 10 = 0.
(c) The end points of the diameter of a
circle have coordinates (8 , 4) and
(2 , 2). Find the equation of the circle
and state its centre and radius.
10. Table R3.2.1 gives corresponding values of
two variables x and y.
x 1.1 1.2 1.3 1.5 1.6
y 0.3 0.5 1.4 3.8 5.2
Table R3.2.1
The variables are known to satisfy an equa-
tion of the form y = ax
3
+ c, where a and c
are constants.
By drawing a graph of y against x
3
, nd the
values of a and c. Hence, write down the
relationship connecting x and y.
11. The variables D and E in Table R3.2.2 are
related in direct variation.
D 3 7 12 25 32
E 210 250 305 440 500
Table R3.2.2
(a) Use the data in the table to draw the
line of best t through the data.
(b) Write down an equation connecting D
and E and use it to nd the value of E
when D = 9.
12. Variables x and y are connected by the
equation y = kx
n
(which may also be written
as log y = n log x + log k), where k and n
are constants. Table R3.2.3 gives corre-
sponding values of log x and log y.
log x 0.301 0.699 1.000 1.301
log y 1.398 0.602 0.000 1.309
Table R3.2.3
By means of a suitable linear graph, nd
the value of k and n. Hence write down the
relationship that connects x and y.
Revision exercise 3.3
1. In Fig. R3.3.1, OV = OP and PX = PQ.
Fig R3.3.1
(a) Given that OP = 12p and OQ = 4q,
express the following in terms of p and
q as simply as possible
(i) PQ (ii) PX (iii) OX (iv) VX
(b) Also, given that VS = aVX and
PS = hp + kq, express h and k in terms
of a.
2. The coordinates of points A and B are
(6 , 5) and (
–
4 ,
–
8) respectively. Find the
coordinates of a point P which divides AB
in the ratio 2 :
–
1.
3. ABCD is a quadrilateral. The midpoints
of AB, BC, CD and DA are P, Q, R and S
respectively. Using vector method, prove
that PQRS is a parallelogram.
202
4. Use binomial expansion to evaluate
(a) (1.01)
10
correct to 3 d.p.
(b) (2.001)
10
correct to 6 s.f.
(c) (0.997)
12
correct to 3 s.f.
(d) (1.998)
8
correct to 2 d.p.
5. Expand and simplify (1 + x)
4
– (1– x)
4
.
Hence use your result to evaluate
(
1 + √5
)
4
–
(
1 – √5
)
4
.
6. All possible two-digit numbers are formed
from the digits 1, 2, 3, 4, 5, 6, 7. If one of
these numbers is chosen at random, what is
the probability that it is divisible by
(a) 5 (b) 3 ?
7. A coin is biased so that the probability of
“Heads” is . Find the probability that
when the coin is tossed three times, it shows
(a) 3 tails (b) 2 heads and 1 tail
(c) no tails
8. Musau is learning how to play darts. The
probability that he hits the mark when he
throws a dart is . If he tries four times,
nd the probability that he
(a) hits the mark four times,
(b) does not hit the mark at all,
(c) hits the mark at least once.
9. (a) Six pumps raise 324 litres of water in
12 minutes. How many litres will eight
pumps raise in 7 minutes.
(b) Seven machines, wrapping sweets at
the rate of 80 a minute, turn out
1 120 kg of sweets in 2 hours 40
minutes. At what rate will 9 machines
turn out 1 080 kg in 2 hours?
10. A grocer mixes 30 sacks of rice costing
sh 4 000 per sack with 50 sacks of another
kind of rice costing sh 3 500 per sack. If
he sells the mixture at a prot of 20%, at
what price does he sell 1 kg, given that
each sack weighs 50 kg?
11. Amina can do a piece of work in 6 days
and Nina can do the same piece of work in
9 days. Amina starts alone and is joined by
Nina after 1 day. How long do they take to
complete the work together?
12. Table R3.3.1 gives values of two variables
x and y.
x 0.05 0.075 0.150 0.250 0.500 0.750
y 1.00 0.660 0.330 0.200 0.100 0.067
Table R3.3.1
(a) Draw a graph to show how varies
with x.
(b) Explain the kind of relationship existing
between x and y.
(c) Given that xy = c, determine the value
of c.
1
2
–
3
4
–
1
10
––
1
y
–
203
MATHEMATICAL FORMULAE AND TABLES
Formulae
1. Plane shapes
Shape Perimeter Area
Square
(side s) 4s s
2
Rectangle
(length l, breadth b) 2(l + b) lb
Triangle
Sum of sides
(base b, height h)
bh
Parallelogram
Sum of sides bh
(base b, height h)
Trapezium (height h,
Sum of sides
(parallels a and b)
(a + b)h
Circle
(radius r)
2πr πr
2
Sector of a circle
2r +
2πr
πr
2
(radius r, angle
θ

)
2. Solid shapes
Shape Area Volume
Cube (edge s) 6s
2
s
3
Cuboid (length l,
breadth b, height h)
2(lb + bh + lh) lbh
Prism (cross-section area Sum of area
ah
a, length h) of all faces
Cylinder (closed)
(radius r, height h)
2πrh + 2πr
2
πr
2
h
Cone (radius r,
slant height l, height h)
πrl + πr
2
πr
2
h
Sphere (radius r) 4πr
2
πr
3
3. Right-angled triangles
In a triangle ABC right-angled at C,
c
2
= a
2
+ b 
tan A = tan B =
sin A = sin B =
cos A = cos B =
θ
360
–––
θ
360
–––
1
2
–
1
2
–
1
2
–
1
2
–
–
b±√b
2
– 4ac
2a
4. Any triangle ABC
Sine rule:
a
=
b
=
c
sin A sin B sin C
Cosine rule: c
2
= a
2
+ b
2
– 2ab cos C
b
2
= a
2
+ c
2
– 2ac cos B
a
2
= b
2
+ c
2
– 2bc cos A
Area = √{s(s – a)(s – b)(s – c)}
where s = (a + b + c)
or Area = bc sin A etc
5. Obtuse angles
sin
θ
= sin (180° –
θ
)
cos
θ
=
–
cos (180° –
θ
)
tan
θ
=
–
tan (180° –
θ
)
6. Quadratic functions
If ax
2
+ bx + c = 0, a ≠ 0 then x =
7. (a) The nth term of an AP is given by
a + (n – 1)d where a is the rst term and d
the common dierence.
(b) The nth term of a GP is given by ar
n–1
where a is the rst term and r the common
ratio.
8. (a) Sum of an arithmetic progression
The sum of n terms is given by
S
n
= n{2a + (n – 1)d} where a is the rst term
and d is the common dierence
(b) Sum of a geometric progression
S
n
= if r < 1, where a is the rst term
and r is the common ratio
or if r > 1
a(1 – r
n
)
1 – r
a(r
n
– 1)
r – 1
a
b
a
c
b
c
a
c
b
c
b
a
c
B
a
A
b
C
1
3
–
4
3
–
1
2
–
B
CA
b
a
c
204
LOGARITHMS BASE 10, x  log
10
x
Add mean dierences
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0 .0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 8 12 17 21 25 29 33 37
1.1 .0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 9 11 15 19 23 27 30 34
1.2 .0792 0828 0864 0899 0934 0969 1004 1039 1072 1106 3 7 10 14 17 21 24 29 31
1.3 .1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 6 10 13 16 19 23 26 29
1.4 .1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 3 6 9 12 15 18 21 24 27
1.5 .1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 3 6 8 11 14 17 20 22 25
1.6 .2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 3 5 8 11 13 16 18 21 24
1.7 .2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 2 5 7 10 12 15 17 20 22
1.8 .2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 2 5 7 9 12 14 16 19 21
1.9 .2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 2 4 7 9 11 13 16 18 20
2.0 .3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2 4 6 8 11 13 15 17 19
2.1 .3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 2 4 6 9 10 12 14 16 18
2.2 .3424 3444 3464 3483 3502 3522 3541 3560 3579 3599 2 4 6 8 10 12 14 15 17
2.3 .3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 2 4 6 7 9 11 13 15 17
2.4 .3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 2 4 5 7 9 11 12 14 16
2.5 .3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2 3 5 7 9 10 12 14 15
2.6 .4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 2 3 5 7 8 10 11 13 15
2.7 .4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 2 3 5 6 8 9 11 13 14
2.8 .4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 2 3 5 6 8 9 11 12 14
2.9 .4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 1 3 4 6 7 9 10 12 13
3.0 .4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1 3 4 6 7 9 10 11 13
3.1 .4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1 3 4 6 7 8 10 11 12
3.2 .5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 1 3 4 5 7 8 9 11 12
3.3 .5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 1 3 4 5 6 8 9 10 12
3.4 .5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 1 3 4 5 6 8 9 10 11
3.5 .5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 1 2 4 5 6 7 9 10 11
3.6 .5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 1 2 4 5 6 7 9 10 11
3.7 .5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 1 2 3 5 6 7 8 9 10
3.8 .5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 1 2 3 5 6 7 9 9 10
3.9 .5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 1 2 3 4 6 7 9 9 10
4.0 .6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 1 2 3 4 5 6 8 9 10
4.1 .6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 1 2 3 4 5 6 7 8 9
4.2 .6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 1 2 3 4 5 6 7 8 9
4.3 .6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 1 2 3 4 5 6 7 8 9
4.4 .6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 1 2 3 4 5 6 7 8 9
4.5 .6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 1 2 3 4 5 6 7 8 9
4.6 .6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 1 2 3 4 5 6 7 7 9
4.7 .6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 1 2 3 4 5 5 6 7 8
4.8 .6812 6821 6830 6839 6948 6857 6866 6875 6884 6893 1 2 3 4 4 5 6 7 8
4.9 .6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 1 2 3 4 4 5 6 7 8
5.0 .6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1 2 3 3 4 5 6 7 8
5.1 .7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 1 2 3 3 4 5 6 7 8
5.2 .7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 1 2 2 3 4 5 6 7 7
5.3 .7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 1 2 2 3 4 5 6 7 7
5.4 .7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 1 2 2 3 4 5 6 6 7
0 1 2 3 4 5 6 7 8 9 1 1 3 4 5 6 7 8 9
205
LOGARITHMS BASE 10, x  log
10
x
Add mean dierences
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
5.5 .7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1 2 2 3 4 5 5 6 7
5.6 .7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 1 2 2 3 4 5 5 6 7
5.7 .7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 1 2 2 3 4 5 5 6 7
5.8 .7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 1 1 2 3 4 4 5 6 7
5.9 .7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 1 1 2 3 4 4 5 6 7
6.0 .7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 1 1 2 3 4 4 5 6 6
6.1 .7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 1 1 2 3 4 4 5 6 6
6.2 .7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 1 1 2 3 3 4 5 6 6
6.3 .7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 1 1 2 3 3 4 5 5 6
6.4 .8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 1 1 2 3 3 4 5 5 6
6.5 .8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 1 1 2 3 3 4 5 5 6
6.6 .8195 8202 8209 8215 8222 8229 8235 8241 8248 8254 1 1 2 3 3 4 5 5 6
6.7 .8261 8267 8274 8280 8287 8293 8299 8306 8312 9319 1 1 2 3 3 4 5 5 6
6.8 .8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 1 1 2 3 3 4 4 5 6
6.9 .8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 1 1 2 3 3 4 4 5 6
7.0 .8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 1 1 2 2 3 4 4 5 6
7.1 .8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 1 1 2 2 3 4 4 5 5
7.2 .8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 1 1 2 2 3 4 4 5 5
7.3 .8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 1 1 2 2 3 4 4 5 5
7.4 .8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 1 1 2 2 3 4 4 5 5
7.5 .8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 1 2 2 3 3 4 5 5
7.6 .8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 1 1 2 2 3 3 4 5 5
7.7 .8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 1 1 2 2 3 3 4 4 5
7.8 .8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 1 1 2 2 3 3 4 4 5
7.9 .8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 1 1 2 2 3 3 4 4 5
8.0 .9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 1 1 2 2 3 3 4 4 5
8.1 .9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 1 1 2 2 3 3 4 4 5
8.2 .9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 1 1 2 2 3 3 4 4 5
8.3 .9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 1 1 2 2 3 3 4 4 5
8.4 .9243 9248 9253 9258 9263 9269 9274 9279 9294 9289 1 1 2 2 3 3 4 4 5
8.5 .9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 1 2 2 3 3 4 4 5
8.6 .9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 1 1 2 2 3 3 4 4 5
8.7 .9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 0 1 1 2 2 3 3 4 4
8.8 .9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 0 1 1 2 2 3 3 4 4
8.9 .9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 0 1 1 2 2 3 3 4 4
9.0 .9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 0 1 1 2 2 3 3 4 4
9.1 .9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 0 1 1 2 2 3 3 4 4
