i
Advancing in
Mathematics
Student’s Book
Form 4
Kinyua Mugo
Lucy Maina
Jared Ondera
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 Road
P.O. Box 1237
Dar es Salaam, Tanzania
© Kinyua Mugo, Lucy Maina, Jared Ondera, 2005
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 2005
Reprinted 2005, 2006
Revised 2008
ISBN 978 9966 49 585 1
Typsetting, design and graphics by Justice Mogaki
Printed by Printwell Ltd., Road ‘A’ o Enterprise Road,
Industrial Area. P.O. Box 5216-00506, Nairobi, Kenya.
iii
Contents
1. Matrices and transformations....... 1
Transformation....................................... 1
Rotation ............................................ 2
Enlargement........................................ 2
Matrices ................................................. 3
Scalar multiplication of matrices........ 3
Multiplication of matrices .................. 3
Inverse and determinant of a matrix ... 4
Matrix of a transformation..................... 5
Identifying the transformation for a
given matrix........................................ 5
Finding the matrix of a given
transformation........................................ 7
Finding the transformation matrix by
calculation .......................................... 7
Finding the transformation matrix
using the unit square ........................... 8
Other transformations ............................ 10
The general rotation............................ 10
Stretch................................................. 11
Two-way stretch.................................. 11
Shear ................................................... 12
Identity transformation ....................... 13
Inverse transformations ......................... 14
Successive transformations.................... 15
Area scale factor and determinant ......... 17
Determinant and area.......................... 17
Isometry .............................................. 18
Application of transformations to
real life situations................................... 19
2. Statistics (2)..................................... 22
Measures of central tendency ................ 22
The mean ............................................... 22
Assumed mean method....................... 22
Scaling down the deviations from the
assumed mean..................................... 25
The median ............................................ 27
Calculation method............................. 27
Cumulative frequency curve (ogive) ..... 28
Median from an ogive......................... 29
Quantiles............................................. 29
Measures of dispersion .......................... 32
Range .................................................. 32
Interquartile range............................... 32
Mean deviation (M.D.) ....................... 32
Variance and standard deviation......... 34
Other methods of nding the
standard deviation............................... 35
Mean and standard deviation of
a distribution with uneven class
intervals .............................................. 37
3. Loci .................................................. 40
Denition of locus ................................. 40
Common types of loci............................ 40
Constant radius locus.......................... 40
Constant distance from a xed line .... 40
The perpendicular bisector locus........ 41
Angle bisector locus ........................... 41
Constant angle locus .............................. 42
90° angle locus.................................... 42
Non-90° angle locus ........................... 43
Intersecting loci ..................................... 47
Loci involving inequalities ................. 49
4. Trigonometry (3)............................. 51
Basic trigonometric ratios...................... 51
The relation cos
2
θ
+ sin
2
θ
= 1 ............... 53
Deriving the relation
cos
2
θ
+ sin
2
θ
= 1 using a unit circle .... 53
Deriving the relation cos
2
θ
+ sin
2
θ
= 1
using a right-angled triangle ............... 54
Proving trigonometric identities using
the relation cos
2
θ
+ sin
2
θ
= 1 ............. 54
Solving trigonometric equations............ 56
Analytical solution.............................. 56
Simple trigonometric graphs .............. 58
The sine graph .................................... 59
The cosine graph................................. 59
Variations of the sine and cosine
curves.................................................. ..59
Graphs of y = tan x and its
variations ............................................ 63
iv
Graphical solution of trigonometric
equations................................................ 65
Revision exercises 1 (Chapters 1–4) ... 68
Revision exercise 1.1 ............................. 68
Revision exercise 1.2 ............................. 69
Revision exercise 1.3 ............................. 70
5. Three dimensional geometry ......... 72
Geometrical properties of common
solids...................................................... 72
Points, lines and planes....................... 72
Skew lines........................................... 73
Determination of a plane .................... 73
Projections and angles ........................... 74
Projections .......................................... 74
Angle between two lines (including
skew lines) .......................................... 74
Angle between a line and a plane ....... 75
Angle between two planes.................. 76
Calculating lengths and angles in solids 77
6. Latitudes and longitudes................ 81
The earth ................................................ 81
Great circles and small circles ............... 81
Latitude and longitude ........................... 82
Locating a point on the surface of
the earth ................................................. 83
Distance along a great circle.................. 86
The nautical mile ................................ 86
Distance along the equator ................. 87
Distance along a meridian .................. 87
Radius of, and distance along, a
small circle.......................................... 88
Longitude and time................................ 91
Speed in knots and kilometres per hour. 92
7. Linear programming...................... 94
Inequalities............................................. 94
Analytical solution of linear
inequalities.......................................... 94
Representing simultaneous linear
inequalities on a number line.............. 94
Graphical solution of linear
inequalities.......................................... 95
Forming linear inequalities .................... 96
Maximising or minimising a function ... 99
Optimisation ..........................................100
Revision exercises 2 (Chapters 5–7) .. 104
Revision exercise 2.1 .............................104
Revision exercise 2.2 .............................105
Revision exercise 2.3 .............................106
8. Dierentiation ................................ 109
Gradient, average and instantaneous
rate of change.........................................109
Gradient of a curve at a point ................110
Gradient function................................ 111
Gradient function of a polynomial...... 112
Delta (δ) notation and the dierential
coecient............................................... 114
Dierentiation by rule.........................116
Equations of tangent and normal to
a curve....................................................117
Stationary points .................................... 118
Curve sketching .....................................120
Application of dierentiation ................121
Velocity and acceleration.................... 121
Maxima and minima..............................123
9. Area approximation ...................... 125
Approximation of areas of irregular
shapes.....................................................125
Area by counting squares ...................125
Area by comparing masses .................126
Area by triangulation .............................127
Finding area under a curve using
trapezium rule ........................................128
Deriving the trapezia..............................129
Using trapezium rule..............................130
Deriving the mid-ordinate rule ..............132
Using the mid-ordinate rule ...................133
10. Integration...................................... 135
Dierentiation........................................ 135
Reverse dierentiation........................... 135
Integration by rule ..............................136
Finding the arbitrary constant............. 137
Area under a curve................................. 138
Denite and indenite integrals.............140
The integral notation..............................142
v
Application of integration in
kinematics ..............................................145
Revision exercises 3 (Chapters 8–10)... 148
Revision exercise 3.1 .............................148
Revision exercise 3.2 .............................149
Revision exercise 3.3 .............................149
11. Model examination papers ........... 151
Examination practice .............................151
Set 1: Model 1.......................................152
Model 2.......................................155
Set 2: Model 1.......................................158
Model 2.......................................161
Set 3: Model 1.......................................164
Model 2.......................................167
Solutions to model examination
papers — Set 1
Model 1...............................................170
Model 2...............................................181
Mathematical formulae and tables......193
vi
1
Transformation
In Book 2, we learnt about the transformations
of translation, reection, rotation and
enlargement. In this chapter, we shall develop
those ideas further and introduce some two
more transformations shear and stretch.
However, we need to rst recall some important
properties of the above transformations.
Translation
We learnt that a translation is a displacement
of an object, without turning it, in a specied
direction. For example, ∆ABC in Fig. 1.1 is
translated to ∆PQR.
Fig. 1.1
Vector AP (=BQ = CR) is a translation
vector. Written as a column vector.
Any translation in the Cartesian plane can
be written as a column vector, i.e. a column
matrix , where x represents a movement
of
x units parallel to the x-axis and y represents
a movement of y units parallel to the y-axis.
Movements to the right and movements
upwards are defined as positive while
movement to the left and movements
downwards are dened as negative.
Under translation, the object and the image
are directly congruent.
Reection
If a point A′ is the image of a point A under a
reection in the mirror line m, then A and A′ are
equidistant from m (Fig. 1.2).
Fig. 1.2
Thus, m is the perpendicular bisector of AA′.
The following are the general properties of
reection.
1. Points which are on the mirror line are
their own images, i.e. they are invariant.
2. The distance of the image from the mirror
line is equal to the distance of the object
from the mirror line.
3. The mirror line m bisects the angle
between any line and its image.
4. Reection preserves lengths and angles
but reverses direction, i.e. an object and
its image are oppositely congruent.
x
y
( )
– – – – – – – – – – – – – – – – –
A′
=
A
•
•
=
m
6
5
4
3
2
0
1 2 3 4 5 6 7
x
y
B
Q
R
P
A
C
1
MATRICES AND TRANSFORMATIONS
1
2
Enlargement
As we learnt, an enlargement is a
transformation which results in an image such
that:
1. All its lengths and the corresponding
lengths on the object bear a constant ratio,
known as the scale factor;
2. Its angles are equal to the corresponding
angles on the object.
In other words, an object and its image are
similar.
Recall, particularly, the following general
properties of enlargement.
1. An object point, its image and the centre
of enlargement are collinear.
2. For any point A on an object, OA′ = k
OA,
where k is the scale factor.
(a) If k > 0, (positive enlargment) the
object and its image lie on the same
side of the centre of enlargement.
(b) If k < 0, (negative enlargment) the
object and its image are on opposite
sides of the centre of enlargement.
(c) If k = ±1, the object and its image are
congruent.
3. The centre of enlargement is the only
point that remains xed, irrespetive of the
scale factor.
4. If the linear scale factor of enlargement
is k, the area scale factor is k
2
. Where the
enlargement is of a solid, the volume scale
factor is k
3
.
Note:
Under translation, reection and rotation, size
and shape are always preserved. However,
under enlargement, it is only the shape that is
always preserved except when the scale factor
is ±1. The size varies depending on the scale
factor used for the enlargement.
Rotation
We learnt that rotation, in a plane, takes place
about a given centre and through a given angle.
A rotation is dened as positive when it is
anticlockwise.
A positive rotation through an angle θ is the
same as a negative rotation through an angle
(360° – θ) about the same centre (Fig. 1.3).
Fig. 1.3
ABCD is the object and A′B′C′D′ is the image.
θ is the angle of rotation with (0 , 0) as the
centre of rotation.
Refer to Fig. 1.3 and note the following general
properties of rotation.
1. The image is directly congruent to the
object.
2. The distance of an image point, equals
the distance of the corresponding object
point from the centre of rotation.
3. The centre of rotation is the only xed
point under a rotation. Every other point
moves along an arc of a circle with this
centre.
4. Each line of, or on, the object turns through
an angle equal to the angle of rotation.
5. The centre of rotation is the intersection
of the perpendicular bisectors of the line
segments AA′, BB′, CC′, and DD′.
–
3
–
2
–
1
0
1 2 3 4
x
y
4
3
2
1
– – – – – – –
– – – – – – –
θ
360° – θ
A
D
C
B
A′
D′
C′
B′
3
( )
–
3
4
1
2
–
( )
3 1
4 5
( )
3 1
4 5
( )
6 2
8 10
Exercise 1.1.
1. A point P is translated by vector to
point Q(6 , 5). Find the coordinates of P.
2. Plot the points A(1 ,
–
1), B(3 ,
–
1), C(3 , 1)
and D(1 , 1).
(a) Find the coordinates of A′, B′, C′ and
D′, the images of A, B, C, and D after
reection in the mirror line y = x.
(b) What is the image of ABCD under a
reection in y = 0?
3. A point is translated by vector . It is
then translated by vector .
(a) What single translation is this equivalent
to?
(b) What are the coordinates of the nal
image of the point (4 ,
–
1) under these
translations?
(c) The nal image of point P under these
translations is Q(4 , 4). What are the
coordinates of P?
4. Points A(4 , 0), B(0 , 3) and C(4 , 3) are the
vertices of a triangle. Draw the triangle on
squared paper. Copy and complete Table
1.1
for the coordinates of A′, B′, and C′, the
images of A, B, and C under enlargement
with centre (0 , 0).
Scale factor A′ B′ C′
(a) 3
(b) 1.5
(c)
–
1
(d)
Table 1.1
5. Under a rotation, the images of points
P(
–
1 , 1) and Q(2 , 4) are P′(3 ,
–
4) and
Q′(0 ,
–
1). Plot these points on squared
paper and nd:
(a) The coordinates of the centre of rotation.
(b) The coordinates of R given that R′
is
(
–
2 , 4).
6. A(
–
1 ,
–
1), B(3 ,
–
1), C(3 , 3), D(
–
1 , 3),
E(
–
3 ,
–
3), F(1 ,
–
3), G(1 , 1) and H(
–
3 , 1)
are the vertices of two squares ABCD and
EFGH. Draw the squares on squared paper.
(a) Find the centre and the angle of rotation
that maps
(i) ABCD onto FGHE,
(ii) EFGH onto DABC.
(b) What is the equation of the mirror line
of the reection that maps ABCD onto
GFEH?
(c) What transformation maps EFGH onto
ABCD ?
Matrices
In Book 3, we learnt about operations on
matrices. In this chapter, we are going to learn
how to describe and perform transformations
using matrices. To be able to do this, we need
to briey recall the procedure of working with
matrices.
Scalar multiplication of matrices.
Given that k is a scalar and A is a matrix, then
kA means that every element of matrix A is
multiplied by k.
For example, 2 means that every element
in the matrix is multiplied by 2, i.e.
2 = .
Multiplication of matrices
Recall that matrices may be multiplied only if
they are compatible, i.e. the number of columns
in the LHS matrix must be equal to the number
of rows in the RHS matrix. We also know that
matrix multiplication is not commutative except
in the case of a matrix and its inverse or its
identity matrix. Thus, if A and B are matrices,
AB ≠ BA.
( )
–
6
1
( )
–
3
0
4
( )
In general:
Given that A = and B =
,
1. kA = k =
2. AB =
=
3. det A = ad – bc
4. A
–1
=
1
and
ad – bc
5. AA
–
1
= A
–
1
A = .
Exercise 1.2
1. Given that A = and B = ,
evaluate
(a) AB (b) BA (c) B
2. (a) Find the inverse of
.
(b) Find the matrix such that
= .
3. Find the values of the unknowns in each of
the following cases.
(a)
= .
(b) = .
4. Find the value of y given that is a
singular matrix.
5. If the determinant of the matrix is 6,
nd the value of p.
( )
4 3
1
–
2
( )
1 5
4
–
6
R
1
C
1
R
2
C
1
R
1
C
2
R
2
C
2
( )
16 2
–
7 17
( )
4
–
3
1
2
( )
2 3
–
1 4
1 5
4
–
6
( )
( )
4 3
1
–
2
1
11
( )
a b
c d
( )
p q
r s
( )
a b
c d
( )
ak bk
ck dk
( )
a b
c d
( )
p q
r s
( )
d
–
b
–
c a
( )
1 0
0 1
( )
ap + br aq + bs
cp + dr cq + ds
( )
–
6 4
1 5
( )
1 4
–
2
–
3
( )
2 6
2 7
1
3
–
( )
a b
c d
( )
1 0
0 1
( )
x
y
( )
2 0
–
1 1
( )
10
2
( )
b c
–
2 0
( )
4 0
1 a
( )
20 12
–
1 d
( )
1
–
2
y
4
( )
p
3
1
2
( )
2 3
1 3
( )
a b
c d
1
11
( )
4
–
3
1
2
( )
2 3
–
1 4
( )
1 0
0 1
( )
4
–
3
1
2
11 0
0 11
( )
( )
2 3
–
1 4
( )
4
–
3
1
2
( )
2 3
–
1 4
( )
1 0
0 1
11 0
0 11
( )
1
11
1
11
1
11
1
11
( )
Remember that when we are multiplying
matrices, we combine row by column, working
from left to right.
Example 1.1
Find the product AB, given that A =
and
B = .
Solution
AB =
=
,
(where R
1
means row 1
and C
1
means column 1)
= 4 × 1 + 3 × 4 4 × 5 + 3 ×
–
6
1 × 1 +
–
2 × 4 1 × 5 +
–
2 ×
–
6
= .
Inverse and determinant of a matrix
We learnt, in Book 3, that to nd the inverse of
a matrix, we must rst nd its determinant. For
example, if
A = , then
det A = |A| = 4 × 2 – 1 × (
–
3) = 11
∴ The inverse of A is denoted as A
–
1
=
=
We also saw that the product of a matrix and its
inverse is the identity matrix. For example,
AA
–
1
= ·
=
=
=
Similarly, A
–
1
A =
= =
R
1
R
1
C
1
C
2
( )
2
11
-1
11
4
11
3
11
5
8
y
6
4
2
2
0
4 6 8 10 12
x
B
A′
B′
C′
A
C
( )
2
–
3
1
–
4
6. X = , Y = . Show that
(a) (XY)
–
1
= Y
–
1
X
–
1
(b) (XY)
–
1
≠ X
–
1
Y
–
1
.
Matrix of a transformation
We have seen that a translation can be
represented by a column vector, which is a
2 × 1 matrix. Let us now look at the
representation of other transformations by
matrices.
Consider triangle ABC with vertices A(1 , 1),
B(1 , 3) and C(4 , 1). What are the position
vectors of the points A, B and C ?
These position vectors are 2 × 1 matrices and
can be premultiplied by any 2 × 2 matrix.
Consider the eect of premultiplying each of
them by the matrix .
A A′
=
B B′
=
C C′
=
Thus, A(1 , 1) is mapped onto A′(5 , 3).
This may be written as:
A(1 , 1) A′(5 , 3).
Likewise,
B(1 , 3) B′(11 , 7) and
C(4 , 1) C′(11 , 6).
The object and image triangles are shown in
Fig. 1.4.
( )
3 4
1
2
( )
2 3
1 2
( )
2 3
1 2
( )
1
3
( )
11
7
( )
2 3
1 2
( )
4
1
( )
11
6
( )
2 3
1 2
( )
1
1
( )
5
3
( )
2 3
1 2
Fig. 1.4
As we can see from Fig. 1.4, ∆ABC has been
transformed to ∆A′B′C′. It is not easy to
describe the transformation in this particular
case. For the moment, however, it is sucient
to notice that a transformation has occurred.
Such a matrix as is called a
transformation matrix or transformation
operator.
Many geometrical transformations can be
represented by 2 × 2 matrices.
Note that to perform a transformation,
the transformation matrix always acts as
a premultiplier. Remember that matrix
multiplication is not commutative.
Identifying the transformation for a
given matrix
In order to nd the transformation for a given
matrix, we need to choose an object gure
and nd out what happens to it under the
transformation represented by that matrix. A
triangle is quite convenient for this purpose.
Example 1.2
Identify the transformation whose matrix is
(a) (b) (c) (d)
Solution
Let us use ∆ABC with vertices A(1 , 1), B(3 , 1)
and C(3 , 4).
( )
1 0
0
–
1
( )
–
1 0
0 1
( )
0 1
1 0
( )
1 0
0 1
6
Instead of premultiplying each of the position
vectors of A, B, and C separately with the
operator, we can set out the working as follows:
(a) A B C A′ B′ C′
=
(b) A B C A′ B′ C′
=
(c) A B C A′ B′ C′
=
(d) A B C A′ B′ C′
=
The object and image, in each case, are shown in
Fig. 1.5(a)–(d).
(a)
Fig. 1.5
From Fig. 1.5, we see that the matrix:
(a) represents reection in the x-axis.
(b) represents reection in the y-axis.
(c) represents reection in the line
y = x.
–
3
–
2
4
3
2
1
A
C′
B′
A′
C
B
A
A′
B′
(c)
A
A′
(d)
( )
1 0
0
–
1
( )
–
1 0
0 1
( )
0 1
1 0
( )
1 3 3
–
1
–
1
–
4
( )
1 3 3
1 1 4
( )
–
1
–
3
–
3
1 1 4
( )
1 3 3
1 1 4
( )
1 0
0 1
( )
1 3 3
1 1 4
( )
1 3 3
1 1 4
( )
1 0
0
–
1
( )
–
1 0
0 1
( )
0 1
1 0
( )
1 1 4
1 3 3
( )
1 3 3
1 1 4
4
3
2
1
0
–
1
–
2
–
3
–
4
1 2 3 4
x
y
A
C
B
A′
B′
C′
x
1 2 3
0
4
4
3
2
1
y
x
y
1 2 3
–
1
0
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
C
B
C′
x
1 2 3
0
4
4
3
2
1
y
C
B
C′
B′
7
( )
1 0
0 1
Matrix Transformation
Reection in x-axis
Reection in the y-axis
Reection in line y = x
Reection in line y =
–
x
“No transformation”
Positive quarter-turn about
the origin (0, 0)
Negative quarter-turn about
the origin (0, 0)
Half-turn about the origin
Enlargement with centre
origin (0, 0) and scale factor k.
Table 1.2.
Important: From Question 4 of Exercise 1.3,
you should have noticed that the origin is
invariant under all the given transformations.
Indeed, it is a fact that it is only transformations
under which the origin is invariant that can
be represented by 2 × 2 matrices. Thus, if the
transformation is a rotation, it must be about the
origin; if it is a reection, the mirror line must
pass through the origin; etc.
In this section, we shall learn how to nd the
matrix of a given transformation. There are two
possible approaches:
(a) Calculation method, and
(b) Using the unit square.
Finding the transformation matrix by
calculation
Example 1.3
Find the matrix whose transformation is a
reection in the line y = 0.
( )
2 1
3 2
( )
1 2
1 3
( )
0
–
1
–
1 0
( )
0
–
1
1 0
( )
0
1
–
1 0
( )
–
1 0
0
–
1
( )
3 0
0 3
( )
–
3 0
0
–
3
( )
2 0
0 2
1
2
–
( )
0
0
(d) represents a transformation that
leaves the object unchanged.
– Maps the object to its image
– Identity transformation
Exercise 1.3
1. Plot the triangles whose vertices are given
by the following points and the transformed
gures under the given matrices.
(a) A(1 , 1), B(1, 3 ), C(4 , 1) ;
(b) P(2 , 2), Q(3 , 2), R(3 , 6) ;
2. Draw triangle LMN with vertices L(2 , 3),
M(2 , 5) and N(6 , 5) and its image under
the transformation represented by each
of the following matrices. In each case,
describe the transformation fully.
(a) (b)
(c) (d)
3. Using a simple object gure of your choice
nd the transformation given by:
(a) (b)
(c) (d)
4. What are the images of the origin under the
transformations given by the matrices in
Questions 1 to 3?
Finding the matrix of a given
transformation
In the previous section, and in Exercise 1.3,
we found transformations that correspond to
given matrices. Table 1.2 is a summary of these
ndings.
1
2
–
( )
0
–
1
–
1 0
( )
1
0
0 1
( )
0
1
–
1 0
( )
–
1 0
0
–
1
( )
k
0
0 k
( )
1 0
0
–
1
( )
–
1
0
0
1
( )
0
1
1 0
( )
0
–
1
1
0
(In the line y = 0)
(In the line x = 0)
Identity transformation
8
Thus, the transformation matrix is ,
conrming our earlier nding.
Finding the transformation matrix using
the unit square
Fig. 1.7 shows a square with vertices O(0 , 0),
I(1 , 0), K(1 , 1) and J(0 , 1).
Fig. 1.7
Since this square is of side 1 unit, it is referred
to as the unit square.
The two points I and J are very useful in nding
the matrices of given transformations, and for
the inverse process of identifying transformations
associated with given 2 × 2 matrices.
Let us nd the images of I and J under the
transformation whose matrix is .
= , and
= .
Thus, and are the images of and
respectively, under the transformation
given by
.
In other words, under the transformation
,
I(1 , 0) → I′(a , c) and
J(0 , 1) → J′(b , d)
If a given transformation maps I(1 , 0) onto
I′(a , c) and J(0 , 1) onto J′(b , d), then the
transformation matrix is
.
( )
1 0
0
–
1
Solution
We need to rst choose a simple object
gure and nd its image under the given
transformation. Let us take ∆PQR with vertices
P(2 , 1), Q(4 , 3) and R(3 , 5) as the object.
Under reection in the x-axis, the image
triangle has vertices P′(2 ,
–
1), Q′(4,
–
3) and
R′(3 ,
–
5) (see Fig. 1.6).
Fig. 1.6
Let the matrix of the transformation be
Then,
P Q R P′ Q′ R′
=
⇒
=
Equating the corresponding elements and solving
the resulting simultaneous equations gives
a = 1, b = 0, c = 0 and d =
–
1. Conrm this!
( )
2 4 3
1 3 5
( )
2a + b 4a + 3b 3a + 5b
2c + d 4c + 3d 3c + 5d
( )
a b
c d
( )
2 4 3
–
1
–
3
–
5
( )
2 4 3
–
1
–
3
–
5
( )
b
d
( )
1
0
( )
a b
c d
( )
a b
c d
( )
0
1
a b
c d
( )
( )
a b
c d
( )
1
0
( )
a
c
( )
a b
c d
( )
0
1
( )
b
d
( )
a b
c d
x
y
J(0 , 1)
I(1 , 0)
K(1 , 1)
O
5
4
3
2
1
0
–
1
–
2
–
3
–
4
–
5
R
Q
P
Q′
R′
P′
1 2 3 4 5
y
x
( )
a b
c d
( )
a
c
9
( )
0
1
–
1 0
( )
–
2.5
0
0
–
2.5
Example 1.4
A triangle with vertices A(1 , 3), B(2 , 1) and
C(3 , 1) is mapped onto another triangle with
vertices A′(
–
3 , 1), B′(
–
1 , 2) and C′(
–
1 , 3).
Describe this transformation and nd its matrix.
Solution
The object and its image are shown in Fig. 1.8.
Fig. 1.8
The transformation is a rotation of +90° about
the origin. Under this transformation,
I(1 , 0) I′( 0, 1) and
J(0 , 1) J′(
–
1 , 0) (see Fig. 1.9).
Fig. 1.9
Thus, the matrix of the transformation is .
Can translation be represented by a 2 × 2
matrix? Choose a simple object gure and any
translation and use them to try this out.
Exercise 1.4
1. By considering the unit square, obtain the
matrix for
(a) enlargement with centre (0 , 0) and
scale factor 1.5.
(b) a rotation of 180° about the origin.
2. By considering the unit square, identify the
transformation represented by the matrix
(a) (b)
3. A triangle with vertices A(
–
3 , 4), B(
–
3 , 1)
and C(
–
1 , 1) is mapped onto triangle
A′(
–
3 ,
–
4), B′(
–
3 ,
–
1) and C′(
–
1 ,
–
1).
(a) Describe this transformation and nd
its matrix.
(b) Find the matrix of transformation
4. The matrix represents a
transformation. Use points I and J to
identify the transformation. What are the
vertices of the object triangle whose image
under this transformation has vertices
A′(
–
2 ,
–
1), B′(
–
2 ,
–
4) and C′(
–
4 ,
–
4)?
5. A triangle has vertices X(3 , 5 ), Y(3 , 2)
and Z(5 , 2). Use calculation to nd the
matrix of the transformation which maps
the triangle onto another triangle with
vertices
(a) X′(3 , 5 ), Y′(3 , 2) and Z′(5 , 2)
(b) X′(
–
3,
–
5 ), Y′(
–
3 ,
–
2) and Z′(
–
5 ,
–
2)
(c) X′(
–
3 , 5) , Y′(
–
3 , 2) and Z′(
–
5 , 2)
(d) X′(1.5 , 2.5), Y′(1.5 , 1) and Z′(2.5 , 1)
6. Under a certain transformation the images
of points (1 , 0) and (0 , 1) are (3 , 2) and
(5 , 6) respetively.
(a) Find the images of point (2 ,
–
3) under
the same transformation.
(b) Find also the point whose image is
(4 , 6) under this transformation.
x
y
1 2 3
–
3
–
2
–
1
0
4
3
2
1
A
C′
B′
A′
C
B
– – – – – – – – – – – – – – –
– – – – – – – – – – – – – – –
x
y
1
0
K′
I′
J′
1
–
1
J
K
I
( )
0
–
1
–
1 0
( )
0
–
1
1 0
10
( )
p q
r s
–
1
2
–
√3
2
––
–
1
2
–
–
Other transformations
The general rotation
Fig. 1.10 shows points I and J, of the unit square,
rotated through the angle
θ
about point (0 , 0).
Fig. 1.10
Let represent the matrix of the
transformation.
Since I(1 , 0) I′(cos
θ
, sin
θ
), then
p = cos
θ
and r = sin
θ
.
Since J(0 , 1) J′(
–
sin
θ
, cos
θ
), then
q =
–
sin
θ
and s = cos
θ
.
Thus:
The general rotation matrix is
, where
θ
represents the
angle of rotation and (0 , 0) is the centre of
rotation.
Example 1.5
A rotation is given by the matrix .
What is the angle of rotation?
Solution
Comparing the matrices;
= , we see
that
cos
θ
=
.
∴
θ
= 45°
Fig. 1.11
From the diagram, the problem clearly involves
a negative rotation.
cos
θ
=
–
But cos 120° =
–
∴ the rotation is
–
120° about the origin.
Exercise 1.5
1. Find the images of I and J under the
transformation given by the matrix
Illustrate this transformation on a sketch
diagram and describe it fully.
2. Point P(2 , 4) is rotated through 60° anti-
clockwise about the origin. Find P′, the
image of point P, leaving the coordinates in
surd form.
3. Find the matrix for a rotation, about the
origin, of (a) 90° (b) 270°.
– – – – – – – – – –
– – – – – – – – – –
sin
θ
cos
θ
J′(
–
sin
θ
,
cos
θ
)
I′(cos
θ
, sin
θ
)
θ
θ
cos
θ
•
I(1 , 0)
sin
θ
J(0 , 1)
•
y
x
0
1
√2
––
1
√2
––
1
√2
––
1
√2
––
( )
–
1
√2
––
( )
√3
2
––
√3
2
––
–
1
2
–
Example 1.6
State the smallest angle of the
rotation represented by the matrix .
Solution
I (1 , 0) I′
(
,
)
J (0 , 1) J′
(
,
)
In working out such a problem, it helps to draw
a sketch, as in Fig. 1.11.
√3
2
––
–
1
2
–
–
( )
.
cos 30°
–
sin 30°
sin 30° cos 30°
–
1
1
–
1
1
•
•
I
(1 , 0)
J
(0 , 1)
y
x
1
2
–
–
√3
2
––
–
I
′
(
,
)
J
′
(
,
)
√3
2
––
1
2
–
–
K
′
K
0
1
√2
––
1
√2
––
1
√2
––
1
√2
––
( )
–
( )
cos
θ
–
sin
θ
sin
θ
cos
θ
( )
cos
θ
–
sin
θ
sin
θ
cos
θ
1
2
–
1
2
–
11
(b)(a)
4. ∆ABC with vertices A(–5 , 6), B(
–
5 , 1)
and C(
–
2 , 1) is mapped onto ∆A′B′C′, with
vertices A′(6 , 5), B′(1 , 5) and C′(1 , 2),
by a clockwise rotation about the origin.
Draw the two triangles and state the angle
of rotation. What is the matrix for this
rotation?
5. Give the smallest angle of rotation
represented by the matrix
(a) (b)
(c) (d)
(e) (f)
Stretch
The enlargement of a plane shape in one
direction only is called a stretch.
A stretch is fully described when we are given
1. its direction, and
2. the scale factor (stretch factor), i.e.
the ratio of corresponding lengths of
sides measured in the direction of the
stretch.
Fig. 1.12(a) and (b) illustrate stretches of the
unit square parallel to the x-axis and to the
y-axis respectively.
In Fig. 1.12(a), I(1 , 0) I′(3 , 0) and
J(0 , 1) J′(0 , 1)
∴ the matrix of the transformation is .
This transformation is a stretch with scale factor
3, parallel to the x-axis. The direction may
also be described as being perpendicular to the
invariant line, the y-axis.
In Fig. 1.12(b), I(1 , 0) I′(1 , 0) and
J(0 , 1) J′(0 , 3)
∴ the matrix of the transformation is .
Describe this transformation fully!
In general:
The matrix of a stretch parallel to the x-axis,
with scale factor c, is . Note that the
y-axis is invariant under this stretch.
The matrix of a stretch parallel to the y-axis,
with scale factor c, is . In this case
the x-axis is invariant.
If the scale factor is negative, the object is
rst turned over about the xed (invariant)
line and then stretched as is appropriate.
Two-way stretch
Sometimes, a stretch may be two-way. This
means that an object is stretched in two
directions as shown in Fig. 1.13.
Fig. 1.13
I(1 , 0) → I′(c , 0) and
J(0 , 1) → J′(0 , d )
Fig. 1.12
( )
3
5
–
4
5
–
( )
0.866
–
0.5
0.5
0.866
( )
0.5
–
0.866
0.866 0.5
( )
–
0.5
–
0.866
0.866
–
0.5
( )
0.866
0.5
–
0.5
0.866
( )
√3
2
––
1
2
–
√3
2
––
–
1
2
–
( )
3 0
0 1
( )
1 0
0 3
( )
1 0
0 c
( )
c 0
0 1
3
5
–
–
4
5
–
x
y
3
2
1
0
1 2 3
pull
J′
J
K′
K
I′
I
– – – – –
– – – – – – – – – –
x
y
3
2
1
0
1 2 3
– – – – – – – – – –
– – – – –
pull
J′
J
K′
K
I′
I
x
y
J′(0 , d)
•
0
pull 2
K′
K
J′(0 , 1)
I′(c , 0)
I(1 , 0)
pull 1
•
12
x
y
•
0
shear, d units
y
Similarly, when the shear is parallel to the
y-axis, the transformation matrix is
, where k is the shear factor.
How do we obtain the shear factor k without
using points I and J?
By the denition of a shear, the value of k is
proportional to the distance of any point on the
object from the invariant line.
Suppose the shear is parallel to the x-axis
(Fig. 1.15)
Fig. 1.15
If d is the number of units by which a point on
the object is displaced, then, by denition of
shear,
d ∝ y, where y is the distance of the point
from the xed line.
i.e. d = ky
⇒ k =
d
y
The shear factor is k. It could be positive or
negative, depending on the direction.
Likewise, if the shear is parallel to the y-axis,
and x is the distance of a point on the object
from the y-axis (Fig. 1.16 ), then
k =
d .
x
Fig. 1.16
( )
c 0
0 d
( )
k 0
0 k
∴ the matrix of a two-way stretch is ,
where c and d are the stretch factors parallel to
the x-axis and to the y-axis, respectively.
If the two stretch factors are equal, i.e. c = d,
then the matrix takes the form . What
transformation does this matrix represent?
Shear
A shear is a transformation which has the
following properties:
1. There is a xed line in the plane.
2. A point which is d units away from the
fixed line is displaced parallel to the
xed line, through a distance which is
proportional to d.
A shear is fully described if we are given
1. the invariant line, and
2. a point and its image not on the invariant
line.
Fig. 1.14(a) and (b) illustrate shears of scale
factor 2, of the unit square, parallel to the x-axis
and to the y-axis, respectively.
Fig. 1.14
When the shear is parallel to the x-axis
I(1 , 0) I′(1 , 0)
J(0 , 1) J′(k , 1)
∴ the transformation matrix is
,
where k is the shear factor.
x
y
3
2
1
0
1 2
shear
(b)
J
K′
K
I
I′
– – – – –
x
y
2
1
0
1 2 3
– – – – –
shear
(a)
J
J′
K′
K
I
( )
1 k
0 1
( )
1 0
k 1
x
y
0
shear, d units
x
•
13
3
2
1
0
1 2 3 4 5 6 7
x
y
D′
B
C
D
A
C′
B′
A′
d
y
d
x
Example 1.7
Under a shear, with the x-axis invariant, a
square with vertices A(1 , 0), B(3 , 0), C(3 , 2)
and D(1 , 2) is mapped onto a parallelogram
with vertices A′(1 , 0), B′(3 , 0), C′(7 , 2) and
D′(5 , 2).
Find the shear factor. Hence, nd the shear
matrix.
Solution
Fig. 1.17 shows the object and its image under
the shear.
Fig. 1.17
A point 2 units away from the invariant line (the
x-axis) is displaced through a distance of 4 units.
∴ the shear factor is k =
4
= 2
2
∴ the matrix of the shear is .
Check
Using points D and C,
D C
=
D′ C′
=
i.e. D(1 , 2) D′(5 , 2) and
C(3 , 2) C′(7 , 2).
In general
1. The matrix for a shear parallel to the
x-axis is , where k is the shear
factor obtained as , d
being the distance
that a point is displaced, y units away from
the x-axis.
2. If the shear is parallel to the y-axis, the
matrix is , where k = , d being
the distance that a point is displaced,
x units away from the y-axis.
3. The sign of d must be included in the
evaluation of k.
4. The invariant line of each of these shears
must pass through the origin.
Identity transformation
Activity 1.1
Consider the following transformations:
1. Enlargement with scale factor 1, centre
(0 , 0).
2. A rotation of one complete revolution
about (0 , 0).
3. A stretch with stretch factor 1.
4. A shear with shear factor 0.
What is the image of the unit square under each
of these transformations?
Write down the matrix of each of the
transformations.
In Activity 1.1, you should have found that
each transformation maps the object onto itself.
Each of these transformations is said to be an
identity transformation. Thus:
An identity transformation is one that maps
an object onto itself, i.e. one that does not
change the position or size of the object.
The matrix of the identity transformation is
.
( )
1 0
k 1
( )
1 2
0 1
( )
1 2
0 1
( )
1 3
2 2
( )
5 7
2 2
( )
1 × 1 + 2 × 2 1 × 3 + 2 × 2
0 × 1 + 1 × 2 0 × 3 + 1 × 2
( )
1 k
0 1
( )
1 0
0 1
14
Now refer back to Example 1.2 part (d).
A B C A′ B′ C′
=
We identied the transformation as one
that leaves the object unchanged, or as “no
transformation”, in Table 1.2. Now we know
what to call it.
Note: Another identity matrix is , but this
applies only to translation.
Exercise 1.6
1. With reference to the unit square, identify
the transformation for each of the following
matrices.
(a) (b)
(c) (d)
2. A rectangle with vertices O(0 , 0), P(0 ,
–
1),
Q(4 ,
–
1) and R(4 , 0) is stretched parallel
to the x-axis with stretch factor 4. Find the
coordinates of the image.
3. The operator represents a shear
used to transform the rhombus with
vertices (0 , 0), (2,
–
1), (3 ,
–
3) and (1 ,
–
2).
(a) What is the distance, from the x-axis,
of the furthest point on the rhombus?
(b) How many units does this point move
under this transformation?
(c) Draw the object and image on the
same diagram.
4. represents a two-way stretch with
scale factor 5 parallel to the x-axis and
scale factor 3 parallel to the y-axis. Find
the value of x.
5. A(0 , 0), B( 2 , 0), C (2 , 1) and D(0 , 1)
are the vertices of a rectangle. The
rectangle undergoes a transformation
represented by the matrix . Describe
this transformation fully. What are the
coordinates of the vertices of the image?
6. is the matrix for a certain
transformation. What are the coordinates
of the vertices of the unit square under
this transformation? Describe fully the
transformation.
7. Under a shear with the x-axis invariant, the
point (2 , 3) is mapped onto the point (4, 3).
Find the matrix representing this shear.
8. A shear with the y-axis invariant maps the
point (3 , 4) onto the point (3 , 10). Find
the matrix representing the shear.
9. ∆O′A′B′ with vertices O′(0 , 0), A′(3 , 0)
and B′(11 , 4) is the image of ∆OAB with
vertices O(0 , 0) , A(3 , 0) and B(3 , 4)
under a shear. Find the matrix of the shear
and dene the shear completely.
10. Rectangle ABCD has vertices A(2 , 4),
B(5 , 4), C(5 , 2) and D(2 , 2). Under a
shear, B(5 , 4) B′(5 , 1) and
C(5 , 2) C’(5 ,
–
1).
(a) What is the invariant axis?
(b) What is the matrix of the shear?
11. Find the value of x for which the matrix
is an identity.
Inverse transformations
Let M be the matrix of a transformation
which maps an object point P onto an image
point P′. The transformation which maps
P′ back onto P is called the inverse of the
transformation M, written as M
–
1
.
( )
1 3 3
1 1 4
( )
1 0
0 1
( )
1 3 3
1 1 4
( )
0
0
( )
3 0
0 2
( )
1 2
0 1
( )
–
2 0
0
–
2
( )
3 0
0 1
( )
1 0
0 5
( )
2x 0
0 3
( )
2 0
0 1
( )
1 4
0 1
( )
2x – 5 x – 3
2x – 6 x – 2
15
For example, if Q represents a positive quarter-
turn about the origin, Q
–
1
must be a negative
quarter-turn about the origin.
From our earlier work (refer to Table 1.2), we
found out that
Q = , so Q
–
1
= .
Work out (a) Q
–
1
Q, and (b) QQ
–1
.
You should have found that Q
–
1
Q = QQ
–
1
= I,
where I is the identity matrix of order 2 × 2.
Thus, given the transformation matrix M and the
image points under the transformation, the object
points can be found using the inverse of M.
Example 1.8
Under the enlargement , the image of
∆ABC has vertices A′(4 , 2), B′(4 , 4) and
C′(8 , 4). Find the coordinates of the vertices
of object ∆ABC.
Solution
Let E = .
det E = 2 × 2 – 0 = 4
∴ E
–
1
= =
i.e. the matrix of the inverse transformation is
∴ the coordinates of the matrices of ∆ABC are
given by:
A′ B′ C′ A B C
=
i.e. the coordinates of the vertices of ∆ABC are
A(2 , 1), B(2 , 2) and C(4 , 2).
Exercise 1.7
1. Find the inverses of the following
transformation matrices.
(a) (b) (c)
(d) (e) (f)
2. State the inverse of each of the following
transformations.
(a) Reection in the mirror line m.
(b) An anticlockwise rotation of 70° about
the point (a , b).
(c) A stretch with scale factor 3 and x-axis
invariant.
(d) A translation with displacement vector
.
3. S = is the matrix for a certain
transformation. This transformation maps
∆ABC onto ∆A′B′C′ with vertices A′(6 , 2),
B′(3 , 1) and C′(6 , 1). Find S
–
1
and the
vertices of ∆ABC.
4. Under the rotation 90° about the origin, the
image of ∆XYZ has vertices X′(4 , 2),
Y′(3 , 2) and Z′( 4 , 6 ). Find the
coordinates of the vertices of ∆XYZ.
5. Under the transformation represented by
S = , the image of ∆PQR has
vertices P′(4 , 1), Q′(6 , 2) and R′(8 , 2).
(a) What transformation does S represent?
(b) Find the coordinates of ∆PQR.
Successive transformations
Let Q represent a positive quarter-turn about the
origin and X represent reection in the x-axis.
Fig. 1.18 shows ∆ABC, vertices A(1 , 1), B(1 , 4)
and C(3 , 4), its image ∆A′B′C′ under Q and
∆A′′B′′C′′, the image of ∆A′B′C′ under X.
What single transformation maps ∆ABC onto
∆A′′B′′C′′?
What is the matrix of this transformation ?
( )
0
–
1
1 0
( )
0 1
–
1 0
( )
2 0
0 2
( )
1 0
0 1
( )
2 0
0 2
1
4
–
( )
2 0
0 2
( )
0
0
1
2
–
1
2
–
( )
4 4 8
2 4 4
( )
0
0
1
2
–
1
2
–
( )
2 2 4
1 2 2
( )
0
0
1
2
–
1
2
–
( )
–
1 0
0
–
1
( )
2 0
0 1
( )
3
–
5
( )
1 1
0
1
2
–
( )
√3
2
––
1
2
–
√3
2
––
( )
3 0
0 2
1
2
–
–
( )
√3
2
––
1
2
–
1
2
–
√3
2
––
–
( )
2 0
0 1
16
–
4
–
3
–
2
–
1
0
–
1
–
2
–
3
–
4
4
3
2
1
1 2 3
A
B
B′
C′
A′
B′′
A′′
C′′
y
x
C
( )
1 0
0
–
1
( )
0 1
1 0
( )
0
0
1
2
–
( )
–
2
–
4
( )
–
1 0
0 1
( )
–
2 0
0
–
2
( )
0
–
1
1 0
( )
1
0
0
–
1
( )
–
1
–
4
–
4
–
1
–
1
–
3
( )
1 1 3
1 4 4
1
2
–
Fig. 1.18.