9.2 .9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 0 1 1 2 2 3 3 4 4
9.3 .9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 0 1 1 2 2 3 3 4 4
9.4 .9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 0 1 1 2 2 3 3 4 4
9.5 .9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 0 1 1 2 2 3 3 4 4
9.6 .9823 9827 9832 9836 9941 9845 9850 9854 9859 9863 0 1 1 2 2 3 3 4 4
9.7 .9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 0 1 1 2 2 3 3 4 4
9.8 .9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 0 1 1 2 2 3 3 4 4
9.9 .9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 0 1 1 2 2 3 3 3 4
0 1 2 3 4 5 6 7 8 9 1 1 3 4 5 6 7 8 9
206
ANTILOGARITHMS x → 10
x
Add Mean Dierences
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
.00 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 0 0 1 1 1 1 2 2 2
.01 1023 1026 1028 1030 1033 1035 1038 1040 1042 1045 0 0 1 1 1 1 2 2 2
.02 1047 1050 1052 1054 1057 1059 1062 1064 1067 1069 0 0 1 1 1 1 2 2 2
.03 1072 1074 1076 1079 1081 1084 1086 1089 1091 1094 0 0 1 1 1 1 2 2 2
.04 1096 1099 1102 1104 1107 1109 1112 1114 1117 1119 0 1 1 1 1 2 2 2 2
.05 1122 1125 1127 1130 1132 1135 1138 1140 1143 1146 0 1 1 1 1 2 2 2 2
.06 1148 1151 1153 1156 1159 1161 1164 1167 1169 1172 0 1 1 1 1 2 2 2 2
.07 1175 1179 1180 1183 1186 1189 1191 1194 1197 1199 0 1 1 1 1 2 2 2 2
.08 1202 1205 1208 1211 1213 1216 1219 1222 1225 1227 0 1 1 1 1 2 2 2 3
.09 1230 1233 1236 1239 1242 1245 1247 1250 1253 1256 0 1 1 1 1 2 2 2 3
.10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 0 1 1 1 1 2 2 2 3
.11 1288 1291 1294 1297 1300 1303 1306 1309 1312 1315 0 1 1 1 2 2 2 2 3
.12 1318 1321 1324 1327 1330 1334 1337 1340 1343 1346 0 1 1 1 2 2 2 2 3
.13 1349 1352 1355 1358 1361 1365 1368 1371 1374 1377 0 1 1 1 2 2 2 2 3
.14 1380 1384 1387 1390 1393 1396 1400 1403 1406 1409 0 1 1 1 2 2 2 2 3
.13 1413 1416 1419 1422 1426 1429 1432 1435 1439 1442 0 1 1 1 2 2 2 3 3
.16 1445 1449 1452 1455 1459 1462 1466 1469 1472 1476 0 1 1 1 2 2 2 3 3
.17 1479 1493 1486 1489 1493 1496 1500 1503 1507 1510 0 1 1 2 2 2 3 3 4
.19 1514 1517 1521 1524 1528 1531 1535 1538 1542 1545 0 1 1 2 2 2 3 3 4
.19 1549 1552 1556 1560 1563 1567 1570 1574 1578 1581 0 1 1 2 2 2 3 3 4
.20 1585 1589 1592 1596 1600 1603 1607 1611 1614 1618 0 1 1 2 2 2 3 3 4
.21 1622 1626 1629 1633 1637 1641 1644 1648 1652 1656 0 1 1 2 2 2 3 3 4
.22 1660 1663 1667 1671 1675 1679 1683 1697 1690 1694 0 1 1 2 2 2 3 3 4
.23 1698 1702 1706 1710 1714 1718 1722 1726 1730 1734 0 1 1 2 2 2 3 3 4
.24 1738 1742 1746 1750 1754 1758 1762 1766 1770 1774 0 1 1 2 2 2 3 3 4
.25 1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 0 1 1 2 2 2 3 3 4
.26 1820 1824 1828 1832 1837 1841 1845 1849 1854 1858 0 1 1 2 2 2 3 3 4
.27 1862 1866 1871 1875 1879 1884 1888 1892 1897 1901 0 1 1 2 2 2 3 3 4
.28 1905 1910 1914 1919 1923 1928 1932 1936 1941 1945 0 1 1 2 2 2 3 3 4
.29 1950 1954 1959 1963 1968 1972 1977 1982 1986 1991 0 1 1 2 2 2 3 3 4
.30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 0 1 1 2 2 3 3 4 4
.31 2042 2046 2051 2056 2061 2065 2070 2075 2080 2084 0 1 1 2 2 3 3 4 4
.32 2089 2094 2099 2104 2109 2113 2118 2123 2128 2133 0 1 1 2 2 3 3 4 4
.33 2138 2143 2148 2153 2158 2163 2168 2173 2178 2183 0 1 1 2 2 3 3 4 4
.34 2198 2193 2198 2203 2208 2213 2218 2223 2228 2234 1 1 2 2 3 3 4 4 5
.35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2296 1 1 2 2 3 3 4 4 5
.36 2291 2296 2301 2307 2312 2317 2323 2328 2333 2339 1 1 2 2 3 3 4 4 5
.37 2344 2350 2355 2360 2366 2371 2377 2382 2388 2393 1 1 2 2 3 3 4 4 5
.38 2399 2404 2410 2415 2421 2427 2432 2438 2443 2449 1 1 2 2 3 3 4 4 5
.39 2455 2460 2466 2472 2477 2483 2489 2495 2500 2506 1 1 2 2 3 3 4 5 5
.40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 1 1 2 2 3 4 4 5 5
.41 2570 2576 2582 2588 2594 2600 2606 2612 2618 2624 1 1 2 2 3 4 4 5 5
.42 2630 2636 2642 2649 2655 2661 2667 2673 2679 2685 1 1 2 2 3 4 4 5 6
.43 2692 2698 2704 2710 2716 2723 2729 2735 2742 2748 1 1 2 3 3 4 4 5 6
.44 2754 2761 2767 2773 2780 2786 2793 2799 2805 2812 1 1 2 3 3 4 4 5 6
.45 2818 2825 2831 2838 2844 2851 2858 2864 2871 2877 1 1 2 3 3 4 5 5 6
.46 2884 2891 2897 2904 2911 2917 2924 2931 2938 2944 1 1 2 3 3 4 5 5 6
.47 2951 2958 2965 2972 2979 2985 2992 2999 3006 3013 1 1 2 3 3 4 5 5 6
.48 3020 3027 3034 3041 3049 3055 3062 3069 3076 3083 1 1 2 3 4 4 5 6 6
.49 3090 3097 3105 3112 3119 3126 3133 3141 3148 3155 1 1 2 3 4 4 5 6 6
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
e
x
= antilog (x log
10
e
) = antilog (0.43429 x)
207
ANTILOGARITHMS x → 10
x
Add Mean Dierences
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
.50 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 1 1 2 3 4 4 5 6 7
.51 3236 3243 3251 3258 3266 3273 3281 3289 3296 3304 1 2 2 3 4 5 5 6 7
.52 3311 3319 3327 3334 3342 3350 3357 3365 3373 3381 1 2 2 3 4 5 5 6 7
.53 3388 3396 3404 3412 3420 3428 3436 3443 3451 3459 1 2 2 3 4 5 6 6 7
.54 3467 3475 3483 3491 3499 3508 3516 3524 3532 3540 1 2 2 3 4 5 6 6 7
.55 3548 3556 3565 3573 3581 3589 3597 3606 3614 3622 1 2 2 3 4 5 6 7 7
.56 3631 3639 3648 3656 3664 3673 3681 3690 3698 3707 1 2 3 3 4 5 6 7 8
.57 3715 3724 3733 3741 3750 3758 3767 3776 3784 3793 1 2 3 4 4 5 6 7 8
.58 3802 3811 3819 3828 3837 3846 3855 3864 3873 3882 1 2 3 4 4 5 6 7 8
.59 3890 3899 3908 3917 3926 3936 3945 3954 3963 3972 1 2 3 4 5 5 6 7 8
.60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1 2 3 4 5 6 6 7 8
.61 4074 4083 4093 4102 4111 4121 4130 4140 4150 4159 1 2 3 4 5 6 7 8 9
.62 4169 4178 4188 4198 4207 4217 4227 4236 4246 4256 1 2 3 4 5 6 7 8 9
.63 4266 4276 4285 4295 4305 4315 4325 4335 4345 4355 1 2 3 4 5 6 7 9 9
.64 4365 4375 4385 4395 4406 4416 4426 4436 4446 4457 1 2 3 4 5 6 7 9 9
.65 4467 4477 4487 4498 4508 4519 4529 4539 4550 4560 1 2 3 4 5 6 7 8 9
.66 4571 4581 4592 4603 4613 4624 4634 4645 4656 4667 1 2 3 4 5 6 7 9 10
.67 4677 4688 4699 4710 4721 4732 4742 4753 4764 4775 1 2 3 4 5 7 8 9 10
.68 4786 4797 4808 4819 4831 4842 4853 4864 4875 4887 1 2 3 4 6 7 9 9 10
.69 4898 4909 4920 4932 4943 4955 4966 4977 4989 5000 1 2 3 5 6 7 9 9 10
.70 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 1 2 4 5 6 7 8 10 11
.71 5129 5140 5152 5164 5176 5188 5200 5212 5224 5236 1 2 4 5 6 7 8 10 11
.72 5248 5260 5272 5284 5297 5309 5321 5333 5346 5358 1 2 4 5 6 7 8 10 11
.73 5370 5383 5395 5408 5420 5433 5445 5458 5470 5483 1 2 4 5 6 7 8 10 11
.74 5495 5508 5521 5534 5546 5559 5572 5595 5598 5610 1 3 4 5 6 8 9 10 12
.75 5623 5636 5649 5662 5675 5689 5702 5715 5728 5741 1 3 4 5 7 8 9 10 12
.76 5754 5768 5791 5794 5908 5921 5834 5948 5961 5875 1 3 4 5 7 9 9 11 12
.77 5998 5902 5916 5929 5943 5957 5970 5994 5998 6012 1 3 4 5 7 9 10 11 12
.75 6026 6039 6053 6067 6091 6095 6109 6124 6138 6152 1 3 4 6 7 9 10 11 13
.79 6166 6190 6194 6209 6223 6237 6252 6266 6281 6295 1 3 4 6 7 9 10 11 13
.80 6310 6324 6339 6353 6369 6383 6387 6412 6427 6442 1 3 4 6 7 9 10 12 13
.81 6457 6471 6486 6501 6516 6531 6546 6561 6577 6592 2 3 5 6 8 9 11 12 14
.82 6607 6622 6637 6653 6668 6683 6699 6714 6730 6745 2 3 5 6 8 9 11 12 14
.83 6761 6776 6792 6808 6823 6839 6855 6871 6887 6902 2 3 5 6 8 10 11 13 14
.84 6918 6934 6950 6966 6982 6998 7015 7031 7047 7063 2 3 5 6 8 10 11 13 14
.85 7079 7096 7112 7129 7145 7161 7178 7194 7211 7229 2 3 5 6 8 10 11 13 14
.86 7244 7261 7278 7295 7311 7328 7345 7362 7379 7396 2 3 5 7 8 10 12 14 15
.87 7413 7430 7447 7464 7482 7499 7516 7534 7551 7568 2 3 5 7 9 10 12 14 16
.88 7586 7603 7621 7638 7656 7674 7691 7709 7727 7745 2 4 5 7 9 11 13 14 16
.89 7762 7780 7798 7816 7834 7852 7870 7889 7907 7925 2 4 5 7 9 11 13 14 16
.90 7943 7962 7980 7998 8017 8035 8054 8072 8091 8110 2 4 5 7 9 11 13 14 16
.91 8128 8147 8166 8185 8204 8222 8241 8260 8279 8299 2 4 6 8 10 11 13 15 17
.92 8318 9337 8356 8375 8395 8414 8433 8453 8472 8492 2 4 6 8 10 11 13 15 17
.93 8511 8531 8551 8570 8590 8610 8630 8650 8670 8690 2 4 6 8 10 12 14 16 18
.94 8710 8730 8750 8770 8790 8810 8831 8851 8872 8892 2 4 6 8 10 12 14 16 18
.95 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 2 4 6 8 10 13 15 17 19
.96 9120 9141 9162 9183 9204 9226 9247 9268 9290 9311 2 4 6 9 11 13 15 17 19
.97 9333 9354 9376 9397 9419 9441 9462 9484 9506 9528 2 4 7 9 11 13 15 18 20
.98 9550 9572 9594 9616 9638 9661 9683 9705 9727 9750 2 4 7 9 11 13 15 18 20
.99 9772 9795 9817 9840 9863 9886 9908 9931 9954 9977 2 5 7 9 11 14 16 18 21
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
e
x
= antilog (x log
10
e
) = antilog (0.43429 x)
208
SINES OF ANGLES
θ
°  sin
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Add mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
0 0.0000 0017 0035 0052 0070 0087 0105 0122 0140 0157 3 6 9 12 15
1 0.0175 0192 0209 0227 0244 0262 0279 0297 0314 0332 3 6 9 11 14
2 0.0349 0366 0384 0401 0419 0436 0454 0471 0488 0506 3 6 9 11 14
3 0.0523 0541 0558 0576 0593 0610 0628 0645 0663 0680 3 6 9 11 14
4 0.0698 0715 0732 0750 0767 0785 0802 0819 0837 0854 3 6 9 11 14
5 0.0872 0889 0906 0924 0941 0958 0976 0993 1011 1028 3 6 9 11 14
6 0.1045 1063 1080 1097 1115 1132 1149 1167 1184 1201 3 6 9 11 14
7 0.1219 1236 1253 1271 1288 1305 1323 1340 1357 1374 3 6 9 11 14
8 0.1392 1409 1426 1444 1461 1478 1495 1513 1530 1547 3 6 9 11 14
9 0.1564 1582 1599 1616 1633 1650 1668 1685 1702 1719 3 6 9 11 14
10 0.1736 1754 1771 1788 1805 1822 1840 1857 1874 1891 3 6 9 11 14
11 0.1908 1925 1942 1959 1977 1994 2011 2028 2045 2062 3 6 9 11 14
12 0.2079 2096 2113 2130 2147 2164 2181 2198 2215 2233 3 6 9 11 14
13 0.2250 2267 2284 2300 2317 2334 2351 2368 2385 2402 3 6 8 11 14
14 0.2419 2436 2453 2470 2487 2504 2521 2538 2554 2571 3 6 8 11 14
15 0.2588 2605 2622 2639 2656 2672 2689 2706 2723 2740 3 6 8 11 14
16 0.2756 2773 2790 2807 2823 2840 2857 2874 2890 2907 3 6 8 11 14
17 0.2924 2940 2957 2974 2990 3007 3024 3040 3057 3074 3 6 8 11 14
18 0.3090 3107 3123 3140 3156 3173 3190 3206 3223 3239 3 6 8 11 14
19 0.3256 3272 3289 3305 3322 3338 3355 3371 3387 3404 3 5 8 11 13
20 0.3420 3437 3453 3469 3486 3502 3518 3535 3551 3567 3 5 8 11 13
21 0.3584 3600 3616 3633 3649 3665 3681 3697 3714 3730 3 5 8 11 13
22 0.3746 3762 3778 3795 3811 3827 3843 3859 3875 3891 3 5 9 11 13
23 0.3907 3923 3939 3955 3971 3987 4003 4019 4035 4051 3 5 8 11 13
24 0.4067 4083 4099 4115 4131 4147 4163 4179 4195 4210 3 5 8 11 13
25 0.4226 4242 4258 4274 4289 4305 4321 4337 4352 4368 3 5 8 11 13
26 0.4384 4399 4415 4431 4446 4462 4478 4493 4509 4524 3 5 8 11 13
27 0.4540 4555 4571 4586 4602 4617 4633 4648 4664 4679 3 5 8 11 13
28 0.4695 4710 4726 4741 4756 4772 4787 4802 4818 4833 3 5 8 10 13
29 0.4848 4863 4879 4894 4909 4924 4939 4955 4970 4985 3 5 8 10 13
30 0.5000 5015 5030 5045 5060 5075 5090 5105 5120 5135 3 5 8 10 13
31 0.5150 5165 5180 5195 5210 5225 5240 5255 5270 5284 2 5 7 10 12
32 0.5299 5314 5329 5344 5358 5373 5388 5402 5417 5432 2 5 7 10 12
33 0.5446 5461 5476 5490 5505 5519 5534 5548 5563 5577 2 5 7 10 12
34 0.5592 5606 5621 5635 5650 5664 5678 5693 5707 5721 2 5 7 9 12
35 0.5736 5750 5764 5779 5793 5807 5821 5835 5850 5864 2 5 7 9 12
36 0.5878 5892 5906 5920 5934 5948 5962 5976 5990 6004 2 5 7 9 12
37 0.6018 6032 6046 6060 6074 6088 6101 6115 6129 6143 2 5 7 9 12
38 0.6157 6170 6184 6198 6211 6225 6239 6252 6266 6280 2 5 7 9 12
39 0.6293 6307 6320 6334 6347 6361 6374 6388 6401 6414 2 5 7 9 12
40 0.6428 6441 6455 6468 6481 6494 6508 6521 6534 6547 2 4 7 9 11
41 0.6561 6574 6587 6600 6613 6626 6639 6652 6665 6678 2 4 7 9 11
42 0.6691 6704 6717 6730 6743 6756 6769 6782 6794 6807 2 4 6 9 11
43 0.6820 6833 6845 6858 6871 6884 6896 6909 6921 6934 2 4 6 9 11
44 0.6947 6959 6972 6984 6997 7009 7022 7034 7046 7059 2 4 6 8 10
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
209
SINES OF ANGLES
θ
°  sin
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Add mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
45 0.7071 7083 7096 7108 7120 7133 7145 7157 7169 7181 2 4 6 8 10
46 0.7193 7206 7218 7230 7242 7254 7266 7278 7290 7302 2 4 6 8 10
47 0.7314 7325 7337 7349 7361 7373 7385 7396 7408 7420 2 4 6 8 10
48 0.7431 7443 7455 7466 7478 7490 7501 7513 7524 7536 2 4 6 8 10
49 0.7547 7559 7570 7581 7593 7604 7615 7627 7638 7649 2 4 6 7 9
50 0.7660 7672 7683 7694 7705 7716 7727 7738 7749 7760 2 4 6 7 9
51 0.7771 7782 7793 7804 7815 7826 7837 7848 7859 7869 2 4 5 7 9
52 0.7880 7891 7902 7912 7923 7934 7944 7955 7965 7976 2 4 5 7 9
53 0.7986 7997 8007 8018 8028 8039 8049 8059 8070 8080 2 3 5 7 8
54 0.8090 8100 8111 8121 8131 8141 8151 8161 8171 8181 2 3 5 7 8
55 0.8192 8202 8211 8221 8231 8241 8251 8261 8271 8281 2 3 5 7 8
56 0.8290 8300 8310 8320 8329 8339 8348 8358 9368 8377 2 3 5 7 8
57 0.8387 8396 9406 8415 9425 8434 8443 8453 9462 8471 2 3 5 6 8
58 0.8480 9490 8499 8508 8517 8526 8536 8545 8554 8563 2 3 5 6 8
59 0.8572 8581 8590 8599 8607 8616 8625 8634 8643 8652 1 3 4 6 7
60 0.8660 8669 9678 8686 8695 8704 8712 8721 8729 8738 1 3 4 6 7
61 0.8746 8755 8763 8771 8780 8788 8796 8805 8813 8821 1 3 4 5 7
62 0.8929 8838 8846 8854 8862 8870 8878 8886 8894 8902 1 3 4 5 7
63 0.8910 8918 8926 8934 8942 8949 8957 8965 8973 8980 1 3 4 5 7
64 0.8988 8996 9003 9011 9018 9026 9033 9041 9048 9056 1 3 4 5 7
65 0.9063 9070 9078 9085 9092 9100 9107 9114 9121 9128 1 2 4 5 6
66 0.9135 9143 9150 9157 9164 9171 9178 9184 9191 9198 1 2 4 5 6
67 0.9205 9212 9219 9225 9232 9239 9245 9252 9259 9265 1 2 3 5 6
68 0.9272 9278 9285 9291 9298 9304 9311 9317 9323 9330 1 2 3 4 5
69 0.9336 9342 9348 9354 9361 9367 9373 9379 9385 9391 1 2 3 4 5
70 0.9397 9403 9409 9415 9421 9426 9432 9438 9444 9449 1 2 3 4 5
71 0.9455 9461 9466 9472 9478 9483 9489 9494 9500 9505 1 2 3 4 5
72 0.9511 9516 9521 9527 9532 9537 9542 9548 9553 9558 1 2 3 3 4
73 0.9563 9568 9573 9578 9583 9588 9593 9598 9603 9608 1 2 3 3 4
74 0.9613 9617 9622 9627 9632 9636 9641 9646 9650 9655 1 2 2 3 4
75 0.9659 9664 9668 9673 9677 9681 9686 9690 9694 9699 1 1 2 3 3
76 0.9703 9707 9711 9715 9720 9724 9728 9732 9736 9740 1 1 2 3 3
77 0.9744 9748 9751 9755 9759 9763 9767 9770 9774 9778 1 1 2 3 3
78 0.9781 9785 9789 9792 9796 9799 9803 9806 9810 9813 1 1 2 3 3
79 0.9816 9820 9823 9826 9829 9833 9836 9839 9942 9845 1 1 2 2 3
80 0.9848 9851 9854 9857 9860 9963 9866 9869 9871 9874 0 1 1 2 2
81 0.9877 9880 9882 9885 9888 9890 9893 9895 9898 9900 0 1 1 2 2
82 0.9903 9905 9907 9910 9912 9914 9917 9919 9921 9923 0 1 1 1 2
83 0.9925 9928 9930 9932 9934 9936 9938 9940 9942 9943 0 1 1 1 2
84 0.9945 9947 9949 9951 9952 9954 9956 9957 9959 9960 0 1 1 1 2
85 0.9962 9963 9965 9966 9968 9969 9971 9972 9973 9974
86 0.9976 9977 9978 9979 9980 9981 9982 9983 9984 9985
87 0.9986 9987 9988 9989 9990 9990 9991 9992 9993 9993
88 0.9994 9995 9995 9996 9996 9997 9997 9997 9998 9998
89 0.9998 9999 9999 9999 9999 1.000 1.000 1.000 1.000 1.000
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
210
COSINES OF ANGLES
θ
°  cos
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Subtract mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
0° 1.0000 1.000 1.000 1.000 1.000 1.000 9999 9999 9999 9999
1 .9998 9998 9998 9997 9997 9997 9996 9996 9995 9995
2 .9994 9993 9993 9992 9991 9990 9990 9989 9988 9987 0 0 0 0 1
3 .9986 9985 9984 9983 9982 9981 9980 9979 9978 9977 0 0 1 1 1
4 .9976 9974 9973 9972 9971 9969 9968 9966 9965 9963 0 0 1 1 1
5 .9962 9960 9959 9957 9956 9954 9952 9951 9949 9947 0 1 1 1 1
6 .9945 9943 9942 9940 9938 9936 9934 9932 9930 9928 0 1 1 1 2
7 .9925 9923 9921 9919 9917 9914 9912 9910 9907 9905 0 1 1 2 2
8 .9903 9900 9898 9895 9893 9890 9888 9885 9882 9880 0 1 1 2 2
9 .9977 9874 9871 9869 9866 9863 9860 9857 9854 9851 0 1 1 2 2
10 .9848 9845 9842 9839 9836 9833 9829 9826 9823 9820 1 1 2 2 3
11 .9816 9813 9810 9806 9803 9799 9796 9792 9789 9785 1 1 2 2 3
12 .9781 9778 9774 9770 9767 9763 9759 9755 9751 9748 1 1 2 3 3
13 .9744 9740 9736 9732 9728 9724 9720 9715 9711 9707 1 1 2 3 3
14 .9703 9699 9694 9690 9686 9681 9677 9673 9668 9664 1 1 2 3 4
15 .9659 9655 9650 9646 9641 9636 9632 9627 9622 9617 1 2 2 3 4
16 .9613 9608 9603 9598 9593 9588 9583 9578 9573 9568 1 2 2 3 4
17 .9563 9558 9553 9548 9542 9537 9532 9527 9521 9516 1 2 3 4 4
18 .9511 9505 9500 9494 9489 9483 9478 9472 9466 9461 1 2 3 4 5
19 .9455 9449 9444 9438 9432 9426 9421 9415 9409 9403 1 2 3 4 5
20 .9397 9391 9385 9379 9373 9367 9361 9354 9348 9342 1 2 3 4 5
21 .9336 9330 9323 9317 9311 9304 9298 9291 9285 9278 1 2 3 4 5
22 .9272 9265 9259 9252 9245 9239 9232 9225 9219 9212 1 2 3 4 6
23 .9205 9198 9191 9184 9178 9171 9164 9157 9150 9143 1 2 3 5 6
24 .9135 9128 9121 9114 9107 9100 9092 9085 9078 9070 1 2 4 5 6
25 .9063 9056 9049 9041 9033 9026 9018 9011 9003 8996 1 3 4 5 6
26 .8988 8980 8973 8965 8957 8949 8942 8934 8926 8918 1 3 4 5 6
27 .8910 8902 8894 8886 8878 8870 8862 8854 8846 8838 1 3 4 5 7
28 .8829 8821 8813 8805 8796 8788 8780 8771 8763 8755 1 3 4 6 7
29 .8746 8738 8729 8721 8712 8704 8695 8686 8678 8669 1 3 4 6 7
30 .8660 8652 8643 8634 8625 8616 8607 8599 8590 8581 1 3 4 6 7
31 .8572 8563 8554 8545 8536 8526 8517 8508 8499 8490 2 3 5 6 8
32 .8480 8471 8462 8453 8443 8434 8425 8415 8406 8396 2 3 5 6 8
33 .8387 8377 8368 8358 8348 8339 8329 8320 8310 8300 2 3 5 6 8
34 .8290 8281 8271 8261 8251 8241 8231 8221 8211 8202 2 3 5 7 8
35 .8192 8181 8171 8161 8151 8141 8131 8121 8111 8100 2 3 5 7 8
36 .8090 8080 8070 8059 8049 8039 8028 8018 8007 7997 2 3 5 7 9
37 .7986 7976 7965 7955 7944 7934 7923 7912 7902 7891 2 4 5 7 9
38 .7880 7869 7859 7848 7837 7826 7815 7804 7793 7782 2 4 5 7 9
39 .7771 7760 7749 7738 7727 7716 7705 7694 7683 7672 2 4 6 7 9
40 .7660 7649 7638 7627 7615 7604 7593 7581 7570 7559 2 4 6 8 9
41 .7547 7536 7524 7513 7501 7490 7478 7466 7455 7443 2 4 6 8 10
42 .7431 7420 7408 7396 7385 7373 7361 7349 7337 7325 2 4 6 8 10
43 .7314 7302 7290 7278 7266 7254 7242 7230 7218 7206 2 4 6 8 10
44 .7193 7181 7169 7157 7145 7133 7120 7108 7096 7083 2 4 6 8 10
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
211
COSINES OF ANGLES
θ
°  cos
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Subtract mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
45° .7071 7059 7046 7034 7022 7009 6997 6984 6972 6959 2 4 6 8 10
46 .6947 6934 6921 6909 6896 6884 6871 6858 6845 6833 2 4 6 8 11
47 .6820 6907 6794 6782 6769 6756 6743 6730 6717 6704 2 4 6 9 11
48 .6691 6678 6665 6652 6639 6626 6613 6600 6587 6574 2 4 7 9 11
49 .6561 6547 6534 6521 6508 6494 6481 6468 6455 6441 2 4 7 9 11
50 .6428 6414 6401 6388 6374 6361 6347 6334 6320 6307 2 4 7 9 11
51 .6293 6280 6266 6252 6239 6225 6211 6198 6194 6170 2 5 7 9 11
52 .6157 6143 6129 6115 6101 6088 6074 6060 6046 6032 2 5 7 9 12
53 .6018 6004 5990 5976 5962 5948 5934 5920 5906 5892 2 5 7 9 12
54 .5878 5864 5850 5835 5821 5807 5793 5779 5764 5750 2 5 7 9 12
55 .5736 5721 5707 5693 5678 5664 5650 5635 5621 5606 2 5 7 10 12
56 .5592 5577 5563 5548 5534 5519 5505 5490 5476 5461 2 5 7 10 12
57 .5446 5432 5417 5402 5388 5373 5358 5344 5329 5314 2 5 7 10 12
58 .5299 5284 5270 5255 5240 5225 5210 5195 5180 5165 2 5 7 10 12
59 .5150 5135 5120 5105 5090 5075 5060 5045 5030 5015 3 5 8 10 13
60 .5000 4985 4970 4955 4939 4924 4909 4894 4879 4863 3 5 8 10 13
61 .4848 4833 4818 4802 4787 4772 4756 4741 4726 4710 3 5 8 10 13
62 .4695 4679 4664 4648 4633 4617 4602 4586 4571 4555 3 5 8 10 13
63 .4540 4524 4509 4493 4478 4462 4446 4431 4415 4399 3 5 8 10 13
64 .4384 4368 4352 4337 4321 4305 4289 4274 4258 4242 3 5 8 11 13
65 .4226 4210 4195 4179 4163 4147 4131 4115 4099 4083 3 5 8 11 13
66 .4067 4051 4035 4019 4003 3987 3971 3955 3939 3923 3 5 8 11 13
67 .3907 3891 3875 3859 3843 3827 3811 3795 3778 3762 3 5 8 11 13
68 .3746 3730 3714 3697 3681 3665 3649 3633 3616 3600 3 5 8 11 14
69 .3584 3567 3551 3535 3518 3502 3486 3469 3453 3437 3 5 8 11 14
70 .3420 3404 3387 3371 3355 3338 3322 3305 3289 3272 3 5 8 11 14
71 .3256 3239 3223 3206 3190 3173 3156 3140 3123 3107 3 6 8 11 14
72 .3090 3074 3057 3040 3024 3007 2990 2974 2957 2940 3 6 8 11 14
73 .2924 2907 2890 2874 2857 2840 2823 2807 2790 2773 3 6 8 11 14
74 .2756 2740 2723 2706 2689 2672 2656 2639 2622 2605 3 6 8 11 14
75 .2588 2571 2554 2538 2521 2504 2487 2470 2453 2436 3 6 8 11 14
76 .2419 2402 2385 2368 2351 2334 2317 2300 2284 2267 3 6 8 11 14
77 .2250 2233 2215 2198 2181 2164 2147 2130 2113 2096 3 6 9 11 14
78 .2079 2062 2045 2028 2011 1994 1977 1959 1942 1925 3 6 9 11 14
79 .1908 1891 1874 1857 1840 1822 1805 1788 1771 1754 3 6 9 11 14
80 .1736 1719 1702 1685 1668 1650 1633 1616 1599 1582 3 6 9 11 14
81 .1564 1547 1530 1513 1495 1478 1461 1444 1426 1409 3 6 9 12 14
82 .1392 1374 1357 1340 1323 1305 1288 1271 1253 1236 3 6 9 12 14
83 .1219 1201 1184 1167 1149 1132 1115 1097 1080 1063 3 6 9 12 14
84 .1045 1028 1011 0993 0976 0958 0941 0924 0906 0889 3 6 9 12 14
85 .0872 0854 0837 0819 0802 0785 0767 0750 0732 0715 3 6 9 12 14
86 .0698 0680 0663 0645 0628 0610 0593 0576 0558 0541 3 6 9 12 15
87 .0523 0506 0488 0471 0454 0436 0419 0401 0384 0366 3 6 9 12 15
88 .0349 0332 0314 0297 0279 0262 0244 0227 0209 0192 3 6 9 12 15
89 .0175 0157 0140 0122 0105 0087 0070 0052 0035 0017 3 6 9 12 15
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
212
TANGENTS OF ANGLES
θ
°  tan
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Add mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
0° 0.0000 0017 0035 0052 0070 0087 0105 0122 0140 0157 3 6 9 12 15
1 0.0175 0192 0209 0227 0244 0262 0279 0297 0314 0332 3 6 9 12 15
2 0.0349 0367 0384 0402 0419 0437 0454 0472 0489 0507 3 6 9 12 15
3 0.0524 0542 0559 0577 0594 0612 0629 0647 0664 0692 3 6 9 12 15
4 0.0699 0717 0734 0752 0769 0787 0805 0822 0840 0857 3 6 9 12 15
5 0.0875 0892 0910 0928 0945 0963 0981 0998 1016 1033 3 6 9 12 15
6 0.1051 1069 1086 1104 1122 1139 1157 1175 1192 1210 3 6 9 12 15
7 0.1228 1246 1263 1281 1299 1317 1334 1352 1370 1388 3 6 9 12 15
8 0.1405 1423 1441 1459 1477 1495 1512 1530 1548 1566 3 6 9 12 15
9 0.1584 1602 1620 1638 1655 1673 1691 1709 1727 1745 3 6 9 12 15
10 0.1763 1781 1799 1817 1835 1853 1871 1890 1908 1926 3 6 9 12 15
11 0.1944 1962 1980 1998 2016 2035 2053 2071 2089 2107 3 6 9 12 15
12 0.2126 2144 2162 2180 2199 2217 2235 2254 2272 2290 3 6 9 12 15
13 0.2309 2327 2345 2364 2382 2401 2419 2438 2456 2475 3 6 9 12 15
14 0.2493 2512 2530 2549 2568 2586 2605 2623 2642 2661 3 6 9 12 16
15 0.2679 2698 2717 2736 2754 2773 2792 2811 2830 2849 3 6 9 13 16
16 0.2867 2886 2905 2924 2943 2962 2981 3000 3019 3038 3 6 9 13 16
17 0.3057 3076 3096 3115 3134 3153 3172 3191 3211 3230 3 6 10 13 16
18 0.3249 3269 3288 3307 3327 3346 3365 3385 3404 3424 3 6 10 13 16
19 0.3443 3463 3482 3502 3522 3541 3561 3581 3600 3620 3 7 10 13 16
20 0.3640 3659 3679 3699 3719 3739 3759 3779 3799 3819 3 7 10 13 17
21 0.3839 3859 3879 3899 3919 3939 3959 3979 4000 4020 3 7 10 13 17
22 0.4040 4061 4081 4101 4122 4142 4163 4183 4204 4224 3 7 10 14 17
23 0.4245 4265 4286 4307 4327 4348 4369 4390 4411 4431 3 7 10 14 17
24 0.4452 4473 4494 4515 4536 4557 4578 4599 4621 4642 4 7 11 14 18
25 0.4663 4684 4706 4727 4748 4770 4791 4813 4834 4856 4 7 11 14 18
26 0.4877 4899 4921 4942 4964 4986 5008 5029 5051 5073 4 7 11 15 19
27 0.5095 5117 5139 5161 5184 5206 5228 5250 5272 5295 4 7 11 15 18
28 0.5317 5340 5362 5384 5407 5430 5452 5475 5498 5520 4 8 11 15 19
29 0.5543 5566 5589 5612 5635 5658 5681 5704 5727 5750 4 8 12 15 19
30 0.5774 5797 5820 5844 5867 5890 5914 5938 5961 5985 4 8 12 16 20
31 0.6009 6032 6056 6080 6104 6128 6152 6176 6200 6224 4 8 12 16 20
32 0.6249 6273 6297 6322 6346 6371 6395 6420 6445 6469 4 8 12 16 20
33 0.6494 6519 6544 6569 6594 6619 6644 6669 6694 6720 4 8 13 17 21
34 0.6745 6771 6796 6822 6847 6873 6899 6924 6950 6976 4 9 13 17 21
35 0.7002 7028 7054 7080 7107 7133 7159 7186 7212 7239 4 9 13 18 22
36 0.7265 7292 7319 7346 7373 7400 7427 7454 7481 7508 5 9 14 18 23
37 0.7536 7563 7590 7618 7646 7673 7701 7729 7757 7785 5 9 14 18 23
38 0.7813 7841 7869 7898 7926 7954 7983 8012 8040 8069 5 9 14 19 24
39 0.8098 8127 8156 8185 8214 8243 8273 8302 8332 8361 5 10 15 20 24
40 0.8391 8421 9451 8481 8511 8541 8571 8601 8632 8662 5 10 15 20 25
41 0.8693 8724 8754 8785 8816 8947 8878 8910 8941 8972 5 10 16 21 26
42 0.9004 9036 9067 9099 9131 9163 9195 9228 9260 9293 5 11 16 21 27
43 0.9325 9358 9391 9424 9457 9490 9523 9556 9590 9623 6 11 17 22 28
44 0.9657 9691 9725 9759 9793 9827 9861 9896 9930 9965 6 11 17 23 29
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
213
TANGENTS OF ANGLES
θ
°  tan
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Add mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
45 1.0000 0035 0070 0105 0141 0176 0212 0247 0283 0319 6 12 18 24 30
46 1.0355 0392 0428 0464 0501 0538 0575 0612 0649 0686 6 12 18 25 31
47 1.0724 0761 0799 0837 0975 0913 0951 0990 1028 1067 6 13 19 25 32
48 1.1106 1145 1184 1224 1263 1303 1343 1383 1423 1463 7 13 20 26 33
49 1.1504 1544 1585 1626 1667 1708 1750 1792 1833 1875 7 14 21 28 34
50 1.1918 1960 2002 2045 2088 2131 2174 2218 2261 2305 7 14 22 29 36