Q = and L = , the matrix of
the position vectors of points A, B and C.
Premultiply L by Q ( i.e. nd QL).
X = .
Now premultiply QL by X.
The product is , the matrix whose
columns give the coordinates of the image of
∆ABC under the combined transformations.
You should have found that the same image
could have been obtained by a single reection
in the line y =
–
x.
Evaluate the matrix product XQ. Compare this
product with the matrix you obtained earlier for
the single transformation.
The eect of the transformation with matrix Q
followed by the transformation with matrix X
is the same as that of the transformation with
matrix XQ.
Is the eect the same as that of the
transformation with matrix QX? (Check this!)
Exercise 1.8
1. ∆ABC has vertices A(2 , 1), B(2 , 2) and
C(4 , 2). X = and U = .
Use matrix multiplication to nd the image
of ∆ABC under the transformation given by
X followed by the transformation given by
U. Identify the single transformation that
maps ∆ABC onto the nal image.
2. The vertices of ∆OPQ are O(0 , 0), P(4 , 2)
and Q(
–
2 , 6). Find the vertices of the image
of ∆OPQ if it is rst enlarged by E and then
translated through T, where
E = and T = .
3. Let Y be the transformation whose matrix is
and H be the transformation whose
matrix is .
(a) Describe each of the transformations
fully.
(b) Use matrix multiplication to nd the
vertices of the image of the unit square
under the successive transformations Y
followed by H.
(c) What single transformation maps the
unit square onto the nal image? State
the matrix of this transformation.
Is Q
–
1
X
–
1
= X
–
1
Q
–
1
? Which of them is the same
as (QX)
–
1
?
We see that :
If X and Q represent transformations, then
XQ means ‘perform Q rst and then X’.
Similarly, if X, Q and R are
transformations, QRX means ‘perform X ,
then R and nally Q, in that order’.
X
2
is the same as XX, i.e. X followed by X.
Note that XQ ≠ QX.
Note also that (QX)
–
1
= X
–
1
Q
–
1
.
17
4. The parallelogram, whose vertices are
(0 , 0), (0,
–
1), (
–
3 ,
–
3) and (
–
3,
–
2), is rst
sheared by the operator , then
reected in the x-axis, and nally translated
by . Find the vertices of the nal image.
5. Let R represent a negative quarter-turn
about the origin, U represent reection in
the line y = x , and T represent translation
. ∆ABC has vertices A(
–
2 , 1), B(
–
2 , 2)
and C(
–
4 , 2). Construct the successive
images of ∆ABC under the transformation
TUR, labelling them as A′B′C′, A′′B′′C′′
and A′′′B′′′C′′′.
Describe the single transformation that
maps ∆A′′′B′′′C′′′ onto ∆ABC. What is the
matrix for this transformation?
6. With reference to the information in
Question 5, nd the nal image of ∆ABC
under the transformation.
(a) U
–
1
UR (b) U
2
R (c) T
2
UR
(d) (T
–
1
)
2
R
4
7.
= + represents a
transformation which maps (x , y) onto
(x′ , y′).
(a) Describe the transformation.
(b) Find the coordinates of A′, the image
of A(2 ,
–
1) under the transformation.
(c) Find the coordinates of a point B,
whose image under this transformation
is B′(
–
4 , 5).
Area scale factor and determinant
Determinant and area.
Consider the transformation of the unit square
shown in Fig. 1.19.
( )
–
1
2
( )
–
2
0
( )
1
–
2
0 1
( )
x′
y′
3
0
0
4
( )
( )
x
y
( )
–
5
–
1
( )
1 1
0 3
( )
a b
c d
( )
a b
c d
( )
0 1 1 0
0 0 1 1
( )
0 a a + b b
0 c c + d d
Fig. 1.19
O(0 , 0) O′(0 , 0), I(1 , 0) I′(1 , 0),
K(1 , 1) K′(2 , 3), and J(0 , 1) J′(1 , 3).
∴ the transformation matrix is
.
What is the determinant of this matrix?
The area of the image OIK′J′ is 3 square units.
(Conrm this!).
We notice that the area of the image is 3 times
that of the object, and that the transformation
matrix has determinant 3.
Thus, the area of the image equals the area of
the object multiplied by the determinant of
the transformation matrix.
Does this result hold true for a transformation
with general matrix ? Let us nd out
using Fig. 1.20 in which the unit square is
transformed, using this transformation as
follows:
O I K J O′ I′ K′ J′
= .
Fig. 1.20
x
y
– – – – –
J
K
I
x
y
– – – – – – – – – – – – – – –
– – – – –
M
A
L
C
I′(a , c)
O
O
J′(b , d)
K′(a+b , c+d)
– – – – – – – – – – – – – – –
c
b
B
D
y
x
3
2
1
0
1 2
J(0 , 1)
K(1 , 1)
J(1 , 0)
J′(1 , 3)
K′(2 , 3)
18
Any transformation that changes the size and/or
shape of an object is said to be non-isometric.
Give examples of such transformations.
Example 1.9
∆ABC with coordinates A(1 , 1), B(1 , 5) and
C(6 , 1) undergoes a transformation represented
by matrix . Find the area of the image
triangle.
Solution
∆ABC is shown in Fig. 1.21.
Fig. 1.21
Area of ∆ABC = × 5 × 4
= 10 square units.
Determinant of is 3 × 2 – 3 × 1 = 3
∴ Area of image ∆A′B′C′ = 3 × 10
= 30 square units.
Exercise 1.9
1. The following matrices represent
transformations. State, in each case, whether
the transformation is an isometry or not.
(a) (b)
(c) (d)
1
2
–
1
2
–
1
2
–
1
2
–
( )
a b
c d
( )
a b
c d
The area of the image is given by:
area of OLK′M – (area A + area B + area C +
area D)
= (a + b)( c + d) – [ bd + (a + b + b)c +
ac + (c + c + d)b].
Conrm that this area simplies to ad – bc.
What is the determinant of the transformation
matrix ?
We see that the area of the image is (ad – bc)
times the area of the object. Thus:
Under any transformation, with matrix
, the area of the image equals
|ad – bc| times the area of the object, where
ad – bc is the determinant of the matrix. Thus,
Area scale factor =
area of image
area of object
= determinant of the
operator.
In case the determinant is negative, what value
do we use?
Isometry
Suppose the determinant of the operator equals 1.
Then,
area of the image = area of the object.
If the shape of the object is unaltered by the
transformation whose matrix has determinant 1,
then the transformation is known as an isometry.
Thus:
An isometry is a transformation in which the
area and shape of the object remain unaltered.
An isometry, therefore, only changes the
position and/or orientation of the object.
Reection is an example of an isometry. Give
other examples.
( )
3 1
3 2
1
2
–
( )
3 1
3 2
( )
3 1
5 2
( )
1
–
1
1 1
( )
–
2 1
–
5 2
( )
–
3 0
0
–
3
x
y
6
5
4
3
2
1
0
1 2 3 4 5 6
B
C
A
19
2. ∆ABC has vertices A(0 , 1), B(2 , 4) and
C(4 , 1). Table 1.3 is about the images of
∆ABC under the transformations whose
matrices are given. Copy and complete the
table.
Matrix
Area of
object
Det. of
matrix
Area scale
factor
Area of
image.
Table 1.3
3. A rectangle with vertices (2 , 0), (6 , 0),
(6 , 2) and (2 , 2) is transformed under the
matrix . Find the area of the image
rectangle.
4. If the area of an object gure is 8 square
units, state the area of the image gure after
a transformation whose matrix is
(a) (b)
(c) (d)
5. Under a certain transformation, the
rectangle with vertices A(1 , 0), B(4 , 0),
C(4 , 2) and D(1 , 2) is mapped onto the
rectangle with vertices A′(1 , 0), B′(4 , 0),
C′(4,
–
4) and D′(1,
–
4 ).
(a) Draw the object and image rectangles
on the same axes.
(b) What transformation maps ABCD onto
A′B′C′D′?
(c) What is the area scale factor under this
transformation?
(d) State the matrix of this transformation.
6. Under the transformation whose matrix is
, the image of ∆ABC has vertices
A′(6 , 2), B′(3 , 1) and C′(6 , 1). What is
the area of ∆ABC?
7. Find the area of the image of ∆ABC,
vertices A(3 , 1), B(3 , 4) and C(6 , 1),
under the transformation whose matrix is
. What is the geometrical
interpretation of your result?
Application of transformations to
real life situations
There are many situations in real life where the
knowledge of transformations is required. For
example, an architect must make a model of a
building in order to illustrate to his client how
the actual building will look like and what its
dimensions will be. The actual building is an
enlargement of the model. In a similar way an
aeroplane builder starts with a prototype (i.e. a
model) on which he perfects all the features
before going into the building of the real plane.
The following example and exercise give more
such situations.
Example 1.10
A company sells tanks for harvesting
rain water. When marketing the tanks, its
salespeople carry with them miniature models
of the tanks and of houses in order to be able to
demonstrate to potential customers, the various
features of the actual tanks. One such model
is in the form of a cylinder and it is a 50 times
reduction of the actual tank.
(a) If the radius of a circular opening on the
top of the model is 0.5 cm and its capacity
is 30.8 ml, nd
( )
2 0
0 2
( )
–
3 0
0 3
( )
0
–
1
1 0
( )
1
–
4
–
1 9
( )
–
1
0
1
1
( )
–
2 1
2 1
( )
2 2
1 3
( )
0
–
2
1 3
( )
0
0
1
2
–
–
1
2
–
–
( )
1 2
2 4
( )
1 1
0
1
2
–
20
(i) the radius of the equivalent opening on
the actual tank;
(ii) the capacity, in litres, of the actual tank.
(b) Given that the height and circumference of
the model are 5 cm and 8.8 cm respectively,
nd the total surface area of the tank in
square metres (assume that the opening
at the top is covered by a lid and that the
overlaps are negligible).
Solution
(a) Linear scale factor (l.s.f) = 1:50.
(i) Let R be the radius of the opening in
the actual tank.
Then, 0.5 : R = 1 : 50
R = 50 × 0.5
R = 25 cm
(ii) Volume scale factor (v.s.f.) = (l.s.f.)
3
= 1 : 125 000
∴ capacity of actual tank = 125 000 × 30.8 ml
= 3 850 000
= 3 850 l
(b) Height of actual tank = 50 × 5
= 250 cm
= 2.5 m
Circumference of actual tank = 50 × 8.8
= 440 cm
= 4.4 m
∴ Area of curved surface of actual tank
= 4.4 × 2.5
= 11 m
2
Radius of actual tank =
C
2π
=
4.4 × 7
2 × 22
= 0.7 m
∴ Area of bottom and top surface = 2 × πr
2
= 2 × × 0.7 × 0.7
= 3.08 m
2
Hence total surface area of actual tank
= 11 + 3.08
= 14.08 m
2
Rainwater harnessing helps
reduce run-off and hence
reduce soil erosion. Thus, it
helps in conserving the top
soil for better crop production.
Also, as fewer and fewer
people draw water from rivers and streams, it
helps save our rivers and hence our ecosystem.
Exercise 1.10
1. A solid cone with base radius 3 cm has a
surface area of 48 cm
2
and a mass of 750 g.
Calculate the surface area and the mass of a
similar cone made of the same material but
with a base radius of 9 cm.
2. A model of a house is made to a scale of
1 : 50.
(a) In the model, the height of the wall is
5 cm. Calculate the height of the wall
of the actual house in metres.
(b) The area of the actual oor is 22 m
2
.
Calculate the corresponding area in the
model, giving your answer in square
centimetres.
3. For a 90°-angle periscope to work, at what
angle must the mirrors be positioned at each
elbow of the periscope?
4. The scale of a map is 1 : 200 000.
(a) If a river is 200 km long, what length
will it be on the map?
(b) On the map, a forest is shown as
occupying an area of 25 cm
2
. What is
the actual area of the forest?
5. I wish to make an indirect measurement of
the height of a ag pole. To do this, I rst
place a mirror on the ground. I then move
back until I can see the top of the agpole
in the mirror. I nd that the distance of the
mirror from the foot of the pole is 20 m and
that my distance from the foot of the pole is
22
7
––
21
24 m. My height, at eye level, is 152 cm.
Make an appropriate sketch and use it to
nd the height of the pole.
6. A restaurant owner nds that the dining
area of her restaurant is a bit small and
crumpled up. To have a semblance of a
more spacious place, she decides to put
mirrors on two opposite parallel walls. She
realises that she gets an extra benet as the
mirrors create multiple images which give
the impression of good patronage.
(a) How many images of herself are formed
if she stands in the dining area?
(b) If she stands 1.2 m facing one of the
mirrors, and given that the distance
between the two walls is 5.2 m, how far
behind the mirror will the second image
that faces her be?
22
Measures of central tendency
In Book 2, we learnt the three common
measures of central tendency. These are the
mean, median and mode. We dened them as
follows.
Mean =
sum of all values
=
∑fx
total number of values
∑f
Median = middle value in a distribution
when the values are arranged in
order.
Mode = most frequent value in a
distribution.
Point of interest
How long do you think you would take to nd
the sum of the numbers from 1 to 100? It must
take some time. However, in 1787 Carl Friedrich
Gauss, then ten-year-old, had a way of solving
this problem within seconds. His method involved
using the mean of the rst and last numbers, i.e.
1 and 100. He reckoned that since the numbers
in the set are evenly distributed, then the mean
of the set of numbers is the same as the mean of
the rst and last of the numbers.
Since the mean of 1 and 100 is 50.5 and there
are 100 numbers, then the sum must be
50.5 × 100 = 5 050.
Gauss later became the foremost mathematician
of his time.
Exercise 2.1
1. Find the mean, median and mode of each of
the following groups of values.
(a) 8, 9, 7, 8, 6, 10, 5, 11, 8, 6, 7
(b) 12, 11, 15, 13, 13, 17, 14, 15, 13, 16
2. Find the mean of each of the following
distributions.
(a) 45, 47, 48, 49, 50, 52, 53, 56
(b) 55, 57, 58, 59, 60, 62, 63, 66
(c) 40, 42, 43, 44, 45, 47, 48, 51
3. Table 2.1 shows the marks obtained by
some students in a quiz.
Marks 33 34 35 36 37 38 39
Frequency 4 6 9 7 6 5 3
Table 2.1
(a) How many students did the quiz?
(b) Find the mean of the distribution.
(c) What is the median mark ?
(d) State the modal mark.
4. Table 2.2 shows the distribution of marks
scored by some 50 students in a mathematics
test.
Marks 21–30 31–40 41–50 51–60
Frequency 1 5 8 10
61–70 71–80 81–90 91–100
11 7 6 2
Table 2.2
(a) Find the mean and median of this
distribution.
(b) State the modal class of the distribution.
The mean
Assumed mean method
Consider the following distributions (Table 2.3).
A: x 45 47 48 49 50
f 1 2 2 3 2
B: x 75 77 78 79 80
f 1 2 2 3 2
C: x 5 7 8 9 10
f 1 2 2 3 2
Table 2.3
STATISTICS (2)
2
23
Conrm that the means of these distributions
are as follows:
Mean of distribution A is 48·2
Mean of distribution B is 78·2
Mean of distribution C is 8·2
Notice that distribution B is obtained by
adding 30 to each of the values in distribution
A. Similarly, distribution C is obtained by
subtracting 40 from each of the values in
distribution A.
Now look at their means.
Adding 30 to the mean of distribution A gives
the mean of distribution B. Subtracting 40 from
the mean of distribution A gives the mean of
distribution C.
You should have obtained similar results in
Question 2 of Exercise 2.1. Conrm this.
In general:
If a constant A is added to or subtracted
from each value in a distribution, the mean
of the new distribution equals the mean of
the old distribution minus or plus the same
constant A. This constant is referred to as a
working mean or an assumed mean.
The assumed mean may be used to make work
easier and quicker when nding the mean of a
distribution, especially if the values are large.
Example 2.1
Find the mean of 105 , 107, 108, 109, 113.
Solution
Step 1:
Choose a reasonable assumed mean. You do
this by looking at the values and checking it
that they range from 105 to 113. The true mean
will lie roughly halfway between these values.
Thus, a reasonable working mean may be 109.
Step 2:
Subtract the assumed mean, 109, from each of
the values to obtain the new distribution
Tabulate the deviation.
x 105 107 108 109 113
x – 10a –4 –2 –1 0 4
This is a distribution of dierences from the
assumed mean known as deviations.
Step 3:
Calculate the mean of the new distribution (i.e.
mean of deviations from the assumed mean).
Mean of deviations =
–
4 +
–
2 +
–
1 +0 + 4
5
=
–
3
5
=
–
0.6
Step 4:
To obtain the true mean, add the assumed mean
to the mean of deviations. Thus:
True mean = 109 +
–
0.6
= 108.4
Check:
Mean of the original values is
105 + 107 + 108 + 109 + 113
=
542
5 5
= 108.4
Example 2.2
A farmer weighed the pigs in his sty and found
their masses to be as in Table 2.4.
Mass (kg) 52 53 54 55 56 57 58 59 60
Frequency 1 2 2 3 5 7 6 3 1
Table 2.4
Using an appropriate assumed mean, nd the
mean mass of the pigs .
Solution
We use a working mean A = 56.
The working is tabulated as in Table 2.5.
24
–
–
– –
Mass, x (kg)
Deviation
f
fd
d = x – A
52
–
4 1
–
4
53
–
3 2
–
6
54
–
2 2
–
4
55
–
1 3
–
3
56 0 5 0
57 1 7 7
58 2 6 12
59 3 3 9
60 4 1 4
∑f = 30 ∑fd = 15
Table 2.5
Mean of deviations =
∑fd
=
15
= 0.5
∑f
30
∴mean mass, x = 56 + 0.5
= 56.5 kg
From Example 2.2, we see that:
Using an assumed mean, A, the formula for
nding the mean, x, of a distribution is
x = A +
∑f(x – A)
or x = A +
∑fd
∑f
∑f
where d = x – A
Recall:
If the given data is in a grouped frequency
distribution, we use mid-interval values, i.e.
class mid-points as the values of x.
Exercise 2.2
1. Using an appropriate assumed mean, nd
the mean of each of the following groups of
values.
(a) 178, 179, 183, 185, 186, 199
(b) 66.4, 67.8, 69.2, 70.0, 71.3
(c) sh 15.40, sh 16.20, sh 17.00, sh 17.80,
sh 19.60, sh 20.40, sh 21.20, sh 22.00.
(d) 221 cm, 229 cm, 227 cm, 226 cm,
220 cm, 221 cm, 228 cm, 225 cm,
220 cm, 223 cm.
2. Table 2.6 shows the marks (out of 50)
obtained by 28 students of a certain school
in an aptitude test.
Marks 38 39 40 41 42 43 44
Frequency 2 4 6 5 5 4 2
Table 2.6
Use the method of working mean to nd the
mean mark.
3. Table 2.7 shows the masses, to the nearest
kilogram, of 40 Form 4 pupils picked at
random.
Mass (kg) 45 – 49 50 – 54 55 – 59 60 – 64
Frequency 2 5 16 9
65 – 69 70 – 74 75 – 79
5 2 1
Table 2.7
Calculate the mean mass, using an
appropriate assumed mean.
4. Table 2.8 shows the grouping by age of
students in a certain polytechnic.
Age group 18 – 19 19 – 20 20 – 21
Number in group 3 6 10
21 – 22 22 – 23 23 – 24
16 13 2
Table 2.8
Calculate the mean age of the students, to
the nearest year.
5. An agricultural researcher measured the
heights of a sample of plants and recorded
them as in Table 2.9. Using an appropriate
working mean, nd the mean height of the
plants.
25
d
4
–
Height
21–30 31–40 41–50 51–60
in cm
Number
2 5 7 9
of plants
61–70 71–80 81–90 91–100
11 8 5 3
Table 2.9
Scaling down the deviations from the
assumed mean.
Consider the following example.
Example 2.3
Use an assumed mean of 120 to nd the mean
of the values:
105, 110, 115, 120, 125, 130
Solution
The working is tabulated as in Table 2.10.
x 105 110 115 120 125 130
d = x – A
–
15
–
10
–
5 0 5 10
Table 2.10
Mean = A +
∑d
,
where n stands for the
n
number of values
= 120 +
–
15
6
= 120 +
–
2.5
= 117.5
Now, let us divide each value of d by any
number, say 5. We obtain.
:
–
3,
–
2,
–
1, 0, 1, 2
∑
=
=
–
0.5
n
Multiplying this value by 5 gives
–
0.5 × 5 =
–
2.5 =
∑d
= mean of deviations
n
∴ mean, x = A +
∑
× 5
n
= 120 +
–
0.5 × 5
= 120 +
–
2.5
= 117.5
Thus:
If each value of a distribution is multiplied
or divided by a constant, c, the mean of the
new distribution equals the mean of the old
distribution multiplied or divided by the
same constant, c. This constant is known as
a scale factor.
This fact may be used to ease working when
nding the mean of a distribution, especially if
it is grouped. In case of a grouped distribution,
with equal class intervals, the best scale factor
to use is the class width. Otherwise, any value
that is a common of the deviations may be used.
Example 2.4
Find the mean of the following distribution
(Table 2.11).
Quantity 101–104 105–108 109–112
Frequency 2 16 41
113–116 117–120 121–124
30 8 3
Table 2.11
Solution
Let A be 110.5, the mid-point of the class
109–112. The working is tabulated as in
Table 2.12.
Quantity
Mid-point
d = x – A t =
f ft
x
101–104 102.5
–
8
–
2 2
–
4
105–108 106.5
–
4
–
1 16
–
16
109–112 110.5 0 0 41 0
113–116 114.5 4 1 30 30
117–120 118.5 8 2 8 16
121–124 122.5 12 3 3 9
∑f = 100 ∑ft = 35
Table 2.12
d
5
–
–
3
6
––
–
d
5
–
d
5
–
26
3. The heights of some 40 students were
grouped as in Table 2.15.
Height
146–150 151–155 156–160 161–165
(cm)
No. of
2 5 16 9
students
166–170 171–175 176–180
5 2 1
Table 2.15
Calculate the mean height.
4. Find the mean of the following distribution
(Table 2.16).
Class
100–105 106–111 112–117 118–123
interval
Frequency 8 12 14 18
124–129 130–135 136–141 142–147
24 22 16 11
148–153 154–159
19 7
Table 2.16
5. Table 2.17 shows the rate of absenteeism in
a year in a certain college.
Days of
1–10 11–20 21–30 31–40 41–50
absenteeism
No. of
2 2 3 9 12
students
51–60 61–70 71–80 81–90 91–100
5 2 3 1 1
Table 2.17
Calculate the mean rate of absenteeism.
6. Table 2.18 shows marks (out of 100) for 40
pupils in a physics test. Make a frequency
distribution with class intervals 20–24,
25–29, ... Estimate the mean mark in the
test.
72 35 49 37 25 25 38 70
63 42 51 40 39 20 35 41
51 39 27 31 38 63 64 72
23 35 46 48 39 56 67 69
31 28 42 51 55 48 37 49
Table 2.18
–
–
Mean, x = A +
∑fd
100
But
∑fd
=
∑f
× 4 =
∑ft
× 4
100 100
100
∴ x = A +
∑ft
× 4
100
= 110.5 +
35
× 4
100
= 110.5 + 0.35 × 4
= 110.5 + 1.4
= 111.9
From Examples 2.3 and 2.4, we see that:
Using an assumed mean A and a scale factor
c, the formula for nding the mean x of a
distribution is
x = A +
∑f
× c, i.e. x = A +
∑f
× c
∑f
∑f
or x = A +
∑ft
× c.
∑f
Exercise 2.3
For each question in this exercise, use an
appropriate working mean and scale factor.
1. Find the mean mark in the following
distribution (Table 2.13).
Mark 15–19 20–24 25–29 30–34 35–39
Frequency 2 10 0 8 2
40–44 45–49
5 1
Table 2.13
2. Table 2.14 shows the percentage scores of
100 students in a mathematics examination.
Score 0–9 10–19 20–29 30–39 40–49
Frequency 1 2 4 18 23
50–59 60–69 70–79 80–89 90–99
24 19 5 3 1
Table 2.14
Find the mean score.
x – A
c
(
––––
)
–
–
d
c
–
–
–
d
4
–
27
( )
( )
The median
Calculation method
As already mentioned, median is the value of
a variable that divides a distribution into two
parts with equal frequencies. For grouped data,
it is determined as illustrated in Example 2.5.
Example 2.5
Estimate the median of the distribution in
Table 2.11 on page 25.
Solution
Recall that:
(1) If N (i.e. the total number of values) is odd,
the median is the (N + 1)
st
value.
(2) If N is even, the median is the average of
the two middle values, i.e. the average of
the N
th
and (N + 2)
nd
values.
In this example, N = 100 (an even number).
Thus, the median is the average of the 50
th
and
51
st
values.
To identify these values, we add a cumulative
frequency (cf) row to the given distribution as in
Table 2.19.
Quantity 101–104 105–108 109–112 113–116
Frequency 2 16 41 30
cf 2 18 59 89
117–120 121–124
8 3
97 100
Table 2.19
From Table 2.19, we see that the 50
th
and 51
st
values are both in the class 109–112, i.e.
109–112 is the median class.
Up to 108.5, there are 18 values. We need 32
more values to get to the 50
th
value, i.e. 32 out
of 41 values (assumed to be evenly distributed
within the class of width 4).
1
2
–
1
2
–
1
2
–
N + 1
2
––––
(
)
∴ 50th value = 108.5 +
50 – 18
× 4
41
= 108.5 +
32
× 4
41
= 108.5 + 3.12
= 111.62
Similarly,
51st value = 108.5 +
51 – 18
× 4
41
= 108.5 +
33
× 4
41
= 108.5 + 3.22 = 111.72
Hence, the median =
111.62 + 111.72
2
= 111.67
Alternatively;
Median value =
50 + 51
th
value
2
=
101
th
value
2
= 50.5
th
value.
Up to 108.5, there are 18 values. We need
50.5 – 18 = 32.5 more values to reach the
50.5
th
value, i.e. 32.5 out of 41 values.
∴ median = 108.5 +
32.5
× 4
41
= 108.5 + 3.171 = 111.67
From Example 2.5, we see that:
Given L = Lower class boundary of the
median class,
N = total frequency of the
distribution,
(∑f )
L
= total of all the frequencies
below the median class,
f
m
= frequency of the median class,
and
c = width of the median class, then
median = L +
– (∑f )
L
c.
f
m
28
Exercise 2.4
1. Calculate the median and state the modal
class of the distribution in Table 2.20.
Mark 0–9 10–19 20–29 30–39 40–49
Frequency 2 2 3 6 10
50–59 60–69 70–79 80–89 90–99
15 6 3 3 1
Table 2.20
2. Calculate the mean and median of the
distribution in Table 2.21.
Class 11–14 15–18 19–22 23–26
Frequency 3 5 7 8
27–30 31–34
5 2
Table 2.21
3. A sample of 40 tree saplings was taken from
a nursery. Their heights were measured and
recorded as in Table 2.22.
Height
145–149 150–154 155–159 160–164
(mm)
Frequency 2 5 16 9
165–169 170–174 175–179
5 2 1
Table 2.22
Estimate the mean and median height.
4. A demographer took a sample of the
population of a town in Kenya and grouped
it by age as in Table 2.23.
Age 0–10 10–20 20–30 30–40 40–50
Frequency 101 93 79 76 67
50–60 60–70 70–80 80–90
67 60 38 12
Table 2.23
Estimate the mean and median age.
5. The times (in seconds) taken by some
athletes who took part in the 400 metres
race during an inter-schools competition
were recorded as in Table 2.24.
56 67 56 55 61 51 52 68
66 59 57 60 46 58 63 52
63 50 64 52 58 53 62 63
47 63 62 68 49 66 55 46
58 45 48 52 55 45 61 65
Table 2.24
(a) Group the data in class intervals of
45–49, 50–54, etc.
(b) Calculate the mean time
(i) when the data is ungrouped,
(ii) when the data is grouped.
(iii) What is the percentage error in
your calculation in (ii)?
(c) Calculate the median time and state the
modal class.
Cumulative frequency curve (ogive)
In Example 2.5, we formed a cumulative
frequency table (Table 2.19).
If we plot each cumulative frequency value
against the upper class boundary of the
corresponding class and join the points, we
obtain a smooth curve called a cumulative
frequency curve (also called an ogive).
The ogive can be used to make deductions
and draw conclusions concerning given data
distributions.
Table 2.19 is reproduced but with the class
limits changed to class boundaries (Table 2.25).
Quantity 100.5–104.5 104.5–108.5 108.5–112.5
Frequency 2 16 41
Cum. freq. 2 18 59
112.5–116.5 116.5–120.5 120.5–124.5
30 8 3
89 97 100
Table 2.25
29
N
2
––
Fig. 2.1 is the corresponding ogive.
frequency in the distribution.
Using Fig. 2.1, the median of the distribution in
Table 2.25 is 111.7.
Quantiles
The median, quartiles, deciles and percentiles,
etc. are collectively known as quantiles.
The median is the value that divides a
distribution into two parts with equal
frequencies.
Quartiles (Q) are values that divide a
distribution into four equal parts. Thus, the
lower quartile (Q
1
) is the value below which
lies a quarter (i.e. 25%) of the distribution.
The middle quartile (Q
2
) or median is the
value below which lies half (i.e. 50%) of the
distribution. The upper quartile (Q
3
) is the
value below which lies (i.e. 75%) of the
distribution.
Deciles (D) are values which divide a
distribution into ten equal parts.
Percentiles (P) divide a distribution into 100
equal parts.
Note that in all cases, the values in the
distribution must be arranged in order of
magnitude for us to locate the required quantile.
Example 2.6
The marks, out of 100, for 100 students were
tabulated as in Table 2.26.
Marks 11–20 21–30 31–40 41–50
Frequency 4 16 27 32
51–60 61–70 71–80
15 4 2
Table 2.26
(a) Estimate, by calculation;
(i) the lower and upper quartile,
(ii) the pass mark if 30% of the students
were to fail,
Fig. 2.1
Note:
1. Each cumulative frequency value is plotted
against the upper class boundary.
2. The lowest class boundary ( in this case
100.5) has a cumulative frequency zero.
This point is plotted for completeness of the
curve.
3. The ogive has a characteristic elongated ‘S’
shape.
Median from the ogive
The median is estimated graphically by locating
the mid-point of a distribution on its cumulative
frequency curve, i.e., we take the value
corresponding with , where N is the total
3
4
–
100
90
80
70
60
50
40
30
20
10
0
y
x
100.5 104.5 108.5 112.5 116.5 120.5 124.5
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – –
median
×
×
×
×
×
×
Cumulative frequency
Quantity
30
( )
3N
4
––
(
)
N
4
––
(
)
30
100
–––
( )
x
100
–––
P
30
= L +
N – (∑f)
P
30
c
f
P
30
The P
30
class is 30.5 – 40.5.
∴ P
30
= 30.5 +
30 – 20
× 10
27
= 30.5 +
10
× 10
27
= 34.20 marks.
Thus, the pass mark is 34 marks.
(iii) For 5 students to get A ⇒ only the top
5% of students get A.
∴we require the 95
th
percentile.
The P
95
class is 60.5 – 70.5.
∴ P
95
= 60.5 +
95 – 94
× 10
4
= 60.5 +
1
× 10
4
= 63 marks.
Thus, the minimum number of marks
for obtaining A is 63.
(iv) Required is the percentile
corresponding with 25 marks. This
mark is in the class 20.5 – 30.5.
∴L = 20.5.
Hence, 20.5 +
N – 4
× 10 = 25
16
i.e. 20.5 +
x – 4
× 10 = 4.5, since
16
N = 100.
⇒
x – 4
× 10 = 4.5
16
∴ x =
4.5 × 16
+ 4
10
= 11.2
≈ 11.
Thus, 11% of the students (11 students)
fail since they get below 25 marks.
Hence, 100 – 11 = 89 students pass.
( )
( )
(iii) the minimum number of marks
required to obtain grade A if only 5
students were to get A,
(iv) how many students were to pass if the
pass mark was set at 25 marks.
(b) Repeat part (a) (i) and (ii) using an ogive.
Solution
To estimate quantiles by calculation, we use
the formula for estimating the median, but with
the appropriate adjustments. Thus, we need a
cumulative frequency table (Table 2.27).
Marks Frequency Cum. freq.
10.5–20.5 4 4
20.5–30.5 16 20
30.5–40.5 27 47
40.5–50.5 32 79
50.5–60.5 15 94
60.5–70.5 4 98
70.5–80.5 2 100
Table 2.27
(a) (i) Q
1
= L +
– (∑f)
Q
1
c
f
Q
1
The Q
1
class is 30.5 – 40.5
∴ Q
1
= 30.5 +
25 – 20
× 10
27
= 30.5 +
5
× 10
27
= 32.35 marks.
Q
3
= L +
– (∑f)
Q
3
c
f
Q
3
The Q
3
class is 40.5 – 50.5
∴ Q
3
= 40.5 +
75 – 47
× 10
32
= 40.5 +
28
× 10
32
= 49.25 marks
(ii) Required is the 30
th
percentile, P
30
.
31
Fig. 2.2
(i) From Fig. 2.2
Q
1
= 32.5 marks, and Q
3
= 49.5 marks
(ii) 30% of the students fail.
We locate the 30
th
student on the ogive
(Fig. 2.2 ) and read the corresponding mark.
This value is 34.5 marks.
Exercise 2.5
1. By calculation, estimate the lower and upper
quartiles of the distribution in Table 2.28.
Class 1– 4 5–8 9–12 13–16
Frequency 3 5 7 8
17–20 21–24
5 2
Table 2.28
(a) Draw a cumulative frequency
curve for this distribution.
(b) From your graph, estimate the
median, the quartiles and the
80
th
percentile.
4. The marks of 50 candidates who sat a history
paper were distributed as in Table 2.31.
Marks 21–30 31–40 41–50 51–60 61–70
No. of
2 5 7 9 11
candidates
71–80 81–90 91–100
8 5 3
Table 2.31
Draw an ogive for this distribution and use
it to determine
(a) the median and the quartiles.
(b) how many candidates passed if the pass
mark was 55 marks.
(b) From Table 2.27, we obtain the ogive in
Fig. 2.2.
x
y
100
90
80
70
60
50
40
30
20
10
0
10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5
×
– – – – – – – – – – – – – – –
– – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – –
– – – – – – – – – – –
(i) Q
1
(ii)
(i) Q
3
Marks
×
×
×
×
Cummulative frequencies
2. Draw an ogive for the distribution in
Table 2.29 and use it to estimate
(a) the 8
th
decile,
(b) the 60
th
percentile,
(c) the median, of the distribution.
Class 11–14 15–18 19–22 23–26
Frequency 3 6 8 12
27–30 31–34 35–38 39–42
5 3 2 1
Table 2.29
3. Table 2.30 shows the distribution of
the number of words per page of a
children’s story book.
No. of words 101–120 121–140 141–160
Frequency 2 6 21
161–180 181–200 201–220
26 18 9
221–240 241–260 261–280
4 3 1
Table 2.30
32
(c) the pass mark if 70% of the candidates
were to pass.
(d) the range of marks obtained by the
middle 80% of candidates.
Measures of dispersion
Consider the distributions in Table 2.32.
A: 50, 50, 50, 50, 50, 50, 50
B: 42, 45, 46, 50, 52, 56, 59
C: 34, 37, 49, 53, 57, 59, 61
Table 2.32
The mean of each distribution is 50.
In distribution A, the values do not vary, while
in distributions B and C, they do. Some of the
values in B and C are above the mean while
others are below. The values show variation or
dispersion. Those of distribution C are more
spread out than those of distribution B. Those
of distribution A have no spread.
It is useful, for statistical purposes, to have a
way of measuring the dispersion (or spread) of
a distribution. Here, we look at such measures.
Range
The range is the difference between the
largest and smallest values in a distribution.
Example 2.7
Find the range of each of the distributions in
Table 2.32.
Solution
Distribution A: Range = 50 – 50 = 0
Distribution B: Range = 59 – 42 = 17
Distribution C: Range = 61 – 34 = 27
Note:
1. The greater the variation of the values in a
distribution, the greater the range.
2. The range is very easy to determine.
However, it is disadvantageous in that it
depends on only two extreme values.
Interquartile range
This is the dierence between the lower and
upper quartiles of a distribution. Half of the
interquartile range is called semi-interquartile
range or quartile deviation.
Thus:
Interquartile range = Q
3
– Q
1
, and
Quartile deviation =
Q
3
– Q
1
.
2
Example 2.8
Determine the quartile deviation of the
distribution in Table 2.26 in Example 2.6.
Solution
Q
1
= 32.35 and Q
3
= 49.25 (found in
Example 2.6).
∴ Quartile deviation =
49.25 – 32.35
2
= 8.45 marks.
Mean deviation (M.D.)
Table 2.33 shows the deviations from the mean
of each of the values in distributions B and C in
Table 2.32.
Deviations from the mean
B: –8, –5, –4, 0, 2, 6, 9
C: –16, –13, –1, 3, 7, 9, 11
Table 2.33
In each case, the sum of the deviations is zero.
For any distribution, the sum of the deviations
is zero. This does not reveal anything about the
dispersion of the values.
Since we are interested only in how far above
or below the mean that the values are, we may
ignore the signs on the deviations and take the
absolute values (i.e sizes of the deviations
33
irrespective of the signs). For distributions B
and C, the absolute deviations are as in Table
2.34.
Absolute deviations from the mean
B: 8, 5, 4, 0, 2, 6, 9
C: 16, 13, 1, 3, 7, 9, 11
Table 2.34
The mean of absolute deviations is called mean
absolute deviation or simply mean deviation
(MD). It tells us how far, on average, the
values are above or below the mean.
For distribution A (Table 2.32),
MD = 0.
This means that every value is equal to the
mean.
For distribution B,
MD =
8 + 5 + 4 + 0 + 2+ 6 + 9
7
=
34
≈ 4.857 (4 s,f,).
7
This means that, on average, the values are 4.9
more or less than the mean.
For distribution C,
MD =
16 + 13 + 1 + 3 + 7 + 9 + 11
7
=
60
≈ 8.571 (4 s.f.).
7
The formula for nding mean deviation is
MD =
∑f |x – x|
.
∑f
Note:
1. The greater the dispersion, the higher the
value of MD.
2. The mean deviation may be calculated from
any other average, e.g. from the median or
mode. However, mean deviation about the
mean is the one most commonly used and
preferred.
3. When the data is grouped, we use the class
mid-values to nd MD.
Exercise 2.6
1. For the following distributions, determine
the range and mean absolute deviation.
(a) 65, 69, 70, 72, 76, 78, 80, 81, 84.
(b) 16, 23, 26, 38, 42, 47, 53, 58, 61,
64, 73, 75, 79, 83, 87.
2. Table 2.35 shows the distribution of shoe
sizes of 100 pupils in a certain school.
Shoe size 4 5 6 7 8 9
No. of pupils 11 26 33 16 10 4
Table 2.35
Find
(a) the range,
(b) the interquartile range,
(c) the mean deviation, of the distribution.
3. Draw an ogive for the data in Table 2.36
and use it to nd
(a) the lower quartile, Q
1
(b) the upper quartile, Q
3
(c) the quartile deviation.
Class 11–19 20–29 30–39 40–49
Frequency 1 3 6 10
50–59 60–69
3 2
Table 2.36
4. Table 2.37 is a frequency distribution of the
mass of coee harvested on a single day by
labourers working on a coee estate.
Mass (kg) 57–59 60–62 63–65 66–68
Frequency 5 6 8 12
69–71 72–74 75–77 78–80
13 8 6 2
Table 2.37
Find the mean, mean deviation, the quartiles
and quartile deviation of this distribution.
–
34
√
Variance and standard deviation
Rather than ignore the signs of the deviations,
we can square each deviation so that we get
only positive values.
The mean of the squares of the deviations
from the mean is called the mean squared
deviation or variance, denoted as s
2
.
Consider the following distribution.
B: 42, 45, 46, 50, 52, 56, 59.
We saw that the mean of this distribution is 50.
The variance of the distribution is worked out
as follows (Table 2.38).
x d = x – x d
2
42
–
8 64
45
–
5 25
46
–
4 16
50 0 0
52 2 4
56 6 36
59 9 81
∑d
2
= 226
Table 2.38
Variance =
∑d
2
=
226
≈ 32.29.
N
7
For a frequency distribution, variance is
given by the formula
s
2
=
∑f(x – x)
2
i.e. s
2
=
∑fd
2
.
∑f
∑f
This is known as the basic formula for nding
the variance.
If the units of the values in the distribution
were centimeters, what would the units of the
variance be?
To be useful, any measure of spread must have
the following properties:
1. Translation along the number line (i.e.
adding a constant A to each value in the
distribution) should not aect it.
2. Enlargement with scale factor c (i.e.
multiplying or dividing each value in the
distribution by c) should multiply or divide
the spread by the same factor c.
3. Multiplying the frequencies by any factor
should not change the spread.
4. All members of the distribution should be
taken into account, but extreme values must
not inuence the spread unduly.
The variance does not satisfy property 2! This
is because we have squared the deviations. To
restore this property, we take the square root of
the variance.
The square root of the variance is known as
root mean squared deviation or standard
deviation (denoted by s). Thus,
s = √variance
⇒ s =
∑f(x – x )
2
i.e. s =
∑fd
2
∑f
∑f
This is the basic formula for nding the
standard deviation.
Example 2.9
Calculate the mean, the variance and the
standard deviation of the distribution in Table
2.39.
x 5 7 9 11 13
f 2 4 8 6 4
Table 2.39
Solution
The working for the mean may be tabulated as
in Table 2.40(a).
–
–
√
–
35
√
–
–
x f fx x f d = x – x d
2
fd
2
5 2 10 5 2
–
4.5 20.25 40.5
7 4 28 7 4
–
2.5 6.25 25.0
9 8 72 9 8
–
0.5 0.25 2.0
11 6 66 11 6 1.5 2.25 13.5
13 4 52 13 4 3.5 12.25 49.0
∑f = 24 ∑fx = 228 ∑f = 24 ∑fd
2
= 130
(a) (b)
Table 2.40
Mean, x =
∑fx
=
228
= 9.5
∑f
24
This value of
x is used to complete Table 2.40(b).
Variance, s
2
=
∑fd
2
=
130
∑f
24
= 5.417 (4 s.f)
Standard deviation, s = √5.417
= 2.327 (4 s.f.)
Note:
1. If s is small, the numbers are closely
grouped about the mean.
2. When the data is grouped, we use the class
mid-values to calculate standard deviation, s.
Exercise 2.7
Calculate the mean and standard deviation of
each of the following distributions, giving your
answer correct to 4 s.f. where appropriate.