51 1.2349 2393 2437 2482 2527 2572 2617 2662 2708 2753 8 15 23 30 38
52 1.2799 2846 2892 2938 2985 3032 3079 3127 3175 3222 8 16 24 31 39
53 1.3270 3319 3367 3416 3465 3514 3564 3613 3663 3713 8 16 25 33 41
54 1.3764 3814 3865 3916 3968 4019 4071 4124 4176 4229 9 17 26 34 43
55 1.4281 4335 4388 4442 4496 4550 4605 4659 4715 4770 9 18 27 36 45
56 1.4826 4882 4938 4994 5051 5108 5166 5224 5282 5340 10 19 29 38 48
57 1.5399 5458 5517 5577 5637 5697 5757 5818 5880 5941 10 20 30 40 50
58 1.6003 6066 6128 6191 6255 6319 6383 6447 6512 6577 11 21 32 43 53
59 1.6643 6709 6775 6842 6909 6977 7045 7113 7182 7251 11 23 34 45 56
60 1.7321 7391 7461 7532 7603 7675 7747 7820 7893 7966 12 24 36 48 60
61 1.8040 9115 8190 8265 8341 8418 8495 8572 8650 8728 13 26 38 51 64
62 1.8807 8887 8967 9047 9128 9210 9292 9375 9458 9542 14 27 41 55 68
63 1.9626 9711 9797 9883 9970 0057 0145 0233 0323 0413 15 29 44 58 73
64 2.0503 0594 0686 0778 0872 0965 1060 1155 1251 1348 16 31 47 63 78
65 2.1445 1543 1642 1742 1842 1943 2045 2148 2251 2355 17 34 51 68 85
66 2.2460 2566 2673 2781 2899 2998 3109 3220 3332 3445 18 37 55 73 91
67 2.3559 3673 3789 3906 4023 4142 4262 4383 4504 4627 20 40 60 79 99
68 2.4751 4876 5002 5129 5257 5386 5517 5649 5782 5916 22 43 65 87 108
69 2.6051 6187 6325 6464 6605 6746 6889 7034 7179 7326 24 47 71 95 119
70 2.7475 7625 7776 7929 8083 8239 8397 8556 8716 8879 26 52 78 104 130
71 2.9042 9208 9375 9544 9714 9887 0061 0237 0415 0595 29 58 97 116 144
72 3.0777 0961 1146 1334 1524 1756 1910 2106 2305 2506 32 64 97 129 161
73 3.2709 2914 3122 3332 3544 3759 3977 4197 4420 4646 36 72 108 144 180
74 3.4874 5105 5339 5576 5816 6059 6305 6554 6806 7062 41 81 122 163 203
75 3.7321 7583 7848 8118 8391 8667 8947 9232 9520 9812 46 93 139 186 232
76 4.0108 0408 0713 1022 1335 1653 1976 2303 2635 2972 53 107 160 214 267
77 4.3315 3662 4015 4373 4737 5107 5483 5864 6252 6646 62 124 186 248 310
78 4.7046 7453 7867 8288 8716 9152 9594 0045 0504 0970 73 146 220 293 366
79 5.1446 1929 2422 2924 3435 3955 4486 5026 5578 6140 87 175 263 350 438
80 5.671 5.730 5.789 5.850 5.912 5.976 6.041 6.107 6.174 6.243
81 6.314 6.386 6.460 6.535 6.612 6.691 6.772 6.855 6.940 7.026
82 7.115 7.207 7.300 7.396 7.495 7.596 7.700 7.806 7.916 8.028
83 8.144 8.264 8.386 8.513 8.643 8.777 8.915 9.058 9.205 9.357
84 9.51 9.68 9.84 10.02 10.20 10.39 10.58 10.78 10.99 11.20
85 11.43 11.66 11.91 12.16 12.43 12.71 13.00 13.30 13.62 13.95
86 14.30 14.67 15.06 15.46 15.89 16.35 16.83 17.34 17.89 18.46
87 19.08 19.74 20.45 21.20 27.02 22.90 23.86 24.90 26.03 27.27
88 28.64 30.14 31.82 33.69 35.80 38.19 40.92 44.07 47.74 52.08
89 57.29 63.66 71.62 81.85 95.49 114.6 143.2 191.0 286.5 573.0
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
214
LOGARITHMS OF SINES
θ
°  log
10
sin
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Add mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
0° – ∞ 3.242 3.543 3.719 3.844 3.941 2.020 2.087 2.145 2.196
1 2.2419 2832 3210 3558 3880 4179 4459 4723 4971 5206
2 2.5428 5640 5842 6035 6220 6397 6567 6731 6889 7041
3 2.7188 7330 7468 7602 7731 7857 7979 8098 8213 8326 21 41 62 83 103
4 2.8436 8543 8647 8749 8849 8946 9042 9135 9226 9315 16 32 48 64 81
5 2.9403 9489 9573 9655 9736 9816 9894 9970 0046 0120 13 26 39 53 66
6 1.0192 0264 0334 0403 0472 0539 0605 0670 0734 0797 11 22 33 44 55
7 1.0859 0920 0981 1040 1099 1157 1214 1271 1326 1391 10 19 29 38 48
8 1.1436 1489 1542 1594 1646 1697 1747 1797 1847 1895 8 17 25 34 42
9 1.1943 1991 2038 2085 2131 2176 2221 2266 2310 2353 8 15 23 30 38
10 1.2397 2439 2482 2524 2565 2606 2647 2687 2727 2767 7 14 20 27 34
11 1.2806 2845 2883 2921 2959 2997 3034 3070 3107 3143 6 12 19 25 31
12 1.3179 3214 3250 3284 3319 3353 3387 3421 3455 3488 6 11 17 23 28
13 1.3521 3554 3586 3618 3650 3682 3713 3745 3775 3806 5 11 16 21 26
14 1.3837 3867 3897 3927 3957 3986 4015 4044 4073 4102 5 10 15 20 24
15 1.4130 4158 4186 4214 4242 4269 4296 4323 4350 4377 5 9 14 18 23
16 1.4403 4430 4456 4482 4508 4533 4559 4584 4609 4634 4 9 13 17 21
17 1.4659 4684 4709 4733 4757 4781 4805 4829 4853 4876 4 8 12 16 20
18 1.4900 4923 4946 4969 4992 5015 5037 5060 5082 5104 4 8 11 15 19
19 1.5126 5148 5170 5192 5213 5235 5256 5278 5299 5320 4 7 11 14 18
20 1.5341 5361 5382 5402 5423 5443 5463 5484 5504 5523 3 7 10 14 17
21 1.5543 5563 5583 5602 5621 5641 5660 5679 5698 5717 3 6 10 13 16
22 1.5736 5754 5773 5792 5810 5828 5847 5865 5883 5901 3 6 9 12 15
23 1.5919 5937 5954 5972 5990 6007 6024 6042 6059 6076 3 6 9 12 15
24 1.6093 6110 6127 6144 6161 6177 6194 6210 6227 6243 3 6 8 11 14
25 1.6259 6276 6292 6308 6324 6340 6356 6371 6387 6403 3 5 8 11 13
26 1.6418 6434 6449 6465 6480 6495 6510 6526 6541 6556 3 5 8 10 13
27 1.6570 6585 6600 6615 6629 6644 6659 6673 6687 6702 2 5 7 10 12
28 1.6716 6730 6744 6759 6773 6787 6801 6814 6828 6842 2 5 7 9 12
29 1.6856 6869 6883 6896 6910 6923 6937 6950 6963 6977 2 4 7 9 11
30 1.6990 7003 7016 7029 7042 7055 7068 7080 7093 7106 2 4 6 9 11
31 1.7118 7131 7144 7156 7168 7181 7193 7205 7218 7230 2 4 6 8 10
32 1.7242 7254 7266 7278 7290 7302 7314 7326 7338 7349 2 4 6 8 10
33 1.7361 7373 7384 7396 7407 7419 7430 7442 7453 7464 2 4 6 8 10
34 1.7476 7487 7498 7509 7520 7531 7542 7553 7564 7575 2 4 6 7 9
35 1.7586 7597 7607 7618 7629 7640 7650 7661 7671 7682 2 4 5 7 9
36 1.7692 7703 7713 7723 7734 7744 7754 7764 7774 7785 2 3 5 7 9
37 1.7795 7805 7815 7825 7835 7844 7854 7864 7874 7884 2 3 5 7 8
38 1.7893 7903 7913 7922 7932 7941 7951 7960 7970 7979 2 3 5 6 8
39 1.7989 7998 8007 8017 8026 8035 8044 8053 8063 8072 2 3 5 6 8
40 1.8081 8090 8099 8108 8117 8125 8134 8143 8152 8161 1 3 4 6 7
41 1.8169 8178 8187 8195 8204 8213 8221 8230 8238 8247 1 3 4 6 7
42 1.8255 8264 8272 8280 8289 8297 8305 8313 8322 8330 1 3 4 6 7
43 1.8338 8346 8354 8362 8370 8378 8386 8394 8402 8410 1 3 4 5 7
44 1.8418 8426 8433 8441 8449 8457 8464 8472 8480 8487 1 3 4 5 6
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
– – – – – – – – –
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
215
LOGARITHMS OF SINES
θ
° log
10
sin
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Add mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
45° 1.8495 8502 8510 8517 8525 8532 8540 8547 8555 8562 1 2 4 5 6
46 1.8569 8577 8584 8591 8598 8606 8613 8620 8627 8634 1 2 4 5 6
47 1.8641 8648 8655 8662 8669 8676 8683 8690 8697 8704 1 2 3 5 6
48 1.8711 8718 8724 8731 8738 8745 8751 8758 8765 8771 1 2 3 4 6
49 1.8778 8784 8791 8797 8804 8810 8817 8823 8830 8836 1 2 3 4 5
50 1.8843 8849 8855 8862 8868 8874 8880 8887 8893 8899 1 2 3 4 5
51 1.8905 8911 8917 8923 8929 8935 8941 8947 8953 8959 1 2 3 4 5
52 1.8965 8971 8977 8983 8989 8995 9000 9006 9012 9018 1 2 3 4 5
53 1.9023 9029 9035 9041 9046 9052 9057 9063 9069 9074 1 2 3 4 5
54 1.9080 9085 9091 9096 9101 9107 9112 9118 9123 9128 1 2 3 4 5
55 1.9134 9139 9144 9149 9155 9160 9165 9170 9175 9181 1 2 3 3 4
56 1.9186 9191 9196 9201 9206 9211 9216 9221 9226 9231 1 2 3 3 4
57 1.9236 9241 9246 9251 9255 9260 9265 9270 9275 9279 1 2 2 3 4
58 1.9284 9289 9294 9298 9303 9308 9312 9317 9322 9326 1 2 2 3 4
59 1.9331 9335 9340 9344 9349 9353 9358 9362 9367 9371 1 1 2 3 4
60 1.9375 9380 9384 9388 9393 9397 9401 9406 9410 9414 1 1 2 3 4
61 1.9418 9422 9427 9431 9435 9439 9443 9447 9451 9455 1 1 2 3 3
62 1.9459 9463 9467 9471 9475 9479 9483 9487 9491 9495 1 1 2 3 3
63 1.9499 9503 9506 9510 9514 9518 9522 9525 9529 9533 1 1 2 3 3
64 1.9537 9540 9544 9548 9551 9555 9558 9562 9566 9569 1 1 2 2 3
65 1.9573 9576 9580 9583 9587 9590 9594 9597 9601 9604 1 1 2 2 3
66 1.9607 9611 9614 9617 9621 9624 9627 9631 9634 9637 1 1 2 2 3
67 1.9640 9643 9647 9650 9653 9656 9659 9662 9666 9669 1 1 2 2 3
68 1.9672 9675 9678 9681 9684 9687 9690 9693 9696 9699 0 1 1 2 2
69 1.9702 9704 9707 9710 9713 9716 9719 9722 9724 9727 0 1 1 2 2
70 1.9730 9733 9735 9738 9741 9743 9746 9749 9751 9754 0 1 1 2 2
71 1.9757 9759 9762 9764 9767 9770 9772 9775 9777 9780 0 1 1 2 2
72 1.9782 9785 9787 9789 9792 9794 9797 9799 9801 9804 0 1 1 2 2
73 1.9806 9808 9811 9813 9815 9817 9820 9822 9824 9826 0 1 1 1 2
74 1.9828 9831 9833 9835 9837 9839 9841 9843 9845 9847 0 1 1 1 2
75 1.9849 9851 9853 9855 9857 9859 9961 9863 9965 9967 0 1 1 1 2
76 1.9869 9871 9873 9875 9876 9878 9880 9882 9884 9885 0 1 1 1 2
77 1.9887 9889 9891 9892 9894 9996 9897 9899 9901 9902 0 1 1 1 1
78 1.9904 9906 9907 9909 9910 9912 9913 9915 9916 9918 0 1 1 1 1
79 1.9919 9921 9922 9924 9925 9927 9928 9929 9931 9932 0 0 1 1 1
80 1.9934 9935 9936 9937 9939 9940 9941 9943 9944 9945 0 0 1 1 1
81 1.9946 9947 9949 9950 9951 9952 9953 9954 9955 9956 0 0 1 1 1
82 1.9958 9959 9960 9961 9962 9963 9964 9965 9966 9967 0 0 0 1 1
83 1.9968 9968 9969 9970 9971 9972 9973 9974 9975 9975
84 1.9976 9977 9978 9978 9979 9980 9981 9981 9982 9983
85 1.9983 9984 9985 9985 9986 9987 9987 9988 9988 9989
86 1.9989 9990 9990 9991 9991 9992 9992 9993 9993 9994
87 1.9994 9994 9995 9995 9996 9996 9996 9996 9997 9997
88 1.9997 9999 9998 9998 9999 9999 9999 9999 9999 9999
89 1.9999 9999 0000 0000 0000 0000 0000 0000 0000 0000
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
–
–
–
–
–
–
–
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–
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–
–
–
–
–
–
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–
–
–
–
–
–
–
–
–
–
–
216
LOGARITHM OF COSINES
θ
° log
10
cos
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Subtract mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
0° 0.0000 0000 0000 0000 0000 0000 0000 0000 0000 9999
1 1.9999 9999 9999 9999 9999 9999 9998 9998 9998 9998
2 1.9997 9997 9997 9996 9996 9996 9996 9995 9995 9994
3 1.9994 9994 9993 9993 9992 9992 9991 9991 9990 9990
4 1.9989 9989 9988 9988 9987 9987 9986 9985 9985 9984
5 1.9983 9983 9982 9981 9981 9980 9979 9978 9978 9977
6 1.9976 9975 9975 9974 9973 9972 9971 9970 9969 9968
7 1.9968 9967 9966 9965 9964 9963 9962 9961 9960 9959 0 0 0 1 1
8 1.9958 9956 9955 9954 9953 9952 9951 9950 9949 9947 0 0 1 1 1
9 1.9946 9945 9944 9943 9941 9940 9939 9937 9936 9935 0 0 1 1 1
10 1.9934 9932 9931 9929 9928 9927 9925 9924 9922 9921 0 0 1 1 1
11 1.9919 9918 9916 9915 9913 9912 9910 9909 9907 9906 0 1 1 1 1
12 1.9904 9902 9901 9899 9897 9896 9894 9892 9891 9889 0 1 1 1 1
13 1.9887 9585 9884 9982 9880 9978 9876 9975 9873 9871 0 1 1 1 2
14 1.9869 9867 9965 9863 9861 9859 9857 9855 9853 9851 0 1 1 1 2
15 1.9849 9847 9945 9843 9841 9839 9837 9835 9833 9831 0 1 1 1 2
16 1.9828 9826 9824 9822 9820 9817 9815 9813 9811 9808 0 1 1 1 2
17 1.9806 9804 9801 9799 9797 9794 9792 9789 9787 9785 0 1 1 2 2
18 1.9782 9780 9777 9775 9772 9770 9767 9764 9762 9759 0 1 1 2 2
19 1.9757 9754 9751 9749 9746 9743 9741 9738 9735 9733 0 1 1 2 2
20 1.9730 9727 9724 9722 9719 9716 9713 9710 9707 9704 0 1 1 2 2
21 1.9702 9699 9696 9693 9690 9687 9684 9681 9678 9675 0 1 1 2 2
22 1.9672 9669 9666 9662 9659 9656 9653 9650 9647 9643 1 1 2 2 3
23 1.9640 9637 9634 9631 9627 9624 9621 9617 9614 9611 1 1 2 2 3
24 1.9607 9604 9601 9597 9594 9590 9587 9583 9580 9576 1 1 2 2 3
25 1.9573 9569 9566 9562 9558 9555 9551 9548 9544 9540 1 1 2 2 3
26 1.9537 9533 9529 9525 9522 9518 9514 9510 9506 9503 1 1 2 3 3
27 1.9499 9495 9491 9487 9483 9479 9475 9471 9467 9463 1 1 2 3 3
28 1.9459 9455 9451 9447 9443 9439 9435 9431 9427 9422 1 1 2 3 3
29 1.9418 9414 9410 9406 9401 9397 9393 9388 9384 9380 1 1 2 3 4
30 1.9375 9371 9367 9362 9358 9353 9349 9344 9340 9335 1 1 2 3 4
31 1.9331 9326 9322 9317 9312 9308 9303 9298 9294 9289 1 2 2 3 4
32 1.9284 9279 9275 9270 9265 9260 9255 9251 9246 9241 1 2 2 3 4
33 1.9236 9231 9226 9221 9216 9211 9206 9201 9196 9191 1 2 3 3 4
34 1.9186 9181 9175 9170 9165 9160 9155 9149 9144 9139 1 2 3 3 4
35 1.9134 9128 9123 9118 9112 9107 9101 9096 9091 9085 1 2 3 4 5
36 1.9080 9074 9069 9063 9057 9052 9046 9041 9035 9029 1 2 3 4 5
37 1.9023 9018 9012 9006 9000 8995 8989 8983 8977 8971 1 2 3 4 5
38 1.8965 8959 8953 8947 8941 8935 8929 8923 8917 8911 1 2 3 4 5
39 1.8905 8899 8893 8887 8880 8874 8868 8862 8855 9849 1 2 3 4 5
40 1.8943 8836 8830 8823 8817 8810 8804 8797 8791 8784 1 2 3 4 5
41 1.8778 8771 9765 8758 8751 8745 8738 8731 8724 8718 1 2 3 4 6
42 1.8711 8704 8697 8690 9683 8676 8669 8662 8655 8649 1 2 3 5 6
43 1.8641 9634 8627 8620 8613 8606 8598 8591 8584 8577 1 2 4 5 6
44 1.8569 8562 8555 8547 8540 8532 8525 8517 8510 8502 1 2 4 5 6
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
217
– – – – – – – – –
–
–
–
–
–