1. 6, 8, 9, 10, 10, 12, 15
2. 34, 37, 49, 53, 57, 59, 61
3. x 2 4 6 8 10
f 1 2 4 3 2
Table 2.41
4. x 1 2 3 4 5 6 7
f 2 3 6 9 4 4 2
Table 2.42
5. Class 11–20 21–30 31–40 41–50
Frequency 5 11 16 9
51–60 61–70
5 4
Table 2.43
6. Class 10–14 15–19 20–24 25–29
Frequency 4 5 8 13
30–34 35–39 40–44
11 6 3
Table 2.44
Othermethodsofndingthestandard
deviation
Computational formula
We have already seen that the variance is given
by the formula
s
2
=
∑f(x – x)
2
.
∑f
Expanding this formula gives:
s
2
=
∑f(x
2
– 2xx + x
2
)
∑f
=
∑fx
2
– 2x∑fx + x
2
∑f
∑f
=
∑fx
2
–
2x∑fx
+
x
2
∑f
∑f
∑f ∑f
=
∑fx
2
– 2x
2
+ x
2
∑f
=
∑fx
2
– x
2
∑f
Thus:
The variance may be found using the
formula
s
2
=
∑fx
2
– x
2
.
∑f
The standard deviation is given by
s =
∑fx
2
– x
2
.
∑f
–
–
–
–
–
–
–
–
–
–
–
36
–
–
√
–
This method is very useful in cases where the
mean is fractional, in which case the working
would be more dicult if we tried to use the
basic formula.
Example 2.10
Use the formula s =
∑fx
2
– x
2
to nd the
∑f
standard deviation of the distribution in
Example 2.9.
Solution
The working may be tabulated as shown in
Table 2.45.
x f fx fx
2
5 2 10 50
7 4 28 196
9 8 72 648
11 6 66 726
13 4 52 676
∑f = 24 ∑fx = 228 ∑fx
2
= 2 296
Table 2.45
x =
∑fx
=
228
= 9.5.
∑f
24
s =
2 296
– 9.5
2
24
= √5.417
= 2.327 (4 s.f.)
Note that using the formula s =
∑fx
2
– x
2
∑f
involves less work, and hence, it is quicker.
This formula is known as the computational
formula for the standard deviation.
Using a working mean
Suppose we add or subtract a constant to/from
each of the values of a distribution. What is the
eect of this on the standard deviation? The
following example will enable us to answer this
question.
Example 2.11
For the distribution in Table 2.39 (Example 2.9),
calculate the standard deviation of d, where d
= x – 9. Compare the value obtained by that of the
standard deviation of x obtained in Example 2.9.
Solution
The working is as shown in Table 2.46.
x f d = x – 9 fd fd
2
5 2
–
4
–
8 32
7 4
–
2
–
8 16
9 8 0 0 0
11 6 2 12 24
13 4 4 16 64
∑f = 24 ∑fd = 12 ∑fd
2
= 136
Table 2.46
Mean of d = d =
∑fd
=
12
= 0.5.
∑f 24
Standard deviation of d =
∑fd
2
– d
2
∑f
=
136
– 0.5
2
24
= √5.667 – 0.25
= √5.417
= 2.327 (4 s.f
).
This value is the same as the standard deviation
of x in Example 2.9. Note that the constant 9
was arbitrarily chosen and any other constant
could have been used.
From Example 2.11, we see that subtracting
or adding a constant from/to each value in
a distribution does not alter the value of
the variance or standard deviation. Hence,
the following may also be used to nd the
standard deviation.
s =
∑fd
2
–
∑fd
2
,
∑f
∑f
where d = x – A and A is a working/
assumed mean.
–
–
√
√
√
√
( )
√
37
d
2
–
[
]
(
)
(
)
The coded method
Let us see what happens to the standard
deviation if each of the values of d is scaled
down by a constant c.
Example 2.12
Repeat Example 2.11, but this time, divide each
value of d by 2.
Solution
Refer to Table 2.47.
x f d = x – 9 t = ft ft
2
5 2
–
4
–
2
–
4 8
7 4
–
2
–
1
–
4 4
9 8 0 0 0 0
11 6 2 1 6 6
13 4 4 2 8 16
∑f = 24 ∑ft = 6 ∑ft
2
= 34
Table 2.47
Mean of t = t =
∑ft
=
6
= 0.25.
∑f 24
Standard deviation of t =
∑ft
2
– t
2
∑f
= 34 – 0.25
2
24
= √1.417 – 0.0625
= √1.3545
= 1.164 (4 s.f).
Note that in order to obtain the standard
deviation of x from that of t, we need to
multiply the standard deviation of t by the
constant 2. To obtain the variance of x from
that of t, we multiply the variance of t by 2
2
.
Thus, variance of x is 1.354 × 4 = 5.416, and
standard deviation of x is 1.164 × 2 = 2.328.
The slight dierences are due to rounding
during calculation.
From Example 2.12, we see that:
s
2
= c
2
∑ft
2
–
∑ft
2
, and
∑f ∑f
–
√
–
√
√
s = c
∑ft
2
–
∑ft
2
, where t =
x – A
,
∑f ∑f
c
A is an assumed mean and c is a constant
(usually class interval size).
These are the coded formulas for nding the
variance and standard deviation.
Mean and standard deviation of a
distribution with uneven class intervals.
In some cases, the given frequency distribution
may have unequal class intervals. When
looking for the mean and standard deviation
of such a distribution, we work in the same
manner as for the case of equal class intervals.
Thus, for each class interval, we rst nd the
mid-interval value to represent the class.
When using the coded method, we scale down
the deviations from the mean by dividing
each deviation by an appropriate factor. If
the classes are of equal size, we usually use
factor c, the class interval size. However, if
the class intervals are unequal, we choose any
appropriate factor, k. Example 2.13 illustrates
this.
Example 2.13
Table 2.48 shows the frequency distribution of
the total marks scored by 100 contestants in a
series of ve mathematics contests.
Marks 50–99 100–149 150–249
Frequency 13 16 38
250–349 350–499
24 9
Table 2.48
Use the coded method to calculate
(a) the mean total mark,
(b) the standard deviation of the marks.
Solution
The working may be arranged as in Table 2.49.
Assumed mean, A = 199.5.
38
( )
3. In a study of the characteristics of
cockroaches, a student of zoology
measured lengths of antennae of a number
of cockroaches and recorded them as in
Table 2.52. Find the mean and standard
deviation of the antenna lengths, using a
working mean of 2.75.
Antenna Number of
length (cm) cockroaches
1.0–1.5 5
1.5–2.0 11
2.0–2.5 25
2.5–3.0 36
3.0–3.5 30
3.5–4.0 20
Table 2.52
4. Using the assumed mean method, nd
the mean and standard deviation of the
distribution of marks scored in a certain
test by a number of students (Table 2.53).
Marks Frequency
70–74 4
75–79 8
80–84 11
85–89 15
90–94 9
95–99 3
Table 2.53
5. Some AIDS suerers were weighed and
their masses were recorded as in Table
2.54. Find the mean and standard deviation
of the masses.
Mass (kg) No. of patients
30–39 4
40–49 26
50–59 40
60–69 26
70–79 2
80–89 2
Table 2.54
d
5
( )
–
Marks f x d = x – A t = ft ft
2
50–99 13 74.5
–
125
–
25
–
325
–
8 125
100–149 16 124.5
–
75
–
15
–
220
–
3 600
150–249 38 199.5 0 0 0 0
250–349 24 299.5 100 20 480 9 600
350–499 9 424.5 225 45 405 18 225
∑
100 340 16 100
Table 2.49
(a) Mean, x = A +
∑ft
× k
= 199.5 +
340
× 5
∑f
100
= 199.5 + 17
= 216.5 marks
(b) Standard deviation, s = k
∑ft
2
–
∑ft
2
∑f ∑f
= 5 16 100 – 340
2
100 100
= 5√161 – 3.40
2
= 5√161 – 11.56
= 5√149.44
= 61.12 marks
Exercise 2.8
1. Find the standard deviation of the
distribution in Table 2.50 using the
computational formula.
x 9 10 11 12 13 14 15
f 1 2 4 4 5 3 1
Table 2.50
2. Using a suitable working mean, nd the
standard deviation of the distribution in
Table 2.51. Check your result using the
computational formula.
Class Frequency
0–10 2
10–20 4
20–30 4
30–40 8
40–50 6
50–60 3
60–70 2
Table 2.51
√
√
39
! CAUTION: AIDS has no cure. However,
there is a sure way of avoiding it: TOTAL
ABSTINENCE before marriage, and once
married, STICK TO YOUR PARTNER!
6. Using an assumed mean of 32 and a
scale factor 5, nd the mean and standard
deviation of the distribution in Table 2.55.
Marks Frequency
15–19 8
20–24 10
25–29 16
30–34 26
35–39 22
40–44 12
45–49 6
Table 2.55
7. Table 2.56 shows the masses of eighty
students in a certain college. Calculate the
mean and standard deviation of the masses.
Mass (kg) Frequency
50–54 12
55–59 14
60–64 24
65–69 15
70–74 8
75–79 7
Table 2.56
8. The frequency distribution in Table 2.57
shows the masses, to the nearest gram, of
some biological specimens.
Mass (g) 0–4 5–14 15–24 25–44 45–75
Frequency, f 14 41 59 70 15
Table 2.57
Calculate
(a) the mean mass,
(b) the standard deviation of the masses.
9. A movie was rated “unsuitable for under
16”. Table 2.58 shows the age distribution
of those who attended one sitting. What is
the mean age and standard deviation of the
ages of the attendants?
Age Frequency
16–23 6
24–33 21
34–43 45
44–53 66
54–73 51
Table 2.58
40
Denitionoflocus
Suppose a particle P moves in a plane so that
its distance from a xed point O is always
the same, say 2 cm. In its motion, P obeys the
condition or law that ‘P is always 2 cm away
from the xed point O’. The motion of P
describes a path which forms the circumference
of a circle centre O, radius 2 cm. Such a path,
as the one traced out by point P, is called the
locus of P.
Literally, the word ‘locus’ means location,
position or place. But the above description of
the motion of particle P demonstrates the simple
denition of locus, which is: the path traced
out by a moving point. However, a more
comprehensive denition of locus is as follows:
A locus (plural: loci) of a point is the set of
all possible positions occupied by a point
which varies its position according to some
given law.
The expressions ‘locus of a point’ and ‘locus
of points’ are often used interchangeably.
Whichever we use depends on how we think of
the problem. If, for example, we are thinking
about a point moving under stated restrictions,
‘locus of a point’ might sound better. If we
think of the problem as a series of points which
satisfy given conditions, then ‘locus of points’
may sound better.
Common types of loci
The following activities are intended to help the
pupil dene and illustrate some common loci.
We will use both geometrical and/or graphical
approach, whichever is appropriate.
Constant radius locus
Activity 3.1
1. On a sheet of paper, mark a point O.
2. Mark at least 10 more points on your sheet
such that each point is 3 cm from O.
3. How many more such points can you mark
on your gure?
4. Draw a smooth curve through the points in
(2) to show all the possible points that are
3 cm away from O.
5. Describe fully the series of points in (4)
above.
6. If O is a point in space, what will be the
locus of P?
From Activity 1, we see that:
In two dimensions, the locus of points d cm
from a xed point, O, is a circle centre O,
radius d cm. In three dimensions, the locus
is a spherical shell centre O, radius d cm.
Locus of a point at a given distance from a
point.
Constantdistancefromaxedline
Activity 3.2
1. In your exercise book draw a line l.
2. Mark several points on either side of l, a
distance of 2 cm from l.
Is it possible to locate some more points
which satisfy the same condition?
LOCI
3
xed point
locus
xed
distance
41
3. Describe the locus of all points that are
2 cm from l.
4. If l is a line in space, what will be the locus
of P?
From Activity 3.2, We see that:
In two dimensions, the locus of points that
are d cm away from a given line l is a pair of
lines, on either side of l, each d cm from l and
therefore parallel to l [Fig. 3.1 (a)].
In three dimensions, the locus is a cylindrical
curved surface of radius d cm, with l as the
central axis. [Fig. 3.1 (b)].
The perpendicular bisector locus
Activity 3.3
1. Draw a line segment AB, 7 cm long.
2. Mark a point P on AB such that AP = PB
(Fig. 3.2).
Fig. 3.2
3. Mark other points not on the line, that are
equidistant fromAand B. Describe the locus
of these points.
4. If AB is a line segment in space what will
be the locus of P?
From Activity 3.3, it is evident that:
The locus of points equidistant from two
given points A and B is the perpendicular
bisector of the line segment joining the two
points [Fig. 3.3 (a)]. In three dimensions,
the locus is a plane surface which bisects line
segment AB perpendicularly [Fig. 3.3 (b)].
Fig. 3.3
Angle bisector locus
Activity 3.4
1. Draw an acute angle AOB (Fig. 3.4)
Fig. 3.4
2. Locate a series of points that are halfway
between OA and OB (Fig. 3.5). Hence
indicate all the points equidistant from OA
and OB.
Fig. 3.5
3. Describe the locus of the points in (2) above.
Activity 3.4 shows that:
Given any angle AOB, the locus of points
that are equidistant from OA and OB is the
bisector of angle AOB.
(b)
Fig. 3.1
d
d
l
(a)
l
– – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – –
d
•• •
BA
P
=
=
A
B
– – – – – – – – – – – –
– – – – – – – – – – –
– – – – – – – – – – – –
– – – – – – – – – – –
– – – – – – – – – – – –
– – – – – – – – – – – –
P
n
P
2
P
1
≡
=
–
≡
=
–
A
B
(a)
(b)
A
B
O
θ
A
P
2
P
1
B
θ
O
AP
1
= BP
1
AP
2
= BP
2
42
Exercise 3.1
1. Given two parallel lines l and m, describe
the locus of points equidistant from l and m.
Sketch the locus.
2. Given lines MN and AB which intersect at
a point O, sketch and describe the locus of
points equidistant from the two lines.
3. Describe, the locus of
(a) the tip of the minute hand of a clock in
a time of 20 minutes.
(b) the pendulum bob, P, of a clock as the
pendulum swings to and fro.
(c) the centre of a round coin as the coin
rolls along a straight line on a table.
(d) the key hole in a door as the door opens
through 90° from its shut position.
4. On the rim of a round coin, there is a small
dent marked A. The coin rolls along a
straight line on a table. If the coin starts
with the dent on the line, sketch the locus of
A as the coin rolls.
5. AB is a xed line, 15 cm long. A particle P
moves in the plane of AB so that it is always
2 cm from AB. Sketch the locus of P.
6. Lay a sheet of paper on a drawing board and
hold it down with two drawing pins
4 cm apart. Make a loop of thread 10 cm
long. Place the loop over the pins (Fig. 3.6).
Draw the thread taut with the tip of a pencil.
Keeping the thread taut, move the pencil
so that the tip traces its locus on the paper.
Sketch the locus of the tip of the pencil.
Fig. 3.6
7. In each of the following cases, describe the
required locus and illustrate your answer
using a sketch diagram.
(a) The locus of points equidistant from
P(0 , 0) and Q(2 , 0).
(b) The locus of points equidistant from
A(
–
1 ,
–
1) and B(1 , 1).
(c) The locus of points equidistant from
points C(0 , 0) and D(3 , 3).
(d) The locus of points equidistant from
the line y = x and y = –x.
8. A point P moves so that it is always 5 cm
from the origin. Describe the locus of P and
give its equation in terms of x and y.
Constant angle locus
Given two xed points A and B, we can nd
the locus of a point P, which moves such that
∠APB is always a constant.
90° angle locus
Activity 3.5
1. Draw a line segment AB, 10 cm long.
2. At point A, draw a series of straight lines at
10° intervals (Fig. 3.7), taking line AB to be
at 0°.
Fig. 3.7
3. From B construct a perpendicular to each
of the lines drawn through A, and label the
points where the perpendiculars meet the
lines as P
1
, P
2
... (Fig. 3.8)
10°
10°
10°
10°
A
B
4 cm
43
Q
R
P
(b)
A
B
(a)
Fig. 3.8
4. Using a smooth curve, join the series of the
points P.
5. Describe the locus of P.
6. On the same Fig. 3.8, locate another
possible series of points P, which satisfy the
condition in 3 above.
Describe the complete locus of P.
Observation:
The locus of the point P is the circumference
of a circle with AB as its diameter. This
locus illustrates that the angle subtended by
a diameter on a semi-circle is always equal
to 90°.
Non-90° angle locus
Activity 3.6
1. On a sheet of paper, draw a line segment
AB, 4 cm long.
2. Mount the paper on a drawing board and
stick a drawing pin through each of the
points, A and B [Fig. 3.9 (a)].
3. Cut out a triangle PQR out of a thin card -
board such that the shortest of its sides is
longer than 4 cm [Fig. 3.9 (b)].
Fig. 3.9
4. Place the triangle on the drawing board so
that PQ rests against A and PR rests against
B (Fig. 3.10).
Fig. 3.10
5. Mark a dot on the paper at the point
occupied by the vertex P.
6. Rotate the triangle to a new position, still
with PQ against A and PR against B and
mark the new position of P with a dot.
7. Repeat this process as many times as you
possibly can, always keeping P on the same
side of AB.
8. What is the locus of P? Do you think the
locus is part of a circle? How would you
test for this? As vertex P occupies dierent
positions, what happens to the size of
∠APB?
9. Repeat steps 4–7 with P on the other side
of AB. Describe the complete locus of P.
Repeat this activity with another angle of
your choice.
Observation:
The locus of P is a pair of major arcs of two
intersecting circles with AB as the common
chord. This locus illustrates that angles
subtended by a chord on the circumference
of a circle on the same side of the chord are
equal.
Example 3.1
AB is a xed line segment 6 cm long. Point P
moves on the plane of AB such that ∠APB is
always equal to 60°. Construct the locus of P.
4 cm
Q
R
P
A
B
– – – – – – –
P
1
P
4
P
3
P
2
B
A
44
B
60°
A
O
P
– – – – – – – – – – – – – –
O′
120°
120°
60°
•
•
•
•
•
•
•
•
•
P
P
P
P
P
P
P
Locus of P
P
The two major arcs in Fig. 3.13 represent the
locus of P such that ∠APB = 60°.
In general
If AB is a xed line segment, and P is a
variable point such that ∠APB = θ°, then the
locus of P is a pair of arcs of two intersecting
circles, with AB as their common chord.
Exercise 3.2
1. A and B are two xed points 5 cm apart.
Sketch and describe the locus of point P
such that
(a) ∠APB = 90° (b) ∠APB > 90°
(c) ∠APB < 90°
2. In Fig. 3.14, A and B are xed points.
If ∠AOB = 60°, what angle does AB
subtend at
(a) O′ (b) P
1
(c) P
2
(d) P
3
(e) P
4
?
Fig. 3.14
Describe the locus of P.
3. ∆ABC is one of a series of right-angled
triangles constructed on either side of a
given base AB as hypotenuse. Sketch and
describe the locus of C.
4. ∆ABP stands on a xed base AB, and has a
constant area. If ∠APB is constant, sketch
the locus of P.
5. ABC is an equilateral triangle. P is a
variable point on the same side of AB as C,
and on the same plane such that ∠APB = 30°.
B
6 cm
30°
120°
30°
O
A
60°
A
O
P
B
Solution
Since AB is xed and ∠APB = 60°, then AB
must be a chord of a circle, and ∠APB = 60° is
an angle subtended by AB on the circumference.
Let O be the centre of the circle.
∴ ∠AOB = 120° (angle subtended by a chord
at the centre is twice the angle subtended on the
circumference).
We locate the centre of the circle by
constructing an isosceles triangle AOB of angles
30°, 30°, 120°, and base 6 cm long (Fig. 3.11).
Fig. 3.11
With O as the centre, and radius OA, we draw
the major arc of the circle (Fig. 3.12).
Fig. 3.12
In order to complete the locus of P on the other
side of AB, we locate another center O′ such
that ∠AO′B = ∠AOB.
With O′ as centre and radius O′A, draw another
major arc of a circle (Fig. 3.13).
Fig. 3.13
B
60°
A
•
P
1
O
P
2
•
P
4
•
P
3
•
•
O′
45
Use a ruler and a pair of compasses only to
construct the locus of P. Describe the locus
of P fully.
6. A, B and O are xed points on the same
plane such that ∠AOB = 100°. A point P
moves in the same plane such that ∠APB is
always a constant. Describe the locus of P,
and illustrate it on a sketch diagram.
7. Draw a line segment AB = 6 cm. Mark a
point C on one side of AB such that ∠BAC
is an acute angle.
(a) Accurately, construct the locus of
points equidistant from AB and AC.
Describe the locus.
(b) (i) A variable point P on the same side
of AB as C is such that ∠APB = 25°.
Construct and describe fully the
locus of P.
(ii) Suppose the locus of P in (b)
(i) was not subject to the given
restriction, what would have been
its description?
8. AB is a tangent to a circle centre, O, radius
5 cm (Fig. 3.15). Given that point B moves
freely along the circumference of the circle,
and that AB has a constant length of 12 cm,
state the locus of A and illustrate it on a
sketch.
Fig. 3.15
9. Using a ruler and a pair of compasses only,
construct triangle ABC such that AB = 7 cm,
BC = 8 cm and ∠ABC = 60°. On the same
diagram, construct
(a) the locus of a point P such that P is
equidistant from A, B and C,
(b) the locus of a point Q such that Q is
equidistant from the locus of P.
Describe fully, the locus of Q.
(c) the locus of points R such that
∠RCB = ∠ACB and ∠BRC = ∠BAC.
10. Construct a parallelogram PQRS such that
PQ = 8 cm, PS = 4.5 cm and ∠QPS = 58°.
On the same diagram, locate
(a) the locus of a point X such that X is
equidistant from P and R.
(b) the locus of a point M such that
∠QMS = 90°. Identify the region
inside the parallelogram in which a
variable point Y lies so that PY > RY
and ∠QYS ≤ 90°.
Loci involving chords
Activity 3.7
1. Draw a circle, centre O, radius 5 cm.
2. On your circle, draw a chord AB = 3 cm.
3. Construct the perpendicular bisector of AB.
4. Describe fully, the locus of points equidistant
from A and B.
Activity 3.8
1. Draw a circle centre O with a radius of your
choice.