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–
–
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–
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–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
LOGARITHM OF COSINES
θ
° log
10
cos
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Subtract mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
45° 1.8495 8487 8480 8472 8464 8457 8449 8441 8433 8426 1 3 4 5 6
46 1.9419 8410 8402 8394 8386 8378 8370 8362 8354 8346 1 3 4 5 7
47 1.8338 8330 8322 8313 8305 8297 8289 8280 8272 8264 1 3 4 6 7
48 1.8255 8247 8238 8230 8221 8213 8204 8195 8187 8178 1 3 4 6 7
49 1.8169 8161 8152 8143 8134 8125 8117 8108 8099 8090 1 3 4 6 7
50 1.8081 8072 8063 8053 8044 8035 8026 8017 8007 7998 2 3 5 6 8
51 1.7989 7979 7970 7960 7951 7941 7932 7922 7913 7903 2 3 5 6 9
52 1.7893 7884 7874 7864 7854 7844 7835 7825 7815 7805 2 3 5 7 8
53 1.7795 7785 7774 7764 7754 7744 7734 7723 7713 7703 2 3 5 7 9
54 1.7692 7682 7671 7661 7650 7640 7629 7618 7607 7597 2 4 5 7 9
55 1.7586 7575 7564 7553 7542 7531 7520 7509 7498 7487 2 4 5 7 9
56 1.7476 7464 7453 7442 7430 7419 7407 7396 7384 7373 2 4 6 9 10
57 1.7361 7349 7338 7326 7314 7302 7290 7278 7266 7254 2 4 6 8 10
58 1.7242 7230 7218 7205 7193 7181 7168 7156 7144 7131 2 4 6 8 10
59 1.7118 7106 7093 7080 7068 7055 7042 7029 7016 7003 2 4 6 9 11
60 1.6990 6977 6963 6950 6937 6923 6910 6896 6883 6869 2 4 7 9 11
61 1.6856 6842 6829 6814 6801 6787 6773 6759 6744 6730 2 5 7 9 12
62 1.6716 6702 6687 6673 6659 6644 6629 6615 6600 6585 2 5 7 10 12
63 1.6570 6556 6541 6526 6510 6495 6480 6465 6449 6434 3 5 8 10 13
64 1.6418 6403 6387 6371 6356 6340 6324 6308 6292 6276 3 5 8 11 13
65 1.6259 6243 6227 6210 6194 6177 6161 6144 6127 6110 3 6 8 11 14
66 1.6093 6076 6059 6042 6024 6007 5990 5972 5954 5937 3 6 9 12 15
67 1.5919 5901 5883 5865 5847 5828 5810 5792 5773 5754 3 6 9 12 15
68 1.5736 5717 5698 5679 5660 5641 5621 5602 5583 5563 3 6 10 13 16
69 1.5543 5523 5504 5484 5463 5443 5423 5402 5382 5361 3 7 10 14 17
70 1.5341 5320 5299 5278 5256 5235 5213 5192 5170 5148 4 7 11 14 18
71 1.5126 5104 5082 5060 5037 5015 4992 4969 4946 4923 4 8 11 15 19
72 1.4900 4876 4853 4829 4805 4781 4757 4733 4709 4684 4 8 12 16 20
73 1.4659 4634 4609 4584 4559 4533 4508 4482 4456 4430 4 9 13 17 21
74 1.4403 4377 4350 4323 4296 4269 4242 4214 4186 4158 5 9 14 19 23
75 1.4130 4102 4073 4044 4015 3986 3957 3927 3897 3867 5 10 15 20 24
76 1.3837 3806 3775 3745 3713 3682 3650 3618 3586 3554 5 11 16 21 26
77 1.3521 3488 3455 3421 3397 3353 3319 3284 3250 3214 6 11 17 23 28
78 1.3179 3143 3107 3070 3034 2997 2959 2921 2883 2845 6 12 19 25 31
79 1.2806 2767 2727 2687 2647 2606 2565 2524 2482 2439 7 14 20 27 34
80 1.2397 2353 2310 2266 2221 2176 2131 2085 2038 1991 8 15 23 30 38
81 1.1943 1895 1847 1797 1747 1697 1646 1594 1542 1489 8 17 25 34 42
82 1.1436 1381 1326 1271 1214 1157 1099 1040 0981 0920 10 19 29 38 48
83 1.0859 0797 0734 0670 0605 0539 0472 0403 0334 0264 11 22 33 44 55
84 1.0192 0120 0046 9970 9894 9816 9736 9655 9573 9489 13 26 39 53 66
85 2.9403 9315 9226 9135 9042 8946 8849 8749 8647 8543 16 32 48 64 81
86 2.8436 8326 8213 8098 7979 7857 7731 7602 7468 7330 21 41 62 83 103
87 2.7188 7041 6889 6731 6567 6397 6220 6035 5842 5640
88 2.5428 5206 4971 4723 4459 4179 3880 3558 3210 2832
89 2.242 2.196 2.145 2.087 2.020 3.941 3.844 3.719 3.543 3.242
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
218
– – – – – – – – –
LOGARITHMS OF TANGENTS
θ
° log
10
tan
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Add mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
0° – ∞ 3.242 3.543 3.719 3.844 3.941 2.020 2.087 2.145 2.196
1 2.2419 2833 3211 3559 3881 4181 4461 4725 4973 5208
2 2.5431 5643 5845 6038 6223 6401 6571 6736 6894 7046
3 2.7194 7337 7475 7609 7739 7865 7988 8107 8223 8336 21 42 63 83 104
4 2.8446 8554 8659 8762 8862 8960 9056 9150 9241 9331 16 32 48 65 81
5 2.9420 9506 9591 9674 9756 9836 9915 9992 0068 0143 13 26 40 53 66
6 1.0216 0289 0360 0430 0499 0567 0633 0699 0764 0828 11 22 34 45 56
7 1.0891 0954 1015 1076 1135 1194 1252 1310 1367 1423 10 20 29 39 49
8 1.1478 1533 1587 1640 1693 1745 1797 1848 1898 1948 9 17 26 35 43
9 1.1997 2046 2094 2142 2189 2236 2282 2328 2374 2419 8 16 23 31 39
10 1.2463 2507 2551 2594 2637 2680 2722 2764 2805 2846 7 14 21 28 35
11 1.2887 2927 2967 3006 3046 3085 3123 3162 3200 3237 6 13 19 26 32
12 1.3275 3312 3349 3385 3422 3458 3493 3529 3564 3599 6 12 18 24 30
13 1.3634 3668 3702 3736 3770 3804 3837 3870 3903 3935 6 11 17 22 28
14 1.3968 4000 4032 4064 4095 4127 4158 4189 4220 4250 5 10 16 21 26
15 1.4281 4311 4341 4371 4400 4430 4459 4488 4517 4546 5 10 15 20 25
16 1.4575 4603 4632 4660 4688 4716 4744 4771 4799 4826 5 9 14 19 23
17 1.4853 4880 4907 4934 4961 4987 5014 5040 5066 5092 4 9 13 18 22
18 1.5118 5143 5169 5195 5220 5245 5270 5295 5320 5345 4 8 13 17 21
19 1.5370 5394 5419 5443 5467 5491 5516 5539 5563 5587 4 8 12 16 20
20 1.5611 5634 5658 5681 5704 5727 5750 5773 5796 5819 4 8 12 15 19
21 1.5842 5864 5887 5909 5932 5954 5976 5998 6020 6042 4 7 11 15 19
22 1.6064 6086 6108 6129 6151 6172 6194 6215 6236 6257 4 7 11 14 18
23 1.6279 6300 6321 6341 6362 6383 6404 6424 6445 6465 3 7 10 14 17
24 1.6486 6506 6527 6547 6567 6587 6607 6627 6647 6667 3 7 10 13 17
25 1.6687 6706 6726 6746 6765 6785 6804 6824 6843 6863 3 7 10 13 16
26 1.6882 6901 6920 6939 6958 6977 6996 7015 7034 7053 3 6 9 13 16
27 1.7072 7090 7109 7128 7146 7165 7183 7202 7220 7238 3 6 9 12 15
28 1.7257 7275 7293 7311 7330 7348 7366 7384 7402 7420 3 6 9 12 15
29 1.7438 7455 7473 7491 7509 7526 7544 7562 7579 7597 3 6 9 12 15
30 1.7614 7632 7649 7667 7684 7701 7719 7736 7753 7771 3 6 9 12 14
31 1.7788 7805 7822 7839 7856 7873 7890 7907 7924 7941 3 6 9 11 14
32 1.7958 7975 7992 8008 8025 8042 8059 8075 8092 8109 3 6 8 11 14
33 1.8125 8142 8158 8175 8191 8208 8224 8241 8257 8274 3 5 8 11 14
34 1.8290 8306 8323 8339 8355 8371 8388 8404 8420 8436 3 5 8 11 14
35 1.8452 8468 8484 8501 8517 8533 8549 8565 8581 8597 3 5 8 11 13
36 1.8613 8629 8644 8660 8676 8692 8708 8724 8740 8755 3 5 8 11 13
37 1.8771 8787 8803 8818 8834 8850 8865 8881 8897 8912 3 5 8 10 13
38 1.8928 8944 8959 8975 8990 9006 9022 9037 9053 9068 3 5 8 10 13
39 1.9084 9099 9115 9130 9146 9161 9176 9192 9207 9223 3 5 8 10 13
40 1.9238 9254 9269 9294 9300 9315 9330 9346 9361 9376 3 5 8 10 13
41 1.9392 9407 9422 9438 9453 9468 9483 9499 9514 9529 3 5 8 10 13
42 1.9544 9560 9575 9590 9605 9621 9636 9651 9666 9681 3 5 8 10 13
43 1.9697 9712 9727 9742 9757 9772 9788 9803 9818 9833 3 5 8 10 13
44 1.9848 9864 9879 9894 9909 9924 9939 9955 9970 9985 3 5 8 10 13
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
–
–
–
–
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–
–
–
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–
–
–
–
–
–
–
–
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–
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–
–
–
–
–
–
219
LOGARITHMS OF TANGENTS
θ
° log
10
tan
θ
°
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Add mean dierences
θ
° 0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
45° 0.0000 0015 0030 0045 0061 0076 0091 0106 0121 0136 3 5 8 10 13
46 0.0152 0167 0182 0197 0212 0228 0243 0258 0273 0288 3 5 8 10 13
47 0.0303 0319 0334 0349 0364 0379 0395 0410 0425 0440 3 5 8 10 13
48 0.0456 0471 0486 0501 0517 0532 0547 0562 0578 0593 3 5 8 10 13
49 0.0608 0624 0639 0654 0670 0685 0700 0716 0731 0746 3 5 8 10 13
50 0.0762 0777 0793 0808 0824 0839 0854 0870 0885 0901 3 5 8 10 13
51 0.0916 0932 0947 0963 0978 0994 1010 1025 1041 1056 3 5 8 10 13
52 0.1072 1088 1103 1119 1135 1150 1166 1182 1197 1213 3 5 8 10 13
53 0.1229 1245 1260 1276 1292 1308 1324 1340 1356 1371 3 5 8 11 13
54 0.1387 1403 1419 1435 1451 1467 1483 1499 1516 1532 3 5 8 11 13
55 0.1548 1564 1580 1596 1612 1629 1645 1661 1677 1694 3 5 8 11 14
56 0.1710 1726 1743 1759 1776 1792 1809 1825 1842 1858 3 5 8 11 14
57 0.1875 1891 1908 1925 1941 1958 1975 1992 2009 2025 3 6 8 11 14
58 0.2042 2059 2076 2093 2110 2127 2144 2161 2178 2195 3 6 9 11 14
59 0.2212 2229 2247 2264 2281 2299 2316 2333 2351 2368 3 6 9 12 14
60 0.2386 2403 2421 2439 2456 2474 2491 2509 2527 2545 3 6 9 12 15
61 0.2562 2580 2598 2616 2634 2652 2670 2689 2707 2725 3 6 9 12 15
62 0.2743 2762 2780 2798 2817 2835 2854 2972 2991 2910 3 6 9 12 15
63 0.2928 2947 2966 2985 3004 3023 3042 3061 3080 3099 3 6 9 13 16
64 0.3118 3137 3157 3176 3196 3215 3235 3254 3274 3294 3 7 10 13 16
65 0.3313 3333 3353 3373 3393 3413 3433 3453 3473 3494 3 7 10 13 17
66 0.3514 3535 3555 3576 3596 3617 3638 3659 3679 3700 3 7 10 14 17
67 0.3721 3743 3764 3785 3806 3828 3849 3871 3892 3914 4 7 11 14 18
68 0.3936 3958 3980 4002 4024 4046 4068 4091 4113 4136 4 7 11 15 19
69 0.4158 4181 4204 4227 4250 4273 4296 4319 4342 4366 4 8 12 15 19
70 0.4389 4413 4437 4461 4484 4509 4533 4557 4581 4606 4 8 12 16 20
71 0.4630 4655 4680 4705 4730 4755 4780 4805 4831 4857 4 8 13 17 21
72 0.4892 4908 4934 4960 4986 5013 5039 5066 5093 5120 4 9 13 18 22
73 0.5147 5174 5201 5229 5256 5284 5312 5340 5368 5397 5 9 14 19 23
74 0.5425 5454 5483 5512 5541 5570 5600 5629 5659 5689 5 10 15 20 25
75 0.5719 5750 5780 5811 5842 5873 5905 5936 5968 6000 5 10 16 21 26
76 0.6032 6065 6097 6130 6163 6196 6230 6264 6298 6332 6 11 17 22 28
77 0.6366 6401 6436 6471 6507 6542 6578 6615 6651 6688 6 12 18 24 30
78 0.6725 6763 6800 6838 6877 6915 6954 6994 7033 7073 6 13 19 26 32
79 0.7113 7154 7195 7236 7278 7320 7363 7406 7449 7493 7 14 21 28 35
80 0.7537 7581 7626 7672 7718 7764 7811 7858 7906 7954 8 16 23 31 39
81 0.8003 8052 8102 8152 8203 8255 8307 8360 8413 8467 9 17 26 35 43
82 0.8522 8577 8633 9690 8748 8806 8865 8924 8985 9046 10 20 29 39 49
83 0.9109 9172 9236 9301 9367 9433 9501 9570 9640 9711 11 22 34 45 56
84 0.9794 9857 9932 0008 0085 0164 0244 0326 0409 0494 13 26 40 53 66
85 1.0580 0669 0759 0850 0944 1040 1138 1238 1341 1446 16 32 48 65 81
86 1.1554 1664 1777 1893 2012 2135 2261 2391 2525 2663 21 42 63 83 104
87 1.2806 2954 3106 3264 3429 3599 3777 3962 4155 4357
88 1.4569 4792 5027 5275 5539 5819 6119 6441 6789 7167
89 1.758 1.804 1.855 1.913 1.980 2.059 2.156 2.281 2.457 2.758
0′ 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′ 54′ 1′ 2′ 3′ 4′ 5′
220
SQUARES OF NUMBERS x x
2
Add mean dierences
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0 1.000 1.020 1.040 1.061 1.082 1.103 1.124 1.145 1.166 1.198 2 4 6 8 10 13 15 17 19
1.1 1.210 1.232 1.254 1.277 1.300 1.323 1.346 1.369 1.392 1.416 2 5 7 9 11 14 16 19 21
1.2 1.440 1.464 1.488 1.513 1.538 1.563 1.588 1.613 1.638 1.664 2 5 7 10 12 15 17 20 22
1.3 1.690 1.716 1.742 1.769 1.796 1.823 1.850 1.877 1.904 1.932 3 5 8 11 13 16 19 22 24
1.4 1.960 1.988 2.016 2.045 2.074 2.103 2.132 2.161 2.190 2.220 3 6 9 12 14 17 20 23 26
1.5 2.250 2.280 2.310 2.341 2.372 2.403 2.434 2.465 2.496 2.528 3 6 9 12 15 19 22 25 28
1.6 2.560 2.592 2.624 2.657 2.690 2.723 2.756 2.789 2.822 2.856 3 7 10 13 16 20 23 26 30
1.7 2.890 2.924 2.958 2993 3.028 3.063 3.098 3.133 3.168 3.204 3 7 10 14 17 21 24 28 31
1.8 3.240 3.276 3.312 3.349 3.386 3.423 3.460 3.497 3.534 3.572 4 7 11 15 18 22 26 30 33
1.9 3.610 3.648 3.686 3.725 3.764 3.803 3.842 3.881 3.920 3.960 4 8 12 16 19 23 27 31 35
2.0 4.000 4.040 4.080 4.121 4.162 4.203 4.244 4.285 4.326 4.368 4 8 12 16 20 25 29 33 37
2.1 4.410 4.452 4.494 4.537 4.580 4.623 4.666 4.709 4.752 4.796 4 9 13 17 21 26 30 34 39
2.2 4.840 4.884 4.928 4.973 5.018 5.063 5.108 5.153 5.198 5.244 4 9 13 18 22 27 31 36 40
2.3 5.290 5.336 5.382 5.429 5.476 5.523 5.570 5.617 5.664 5.712 5 9 14 19 23 28 33 38 42
2.4 5.760 5.808 5.856 5.905 5.954 6.003 6.052 6.101 6.150 6.200 5 10 15 20 24 29 34 39 44
2.5 6.250 6.300 6.350 6.401 6.452 6.503 6.554 6.605 6.656 6.708 5 10 15 20 25 31 36 41 46
2.6 6.760 6.812 6.864 6.917 6.970 7.023 7.076 7.129 7.182 7.236 5 11 16 21 26 32 37 42 48
2.7 7.290 7.344 7.398 7.453 7.508 7.563 7.618 7.673 7.728 7.784 5 11 16 22 27 33 38 44 49
2.8 7.840 7.896 7.952 8.009 8.066 8.123 8.180 8.237 8.294 8.352 6 11 17 23 28 34 40 46 51
2.9 8.410 8.468 8.526 8.585 8.644 8.703 8.762 8.821 8.880 8.940 6 12 18 24 29 35 41 47 53
3.0 9.000 9.060 9.120 9.181 9.242 9.303 9.364 9.425 9.486 9.548 6 12 18 24 30 37 43 49 55
3.1 9.610 9.672 9.734 9.797 9.860 9.923 9.986 10.049 10.112 10.176 6 13 19 25 31 38 44 50 57
3.2 10.240 10.304 10.368 10.433 10.498 10.563 10.628 10.693 10.758 10.824 6 13 19 26 32 39 45 52 58