2. Draw two equal chords AB and CD, not
parallel.
3. Construct and describe the locus of points
equidistant from
(a) A and B, (b) C and D.
4. Comment on the two loci in (3) above.
5. What would happen if the two chords AB
and CD were parallel?
Activity 3.9
1. Draw a circle radius 6 cm.
2. On the circle, draw chords AB and XY to
intersect at a point P (not the centre).
A
B
•
5 cm
12 cm
O
1
2
–
46
There are two possible positions for the two
chords:
1. they could be on the same side of the
centre O [Fig. 3.16(a)], or
2. they could be on opposite sides of the
centre O [Fig. 3.16(b)].
In case 1, we begin by locating the locus of
points equidistant from A and B and also from
C and D.
Since AB is parallel to CD, their perpendicular
bisectors coincide. Using Fig. 3.16(a), MN
represents the distance between the chords
Thus, MN = OM – ON.
Using ∆OAM,
OM
2
= 6.5
2
– 5.25
2
(Fig. 3.17)
∴ OM = √42.25 – 27.562 5
= √14.687 5
= 3.832 cm (4 s.f.).
Similarly,
Using ∆OCN (Fig. 3.18)
ON = √6.5
2
– 6
2
= √42.25 – 36
= √6.25 = 2.5
MN = OM – ON
= 3.832 – 2.5
= 1.332
∴ The distance between AB and CD in Fig.
3.16(a) is 1.332 cm.
Fig. 3.16(b) shows chords AB and CD on
opposite sides of the centre, O.
As in case 1, the locus of points equidistant
from A and B and also from C and D coincide.
The required distance is MN
i.e. MN = NO + OM
= 2.5 + 3.832
= 6.332
∴ the distance between AB and CD in Fig.
3.16(b) is 6.332 cm.
(b)
Fig. 3.16
3. Carefully measure the lengths of PA, PB,
PX and PY.
4. Evaluate (a) PA × PB (b) PY × PX.
Comment on the products.
5. Suppose that P is the only xed point in this
activity, and all the other points move along
the circumference of the circle in the same
direction. Does condition (4) above always
hold?
Observations:
1. The locus of points equidistant from the
end points of a chord of a circle is a line
passing through the centre of the circle,
i.e. diameter.
2. The locus of points equidistant from two
equal chords is the centre of the circle
to which the chords belong. Thus, equal
chords of the same circle are equidistant
from the centre of the circle.
3. If two chords are parallel, their
perpendicular bisectors coincide.
4. If chords AB and XY intersect inside the
circle at a point P, then PA · PB = PY · PX
Example 3.2
AB and CD are two parallel chords of a circle,
centre O, radius 6.5 cm. Given that
AB = 10.5 cm and CD = 12 cm, calculate the
distance between them.
Solution
We start by illustrating the given information in
a sketch (Fig. 3.16).
O
C
N
Fig. 3.18
O
A
M
Fig. 3.17
(a)
– – – – – – –
– – – – – – –
N
•
C
D
O
A
B
M
– – – – – – – – – – – – – –
– – – – – – – – – – – – – –
N
•
C
D
O
A
B
M
47
– – – – – – – – – –
•
K
•
N
l
Remember that, distance between two lines
means perpendicular distance hence the need
to use the mediators of the chords.
Exercise 3.3
1. AB is a chord of a circle, centre O, radius
10 cm. N is a point on chord AB such that
ON is perpendicular to AB. Given that ON
is 8 cm long, nd the length of AB.
2. The diameter AB and a chord CD intersect
at right angles at a point P. Given that O is
the centre of the circle, and that AB = 13 cm
while CD = 12 cm, nd the length of OP.
3. AB is a chord of a circle centre O. The tangent
at point B meets AO produced at a point T.
Given that ∠BTA = 52°, calculate ∠B AT.
4. A is a xed point on the circumference of a
xed circle centre O. M is the midpoint of
a variable chord AB. State the locus of M,
and illustrate it on a sketch.
5. Two intersecting circles meet at points A
and B. A line CF meets AB at point X, one
circle at C and D, and the other at E and F
(Fig. 3.19).
Fig. 3.19
Given that CF passes through the centre of
the two circles, and that CE = 9 cm,
EX = 3 cm, XD = 2 cm and DF = 4 cm,
calculate the length of
(a) XB (b) BF.
6. Fig. 3.20 shows a line l and a point N, 4
cm from l. K is a point on l, 6 cm from N.
Construct a circle through N, whose tangent
at K is l. State the radius of the circle.
Fig. 3.20
7. A, B, C, D, and O are variable points such
that at any time, AB and CD intersect at O.
If OA × OB = OC × OD, describe the locus
of points A, B, C, and D. What is the locus
of O?
8. AB and DC are two chords equidistant from
the centre of a circle. Given that BD is a
diameter of the same circle, describe the
locus of a point moving along BD.
9. Construct triangle ABC in which
AB = 9 cm, BC = 5.5 cm, and ∠ABC = 40°.
Locate a point O, equidistant from A, B,
and C. Hence, nd and describe the locus
of a point P which moves such that
∠BPC = ∠BAC.
10. PQR is a triangle in which QR = 9 cm,
PR = 5 cm and PQ = 7 cm. Find, by
construction, the locus of points S,
equidistant from the three sides of the
triangle.
Intersecting loci
Sometimes the solution to a construction
problem depends on nding a point or points
that satisfy given conditions. Finding such a
point requires the use of locus, although the
word locus may not be used.
Example 3.3
Three boys, Juma, Soi and Makau, play a
certain game as follows. The three boys cluster
around a fourth one, Kim. As soon as Kim
blows a whistle, the other three boys in the
game are supposed to run in dierent directions
at the same speed and in straight lines. After 30
B
E
C
D
X
F
A
48
In this example, the locus is a single point.
Generally:
Given any three points, not on a line, the locus
of points equidistant from each one of them is
the intersection of the perpendicular bisectors
of any two of the three line segments dened
by the points.
Example 3.4
Fig. 3.23 shows Mr. Kombo’s house and its
backyard.
Fig. 3.23
Mr. Kombo wishes to plant an orange tree in the
garden. It must be at least 40 m from the house,
at least 30 m from an existing mango tree T,
at least 20 m from the back fence and not less
than 14 m from either side boundary. Draw a
diagram showing where he should plant it.
Solution
The set of all possible positions of the tree
makes up the locus of points satisfying the given
conditions.
1. The tree must be a minimum of 40 m from
the house.
Therefore, using an appropriate scale,
say 1 cm : 10 m, draw the back garden,
mark the position of the house and draw a
line parallel to the house, 40 m away (Fig.
3.24). The tree must be 40 m or more from
the house. Therefore we shade the region
‘<40 m’.
2. The tree must be a minimum of 30 m from T.
Therefore, show the locus of points 30 m
from T and shade the region ‘<30 m’.
*
*
*
Juma
Makau
Soi
J
M
S
–
–
=
=
Locus of points
equidistant fr om
Juma and Soi
Locus of points
equidistant fr om
Makau and Soi
K
– – – – –
– – – – – – – – – –
T
•
10 m
Back
fence
52 m
72 m
12 m
56 m
House
seconds, their positions may be as shown in
Fig. 3.21.
Fig. 3.21
Make a copy of Fig. 3.21 and on it locate the
exact position where Kim would be standing as
he blows the whistle.
Solution
Since the boys are running at the same speed,
each boy will be the same distance away from
Kim after 30 seconds. We therefore require to
nd two loci as follows.
1. Locus of points equidistant from Juma (J)
and Soi (S),
2. Locus of points equidistant from either,
Juma (J) and Makau (M) or Soi (S) and
Makaku (M).
The intersection of the two loci will represent
the position of Kim (K) at the time he blew the
whistle.
The procedure is as follows.
1. Copy Fig. 3.21.
2. Construct the perpendicular bisector of the
line segment MS.
3. Construct the perpendicular bisector of JS.
Let the perpendicular bisectors in 2 and 3
intersect at a point K (Fig. 3.22)
Fig. 3.22
The point K is the intersection of the two loci.
It is the position where Kim stood when he blew
the whistle.
49
3. Construct the locus of points 20 m from the
back fence. The tree must be a minimum
of 20 m from the back fence. Shade the
region ‘<20 m’.
4. The tree must not be less than 14 m from
any boundary.
Therefore, show the locus of points 14 m
from boundary CD and shade the region
‘<14 m’.
Fig. 3.24
The unshaded region, marked M, satises all
the four conditions. It is the intersection of the
four loci. The new tree can be planted anywhere
in that region.
Why are the boundaries in Example 3.4 solid,
and not broken?
Loci involving inequalities
Sometimes, loci may be constructed on the
Cartesian plane. In such cases, the conditions
are often in the form of inequalities and the loci
are regions.
Thus:
The locus of points that satisfy a given
inequality is a region in the x, y-plane.
The locus of points that satisfy two or
more inequalities is the intersection of the
corresponding regions in the x, y-plane.
Example 3.5
Draw the locus of points (x, y) such that:
y + x < 2 and y ≥ x
2
.
– – – – –
T
•
10 m
30 m
fr om T
52 m
House
14 m fr om CD
20 m fr om the
back fence
30 m
A
D
C
M
72 m
40 m fr om the house
B
Solution
Required is the intersection of the regions which
satisfy the given inequalities. This is shown
in Fig. 3.25 as the region marked R. Note that
the unwanted region is shaded, this being the
convention most commonly used.
Fig. 3.25
Exercise 3.4
1. Given that A and B are two xed points
4.5 cm apart, show in a sketch diagram, the
locus of all points equidistant from A and B.
Hence, show the locus of all points nearer
to A than to B. Shade the unwanted region.
2. A, O and B are three points such that ∠AOB
is an acute angle. Draw a diagram to show
the locus of points nearer to OA than to OB.
Shade the unwanted region.
3. Each of the inequalities below denes a
locus of points. Illustrate each one of them
on a separate diagram.
(a) y > x + 2 (b) y < x + 5
(c) 1 ≤ x < 3 (d) 2 ≤ y ≤ 5.
4. O is a xed point. Show, by shading the
unwanted region, the locus of all points P
such that OP < 4 cm.
10
8
6
4
2
–
3
–
2
2
x
y
– – – – – – – – – – – – – – – – – – – – – – – – – – – – –
R
y + x = 2
y = x
2
–
2
–
1
0
1 3
50
3 cm
5 cm
O
Fig. 3.26
5. P is the locus of points equidistant from the
y-axis and the line x = 3. Q is the locus of
points equidistant from the x-axis and the
line y = 5. Sketch the two loci and describe
their intersection.
6. ABCD is a rectangle in which AB = 8 cm
and BC = 4 cm. Show the locus of a point
P, within the rectangle, equidistant from AB
and BC. How many points satisfy the locus
if
(a) AP = 4 cm (b) CP = 4 cm
(c) PD = 5 cm?
7. ABC is a triangle of area 24 cm
2
. If AB is the
xed base of the triangle and it is 8 cm long,
sketch and describe the locus of point C.
8. Fig. 3.26 shows two
concentric circles, centre
O, radii 3 cm and 5 cm
respectively. Describe the
locus of the points in the
shaded region.
9. ABC is a triangle. Sketch and describe the
locus of points equidistant from
(a) the three vertices of the triangle,
(b) the three sides of the triangle.
10. Points P and Q are 20 m apart. A tree is
10 m from P and 24 m from Q. Find a locus
of points 18 m from the tree and equidistant
from P and Q. Describe the locus if the
distance from Q to the tree is 27.4 m and
the other facts remain unchanged.
11. Draw the locus of points (x , y) such that
x ≤ 2, y + x < 3 and y – x < 4.
12. Show on the Cartesian plane the locus of
points that satisfy the inequality x
2
+ y
2
≤ 25
and y ≤ x.
13. Draw the locus of points (x, y) that satisfy
the conditions:
(x – 1)
2
+ (y – 2)
2
≤ 25, x < 3, y ≥ 0.
14. Illustrate, on the x, y-plane, the locus of
points such that y > x
2
and y ≤ –2x
2
+ 5.
51
A
θ
C
B
c
b
a
Fig. 4.1
Basic trigonometric ratios
In Book 2, we dened the sine, cosine and
tangent of an acute angle, using a right-angled
triangle.
Given acute angle
θ
in a triangle ABC (Fig. 4.1)
and using the usual notation we recall that:
sin
θ
= cos(90° –
θ
) =
a
b
cos
θ
= sin(90° –
θ
) =
c
b
tan
θ
=
sin
θ
=
a
cos
θ
c
In Book 3, we extended the denitions of sine,
cosine and tangent to cover angles greater than
90° and angles less than 0°, by relating these
angles to appropriate acute angles, using the
unit circle.
Consider Fig 4.2.
(a) (b)
Fig. 4.2
If an angle
θ
lies in the second quadrant
[Fig. 4.2(a)], then it lies between 90° and 180°,
i.e. it is obtuse. In order to evaluate the ratios
for this angle, we must rst express it in the
form
θ
=180° – α, where α is an ‘acute angle’.
Then we work as in Example 4.1.
Example 4.1
Evaluate the sine , cosine and tangent of 130°.
Solution
130° = 180° – 50°
∴ sin 130° = sin(180° – 50°).
The numerical value of sin 130°
is the same as
that of sin 50°.
Since 130° is an angle in the second quadrant,
its sine must be positive.
∴ sin 130° = sin 50°
= 0.766 0
Also,
cos 130° = cos(180° – 50°).
The numerical value of cos 130° is the same as
that of cos 50°.
But since 130° is in the second quadrant, its
cosine must be negative.
∴ cos 130° =
–
cos 50°
=
–
0.642 8
Similarly,
tan 130° = tan(180° – 50°).
Since 130° is in the second quadrant, its
tangent must be negative.
∴ tan 130° =
–
tan 50°
=
–
1.191 8
Alternatively,
tan 130° =
sin 130°
=
sin 50°
cos 130°
–
cos 50°
O
y
x
A
θ
P(x, y)
– – – – – – – – – –
(x, y)Q
θ
x
B
R(x, y)
O
y
TRIGONOMETRY (3)
4
2
nd
3
rd
4
th
180 +
θ
180 –
θ
360 –
θ
θ
1
st
S
T C
A
52
=
0.766 0
=
–
1.191 7
–
0.642 8
Note: The small dierence in the value of
tan 130° occurs because the values of sine and
cosine used are already approximations to 4 d.p.
If
θ
is in the third quadrant, to evaluate its ratios,
we must express it in the form
θ
= 180° + α, where
α is an acute angle (Fig. 4.2(b);
θ
= ∠BOQ).
Example 4.2
Find the trigonometric ratio of 220°.
Solution
220° = 180° + 40°
∴ sin 220° = sin(180° + 40°).
We know that the numerical value of sin 220° is
the same as that of sin 40°.
Since 220° is in the third quadrant, its sine must
be negative.
∴ sin 220° =
–
sin 40°
=
–
0.642 8
220° = 180° + 40°
∴ cos 220° = cos (180° + 40°)
=
–
cos 40°
=
–
0.766 0
tan 220° = tan (180° + 40°)
= tan 40°
= 0.839 1
If
θ
is in the fourth quadrant, i.e. angle BOR in
Fig. 4.2(b), we express it in the form
θ
= 360° – α,
where α is an acute angle.
Example 4.3
Find the trigonometric ratios of 290°.
Solution
290° = 360° – 70°
∴sin 290° = sin(360° – 70°)
= sin(
–
70°)
=
–
sin 70° (since
–
70° is in the
fourth quadrant)
=
–
0.939 7
Similarly, cos 290° = cos (360° – 70°)
= cos (–70°)
= cos 70° (
–
70° is in the fourth
quadrant)
= 0.342 0
tan 290° = tan (360° – 70°)
= tan (
–
70°)
=
–
tan 70°
=
–
2.747 5
In general, if
θ
is an acute angle, then we recall
that:
sin (180° –
θ
) = sin
θ
and
sin (180° +
θ
) =
–
sin
θ
sin (360° –
θ
) =
–
sin
θ
and
sin (360° +
θ
) = sin
θ
Also
sin (360°n +
θ
) = sin
θ
and
sin (360°n –
θ
) =
–
sin
θ
, where n is an
integer.
cos (180° –
θ
) =
–
cos
θ
and
cos (180° +
θ
) =
–
cos
θ
cos (360° ±
θ
) = cos
θ
cos (360°n ±
θ
) = cos
θ
tan (180° +
θ
) = tan
θ
and
tan (180° –
θ
) =
–
tan
θ
tan (360° +
θ
) = tan
θ
and
tan (360° –
θ
) =
–
tan
θ
tan (360°n +
θ
) = tan
θ
and
tan (360°n –
θ
) =
–
tan
θ
.
Exercise 4.1
1. If cos
θ
is
(a) negative, in which quadrant must
θ
be?
(b) positive, in which quadrant must
θ
be?
2. Repeat Question 1 for:
(i) sin
θ
(ii) tan
θ
3. Express each of the following in terms of
an acute angle.
(a) (i) sin 100° (ii) sin 205°
(iii) sin 280°
53
(e)
cos(90° –
θ
)
(f) cos 70° – sin 160°
cos(90° +
θ
)
12. Given that x is an acute angle, and that
cos x =
m
, nd in terms of m and n
n
(a) sin x, (b) tan x.
13. Given that
θ
is an obtuse angle, and that
cos
θ
=
–
0.6, state sin
θ
and tan
θ
.
14. If cos
θ
= and 0° ≤
θ
≤ 360°, nd the
values of the other ratios giving your
answers in surd form.
15. Given that
–
180° ≤
θ
≤ 180° and that
tan
θ
= 0.75, nd without using
trigonometric tables
(a) sin
θ
, (b) cos
θ
.
16. Given that x is an acute angle, simplify:
(a) cos (180° + x) (b) tan (360° – x)
(c) sin (90° + x) (d) tan (270°
+ x).
The relation cos
2
θ
+ sin
2
θ
= 1
The trigonometric identity cos
2
θ
+ sin
2
θ
= 1 is a
Pythagorean relation, connecting three sides of
a triangle with reference to the unit circle and
an acute angle formed at the centre of the circle.
Deriving the relation cos
2
θ
+ sin
2
θ
= 1
using a unit circle
If P is a point on the unit circle (Fig. 4.3) such
that ∠NOP = x, then
the coordinates of P
are (cos x , sin x).
But ON
2
+ NP
2
= OP
2
.
(b) (i) cos 105° (ii) cos 250°
(iii) cos 305°
(c) (i) tan 95° (ii) tan 160°
(iii) tan 355°
(d) (i) sin 400° (ii) cos 490°
(iii) tan 840°
4. Name another positive angle less than 360°
which has the same
(a) sine as 70°, (b) cosine as (
–
50°),
(c) tangent as 235°.
5. List all the positive angles less than 360°
for which
(a) sine is
, (b) tangent is 1,
(c) cosine is
–
, (d) sine is
–
,
(e) cosine is
–
, (f) tangent is .
6. List all the angles between
–
180° and 180°
inclusive for which
(a) cosine is 0, (b) tangent is
–
,
(c) sine is 1, (d) cosine is
–
1.5,
(e) sine is
–
, (f) tangent is √3.
7. Without using tables, simplify:
(a)
sin 20°
(b)
cos 10°
cos 70° sin 80°
(c) tan 72° sin 18°
8. Given that x is an acute angle such that
sin x =
a
, nd the value of tan x in terms
b
of a and b.
9. Given that sin x = cos 24°, and x is an acute
angle, nd the value of x.
10. Given that x is acute, simplify
tan x × sin(90° – x).
11. Simplify:
(a) sin(180° + x) (b) cos(90° + x)
(c) tan(360° – x) (d)
sin(90° +
θ
)
cos
θ
1
2
–
1
2
–
√3
2
––
√3
2
––
4
3
–
1
2
–
1
√3
––
1
3
–
y
x
x
– – – – – – – – –
O
– – – – – – –
P(cos x , sin x)
N
Fig. 4.3
sin x
cos x
54
( )
C
B
A
θ
Since OP = 1,
ON = cos x, and
NP = sin x,
then ON
2
+ NP
2
= OP
2
becomes
(cos x)
2
+ (sin x)
2
= 1
2
.
(cos x)
2
is usually written as cos
2
x and (sin x)
2
as sin
2
x. Thus,
cos
2
x + sin
2
x = 1
From this we have:
cos
2
x = 1 – sin
2
x
sin
2
x = 1 – cos
2
x
This identity is true for angles of any
magnitude.
Deriving the relation cos
2
θ
+ sin
2
θ
= 1
using a right-angled triangle
Using any right-angled triangle, we can derive
the relation cos
2
θ
+ sin
2
θ
= 1.
Consider ∆ABC (Fig. 4.4).
Fig. 4.4
Since ∆ABC is right-angled at B,
AB
2
+ BC
2
= AC
2
AB
2
+
BC
2
=
AC
2
(dividing both sides by AC
2
)
AC
2
AC
2
AC
2
⇒
AB
2
+ BC
2
= 1
AC
2
AC
2
But
AB
= cos
θ
and
BC
= sin
θ
(by denition)
AC AC
∴
AB
2
= (cos
θ
)
2
and
BC
2
= (sin
θ
)
2
AC
2
AC
2
∴ (cos
θ
)
2
+ (sin
θ
)
2
= 1
i.e. cos
2
θ
+ sin
2
θ
= 1
Proving trigonometric identities using
the relation cos
2
θ
+ sin
2
θ
= 1
The trigonometric identity cos
2
θ
+ sin
2
θ
= 1
may be used to prove other given identities.
In order to prove a given trigonometric
relation:
1. We start by looking for a direct substi-
tution. If none is obvious, an identity
can often be established by expressing
all given quantities in terms of sine and/
or cosine.
2. We must resist the temptation of
assuming, at the start, the truth of what
we want to prove.
3. Normally, we start with one side of the
identity and transform it systematically
until both sides are identical, i.e. we
reduce one side to the other side or
both sides, one at a time, to the same
expression.