3.3 10.890 10.956 11.022 11.089 11.156 11.223 11.290 11.357 11.424 11.492 7 13 20 27 33 40 47 54 60
3.4 11.560 11.628 11.696 11.765 11.834 11.903 11.972 12.041 12.110 12.180 7 14 21 28 34 41 48 55 62
3.5 12.250 12.320 12.390 12.461 12.532 12.603 12.674 12.745 12.816 12.888 7 14 21 28 35 43 50 57 64
3.6 12.960 13.032 13.104 13.177 13.250 13.323 13.396 13.469 13.542 13.616 7 15 22 29 36 44 51 58 66
3.7 13.690 13.764 13.838 13.913 13.988 14.063 14.138 14.213 14.288 14.364 7 15 22 30 37 45 52 60 67
3.8 14.440 14.516 14.592 14.669 14.746 14.823 14.900 14.977 15.054 15.132 8 15 23 31 38 46 54 62 69
3.9 15.210 15.288 15.366 15.445 15.524 15.603 15.682 15.761 15.840 15.920 8 16 24 32 39 47 55 63 71
4.0 16.000 16.080 16.160 16.241 16.322 16.403 16.484 16.565 16.646 16.728 8 16 24 32 40 49 57 65 73
4.1 16.810 16.892 16.974 17.057 17.140 17.223 17.306 17.389 17.472 17.556 8 17 25 33 41 50 58 66 75
4.2 17.640 17.724 17.808 17.893 17.978 18.063 18.148 18.233 18.318 18.404 8 17 25 34 42 51 59 68 76
4.3 18.490 18.576 18.662 18.749 18.836 18.923 19.010 19.097 19.184 19.272 9 17 26 35 43 52 61 70 78
4.4 19.360 19.448 19.536 19.625 19.714 19.803 19.892 19.981 20.070 20.160 9 18 27 36 44 53 62 71 80
4.5 20.250 20.340 20.430 20.521 20.612 20.703 20.794 20.885 20.976 21.068 9 18 Z7 36 45 55 64 73 82
4.6 21.160 21.252 21.344 21.437 21.530 21.623 21.716 21.809 21.902 21.996 9 19 28 37 46 56 65 74 84
4.7 22.090 22.184 22.278 22.373 22.468 22.563 22.658 22.753 22.848 22.944 9 19 28 38 47 57 66 76 85
4.8 23.040 23.136 23.232 23.329 23.426 23.523 23.620 23.717 23.814 23.912 10 19 29 39 48 58 68 78 87
4.9 24.010 24.108 24.206 24.305 24.404 24.503 24.602 24.701 24.800 24.900 10 20 30 40 49 59 69 79 89
5.0 25.000 25.100 25.200 25.301 25.402 25.503 25.604 25.705 25.806 25.908 10 20 30 40 50 61 71 81 91
5.1 26.010 26.112 26.214 26.317 26.420 26.523 26.626 26.729 26.832 26.936 10 21 31 41 51 62 72 82 93
5.2 27.040 27.144 27.248 27.353 27.458 27.563 27.668 27.773 27.878 77.984 10 21 31 42 52 63 73 84 94
5.3 28.090 28.196 28.302 28.409 28.516 28.623 28.730 28.837 28.944 29.052 11 21 32 43 53 64 75 86 96
5.4 29.160 29.268 29.376 29.485 29.594 29.703 29.812 29.921 30.030 30.140 11 22 33 44 54 65 76 87 98
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
221
SQUARES OF NUMBERS x x
2
Add mean dierences
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
5.5 30.25 30.36 30.47 30.58 30.69 30.80 30.91 31.02 31.14 31.25 1 2 3 4 6 7 9 9 10
5.6 31.36 31.47 31.58 31.70 31.81 31.92 32.04 32.15 31.26 37.38 1 2 3 5 6 7 8 9 10
5.7 32.49 32.60 32.72 32.83 32.95 33.06 33.18 33.29 33.41 33.52 1 2 3 5 6 7 8 9 10
5.8 33.64 33.76 33.87 33.99 34.11 34.22 34.34 34.46 34.57 34.69 1 2 4 5 6 7 8 9 11
5.9 34.81 34.93 35.05 35.16 35.28 35.40 35.52 35.64 35.76 35.88 1 2 4 5 6 7 8 10 11
6.0 36.00 36.12 36.24 36.36 36.48 36.60 36.72 36.84 36.97 37.09 1 2 4 5 6 7 8 10 11
6.1 37.21 37.33 37.45 37.58 37.70 37.82 37.95 38.07 38.19 39.32 1 2 4 5 6 7 9 10 11
6.2 39.44 38.56 38.69 38.81 38.94 39.06 39.19 39.31 39.44 39.56 1 2 4 5 6 7 9 10 11
6.3 39.69 39.82 39.94 40.07 40.20 40.32 40.45 40.58 40.70 40.83 1 3 4 S 6 8 9 10 11
6.4 40.96 41.09 41.22 41.34 41.47 41.60 41.73 41.86 41.99 42.12 1 3 4 5 6 8 9 10 12
6.5 42.25 42.38 42.51 41.64 42.77 41.90 43.03 43.16 43.30 43.43 1 3 4 5 7 8 9 10 12
6.6 43.56 43.69 43.82 43.96 44.09 44.22 44.36 44.49 44.62 44.76 1 3 4 5 7 8 9 11 12
6.7 44.89 45.02 45.16 45.29 45.43 45.56 45.70 45.83 45.97 46.10 1 3 4 5 7 8 9 11 12
6.8 46.24 46.38 46.51 46.65 46.79 46.92 47.06 47.20 47.33 47.47 1 3 4 5 7 8 10 11 12
6.9 47.61 47.75 47.89 48.02 48.16 48.30 48.44 48.58 49.72 48.86 1 3 4 6 7 8 10 11 13
7.0 49.00 49.14 49.28 49.42 49.56 49.70 49.84 49.98 50.13 50.27 1 3 4 6 7 8 10 11 13
7.1 50.41 50.55 50.69 50.84 50.98 51.12 51.27 51.41 51.55 51.70 1 3 4 6 7 9 10 11 13
7.2 51.84 51.98 52.13 52.27 52.42 52.56 52.71 52.85 53.00 53.14 1 3 4 6 7 9 10 11 13
7.3 53.29 53.44 53.58 53.73 53.88 54.02 54.17 54.32 54.46 54.61 1 3 4 6 7 9 10 12 13
7.4 54.76 54.91 55.06 55.20 55.35 55.50 55.65 55.80 55.95 56.10 1 3 4 6 7 9 10 12 13
7.5 56.25 56.40 56.55 56.70 56.85 57.00 57.15 57.30 57.46 57.61 2 3 5 6 8 9 11 12 14
7.6 57.76 57.91 58.06 58.22 58.37 58.52 58.68 58.83 58.98 59.14 2 3 5 6 8 9 11 12 14
7.7 59.29 59.44 59.60 59.75 59.91 60.06 60.22 60.37 60.53 60.68 2 3 5 6 8 9 11 12 14
7.8 60.84 61.00 61.15 61.31 61.47 61.62 61.78 61.94 62.09 62.25 2 3 5 6 8 9 11 13 14
7.9 62.41 62.57 61.73 62.88 63.04 63.20 63.36 63.52 63.68 63.84 2 3 5 6 8 10 11 13 14
8.0 64.00 64.16 64.32 64.48 64.64 64.80 64.96 65.12 65.29 65.45 2 3 5 6 8 10 11 13 14
8.1 65.61 65.77 65.93 66.10 66.26 66.42 66.59 66.75 66.91 67.08 2 3 5 7 8 10 11 13 15
8.2 67.24 61.40 67.57 67.73 67.90 68.06 68.23 68.39 68.56 68.72 2 3 5 7 8 10 12 13 15
8.3 68.89 69.06 69.22 69.39 69.56 69.72 69.89 70.06 70.22 70.39 2 3 5 7 8 10 12 13 15
8.4 70.56 70.73 70.90 71.06 71.23 71.40 71.57 71.74 71.91 72.08 2 3 5 7 8 10 12 14 15
8.5 77.25 72.42 71.59 71.76 72.93 73.10 73.27 73.44 73.62 73.79 2 3 5 7 9 10 12 14 15
8.6 73.96 74.13 74.30 74.48 74.65 74.82 75.00 75.17 75.34 75.52 2 3 5 7 9 10 12 14 16
8.7 75.69 75.86 76.04 76.21 76.39 76.56 76.74 76.91 77.09 77.26 2 3 5 7 9 10 12 14 16
8.8 77.44 77.62 77.79 77.97 78.15 78.32 78.50 78.68 78.85 79.03 2 4 5 7 9 11 12 14 16
8.9 79.21 79.39 79.57 79.74 79.92 80.10 80.28 80.46 80.64 80.82 2 4 5 7 9 11 13 14 16
9.0 81.00 81.18 81.36 81.54 81.72 81.90 82.08 81.26 81.45 82.63 2 4 5 7 9 11 13 14 16
9.1 82.81 82.99 83.17 83.36 83.54 83.72 83.91 84.09 84.27 84.46 2 4 5 7 9 11 13 15 16
9.2 84.64 84.82 85.01 85.19 85.38 85.56 85.75 85.93 86.12 86.30 2 4 6 7 9 11 13 15 17
9.3 86.49 86.68 86.86 87.05 87.24 87.42 87.61 87.80 87.98 88.17 2 4 6 7 9 11 13 15 17
9.4 88.36 88.55 88.74 88.92 89.11 89.30 89.49 89.68 89.87 90.06 2 4 6 8 9 11 13 15 17
9.5 90.25 90.44 90.63 90.82 91.01 91.20 91.39 91.58 91.78 91.97 2 4 6 8 10 11 13 15 17
9.6 92.16 92.35 92.54 91.74 92.93 93.12 93.32 93.51 93.70 93.90 2 4 6 8 10 12 14 15 17
9.7 94.09 94.28 94.48 94.67 94.87 95.06 95.26 95.45 95.65 95.84 2 4 6 8 10 12 14 16 18
9.8 96.04 96.24 96.43 96.63 96.83 97.02 97.22 97.42 97.61 97.81 2 4 6 8 10 12 14 16 18
9.9 98.01 98.21 98.41 98.60 98.80 99.00 99.20 99.40 99.60 99.80 2 4 6 8 10 12 14 16 18
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
222
SQUARE ROOTS OF NUMBERS x √x : (1 ≤ x < 10)
Add mean dierences
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0 1.0000 0050 0100 0149 0198 0247 0296 0344 0392 0440 5 10 15 20 24 29 34 39 44
1.1 1.0488 0536 0583 0630 0677 0724 0770 0817 0863 0909 5 9 14 19 23 28 33 37 42
1.2 1.0954 1000 1045 1091 1136 1180 1225 1269 1314 1358 4 9 13 18 22 27 31 36 40
1.3 1.1402 1446 1489 1533 1576 1619 1662 1705 1747 1790 4 9 13 17 22 26 30 34 39
1.4 1.1832 1874 1916 1958 2000 2042 2083 2124 2166 2207 4 8 12 17 21 25 29 33 37
1.5 1.2247 2288 2329 2369 2410 2450 2490 2530 2570 2610 4 8 12 16 20 24 28 32 36
1.6 1.2649 2689 2728 2267 2806 2845 2984 2923 2961 3000 4 8 12 16 19 23 27 31 35
1.7 1.3038 3077 3115 3153 3191 3229 3266 3304 3342 3379 4 8 11 15 19 23 26 30 34
1.8 1.3416 3454 3491 3528 3565 3601 3638 3675 3711 3748 4 7 11 15 18 22 26 29 33
1.9 1.3784 3820 3856 3892 3928 3964 4000 4036 4071 4107 4 7 11 14 18 22 25 29 32
2.0 1.4142 4177 4213 4248 4283 4318 4353 4387 4422 4457 3 7 10 14 17 21 24 28 31
2.1 1.4491 4526 4560 4595 4629 4663 4697 4731 4765 4799 3 7 10 14 17 20 24 27 31
2.2 1.4832 4866 4900 4933 4967 5000 5033 5067 5100 5133 3 7 10 13 17 20 23 27 30
2.3 1.5166 5199 5232 5264 5297 5330 5362 5395 5427 5460 3 7 10 13 16 20 23 26 29
2.4 1.5492 5524 5556 5588 5620 5652 5684 5716 5748 5780 3 6 10 13 16 19 22 26 29
2.5 1.5811 5843 5875 5906 5937 5969 6000 6031 6062 6093 3 6 9 13 16 19 22 25 28
2.6 1.6125 6155 6186 6217 6248 6279 6310 6340 6371 6401 3 6 9 12 15 18 22 25 28
2.7 1.6432 6462 6492 6523 6553 6583 6613 6643 6673 6703 3 6 9 12 15 18 21 24 27
2.8 1.6733 6763 6793 6823 6952 6882 6912 6941 6971 7000 3 6 9 12 15 18 21 24 27
2.9 1.7029 7059 7088 7117 7146 7176 7205 7234 7263 7292 3 6 9 12 15 17 20 23 26
3.0 1.7321 7349 7378 7407 7436 7464 7493 7521 7550 7578 3 6 9 11 14 17 20 23 26
3.1 1.7607 7635 7664 7692 7720 7748 7776 7804 7833 7861 3 6 8 11 14 17 20 23 25
3.2 1.7889 7916 7944 7972 8000 8028 8055 8083 8111 8138 3 6 8 11 14 17 19 22 25
3.3 1.8166 8193 8221 8248 8276 8303 8330 8358 8385 8412 3 5 8 11 14 16 19 22 25
3.4 1.8439 8466 8493 8520 8547 8574 8601 8628 8655 8682 3 5 8 11 13 16 19 22 24
3.5 1.8708 8735 8762 8788 8815 8841 8868 8894 8921 8947 3 5 8 11 13 16 19 21 24
3.6 1.8974 9000 9026 9053 9079 9105 9131 9157 9183 9209 3 5 8 10 13 16 18 21 24
3.7 1.9235 9261 9287 9313 9339 9365 9391 9416 9442 9468 3 5 8 10 13 16 18 21 23
3.8 1.9494 9519 9545 9570 9596 9621 9647 9672 9698 9723 3 5 8 10 13 15 18 20 23
3.9 1.9748 9774 9799 9824 9849 9875 9900 9925 9950 9975 3 5 8 10 13 15 18 20 23
4.0 2.0000 0025 0050 0075 0100 0125 0149 0174 0199 0224 2 5 7 10 12 15 17 20 22
4.1 7.0248 0273 0298 0322 0347 0372 0396 0421 0445 0469 2 5 7 10 12 15 17 20 22
4.2 2.0494 0518 0543 0567 0591 0616 0640 0664 0688 0712 2 5 7 10 12 15 17 19 22
4.3 2.0736 0761 0785 0809 0833 0857 0881 0905 0928 0952 2 5 7 10 12 14 17 19 22
4.4 2.0976 1000 1024 1048 1071 1095 1119 1142 1166 1190 2 5 7 9 12 14 17 19 21
4.5 7.1213 1237 1260 1284 1307 1331 1354 1378 1401 1424 2 5 7 9 12 14 16 19 21
4.6 2.1448 1471 1494 1517 1541 1564 1587 1610 1633 1656 2 5 7 9 12 14 16 19 21
4.7 2.1679 1703 1726 1749 1772 1794 1817 1840 1863 1886 2 5 7 9 11 14 16 18 21
4.8 2.1909 1932 1954 1977 2000 2023 2045 2068 2091 2113 2 5 7 9 11 14 16 18 20
4.9 2.2136 2159 2181 2204 2226 2249 2271 2293 2316 2338 2 4 7 9 11 13 16 19 20
5.0 2.2361 2383 2405 2428 2450 2472 2494 2517 2539 2561 2 4 7 9 11 13 16 18 20
5.1 2.2583 2605 2627 2650 2672 2694 2716 2738 2760 2782 2 4 7 9 11 13 15 18 20
5.2 7.2804 2825 2947 2869 2891 2913 2935 2956 2978 3000 2 4 7 9 11 13 15 17 20
5.3 2.3022 3043 3065 3087 3108 3130 3152 3173 3195 3216 2 4 6 9 11 13 15 17 19
5.4 2.3238 3259 3281 3302 3324 3345 3367 3388 3409 3431 2 4 6 9 11 13 15 17 19
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
223
SQUARE ROOTS OF NUMBERS x √x : (1 ≤ x < 10)
Add mean dierences
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
5.5 2.3452 3473 3495 3516 3537 3558 3580 3601 3622 3643 2 4 6 8 11 13 15 17 19
5.6 2.3664 3685 3707 3728 3749 3770 3791 3812 3833 3854 2 4 6 8 11 13 15 17 19
5.7 2.3875 3896 3917 3937 3958 3979 4000 4021 4042 4062 2 4 6 8 10 13 15 17 19
5.8 2.4083 4104 4125 4145 4166 4187 4207 4228 4249 4269 2 4 6 8 10 12 14 17 19
5.9 2.4290 4310 4331 4352 4372 4393 4413 4434 4454 4474 2 4 6 8 10 12 14 16 18
6.0 2.4495 4515 4536 4556 4576 4597 4617 4637 4658 4678 2 4 6 8 10 12 14 16 18
6.1 2.4698 4718 4739 4759 4779 4799 4819 4839 4860 4980 2 4 6 8 10 12 14 16 18
6.2 2.4900 4920 4940 4960 4980 5000 5020 5040 5060 5080 2 4 6 8 10 12 14 16 18
6.3 2.5100 5120 5140 5159 5179 5199 5219 5239 5259 5278 2 4 6 8 10 12 14 16 13
6.4 2.5298 5318 5338 5357 5377 5397 5417 5436 5456 5475 2 4 6 8 10 12 14 16 18
6.5 2.5495 5515 5534 5554 5573 5593 5612 5632 5652 5671 2 4 6 8 10 12 14 16 18
6.6 2.5690 5710 5729 5749 5768 5788 5807 5826 5846 5865 2 4 6 8 10 12 14 16 17
6.7 2.5884 5904 5923 5942 5962 5981 6000 6019 6038 6058 2 4 6 8 10 12 13 15 17
6.8 2.6077 6096 6115 6134 6153 6173 6192 6211 6230 6249 2 4 6 8 10 11 13 15 17
6.9 2.6268 6287 6306 6325 6344 6363 6382 6401 6420 6439 2 4 6 8 9 11 13 15 17
7.0 2.6458 6476 6495 6514 6533 6552 6571 6589 6608 6627 2 4 6 8 9 11 13 15 17
7.1 2.6646 6665 6683 6702 6721 6739 6758 6777 6796 6814 2 4 6 7 9 11 13 15 17
7.2 2.6833 6951 6870 6889 6907 6926 6944 6963 6981 7000 2 4 6 7 9 11 13 15 17
7.3 2.7019 7037 7055 7074 7092 7111 7129 7148 7166 7185 2 4 6 7 9 11 13 15 17
7.4 2.7203 7221 7240 7258 7276 7295 7313 7331 7350 7368 2 4 5 7 9 11 13 15 16
7.5 2.7386 7404 7423 7441 7459 7477 7495 7514 7532 7550 2 4 5 7 9 11 13 15 16
7.6 2.7568 7586 7604 7622 7641 7659 7677 7695 7713 7731 2 4 5 7 9 11 13 14 16
7.7 7.7749 7767 7785 7803 7821 7839 7857 7875 7893 7911 2 4 5 7 9 11 13 14 16
7.8 2.7928 7946 7964 7982 8000 8018 8036 8054 8071 8089 2 4 5 7 9 11 12 14 16
7.9 2.8107 8125 9142 8160 8178 9196 8213 8231 8249 8267 2 4 5 7 9 11 12 14 16
8.0 2.8284 8302 8320 8337 8355 8373 8390 8408 8425 8443 2 4 5 7 9 11 12 14 16
8.1 2.8460 8478 8496 8513 8531 8548 8566 8583 8601 8618 2 4 5 7 9 11 12 14 16
8.2 2.8636 8653 8671 8688 8705 8723 8740 8758 8775 8792 2 3 5 7 9 10 12 14 16
8.3 2.8810 8827 8844 8862 8879 8896 8914 8931 8948 8965 2 3 5 7 9 10 12 14 16
8.4 2.9983 9000 9017 9034 9052 9069 9086 9103 9120 9138 2 3 5 7 9 10 12 14 15
8.5 2.9155 9172 9189 9206 9223 9240 9257 9275 9292 9309 2 3 5 7 9 10 12 14 15
8.6 2.9326 9343 9360 9377 9394 9411 9428 9445 9462 9479 2 3 5 7 9 10 12 14 15
8.7 2.9496 9513 9530 9547 9563 9580 9597 9614 9631 9648 2 3 5 7 8 10 12 14 15
8.8 2.9665 9682 9698 9715 9732 9749 9766 9783 9799 9816 2 3 5 7 8 10 12 13 15
8.9 2.9833 9850 9866 9883 9900 9917 9933 9950 9967 9993 2 3 5 7 8 10 12 13 15
9.0 3.0000 0017 0033 0050 0067 0083 0100 0116 0133 0150 2 3 5 7 8 10 12 13 15
9.1 3.0166 0183 0199 0216 0232 0249 0265 0282 0299 0315 2 3 5 7 8 10 12 13 15
9.2 3.0332 0348 0364 0381 0397 0414 0430 0447 0463 0480 2 3 5 7 8 10 12 13 15
9.3 3.0496 0512 0529 0545 0561 0578 0594 0610 0627 0643 2 3 5 7 8 10 11 13 15
9.4 3.0659 0676 0692 0708 0725 0741 0757 0773 0790 0806 2 3 5 7 8 10 11 13 15
9.5 3.0822 0838 0854 0871 0887 0903 0919 0935 0952 0968 2 3 5 6 8 10 11 13 15
9.6 3.0984 1000 1016 1032 1048 1064 1081 1097 1113 1129 2 3 5 6 8 10 11 13 14
9.7 3.1145 1161 1177 1193 1209 1225 1241 1257 I273 1289 2 3 5 6 8 10 11 13 14
9.8 3.1305 1321 1337 1353 1369 1385 1401 1417 1432 1448 2 3 5 6 8 10 11 13 14
9.9 3.1464 1480 1496 1512 1528 1544 1559 1575 1591 1607 2 3 5 6 8 10 11 13 14
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
224
SQUARE ROOTS OF NUMBERS x √x : (10 ≤ x < 100)
Add mean dierences
x .0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 2 3 4 5 6 7 8 9
10 3.1623 1780 1937 2094 2249 2404 2558 2711 2863 3015 15 31 46 62 77 93 108 124 139
11 3.3166 3317 3466 3615 3764 3912 4059 4205 4351 4496 15 30 44 59 74 89 103 118 133
12 3.4641 4785 4928 5071 5214 5355 5496 5637 5777 5917 14 28 43 57 71 85 99 113 128
13 3.6056 6194 6332 6469 6606 6742 6879 7014 7148 7283 14 27 41 55 68 82 95 109 123
14 3.7417 7550 7683 7815 7947 8079 8210 8341 8471 8601 13 26 39 53 66 79 92 105 118
15 3.8730 8859 8987 9115 9243 9370 9497 9623 9749 9875 13 25 38 51 64 76 89 102 114
16 4.0000 0125 0249 0373 0497 0620 0743 0866 0988 1110 12 25 37 49 62 74 86 99 111
17 4.1231 1352 1473 1593 1713 1833 1952 2071 2190 2308 12 24 36 48 60 72 84 96 108
18 4.2426 2544 2661 2778 2895 3012 3128 3243 3359 3474 12 23 35 47 58 70 81 93 105
19 4.3589 3704 3818 3932 4045 4159 4272 4385 4497 4609 11 23 34 45 57 68 79 91 102
20 4.4721 4833 4944 5056 5166 5277 5387 5497 5607 5717 11 22 33 44 55 66 77 88 100
21 4.5826 5935 6043 6152 6260 6368 6476 6583 6690 6797 11 22 32 43 54 65 76 86 97
22 4.6904 7011 7117 7223 7329 7434 7539 7645 7749 7854 11 21 32 42 53 63 74 84 95
23 4.7958 8062 8166 8270 8374 8477 8580 8683 8785 8888 10 21 31 41 52 62 72 83 93
24 4.9990 9092 9193 9295 9396 9497 9598 9699 9800 9900 10 20 30 40 51 61 71 81 91
25 5.0000 0100 0200 0299 0398 0498 0596 0695 0794 0892 10 20 30 40 50 59 69 79 89
26 5.0990 1088 1186 1284 1381 1478 1575 1672 1769 1865 10 19 29 39 49 58 68 78 88
27 5.1962 2058 2154 2249 2345 2440 2536 2631 2726 2820 10 19 29 38 48 57 67 76 86
28 5.2915 3009 3104 3198 3292 3385 3479 3572 3666 3759 9 19 28 37 47 56 66 75 94
29 5.3852 3944 4037 4129 4222 4314 4406 4498 4589 4681 9 18 28 37 46 55 64 74 93
30 5.4772 4863 4955 5045 5136 5227 5317 5408 5498 5588 9 18 27 36 45 54 63 72 82
31 5.5678 5767 5857 5946 6036 6125 6214 6303 6391 6480 9 18 27 36 45 53 62 71 80
32 5.6569 6657 6745 6833 6921 7009 7096 7194 7271 7359 9 18 26 35 44 53 61 70 79
33 5.7446 7533 7619 7706 7793 7879 7966 8052 8138 8224 9 17 26 35 43 52 61 69 78
34 5.8310 8395 8481 8566 8652 8737 8822 8907 8992 9076 9 17 26 34 43 51 60 68 77
35 5.9161 9245 9330 9414 9498 9582 9666 9749 9833 9917 8 17 25 34 42 50 59 67 76
36 6.0000 0083 0166 0249 0332 0415 0498 0581 0663 0745 8 17 25 33 41 50 58 66 75
37 6.0828 0910 0992 1074 1156 1237 1319 1400 1492 1563 8 16 25 33 41 49 57 65 74
38 6.1644 1725 1806 1887 1968 2048 2129 2209 2290 2370 8 16 24 32 40 48 56 65 73
39 6.2450 2530 2610 2690 2769 2849 2929 3008 3087 3166 8 16 24 32 40 48 56 64 72
40 6.3246 3325 3403 3482 3561 3640 3718 3797 3875 3953 8 16 24 31 39 47 55 63 71
41 6.4031 4109 4187 4265 4343 4420 4498 4576 4653 4730 8 16 23 31 39 47 54 62 70
42 6.4807 4885 4962 5038 5115 5192 5269 5345 5422 5499 8 15 23 31 38 46 54 61 69
43 6.5574 5651 5727 5803 5879 5955 6030 6106 6182 6257 8 15 23 30 38 46 53 61 68
44 6.6332 6408 6483 6558 6633 6708 6783 6858 6933 7007 7 15 22 30 37 45 52 60 67
45 6.7082 7157 7231 7305 7380 7454 7528 7602 7676 7750 7 15 22 30 37 44 52 59 67
46 6.7823 7897 7971 8044 8118 8191 8264 8337 8411 8484 7 15 22 29 37 44 51 59 66
47 6.8557 8629 8702 8775 8848 8920 8993 9065 9138 9210 7 15 22 29 36 44 51 58 65
48 6.9282 9354 9426 9498 9570 9642 9714 9785 9857 9929 7 14 22 29 36 43 50 57 65
49 7.0000 0071 0143 0214 0285 0356 0427 0498 0569 0640 7 14 21 28 36 43 50 57 64
50 7.0711 0781 0852 0922 0993 1063 1134 1204 1274 1344 7 14 21 28 35 42 49 56 63
51 7.1414 1484 1554 1624 1694 1764 1833 1903 1972 2042 7 14 21 28 35 42 49 56 63
52 7.2111 2180 2250 2319 2388 2457 2526 2595 2664 2732 7 14 21 28 35 41 49 55 62
53 7.2801 2870 2938 3007 3075 3144 3212 3280 3348 3417 7 14 21 27 34 41 48 55 62
54 7.3485 3553 3621 3689 3756 3824 3892 3959 4027 4095 7 14 20 27 34 41 47 54 61
.0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 2 3 4 5 6 7 8 9
225
SQUARE ROOTS OF NUMBERS x √x : (10 ≤ x < 100)
Add mean dierences
x .0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 2 3 4 5 6 7 8 99
55 7.4162 4229 4297 4364 4431 4498 4565 4632 4699 4766 7 13 20 27 34 40 47 54 60
56 7.4833 4900 4967 5033 5100 5166 5233 5299 5366 5432 7 13 20 27 33 40 47 53 60
57 7.5498 5565 5631 5697 5763 5829 5895 5961 6026 6092 7 13 20 26 33 40 46 53 59
58 7.6158 6223 6289 6354 6420 6485 6551 6616 6681 6746 7 13 20 26 33 39 46 52 59
59 7.6811 6877 6942 7006 7071 7136 7201 7266 7330 7395 6 13 19 26 32 39 45 52 58
60 7.7460 7524 7589 7653 7717 7782 7846 7910 7974 8039 6 13 19 26 32 39 45 51 58
61 7.8102 8166 8230 8294 8358 9422 8486 8549 8613 8677 6 13 19 26 32 38 45 51 57
62 7.9740 8804 8867 8930 8994 9057 9120 9183 9246 9310 6 13 19 25 32 38 44 51 57
63 7.9373 9436 9498 9561 9624 9687 9750 9812 9875 9937 6 13 19 25 31 38 44 50 56
64 8.0000 0062 0125 0187 0250 0312 0374 0436 0498 0561 6 12 19 25 31 37 44 50 56
65 8.0623 0685 0747 0808 0870 0932 0994 1056 1117 1179 6 12 19 25 31 37 43 49 56
66 8.1240 1302 1363 1425 1486 1548 1609 1670 1731 1792 6 12 18 25 31 37 43 49 55
67 8.1854 1915 1976 2037 2098 2158 2219 2280 2341 2401 6 12 18 24 30 37 43 49 55
68 8.2462 2523 2583 2644 2704 2765 2825 2885 2946 3006 6 12 18 24 30 36 42 48 54
69 8.3066 3126 3187 3247 3307 3367 3427 3487 3546 3606 6 12 18 24 30 36 42 48 54
70 8.3666 3726 3785 3845 3905 3964 4024 4083 4143 4202 6 12 18 24 30 36 42 48 54
71 8.4261 4321 4380 4439 4499 4558 4617 4676 4735 4794 6 12 18 24 30 35 41 47 53
72 8.4853 4912 4971 5029 5088 5147 5206 5264 5323 5381 6 12 18 23 29 35 41 47 53
73 8.5440 5499 5557 5615 5674 5732 5790 5849 5907 5965 6 12 18 23 29 35 41 47 53
74 8.6023 6081 6139 6197 6255 6313 6371 6429 6487 6545 6 12 17 23 29 35 41 46 52
75 8.6603 6660 6718 6776 6833 6891 6948 7006 7063 7121 6 12 17 23 29 35 40 46 52
76 8.7178 7235 7293 7350 7407 7464 7521 7579 7636 7693 6 11 17 23 29 34 40 46 51
77 8.7750 7807 7864 7920 7977 8034 8091 8148 8204 8261 6 11 17 23 28 34 40 45 51
78 8.8318 8374 8431 8487 8544 8600 8657 8713 8769 8826 6 11 17 23 28 34 40 45 51
79 8.8882 8938 8994 9051 9107 9163 9219 9275 9331 9387 6 11 17 22 28 34 39 45 50
80 8.9443 9499 9554 9610 9666 9722 9778 9833 9889 9944 6 11 17 22 28 33 39 45 50
81 9.0000 0056 0111 0167 0222 O277 0333 0388 0443 0499 6 11 17 22 28 33 39 44 50
82 9.0554 0609 0664 0719 0774 0830 0885 0940 0995 1049 6 11 17 22 28 33 39 44 50
83 9.1104 1159 1214 1269 1324 1378 1433 1488 1542 1597 5 11 16 22 27 33 38 44 49
84 9.1652 1706 1761 1815 1869 1924 1978 2033 2087 2141 5 11 16 22 27 33 38 44 49
85 9.2195 2250 2304 2358 2412 2466 2520 2574 2628 2682 5 11 16 22 27 32 38 43 49
86 9.2736 2790 2844 2898 2952 3005 3059 3113 3167 3220 5 11 16 22 27 32 38 43 48
87 9.3274 3327 3381 3434 3488 3541 3595 3648 3702 3755 5 11 16 21 27 32 37 43 48
88 9.3808 3862 3915 3968 4021 4074 4128 4181 4234 4287 5 11 16 21 27 32 37 43 48
89 9.4340 4393 4446 4499 4552 4604 4657 4710 4763 4816 5 11 16 21 26 32 37 42 48
90 9.4868 4921 4974 5026 5079 5131 5184 5237 5289 5341 5 11 16 21 26 32 37 42 47
91 9.5394 5446 5499 5551 5603 5656 5708 5760 5812 5864 5 10 16 21 26 31 37 42 47
92 9.5917 5969 6021 6073 6125 6177 6229 6281 6333 6385 5 10 16 21 26 31 36 42 47
93 9.6437 6488 6540 6592 6644 6695 6747 6799 6850 6902 5 10 16 21 26 31 36 41 47
94 9.6954 7005 7057 7108 7160 7211 7263 7314 7365 7417 5 10 15 21 26 31 36 41 46
95 9.7468 7519 7570 7622 7673 7724 7775 7826 7877 7929 5 10 15 20 26 31 36 41 46
96 9.7980 8031 8082 8133 8184 8234 8285 8336 8387 8438 5 10 15 20 25 31 36 41 46
97 9.8489 8539 8590 8641 8691 8742 8793 8843 8894 8944 5 10 15 20 25 30 35 41 46
98 9.8995 9045 9096 9146 9197 9247 9298 9348 9398 9448 5 10 15 20 25 30 35 40 45
99 9.9499 9549 9599 9649 9700 9750 9800 9850 9900 9950 5 10 15 20 25 30 35 40 45
0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 2 3 4 5 6 7 8 9
226
CUBES OF NUMBERS x → x
3
Add Mean Dierences
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0 1.000 1.030 1.061 1.093 1.125 1.158 1.191 1.225 1.260 1.295 3 7 10 13 16 20 23 26 30
1.1 1.331 1.368 1.405 1.443 1.482 1.521 1.561 1.602 1.643 1.685 4 8 12 16 20 24 28 31 35
1.2 1.728 1.772 1.816 1.861 1.907 1.953 2000 2.048 2.097 2.147 5 9 14 19 23 28 33 37 42
1.3 2.197 2.248 2.300 2.353 2.406 2.460 2.515 2.571 2.628 2.686 5 11 16 22 27 33 38 43 49
1.4 2.744 2.803 2.863 2.924 2.986 3.049 3.112 3.177 3.242 3.308 6 13 19 25 31 38 44 50 56
1.5 3.375 3.443 3.512 3.582 3.652 3.724 3.796 3.870 3.944 4.020 7 14 21 29 36 43 50 57 64
1.6 4.096 4.173 4.252 4.331 4.411 4.492 4.574 4.657 4.742 4.827 8 16 24 32 41 49 57 65 73
1.7 4.913 5.000 5.088 5.178 5.268 5.359 5.452 5.545 5.640 5.735 9 18 27 37 46 55 64 73 82
1.8 5.832 5.930 6.029 6.128 6.230 6.332 6.435 6.539 6.645 6.751 10 20 31 41 51 61 71 82 92
1.9 6.859 6.968 7.078 7.189 7.301 7.415 7.530 7.645 7.762 7.881 11 23 34 45 57 68 79 91 102
2.0 8.000 8.121 8.242 8.365 8.490 8.615 8.742 8.870 8.999 9.129 13 25 38 50 63 75 88 100 113
2.1 9.261 9.394 9.528 9.664 9.800 9.938 10.078 10.218 10.360 10.503 14 28 41 55 69 83 97 110 124
2.2 10.648 10.794 10.941 11.090 11.239 11.391 11.543 11.697 11.852 12.009 15 30 45 60 76 91 106 121 136
2.3 11.167 12.326 12.487 11.649 12.813 12.978 13.144 13.312 13.481 13.652 16 33 49 66 82 99 115 132 148
2.4 13.824 13.998 14.172 14.349 14.527 14.706 14.887 15.069 15.253 15.438 18 36 54 72 90 108 126 143 161
2.5 15.625 15.813 16.003 16.194 16.387 16.581 16.777 16.975 17.174 17.374 19 39 58 78 97 117 136 155 175
2.6 17.576 17.780 17.985 18.191 18.400 18.610 18.821 19.034 19.249 19.465 21 42 63 84 105 126 147 168 189
2.7 19.683 19.903 20.124 20.346 20.571 20.797 21.025 21.254 21.485 21.718 23 45 68 90 113 136 158 181 203
2.8 21.952 21.188 22.426 22.665 22.906 23.149 23.394 23.640 23.888 24.138 24 49 73 97 121 146 170 194 219
2.9 24.389 24.642 24.897 25.154 25.412 25.672 25.934 26.198 26.464 26.731 26 52 78 104 130 156 182 208 234
3.0 27.000 27.271 27.544 27.818 28.094 28.373 28.653 28.934 29.218 29.504 28 56 83 111 139 167 195 223 250
3.1 29.791 30.080 30.371 30.664 30.959 31.256 31.554 31.855 32.157 32.462 30 59 89 119 148 178 208 237 267
3.2 32.768 33.076 33.386 33.698 34.012 34.328 34.646 34.966 35.288 35.611 32 63 95 126 158 190 221 253 284
3.3 35.937 36.265 36.594 36.926 37.260 37.595 37.933 38.273 38.614 38.958 34 67 101 134 168 201 235 269 302
3.4 39.304 39.652 40.002 40.354 40.708 41.064 41.422 41.782 42.144 42.509 36 71 107 142 178 214 249 285 320
3.5 42.88 43.24 43.61 43.99 44.36 44.74 45.12 45.50 45.88 46.27 4 8 11 15 19 23 26 30 34
3.6 46.66 47.05 47.44 47.83 48.23 48.63 49.03 49.43 49.84 50.24 4 8 12 16 20 24 28 32 36
3.7 50.65 51.06 51.48 51.90 52.31 52.73 53.16 53.58 54.01 54.44 4 8 13 17 21 25 29 34 38
3.8 54.87 55.31 55.74 56.18 56.62 57.07 57.51 57.96 58.41 58.86 4 9 13 18 22 27 31 35 40
3.9 59.32 59.78 60.24 60.70 61.16 61.63 62.10 62.57 63.04 63.52 5 9 14 19 23 28 33 37 42
4.0 64.00 64.48 64.96 65.45 65.94 66.43 66.92 67.42 67.92 68.42 5 10 15 20 25 29 34 39 44
4.1 68.92 69.43 69.93 70.44 70.96 71.47 71.99 72.51 73.03 73.56 5 10 15 21 26 31 36 41 46