Examples 4.4 and 4.5 demonstrate how to
use these tips. The ability to manipulate
given identities comes in handy when solving
trigonometric equations.
Example 4.4
Use the relation cos
2
θ
+ sin
2
θ
= 1 to show that
tan
θ
+
cos
θ
2
=
1
+
1
sin
θ
cos
2
θ
sin
2
θ
Solution
Consider the expression on the left hand side.
L.H.S. = tan
θ
+
cos
θ
2
sin
θ
= tan
2
θ
+ 2tan
θ
·
cos
θ
+
cos
2
θ
sin
θ
sin
2
θ
=
sin
2
θ
+ 2
sin
θ
·
cos
θ
+
cos
2
θ
cos
2
θ
cos
θ
sin
θ
sin
2
θ
( )
55
( )
{
{
( ) ( )
4. Given that
θ
is an obtuse angle, and that
sin
θ
= 0.8, use the relation cos
2
θ
+ sin
2
θ
= 1
to evaluate
(a) cos
θ
, (b) tan
θ
.
5. If S denotes sin x, express each of the
following in terms S, and simplify where
possible.
(a) 2 cos
2
x – 3sin x (b) 2sin x +
cos
2
x
sin
2
x
(c)
3
–
1
sin x cos x
(d) cos
2
x – sin x +
1
sin x
6. (a) Using the relation cos
2
x + sin
2
x = 1,
show that
sin
2
x
= 1 – cos x.
1 + cos x
(b) Express tan
2
θ
in terms of sin
θ
and
cos
θ
and simplify the left hand side of
tan
2
θ
=
sin
2
θ
+ 1.
sin
2
θ
cos
2
θ
Hence, show that the LHS of the
equation is equal to the right hand side.
7. Use the identity cos
2
θ
+ sin
2
θ
= 1 to
express the left hand side of the given
relation in terms of cos
θ
. Hence show that
sin
2
θ
= 1 – cos
θ
.
1 + cos
θ
8. (a) Show that 3cos
2
θ
+ 5sin
2
θ
= 3 + 2sin
2
θ
.
(b) Show that 1 + tan
2
θ
=
1
.
cos
2
θ
9. Show that
(a) tan
θ
+
cos
θ
=
1
,
sin
θ
sin
θ
cos
θ
(b) 1
+ tan
θ
1
– tan
θ
= 1.
cos
θ
cos
θ
10. Show that
(a) 2cos
2
θ
– 1 = 1 – 2sin
2
θ
.
(b) 1 – 2sin
2
θ
= cos
2
θ
– sin
2
θ
.
(c) 2cos
2
θ
– 1 = cos
2
θ
– sin
2
θ
.
=
sin
2
θ
+ 1
+
1 +
cos
2
θ
cos
2
θ
sin
2
θ
=
sin
2
θ
+ cos
2
θ
+
sin
2
θ
+ cos
2
θ
cos
2
θ
sin
2
θ
=
1
+
1
(cos
2
θ
+ sin
2
θ
= 1)
cos
2
θ
sin
2
θ
But RHS =
1
+
1
cos
2
θ
sin
2
θ
∴ LHS = RHS
i.e. tan
θ
+
cos
θ
2
=
1
+
1
sin
θ
cos
2
θ
sin
2
θ
Example 4.5
Use the relation sin
2
x + cos
2
x = 1 to show that
(cos x + sin x)
2
+ (cos x – sin x)
2
= 2.
Solution
LHS = (cos x + sin x)
2
+ (cos x – sin x)
2
= cos
2
x + 2 sin x cos x + sin
2
x + cos
2
x
– 2 cos x sin x + sin
2
x
= cos
2
x + sin
2
x + cos
2
x + sin
2
x
= 1 + 1 (since sin
2
x + cos
2
x = 1)
= 2.
Exercise 4.2
1. Simplify the following.
(a) cos
2
θ
+ cos
2
(90° –
θ
)
(b) sin
2
(180° –
θ
) + cos
2
(360° –
θ
) given
that 0 <
θ
< 90°.
2. Simplify:
(a)
√1 – cos
2
θ
(b)
√1 – cos
2
θ
cos
θ
(c)
1 – cos
2
θ
cos
2
θ
3. Simplify the following.
(a)
1 – cos
2
θ
(b)
sin
θ
cos
θ
1 – sin
2
θ
√1 – sin
2
θ
(c)
cos
2
θ
– 1
(d)
1
+ tan
θ
sin
θ
tan
θ
( )( )
56
or sin
θ
= sin(180° – 60°)
⇒
θ
= 60° or 120°
(b) cos
θ
=
–
0.8.
Since cos
θ
is negative, and
θ
is between
–
180° and 180°,
θ
must be in the second or
third quadrant, giving two possible values
(Fig. 4.6).
Fig. 4.6
Now, cos 36° 52′ = 0.8 (from tables)
Then, cos(180° ± 36° 52′) =
–
0.8
∴
θ
= 180° ± 36° 52′
= 143° 8′ or 216° 52′
216° 52′ is in the third quadrant and so
invalid since
–
180° ≤
θ
≤ 180°. But this
angle can be taken as
–
180° + 36° 52′ =
–
143° 8′, which is between
–
180° and 180°,
as required.
∴
θ
=
–
143° 8′ or 143° 8′
(c) tan
θ
= √3.
Since tan
θ
is positive, and
–
360° ≤
θ
≤ 360°,
then
θ
must be in the rst or third quadrant
(Fig. 4.7).
Fig. 4.7
Solving trigonometric equations
An equation involving trigonometric ratios of
an unknown angle is called a trigonometric
equation. When solving a trigonometric
equation we often restrict the solution to a
particular interval.
The equation is completely solved once we
obtain all the angles which satisfy it within the
given range. Trigonometric equations can be
solved either analytically or graphically.
Analytical solution
Example 4.6
Solve the equation
(a) sin
θ
=
√3
for 0° ≤
θ
≤ 360°,
2
(b) cos
θ
=
–
0.8 for
–
180° ≤
θ
≤ 180°,
(c) tan
θ
= √3 for
–
360° ≤
θ
≤ 360°.
Solution
(a) sin
θ
= √3 .
2
Since
√3
is positive, and
θ
is between 0° and
2
360°, there are two angles, one acute
(rst quadrant) and the other obtuse
(second quadrant) that satisfy the equation
(Fig. 4.5).
Fig. 4.5
Now, sin 60° =
√3
2
∴sin
θ
= sin 60°
90°
180°
270°
0°
O
90°
180°
270°
0°
O
90°
180°
270°
0°
O
θ
= sin
–1
√3
2
θ
= 60
57
( )
( )
( )
( )
⇒ 2x =
2π
c
3
i.e. x =
π
c
3
Also, 2x +
π
c
= 2π
c
+
π
c
,
since adding 2π
c
6
6
does not change the
value or sign of the
sine ratio.
⇒ 2x
= 2π
c
i.e. x = π
c
Likewise,
2x +
π
c
= 2π
c
+
5π
c
(adding 2π
c
)
6
6
⇒ 2x = 2π
c
+
2π
c
3
⇒ 2x =
8π
c
3
i.e. x =
4π
c
3
∴ x = 0
c
,
π
c
,
π
c
or
4π
c
.
3
3
Note that we could obtain more values of x by
adding other multiples of 2π
c
(360°) to
π
c
and
5π
c
.
6 6
However, such values would be outside the
specied range, so, the above values constitute
the required solution.
Example 4.8
Find all the positive angles, not greater than
180°, which satisfy the equation
sin
2
θ
– 2tan
θ
= 0.
cos
θ
Solution
If the equation involves more than one ratio, we
use trigonometric relations or basic denitions
of trigonometric ratios to change it to an
equation that involves only one ratio. Thus
sin
2
θ
– 2tan
θ
= 0 can be written as
cos
θ
tan 60° = √3
∴ tan
θ
= tan 60°
or tan
θ
= tan(180° + 60°)
Thus,
θ
= 60° or 240°
But these angles could also be measured in
the negative direction to give
θ
=
–
120° or
–
300°, which are also in the specied range.
∴
θ
=
–
300°,
–
120°, 60° or 240°.
Example 4.7
Solve the equation 4sin 2x +
π
c
= 2, for
6
0
c
≤ x ≤ 2π
c
.
Solution
We rst write
2x +
π
c
as a single quantity,
6
say α.
Thus, 4sin 2x +
π
c
= 2 becomes
6
4 sin α
= 2
⇒ sin α = ⇒ α is in the rst or second
quadrant.
But
π
c
= 30° and sin 30° =
6
∴ sin
π
c
=
6
∴ sin α = sin
π
c
…………… rst quadrant
6
or sin α = sin π
c
–
π
c
………second quadrant
6
∴ α =
π
c
or
5π
c
.
6 6
If α =
π
c
,
6
then 2x +
π
c
=
π
c
6 6
⇒ 2x = 0
c
i.e. x = 0
c
If α =
5π
c
,
6
then 2x +
π
c
=
5π
c
6 6
1
2
–
1
2
–
1
2
–
58
( )
1
3
–
1
3
–
3
5
–
15
17
––
5
12
––
sin
2
θ
– 2
sin
θ
= 0.
cos
θ
cos
θ
Multiplying both sides by cos
θ
gives
sin
2
θ
– 2sin
θ
= 0.
Note that sin
2
θ
– 2sin
θ
= 0 is of the form
x
2
– 2x = 0 Let sin
θ
= x
⇒ x(x – 2) = 0
⇒ either x = 0 or x = 2
Substituting back:
sin
θ
= 0
θ =
sin
–1
⇒
θ
= 0° or 180°
or sin
θ
= 2, which is invalid since sine
cannot be greater than 1.
∴ the required solution is
θ
= 0° or 180°.
Exercise 4.3
1. If angle α is known to be positive and less
than 360°, nd the possible values of α if:
(a) tan α = √3 (b) tan
α
=
–
1
2
(c) tan
2
α = (d) cos α =
–
2. Given that 0° ≤
θ
≤ 360°, solve the
following equations completely.
(a) 3cos
θ
+ 1 = 0 (b) 4sin
θ
–1 = 0
(c) 3sin
θ
+ 4 = 0 (d) 4tan
2
θ
= 1
3. Solve for x, given that
–
180° ≤ x ≤ 180°.
(a) 3cos
2
x – 2cos x = 0
(b) 2tan
2
x = 1 + tan x
4. Solve the following equations given that
0° ≤
θ
≤ 360°.
(a)
2 – cos
θ
=
1
(b) 3tan
θ
=
cos
θ
sin
θ
tan
θ
sin
θ
5. Solve the following equations for x
between
–
180° and 180°.
(a) cos
2
x + 3sin
2
x – 2 = 0
(b) tan x –
3cos x
= 0
sin x
(c) 2tan x =
1
cos
2
x
(d) tan
2
x –
3
+ 3 = 0.
cos x
6. (a) Given that 0° ≤
θ
≤ 90°, solve the
equation:
(i) sin 2
θ
= cos 3
θ
(ii) 4cos
2
θ
– 4cos
θ
+ 1 = sin
2
θ
(b) Express the equation
sin
2
θ
+ 2cos
2
θ
= 3sin
θ
cos
θ
in terms
of tan
θ
. Hence solve for
θ
, given that
θ
is an acute angle.
7. If sin
θ
= , nd, without using tables, the
value of (a) cos
θ
(b) tan
θ
.
8. Given that sin
θ
= , nd without using
tables, the value of
(a) cos
θ
, (b) tan
θ
, for 0° <
θ
< 360°.
9. If tan
θ
=
–
and
θ
is such that
0° <
θ
< 360°, nd the possible value of
(a) sin
θ
, (b) cos
θ
.
Hence, state the possible value of
θ
.
10. Solve the following equations for values of
θ
from 0° to 360°.
(a) 1 – 2sin
2
θ
+ cos
θ
= 0
(b) 3tan
2
θ
–
7
+ 5 = 0
cos
θ
(c) 3cos
2
θ
– 7cos
θ
= 6
(d) sin (
θ
+ 30°) = 0.7
Simple trigonometric graphs
In Book 3, we drew graphs of simple
trigonometric functions such as y = sin x,
y = cos x and y = tan x, where angles were
expressed in degrees or in radians.
In this section, we are going to learn more about
the behaviour and properties of the sine and
cosine functions and graphs.
59
The sine graph
It is important that we realize that all sine
curves have the same general shape and
properties. For example, the graph of sin x
(Fig. 4.8) gives a curve in the form of a wave
which repeats itself every 360° (2π
c
). Hence,
the sine curve is also called the sine wave.
Note: Both degrees and radians are used in this
graph to illustrate the equivalence of the
specied values.
Fig. 4.8
Between the origin and the point B on the x-axis
(F ig. 4.8), the curve has completed one cycle and
is about to start the same behaviour all over again.
This graph is therefore said to be a periodic
curve, whose period is 360° (= 2π radians).
The wave oscillates between a maximum of 1
unit and a minimum of
–
1 unit about the x-axis
(the mean line). The vertical distance between
a maximum or minimum point on the curve
and the x-axis is called the amplitude of the
curve. The horizontal distance between two
corresponding points on a wave, e.g. the origin
and B in Fig. 4.8, is called the wavelength of
the function.
The cosine graph
The cosine graph (Fig. 4.8) is also a wave with
a period of 360°, and an amplitude of 1. It is
important to note that a cosine curve is the same
as the sine curve translated through 90°
along
the x-axis.
Variations of the sine and cosine curves
Graphs of the form y = asin x and y = bcos x
For the purpose of illustrating these graphs, we
will use the specic functions y = 3sin x and
y = 2cos x. We start by making a table of values
to help us draw reasonably accurate graphs for
0°≤ x ≤ 360°.
x 0° 30° 60° 90° 120° 150°
sin x 0 0.5 0.866 1 0.866 0.5
3sin x 0 1.5 2.598 3 2.598 1.5
cos x 1 0.866 0.5 0
–
0.5
–
0.866
2cos x 2 1.732 1.0 0
–
1.0
–
1.732
180° 210° 240° 270° 300° 330° 360°
0
–
0.5
–
0.866
–
1
–
0.866
–
0.5 0
0
–
1.5
–
2.598
–
3
–
2.598
–
1.5 0
–
1
–
0.866
–
0.5 0 0.5 0.866 1
–
2
–
1.732
–
1.0 0 1.0 1.732 2
Table 4.1
The graphs in Fig 4.9 represent waves similar
to the basic sine and cosine graphs in Fig 4.8.
The only dierence is in their amplitudes.
The amplitude of y = 3sin x is 3 while that of
y = 2cos x is 2.
Fig. 4.9
In general, if y = a sin x, the curve gives a wave
whose amplitude is a. Similarly, y = b cos x
gives a wave whose amplitude is b. Both waves
have a period of 360° or 2π
c
.
3
2
1
0
–
1
–
2
–
3
60° 120° 180° 240° 300° 360°
x
y
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
y = 3sin x
y = 2cos x
3
2
–
5
2
0
1
–
1
amplitude = 1 unit
– – – – – – – – – – – – – – – – – –
B
y
x
wavelength
π
c
2
–
π
c
2π
c
π
c
3π
c
90° 180° 270° 360° 450° 540°
– – – – – – – – – – – –
– – – – – – – – – –
y = cos x
y = sin x
–
}
π
c
60
1
2
–
y = 3sin x
1
2
–
y = 2 cos 2x
radians. This period may be found analytically
as follows:
Period =
360°
= 720° or 4π
radians
Similarly the graph of y = cos 2x has a period of
360°
= 180° or π radians.
2
In general, the graph of y = sin ax, where a is a
constant, has a period equal to
360°
or
2π
c
.
a
a
Also, the graph of y = cos bx has a period of
360°
or
2π
c
.
b b
Note: The graph of y = sin ax is the image of
the graph of y = sin x under a stretch factor
1
a
parallel to the x-axis. Describe the transformation
that maps y = cos x onto y = cos bx.
Graphs of the form y = a sin bx and
y = a cos bx
Now consider the graphs in Fig. 4.11, i.e.
y = 3 sin x and y = 2cos 2x.
Fig. 4.11
In the case of y = 3sin x, the curve repeats
itself after every 4π
c
(= 720°) and the maximum
value of y is 3. Hence, the curve has a period of
4π
c
and an amplitude of 3.
1
2
–
1
2
–
1
2
–
Note: The graph of y = asin x is the image of
the graph of y = sin x under a stretch factor a
parallel to the y-axis. Similarly, the curve of
y = bcos x is the image of y = cos x under a
stretch factor b parallel to the y-axis.
Graphs of the form y = sin ax and
y = cos bx
To illustrate these graphs, we will use the
function y = sin x and y = cos 2x, for
0≤ x ≤ 360°.
x 0° 30° 60° 90° 120° 150° 180°
x 0° 15° 30° 45° 60° 75° 90°
sin x 0 0.2588 0.5 0.7071 0.866 0.9659 1
2x 0° 60° 120° 180° 240° 300° 360°
cos 2x 1 0.5
–
0.5
–
1
–
0.5 0.5 1
210° 240° 270° 300° 330° 360°
105° 120° 135° 150° 165° 180°
0.9659 0.866 0.7071 0.5 0.2588 0
420° 480° 540° 600° 660° 720°
0.5
–
0.5
–
1
–
0.5 0.5 1
Table 4.2
When making tables of values for such functions
as y = sin x and y = cos 2x, a step by step method
is the best. It minimises errors in the calculations.
It is important to note that for y = sin x, we
draw a graph of y against x and not against
x. Similarly, for y = cos 2x, we draw a graph
of y against x and not against 2x (Fig. 4.10).
Fig. 4.10
The graph of y = sin x repeats itself every
720°. Therefore, its period is 720° or 4π
×
y
1
0.5
0
–
0.5
–
1
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
y
= cos 2
x
y
= sin
x
1
2
–
x
120° 180° 240° 300° 360°60°
1
2
–
y
x
π
c
0
3
1
2
–
1
–
3
–
2
2π
c
3π
c
4π
c
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
61
For y = 2cos 2x, the period is π
c
and the
amplitude is 2.
In general:
For the function of the form y = a sin bx or
y = a cos bx, where a and b are constants, the
value a represents the amplitude and
360°
or
2π
c
represents the period.
b
b
Note: The transformation involved in y = a sin bx
or y = acos bx is a two way stretch. Fully
describe it.
Graphs of the form y = sin (x ± b) and
y = cos (x ± b)
You are now familiar with the process
of making tables of values for various
trigonometric functions. Hence, carry out the
following activity to establish the properties
of graphs of the form y = sin (x ± b) and
y = cos (x ± b).
Activity 4.1
1. Using the functions y = sin (x + 60°) and
y = cos (x – 60°), make the respective tables
of values for 0° ≤ x ≤ 360°.
2. Using your tables of values, and on the
same axes, draw the graphs of
y = sin (x + 60°) and y = cos (x + 60°),
labelling them clearly.
3. Describe the similarities between
(a) y = sin (x + 60°) and y = sin x, and
(b) y = cos (x + 60°) and y = cos x.
4. Describe the dierence between the two
pairs of graphs, and describe the symmetry
of each graph.
Now consider the graph in Fig. 4.12.
Fig. 4.12
The addition of 60° in y = 4sin (x + 60°) does
not aect the period or the amplitude of the
graph. It only shifts the curve by 60° to the left
along the x-axis. This shift is called the phase
angle. In this case the phase angle is +60°.
If we had y = 4sin (x + 60°), the curve would
have been shifted by 60° to the right. Conrm
this by drawing the curve of y = 4sin (x + 60°)!
The same properties that are true for a sine
function always hold for a similar cosine
function.
In general:
The curve of y = asin (bx + c) or
y = acos (bx + c) is like that of y = sin x
or y = cos x, respectively,
1. with the amplitude multiplied by |a|,
2. with the period divided by |b|,
3. shifted by
c
along the x-axis (to the left
b
if
c
is positive and to the right if
c
is
b b
negative) and
4. inverted if a is negative, i.e. a reection
in the x-axis.
y
x
4
0
–
4
–
90°
90° 180° 270° 360°
y = 4sin (x + 60°)
62
Points to remember:
1. All sine curves have the same general shape
and properties.
2. A cosine curve is the same as the sine curve
translated through
π
radians along the x-axis
(Fig. 4.8).
2
3. The sine and cosine curves have bilateral
symmetry about lines parallel to the vertical
axis, through maximum or minimum points
on the curves.
4. Sine and cosine curves have rotational
symmetry of order two about the
intersections of the horizontal axis and
the curves.
Example 4.6
On the same axes, sketch th graph of the
function y = sin x and y = sin(x + 60º) in the
domain 0 ≤ x ≤ 360º
Describe the transformation that maps y = sin x
onto y = sin(x + 60º)
Solution
Translation, translation vector
-60
0
Exercise 4.4
1. Copy and complete Table 4.3.
x 0 30 60 90 120 150 180
sin 2x 0 0.866
–
0.866
3sin 2x 2.598
–
2.598
4cos x 4 2
–
2
x 210 240 270 300 330 360
sin 2x
–
0.866
3sin 2x 2.598 0
4cos x
–
3.464
Table 4.3
(a) On the same axes, draw the graphs of
(i) y = 3sin 2x, and (ii) y = 4cos x.
(Retain your graphs: You will need
them later)
(b) State
(i) the amplitudes of the two graphs,
(ii) the periods of the two graphs.
(c) Describe fully the transformation that
maps the graph of
(i) y = cos x onto the graph of
y = 4cos x.
(ii) y = sin x onto the graph of
y = 3sin 2x.
2. Copy and complete Table 4.4 for
y = cos (2x + 30°).
x 0 30 60 90 120 150 180
2x + 30 30 210
cos (2x + 30) 0 0.866
y 0 1.299
x 210 240 270 300 330 360
2x + 30 450 690
cos (2x + 30)
–
0.866 0.866
y
–
1.299 1.299
Table 4.4
(a) Draw the graph of y = cos (2 x + 30).