4.2 74.09 74.62 75.15 75.69 76.23 76.77 77.31 77.85 78.40 78.95 5 11 16 22 27 32 38 43 49
4.3 79.51 80.06 80.62 81.18 81.75 82.31 82.88 83.45 84.03 84.60 6 11 17 23 28 34 40 45 51
4.4 85.18 85.77 86.35 86.94 87.53 88.12 88.72 89.31 89.92 90.52 6 12 18 24 30 36 41 47 53
4.5 91.13 91.73 92.35 92.96 93.58 94.20 94.82 95.44 96.07 96.70 6 12 19 25 31 37 43 50 56
4.6 97.34 97.97 98.61 99.25 99.90 100.54 101.19 101.85 102.50 103.16 6 13 19 26 32 39 45 52 58
4.7 103.82 104.49 105.15 105.82 106.50 107.17 107.85 108.53 109.22 109.90 7 14 20 27 34 41 47 54 61
4.8 110.59 111.28 111.98 112.68 113.38 114.08 114.79 115.50 116.21 116.93 7 14 21 28 35 42 49 56 63
4.9 117.65 118.37 119.10 119.82 120.55 121.29 122.02 122.76 123.51 124.25 7 15 22 29 37 44 51 59 66
5.0 125.00 125.75 126.51 127.26 128.02 128.79 129.55 130.32 131.10 131.87 8 15 23 31 38 46 53 61 69
5.1 132.65 133.43 134.22 135.01 135.80 136.59 137.39 138.19 138.99 139.80 8 16 24 32 40 48 56 64 71
5.2 140.61 141.42 142.24 143.06 143.88 144.70 145.53 146.36 147.20 148.04 8 17 25 33 41 50 58 66 74
5.3 148.88 149.72 150.57 151.42 152.27 153.13 153.99 154.85 155.72 156.59 9 17 26 34 43 51 60 69 77
5.4 157.46 158.34 159.22 160.10 160.99 161.88 167-77 163.67 164.57 165.47 9 18 27 36 44 53 62 71 80
0 1 2 3 4 5 6 7 8 9 1 1 3 4 5 6 7 8 9
227
CUBES OF NUMBERS x → x
3
Add Mean Dierences
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
5.5 166.4 167.3 168.2 169.1 170.0 171.0 171.9 172.8 173.7 174.7 1 2 3 4 5 6 6 7 8
5.6 175.6 176.6 177.5 178.5 179.4 180.4 181.3 182.3 183.3 184.2 1 2 3 4 5 6 7 8 9
5.7 185.2 196.2 187.1 188.1 189.1 190.1 191.1 192.1 193.1 194.1 1 2 3 4 5 6 7 8 9
5.8 195.1 196.1 197.1 198.2 199.2 200.2 201.2 202.3 203.3 204.3 1 2 3 4 5 6 7 8 9
5.9 205.4 206.4 207.5 208.5 209.6 210.6 211.7 212.8 213.8 214.9 1 2 3 4 5 6 7 8 10
6.0 216.0 217.1 218.2 219.3 220.3 221.4 222.5 223.6 224.8 225.9 1 2 3 4 5 7 9 9 10
6.1 227.0 228.1 229.2 230.3 231.5 232-6 233.7 234.9 236.0 237.2 1 2 3 5 6 7 8 9 10
6.2 238.3 239.5 240.6 241.8 243.0 244.1 245.3 246.5 247.7 248.9 1 2 4 5 6 7 8 9 11
6.3 250.0 251.2 252.4 253.6 254.8 256.0 257.3 258.5 259.7 260.9 1 2 4 5 6 7 8 10 11
6.4 262.1 263.4 264.6 265.8 267.1 268.3 269.6 270.8 272.1 273.4 1 2 4 5 6 7 9 10 11
6.5 274.6 275.9 277.2 278.4 279.7 281.0 282.3 283.6 284.9 286.2 1 3 4 5 6 9 9 10 12
6.6 297.5 288.8 290.1 291.4 292.8 294.1 295.4 296.7 298.1 299.4 1 3 4 5 7 8 9 11 12
6.7 300.8 302.1 303.5 304.8 306.2 307.5 308.9 310.3 311.7 313.0 1 3 4 5 7 8 10 11 12
6.8 314.4 315.8 317.2 318.6 320.0 321.4 322.8 324.2 325.7 327.1 1 3 4 6 7 8 10 11 13
6.9 328.5 329.9 331.4 332.8 334.3 335.7 337.2 338.6 340.1 341.5 1 3 4 6 7 9 10 12 13
7.0 343.0 344.5 345.9 347.4 348.9 350.4 351.9 353.4 354.9 356.4 1 3 4 6 7 9 10 12 13
7.1 357.9 359.4 360.9 362.5 364.0 365.5 367.1 368.6 370.1 371.7 2 3 5 6 8 9 11 12 14
7.2 373.2 374.8 376.4 377.9 379.5 381.1 382.7 384.2 385.8 387.4 2 3 5 6 8 9 11 13 14
7.3 389.0 390.6 392.2 393.8 395.4 397.1 398.7 400.3 401.9 403.6 2 3 5 6 8 10 11 13 15
7.4 405.2 406.9 408.5 410.2 411.8 413.5 415.2 416.8 418.5 420.2 2 3 5 7 8 10 12 13 15
7.5 421.9 423.6 425.3 427.0 428.7 430.4 432.1 433.8 435.5 437.2 2 3 5 7 9 10 12 14 15
7.6 439.0 440.7 442.5 444.2 445.9 447.7 449.5 451.2 453.0 454.8 2 4 5 7 9 11 12 14 16
7.7 456.5 458.3 460.1 461.9 463.7 465.5 467.3 469.1 470.9 472.7 2 4 5 7 9 11 13 14 16
7.8 474.6 476.4 478.2 480.0 481.9 483.7 485.6 487.4 489.3 491.2 2 4 6 7 9 11 13 15 17
7.9 493.0 494.9 496.8 498.7 500.6 502.5 504.4 506.3 508.2 510.1 2 4 6 8 9 11 13 15 17
8.0 512.0 513.9 515.8 517.8 519.7 521.7 523.6 525.6 527.5 529.5 2 4 6 8 10 12 14 16 17
8.1 531.4 533.4 535.4 537.4 539.4 541.3 543.3 545.3 547.3 549.4 2 4 6 8 10 12 14 16 19
8.2 551.4 553.4 555.4 557.4 559.5 561.5 563.6 565.6 567.7 569.7 2 4 6 8 10 12 14 16 18
8.3 571.8 573.9 575.9 578.0 580.1 582.2 584.3 586.4 588.5 590.6 2 4 6 8 10 13 15 17 19
8.4 592.7 594.8 596.9 599.1 601.2 603.4 605.5 607.6 609.8 612.0 2 4 6 9 11 13 15 17 19
8.5 614.1 616.3 618.5 620.7 622.8 625.0 627.2 629.4 631.6 633.8 2 4 7 9 11 13 15 18 20
8.6 636.1 638.3 640.5 642.7 645.0 647.2 649.5 651.7 654.0 656.2 2 4 7 9 11 13 16 18 20
8.7 658.5 660.8 663.1 665.3 667.6 669.9 672.2 674.5 676.8 679.2 2 5 7 9 11 14 16 18 21
8.8 681.5 683.8 686.1 688.5 690.8 693.2 695.5 697.9 700.2 702.6 2 5 7 9 12 14 16 19 21
8.9 705.0 707.3 709.7 712.1 714.5 716.9 719.3 721.7 724.2 726.6 2 5 7 10 12 14 17 19 22
9.0 729.0 731.4 733.9 736.3 738.8 741.2 743.7 746.1 748.6 751.1 2 5 7 10 12 15 17 20 22
9.1 753.6 756.1 758.6 761.0 763.6 766.1 768.6 771.1 773.6 776.2 3 5 8 10 13 15 18 20 23
9.2 778.7 781.2 783.8 786.3 788.9 791.5 794.0 796.6 799.2 801.8 3 5 8 10 13 15 18 21 23
9.3 804.4 807.0 809.6 812.2 814.8 817.4 820.0 822.7 825.3 827.9 3 5 8 10 13 16 18 21 24
9.4 830.6 833.2 835.9 838.6 841.2 843.9 846.6 849.3 852.0 854.7 3 5 8 11 13 16 19 21 24
9.5 857.4 860.1 862.8 865.5 868.3 871.0 873.7 876.5 879.2 882.0 3 5 8 11 14 16 19 22 25
9.6 884.7 887.5 890.3 893.1 895.8 898.6 901.4 904.2 907.0 909.9 3 6 8 11 14 17 20 22 25
9.7 912.7 915.5 918.3 921.2 924.0 926.9 929.7 932.6 935.4 938.3 3 6 9 11 14 17 20 23 26
9.8 941.2 944.1 947.0 949.9 952.8 955.7 958.6 961.5 964.4 967.4 3 6 9 12 15 17 20 23 26
9.9 970.3 973.2 976.2 979.1 982.1 985.1 988.0 991.0 994.0 997.0 3 6 9 12 15 18 21 24 27
0 1 2 3 4 5 6 7 8 9 1 1 3 4 5 6 7 8 9
228
RECIPROCAL OF NUMBERS x 
Subtract mean dierences
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0 1.0000 9901 9804 9709 9615 9524 9434 9346 9259 9174 9 18 28 37 46 55 64 73 83
1.1 0.9091 9009 8929 8850 8772 8696 8621 8547 8475 8403 8 15 23 31 38 46 53 61 69
1.2 0.8333 8264 8197 8130 8065 8000 7937 7874 7813 7752 6 13 19 26 32 39 45 52 58
1.3 0.7692 7634 7576 7519 7463 7407 7353 7299 7246 7194 6 11 17 22 28 33 39 44 50
1.4 0.7143 7092 7042 6993 6944 6897 6849 6803 6757 6711 5 10 14 19 24 29 34 38 43
1.5 0.6667 6623 6579 6536 6494 6452 6410 6369 6329 6289 4 8 13 17 21 25 29 34 38
1.6 0.6250 6211 6173 6135 6098 6061 6024 5988 5952 5917 4 7 11 15 18 22 26 30 33
1.7 0.5882 5848 5814 5780 5747 5714 5682 5650 5618 5587 3 7 10 13 16 20 23 26 30
1.8 0.5556 5525 5495 5464 5435 5405 5376 5348 5319 5291 3 6 9 12 15 18 21 24 26
1.9 0.5263 5236 5208 5181 5155 5128 5102 5076 5051 5025 3 5 8 11 13 16 19 21 24
2.0 0.5000 4975 4950 4926 4902 4878 4854 4831 4808 4785 2 5 7 10 12 14 17 19 22
2.1 0.4762 4739 4717 4695 4673 4651 4630 4608 4587 4566 2 4 7 9 11 13 15 17 20
2.2 0.4545 4525 4505 4484 4464 4444 4425 4405 4386 4367 2 4 6 8 10 12 14 16 18
2.3 0.4348 4329 4310 4292 4274 4255 4237 4219 4202 4184 2 4 5 7 9 11 13 15 16
2.4 0.4167 4149 4132 4115 4098 4082 4065 4049 4032 4016 2 3 5 7 8 10 12 13 15
2.5 0.4000 3984 3968 3953 3937 3922 3906 3891 3876 3861 2 3 5 6 8 9 11 12 14
2.6 0.3846 3831 3817 3802 3788 3774 3759 3745 3731 3717 1 3 4 6 7 9 10 11 13
2.7 0.3704 3690 3676 3663 3650 3636 3623 3610 3597 3584 1 3 4 5 7 8 9 11 12
2.8 0.3571 3559 3546 3534 3521 3509 3497 3484 3472 3460 1 2 4 5 6 7 9 10 11
2.9 0.3448 3436 3425 3413 3401 3390 3378 3367 3356 3.344 1 2 3 5 6 7 8 9 10
3.0 0.3333 3322 3311 3300 3289 3279 3268 3257 3247 3236 1 2 3 4 5 6 8 9 10
3.1 0.3226 3215 3205 3195 3185 3175 3165 3155 3145 3135 1 2 3 4 5 6 7 8 9
3.2 0.3125 3115 3106 3096 3086 3077 3067 3058 3049 3040 1 2 3 4 5 6 7 8 9
3.3 0.3030 3021 3012 3003 2994 2985 2976 2967 2959 2950 1 2 3 4 4 5 6 7 8
3.4 0.2941 2933 2924 2915 2907 2899 2890 2882 2874 2865 1 2 3 3 4 5 6 7 8
3.5 0.2857 2849 2841 2833 2825 2817 2809 2801 2793 2786 1 2 2 3 4 5 6 6 7
3.6 0.2778 2770 2762 2755 2747 2740 2732 2725 2717 2710 1 2 2 3 4 5 5 6 7
3.7 0.2703 2695 2688 2681 2674 2667 2660 2653 2646 2639 1 1 2 3 4 4 5 6 6
3.8 0.2632 2625 2618 2611 2604 2597 2591 2584 2577 2571 1 1 2 3 3 4 5 5 6
3.9 0.2564 2558 2551 2545 2538 2532 2525 2519 2513 2506 1 1 2 3 3 4 4 5 6
4.0 0.2500 2494 2488 2481 2475 2469 2463 2457 2451 2445 1 1 2 2 3 4 4 5 6
4.1 0.2439 2433 2427 2421 2415 2410 2404 2398 2392 2387 1 1 2 2 3 3 4 5 5
4.2 0.2381 2375 2370 2364 2358 2353 2347 2342 2336 2331 1 1 2 2 3 3 4 4 5
4.3 0.2326 2320 2315 2309 2304 2299 2294 2288 2283 2278 1 1 2 2 3 3 4 4 5
4.4 0.2273 2268 2262 2257 2252 2247 2242 2237 2232 2227 1 1 2 2 3 3 4 4 5
4.5 0.2222 2217 2212 2208 2203 2198 2193 2188 2183 2179 0 1 1 2 2 3 3 4 4
4.6 0.2174 2169 2165 2160 2155 2151 2146 2141 2137 2132 0 1 1 2 2 3 3 4 4
4.7 0.2128 2123 2119 2114 2110 2105 2101 2096 2092 2088 0 1 1 2 2 3 3 4 4
4.8 0.2083 2079 2075 2070 2066 2062 2058 2053 2049 2045 0 1 1 2 2 3 3 3 4
4.9 0.2041 2037 2033 2028 2024 2020 2016 2012 2008 2004 0 1 1 2 2 2 3 3 4
5.0 0.2000 1996 1992 1988 1984 1980 1976 1972 1969 1965 0 1 1 2 2 2 3 3 4
5.1 0.1961 1957 1953 1949 1946 1942 1938 1934 1931 1927 0 1 1 2 2 2 3 3 3
5.2 0.1923 1919 1916 1912 1908 1905 1901 1898 1894 1890 0 1 1 1 2 2 3 3 3
5.3 0.1887 1883 1880 1876 1973 1969 1866 1862 1859 1855 0 1 1 1 2 2 2 3 3
5.4 0.1852 1848 1845 1842 1938 1835 1832 1828 1825 1821 0 1 1 1 2 2 2 3 3
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1
x
229
RECIPROCAL OF NUMBERS x 
Subtract mean dierences
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
5.5 0.1818 1815 1812 1808 1805 1802 1799 1795 1792 1789 0 1 1 1 2 2 2 3 3
5.6 0.1786 1783 1779. 1776 1773 1770 1767 1764 1761 1757 0 1 1 1 2 2 2 3 3
5.7 0.1754 1751 1748 1745 1742 1739 1736 1733 1730 1727 0 1 1 1 2 2 2 2 3
5.8 0.1724 1721 1718 1715 1712 1709 1706 1704 1701 1698 0 1 1 1 1 2 2 2 3
5.9 0.1695 1692 1689 1686 1684 1681 1678 1675 1672 1669 0 1 1 1 1 2 2 2 3
6.0 0.1667 1664 1661 1658 1656 1653 1650 1647 1645 1642 0 1 1 1 1 2 2 2 2
6.1 0.1639 1637 1634 1631 1629 1626 1623 1621 1618 1616 0 1 1 1 1 2 2 2 2
6.2 0.1613 1610 1608 1605 1603 1600 1597 1595 1592 1590 0 1 1 1 1 2 2 2 2
6.3 0.1587 1585 1582 1580 1577 1575 1572 1570 1567 1565 0 0 1 1 1 1 2 2 2
6.4 0.1563 1560 1558 1555 1553 1550 1548 1546 1543 1541 0 0 1 1 1 1 2 2 2
6.5 0.1538 1536 1534 1531 1529 1527 1524 1522 1520 1517 0 0 1 1 1 1 2 2 2
6.6 0.1515 1513 1511 1508 1506 1504 1502 1499 1497 1495 0 0 1 1 1 1 2 2 2
6.7 0.1493 1490 1488 1486 1484 1481 1479 1477 1475 1473 0 0 1 1 1 1 2 2 2
6.8 0.1471 1468 1466 1464 1462 1460 1458 1456 1453 1451 0 0 1 1 1 1 1 2 2
6.9 0.1449 1447 1445 1443 1441 1439 1437 1435 1433 1431 0 0 1 1 1 1 1 2 2
7.0 0.1429 1427 1425 1422 1420 1418 1416 1414 1412 1410 0 0 1 1 1 1 1 2 2
7.1 0.1408 1406 1404 1403 1401 1399 1397 1395 1393 1391 0 0 1 1 1 1 1 2 2
7.2 0.1389 1387 1385 1383 1381 1379 1377 1376 1374 1372 0 0 1 1 1 1 1 2 2
7.3 0.1370 1368 1366 1364 1362 1361 1359 1357 1355 1353 0 0 1 1 1 1 1 1 2
7.4 0.1351 1350 1348 1346 1344 1342 1340 1339 1337 1335 0 0 1 1 1 1 1 1 2
7.5 0.1333 1332 1330 1328 1326 1325 1323 1321 1319 1318 0 0 1 1 1 1 1 1 2
7.6 0.1316 1314 1312 1311 1309 1307 1305 1304 1302 1300 0 0 1 1 1 1 1 1 2
7.7 0.1299 1297 1295 1294 1292 1290 1289 1287 1285 1284 0 0 1 1 1 1 1 1 2
7.8 0.1282 1280 1279 1277 1276 1274 1272 1271 1269 1267 0 0 0 1 1 1 1 1 1
7.9 0.1266 1264 1263 1261 1259 1258 1256 1255 1253 1252 0 0 0 1 1 1 1 1 1
8.0 0.1250 1248 1247 1245 1244 1242 1241 1239 1238 1236 0 0 0 1 1 1 1 1 1
8.1 0.1235 1233 1232 1230 1229 1227 1225 1224 1222 1221 0 0 0 1 1 1 1 1 1
8.2 0.1220 1218 1217 1215 1214 1212 1211 1209 1208 1206 0 0 0 1 1 1 1 1 1
8.3 0.1205 1203 1202 1200 1199 1198 1196 1195 1193 1192 0 0 0 1 1 1 1 1 1
8.4 0.1190 1189 1188 1186 1185 1183 1182 1181 1179 1178 0 0 0 1 1 1 1 1 1
8.5 0.1176 1175 1174 1172 1171 1170 1168 1167 1166 1164 0 0 0 1 1 1 1 1 1
8.6 0.1163 1161 1160 1159 1157 1156 1155 1153 1152 1151 0 0 0 1 1 1 1 1 1
8.7 0.1149 1148 1147 1145 1144 1143 1142 1140 1139 1138 0 0 0 1 1 1 1 1 1
8.8 0.1136 1135 1134 1133 1131 1130 1129 1127 1126 1125 0 0 0 1 1 1 1 1 1
8.9 0.1124 1122 1121 1120 1119 1117 1116 1115 1114 1112 0 0 0 0 1 1 1 1 1
9.0 0.1111 1110 1109 1107 1106 1105 1104 1103 1101 1100 0 0 0 0 1 1 1 1 1
9.1 0.1099 1098 1096 1095 1094 1093 1092 1091 1089 1088 0 0 0 0 1 1 1 1 1
9.2 0.1087 1086 1085 1083 1082 1081 1080 1079 1078 1076 0 0 0 0 1 1 1 1 1
9.3 0.1075 1074 1073 1072 1071 1070 1068 1067 1066 1065 0 0 0 0 1 1 1 1 1
9.4 0.1064 1063 1062 1060 1059 1058 1057 1056 1055 1054 0 0 0 0 1 1 1 1 1
9.5 0.1053 1052 1050 1049 1048 1047 1046 1045 1044 1043 0 0 0 0 1 1 1 1 1
9.6 0.1042 1041 1040 1038 1037 1036 1035 1034 1033 1032 0 0 0 0 1 1 1 1 1
9.7 0.1031 1030 1029 1028 1027 1026 1025 1024 1022 1021 0 0 0 0 1 1 1 1 1
9.8 0.1020 1019 1018 1017 1016 1015 1014 1013 1012 1011 0 0 0 0 1 1 1 1 1
9.9 0.1010 1009 1008 1007 1006 1005 1004 1003 1002 1001 0 0 0 0 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1
x