(You will need this graph later).
(b) State
(i) the amplitude,
(ii) the period, and
(iii) the shift angle (left or right).
(c) Describe fully the transformation that
maps the graph of y = cos x onto the
graph of y =
cos (2x + 30°).
3
2
–
3
2
–
y
x
0
1
–
0.5
60° 120° 180° 240°
300°
–
1
0.5
360°
63
1
2
–
1
2
–
3. Copy and complete Table 4.5.
Function Amplitude Period
Shift
Angle Left/Right
sin x
cos 2x
–
2cos x
sin x
5sin (x + 30°)
cos (2x – 60°)
a sin(px +α)
–
b cos(πx –
θ
)
Table 4.5
In Questions 4 to 8, use the given graphs
(Fig 4.13 to Fig 4.17) to state:
(a) the amplitude,
(b) the period, and
(c) the shift angle (left/right).
In each case, suggest a possible function for the
curve.
4.
Fig. 4.13
5.
y
x
0
1
–
1
30° 60° 90° 120°
150°
y
x
0
2
–
2
90° 180° 270° 360°
450°
–
1
1
6.
Fig. 4.15
7.
Fig. 4.16
8.
Fig. 4.17
Graphs of y = tan x and its variations
The graphs of y = tan x and its variations are
dierent from the graphs we have dealt with
so far. However, they are periodic and have a
shape which is best described by a graphical
illustration. For the purpose of this illustration,
we will use two simple functions y = tan x and
y = tan 2x, for 0° ≤ x ≤ 360°.
y
x
0
4
–
4
180° 360° 540° 720°
900°
Fig. 4.14
y
x
0
1
–
1
60° 120° 180° 240°
300°
360°
y
x
0
1
–
1
45° 90° 135° 180°
225°
64
Table 4.6 gives the corresponding values of x
and y in each case.
x 0° 30° 60° 90° 120° 150° 180°
tan x 0 0.5773 1.732 ∞
–
1.732
–
0.5773 0
210° 240° 270° 300° 330° 360°
0.5773 1.732 ∞
–
1.732
–
0.5773 0
2x 0° 60° 120° 180° 240° 300° 360°
tan 2x 0 1.732
–
1.732 0 1.732
–
1.732 0
420° 480° 540° 600° 660° 720°
1.732
–
1.732 0 1.732
–
1.732 0
Table 4.6
Fig. 4.18 represents the graph of y = tan x
whose period is 360°. Though this graph can be
extended innitely, it is discontinuous at certain
points, i.e. when x = 90° and x = 270°. At these
points, the values of y are undened and
therefore the curve does not cross the x = 90°
and x = 270° lines.
Fig. 4.18
Remember that:
tan x =
sin x
cos x
and sin 90° = 1 while cos 90° = 0
x
4
y
3
2
1
0
–
1
–
2
–
3
–
4
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
60° 120° 180° 240° 300° 360°
Fig. 4.19
4
3
2
1
135° 235°45°
315°
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
x
120° 180° 240° 300° 360°60°
–
1
0
–
2
–
3
–
4
y
⇒ tan 90° =
sin 90°
cos 90°
=
1
(division by 0 is undened)
0
∴ tan 90° is undened.
Similarly, tan 270° =
sin 270°
cos 270°
=
–
1
(undened)
0
∴ the graph of y = tan x only approaches the
lines x = 90° and x = 270° and never crosses
them. Such lines are known as asymptotes.
The graph of y = tan x has rotational symmetry
of order two about the point (180°, 0). If this
graph were extended in both directions innitely,
every crossing point on the x-axis would be a
centre of rotation of order two.
Fig. 4.19 shows the graph of y = tan 2x for
0° ≤ x ≤ 360°. The graph is discontinuous at the
point where x = 45°, 135°, 225° and 315°.
The graph in Fig. 4.19 has rotational symmetry
of order 2 about (180°, 0).
65
Note that this graph has the same properties
as that of y = tan x, and that x can be given in
terms of radians or degrees.
Now, consider the function y =
tan (x + 30°).
Copy and complete Table 4.7. Round o the
nal values of the function to 1 d.p., and use
them to draw the graph of y =
tan (x + 30°)
for 0 ≤ x ≤ 360°.
x 0° 30° 60° 90° 120° 150° 180°
x + 30 30° 60° 90° 120° 150° 180° 210°
tan (x + 30) 0.5773
–
0.5773
tan(x + 30) 0.2887
210° 240° 270° 300° 330° 360°
240° 270° 300° 330° 360° 390°
0.5773
–
0.866
Table 4.7
Use your graph to answer the following
questions.
1. State the period and the symmetry of your
graph.
2. Does your graph have asymptotes? If yes,
how many? State their equations.
3. How does your graph compare with the
graph of y = tan x for the same interval?
Exercise 4.5
In each of the questions in this exercise, make a
table of values for the stated interval and use it
to draw the graph of the given function. Clearly,
mark and label the asymptotes, state the period
of the graph, and describe the symmetry of the
graph within the given interval.
1. y = tan x ; 0
c
≤ x ≤ 2π
c
2. y = tan x ; 0° ≤ x ≤ 360°
3. y = 3tan x ; 0
c
≤ x ≤ 3π
c
4. y = tan (x – 30°) ; 0° ≤ x ≤ 360°
5. y = tan (2x + 60°) ; 0° ≤ x ≤ 360°
1
2
–
1
2
–
6. y = 2tan (x – ) ; 0
c
≤ x ≤ 2π
c
7. y = 3tan (2x + 30°) ; 0° ≤ x ≤ 360°
Graphical solution of trigonometric
equations
We have already solved trigonometric equations
by analytical method. Graphical methods can
also be used for such equations.
Graphical method of solving equations
involves drawing two independent graphs on
the same axes, and deriving the roots from the
intersection of the graphs.
Example 4.9
On the same axes, draw the graphs of
y = 2cos x and y = sin x for 0° ≤ x ≤ 360°.
Hence nd the values of x that satisfy the
equation 2cos x = sin x.
Solution
Table 4.8 is the required table of values.
x 0° 30° 60° 90° 120° 150° 180°
x 0° 15° 30° 45° 60° 75° 90°
cos x 1 0.9659 0.866 0.7071 0.5 0.2588 0
2cos x 2 1.9318 1.732 1.4142 1 0.5176 0
sin x 0 0.866 0.5 0 0.5 0.866 0
x 210° 240° 270° 300° 330° 360°
x 105° 120° 135° 150° 165° 180°
cos x
–
0.2588
–
0.5
–
0.7071
–
0.866
–
0.9659
–
1
2cos x
–
0.5176
–
1
–
1.4142
–
1.732
–
1.9318
–
2
sin x
–
0.5
–
0.866
–
1
–
0.866
–
0.5 0
Table 4.8
Using a scale of, say, 1 cm to 60° on the
horizontal axis, and 1 cm to 1 unit on the
vertical axis, we plot the two graphs on the
same axis (Fig. 4.20).
1
2
–
1
2
–
1
3
–
1
2
–
1
4
–
π
c
3
––
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
66
Fig. 4.20
The graphs in Fig. 4.20 intersect at only one
point, in the given interval, where the value of x
is 180°.
Therefore the root of the equation 2cos x = sin x
is 180°.
Note that the same result could have been
achieved by solving the equation
2cos x – sin x = 0. This is done by plotting the
graph of y = 2cos x – sin x for 0° ≤ x ≤ 360°.
The root is then given by the value of x at the
intersection of this graph and the line y = 0.
Example 4.10
Solve graphically, the equation 2x + tan x – 5 = 0
for x from 0
c
to 2π
c
.
Solution
If 2x + tan x – 5 = 0, then tan x = 5 – 2x.
Hence, the roots of the equation
2x + tan x – 5 = 0 will be given by the
intersection of the two graphs, y = tan x and
y =
–
2x + 5.
To plot the graph of y =
–
2x + 5, we make a
table of values as Table 4.9.
x 0
c
π
c
π
c
2
–
2x 0
–
3.14
–
6.28
+
5
+
5
+
5
+
5
y 5 1.86
–
1.28
Table 4.9
y
x
0
2
–
2
60° 120° 180° 240°
–
1
1
300° 360°
y
= sin
x
y = 2cos x
1
2
–
– – – – – – – – – –
To plot the graph of y = tan x, we assign values
to x, in terms of π radians, and make a table of
values for tan x as in Table 4.10.
x 0
c
π
c
π
c
π
c
2π
c
5π
c
π
c
6 3 2 3 6
tan x 0 0.577 1.732 ∞
–
1.732
–
0.577 0
7π
c
4π
c
3π
c
5π
c
11 π
c
2π
c
6 3 2 6 6
0.577 1.732 ∞
–
1.732
–
0.577 0
Table 4.10
On the same axis, and using a scale of, say,
1 cm to represent
π
radians on the horizontal
3
axis, and 1 cm to represent 2 units on the
vertical axis, we draw the graphs of y = tan x
and y = 5 – 2x as in Fig. 4.21.
Fig 4.21
Fig 4.21 shows that line y = 5 – 2x meets
the curve y = tan x where the values of x are
approximately
23π
c
(≈ 1.19
c
),
53π
c
(≈ 2.78
c
) and
60 60
31π
c
(≈ 4.87
c
).
20
1
2
–
1
2
–
1
2
–
y
6
4
2
0
–
2
–
4
–
6
π
c
3
π
c
2π
c
x
y = tan x
y = 5 – 2x
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
2
π
c
3
4
π
c
3
5
π
c
3
67
Thus, the roots of the equation 2x + tan x – 5 =
0 are x ≈ 1.19
c
, 2.78
c
or 4.87
c
.
These values can be converted into degrees if so
required.
Exercise 4.6
1. Use the graphs that you drew in Question 1
of Exercise 4.4 to solve the equation
3sin 2x = 4 cos x.
2. Use the graph that you drew in Question 2
of Exercise 4.4 to solve the equation
(a) cos (2x + 30°) = 0
(b) cos (2x + 30°) =
–
1
3. Draw the graphs of y = 3sin x and
y = sin (50° – x) for 0° ≤ x ≤ 90°.
Use your graphs to solve the equation
3sin x = sin (50° – x).
4. Draw the graphs of y = 1 + cos 3x and
y = sin 2x from
–
90° to + 90°. Hence,
determine the solutions of the equation
1 + cos 3x = sin 2x.
5. Draw the graph of y = 2sin x + 3cos 2x from
x = 0° to x = 180°.
Use your graph to solve the equation
2sin x + 3cos 2x = 1.4.
3
2
–
3
2
–
6. On the same axes, draw the graphs of
y = cos (x + 30°) and y = sin x for
0° ≤ x ≤ 360°.
Use your graphs to solve the equation
cos (x + 30°) = sin x.
7. Using the same scale and axes, draw the
graphs of y = tan 2x and y = 4cos x – 1 for
values of x from 10° to 35° at intervals of 5°.
Use your graphs to solve the equation
4cos x – tan 2x = 1
8. Using the same axes draw the graphs of
y = sin (x + 30°) and y = cos (x – 15°) for
–
30° ≤ x ≤ 240°, using 30° intervals.
Use your graphs to solve the equation
cos (x – 30°) = sin (x + 30°).
9. On the same axes, using a scale of 1 cm for
30° on the horizontal axis and 2 cm for 1
unit on the vertical axis, draw the graphs of
y = 4cos x and y = 3 sin 2x for 0° ≤ x ≤ 360°.
Use your graphs to solve the equation
3sin 2x – 4cos x = 0.
10. On the same axes draw the graphs of
y
1
= 3cos (2x + ) and y
2
= cos x for
0
c
≤ x ≤ 2π
c
.
State the period and amplitude of
(i) y
1
(ii) y
2
.
1
3
–
1
3
–
π
c
3
68
( )
( )
1
2
–
(b) the locus of points 3 cm from the
midpoint of AB.
Measure the distance between the
intersections of the two loci.
6. A rectangle has vertices O(0 , 0), A(6 , 0),
B(6 , 4) and C(0 , 4). Find the image of
the rectangle under the transformation
represented by
(a) S
1
=
1 0
(b) S
2
=
1 0
0 0 0
In each case, describe the transformation
fully.
7. Using a suitable working mean, calculate
the mean of the values:
166, 171, 163, 169, 174, 172, 175,
168, 171
8. Solve the equation:
(a) sin x = 0.5 for 0° ≤ x ≤ 180°,
(b) sin
2
x = 0.866 for values of x such that
–
180° ≤ x ≤ 180°.
9. Construct ∆PQR with PQ = 6 cm,
QR = 4.4 cm and PR = 5.2 cm. On the
same diagram, construct
(a) the locus of points equidistant from RP
and QP,
(b) the locus of points 2 cm from RQ.
There are two points A and B that satisfy
the two loci. Identify A and B and state the
distance between them.
10. Solve the following equation for
0° ≤ x ≤ 360°.
(a) 1 + cos x = 2 sin
2
x
(b) 2 sin
2
x – cos
2
x = 2
11. Two hundred Form 1 pupils were weighed
and their masses recorded as in Table R1.2.
1
2
–
( )
Revision exercise 1.1
1. A quadrilateral has vertices A(
–
4 , 0),
B(4 , 2), C(2 , 0) and D (4 ,
–
2). On the same
diagram, draw the quadrilateral and its image
under the transformation with matrix
–
2 0
.
0
–
2
2. In each of the following cases, state and
illustrate the locus of P.
(a) P is the locus of points 2 cm from a
xed line of innite length.
(b) P is 4 cm from a xed point A.
(c) P is equidistant from two xed points B
and C, on the same plane.
(d) P is equidistant from two xed line
segments DE and DF.
(e) P is equidistant from two xed parallel
lines AB and CD.
(f) G and H are xed points and the area of
triangle GPH is constant.
(g) M and N are xed points and ∠MPN is
constant.
3. Solve the following equations for x such
that 0° ≤ x ≤ 360°.
(a) sin x = 0.8
(b) tan x = 2.5
(c) cos 2x = 0.314 9 (d) sin x =
–
0.953 7
4. Find the mean of the following distribution
of marks (Table R1.1).
Mark 0–9 10–19 20–29 30–39 40–49
Frequency 2 6 6 7 10
Mark 50–59 60–69 70–79 80–89
Frequency 11 6 8 4
Table R1.1
5. Construct ∆ABC with AB = 8 cm, BC = 6 cm
and AC = 7cm. On the same diagram
construct
(a) the locus of P equidistant from B and C.
REVISION EXERCISES 1 (CHAPTERS 1–4)
1-4
69
( ) ( )
Mass (kg) 35–39 40–44 45–49 50–54 55–59
No. of
4 0 13 54 58
pupils
Mass (kg) 60–64 65–69 70–74
No. of
50 13 8
pupils
Table R1.2
Using an appropriate working mean and
scale factor, use the coded method to nd
the mean mass of the pupils.
12. The transformation with matrix
1
–
1
maps
1 1
the square WXYZ with W(2 , 0), X(0 , 2),
Y(
–
2, 0) and Z(0 ,
–
2) onto W′X′Y′Z′. Plot
WXYZ and W′X′Y′Z′ on the same axes and
describe the transformation fully.
Revision exercise 1.2
1. Draw a line segment AB = 6 cm. Construct
(a) the locus of points equidistant from A
and B,
(b) the locus of points P such that the area
of ∆APB = 12 cm
2
,
(c) the locus of points C such that ABC is
an isosceles triangle of area 12cm
2
.
2. Given that sin 2x = 2 sin x cos x, show that
(sin x + cos x)
2
= 1 + sin 2x.
3. Calculate the mean and standard deviation
of the values:
87, 89, 90, 93, 95, 96, 98, 99, 101, 102.
4. A(0 , 1), B(1 , 1), C(1 , 2) and D(0 , 2) are
the vertices of a square. A′(2 , 1), B′(3 , 1),
C′(5, 2) and D′(4 , 2) are the vertices of its
image under a shear. Draw the object and
its image on the same axes and use your
diagram to nd the shear factor and hence
the matrix of the shear.
5. A certain transformation maps the
parallelogram with vertices O(0 , 0),
M(1 , 0), N(1.5 , 1) and P(0.5 , 1) onto
another one with vertices O′(0 , 0), M′(1 , 0),
N′(
–
1 , 1) and P′(
–
2 , 1). Write down the
matrix of the transformation and describe
the transformation fully.
6. PQ is a line segment of length 8 cm, and R
is the midpoint of PQ. State the locus of the
centre of a circle such that
(a) its radius is 2 cm, and PQ is a tangent
to the circle,
(b) the circle touches the line PQ at R and
has a radius of any length,
(c) PQ is a chord of the circle and the
circle has a radius of any length.
Illustrate each of the loci by means of a
sketch.
7. Calculate the mean and mean deviation of
the data in Table R1.3.
x 70 71 72 73 74 75 76 77 78 79 80
f 2 5 10 11 15 17 14 12 8 5 1
Table R1.3
8. Show that
(a) tan x +
1
=
1
tan x cos x sin x
(b) 2 – 3 cos
2
x = 3sin
2
x – 1
(c)
sin x
+
1 + cos x
=
2
1 + cos x sin x sin x
9. A is a xed point on the circumference of
a circle centre O. M is the midpoint of a
variable chord AB. Given that O is a xed
point, illustrate and state the locus of M.
10. S =
1 0
and E =
2 0
.
2 1 0 2
(a) What transformations are represented
by S and E?
(b) What is the single transformation
matrix equivalent to (i) SE (ii) ES?
11. Choose a suitable working mean and use it
to nd the mean and standard deviation of
the distribution in Table R1.4.
( )
70
( )
Class 10–20 20–30 30–40 40–50 50–60
Frequency 2 4 4 8 6
Class 60–70 70–80
Frequency 3 2
Table R1.4
12. Copy and complete Table R1.5 given below.
x 0° 30° 60° 90° 120°
2 cos x 2 0
sin (x – 15°) 0.258 8 0.965 9
x 150° 180° 210° 240° 270°
2 cos x
–
1.732 0
sin (x – 15°)
0.258 8
–
0.707 1
x 300° 330° 360°
2 cos x 2
sin (x – 15°)
–
0.707 1
Table R1.5
Using the same axes, draw the graphs of
y = sin (x – 15°) and y = 2cos x for
0° ≤ x ≤ 360°.
Use your graphs to solve the equation
(a) sin (x – 15) – 0.5 = 0,
(b) 2cos x – 1 = 0.5,
(c) sin (x – 15) – 2cos x = 0.
Revision exercise 1.3
1. The vertices of a quadrilateral are (0 , 2),
(2 , 1), (5 , 3) and (3 , 4).
(a) What shape is the quadrilateral?
(b) Find its area.
(c) Find the area of its image under the
transformation given by
–
2 1
.
4
–
5
2. AB is a tangent to a circle at B. If AB is
12 cm long, and the circle has a radius 5 cm
long, illustrate and state the locus of A.
3. Table R1.6 shows the distribution of marks
obtained in a test by some students.
Marks 0–20 20–30 30–40 40–50 50–60 60–70 70–80
No. of
3 8 17 30 23 10 9
students
Table R1.6
(a) Construct a cumulative frequency
curve for this distribution.
(b) Using your graph, estimate
(i) the median mark,
(ii) the number of students who got
65 marks and above,
(iii) the pass mark if only 10% of the
students passed.
4. Using the same axes, draw the graphs of
y = cos (x – 15°) and y = 2 sin x for
0° ≤ x ≤ 360°.
Use your graphs to solve the equation
(a) cos (x – 15°) – 0.5 = 0,
(b) 2 sin x – 1 = 0.5,
(c) cos (x – 15°) – 2 sin x = 0.
5. A stretch with the x-axis invariant maps a
point P(5 , 2) onto P′(5 , 6). Find
(a) the scale factor of the stretch,
(b) the operator of the stretch,
(c) the vertices of the image A′B′C′D′ of
rectangle ABCD with vertices A(0 , 1),
B(4 , 1), C(4 , 4) and D(0 , 4),
(d) the area of the image rectangle.
6. The data in Table R1.7 refer to the marks
obtained by some students in a mathematics
test.
63 61 73 49 36 61 42 52 58 39 46 74
56 73 43 51 51 82 42 69 60 84 66 39
76 21 67 64 73 37 31 71 47 66 48 67
89 57 49 30 57 63 31 37 48 77 20 53
63 74 59 59 56 71 71 42 51 76 61 40
61 74 69 99 56 57 69 22 81 42 63 47
Table R1.7
(a) Construct a frequency distribution of
this data with class intervals of 10,
starting from 20–29.
(b) Draw an ogive for the distribution
(c) Use your ogive to determine the lower
quartile, the median and the upper
quartile of the distribution. Hence, nd
the quartile deviation.
71
7. (a) Draw a line segment AB = 10 cm. On
AB, construct ∆ABC of area 20 cm
2
and ∠ACB = 90°.
(b) On the ∆ABC, construct the locus of P
such that ∠APB = 45°.
(c) Construct also, the locus of Q such
that ∆AQB has maximum area and
calculate this area.
8. (a) Copy and complete Table R1.8.
x 0° 15° 30° 45° 60° 75° 90°
cos (4x – 60°) 0.5
–
0.5
–
0.5
x 105° 120° 135° 150° 165° 180°
cos (4x – 60°)
0.5
–
1 0.5
Table R1.8
(b) Draw the graph of y = cos (4x – 60°)
for 0° ≤ x ≤ 180° using a scale of 1 cm
for 15° on the x-axis and 4 cm to
represent 1 unit on the y-axis.
(c) State the period of the wave.
(d) Use your graph to solve the equation
cos (4x – 60°) = 0.2.
9. ∆PQR with vertices P(2 , 2), Q(4 , 2) and
R(3 , 4) is enlarged by a scale factor of 2
and centre (0 , 0) to ∆P′Q′R′. It is then
given a positive quarter-turn about (0 , 0) to
map onto ∆P′′Q′′R′′.
(a) Write down the matrix of each of these
transformations.
(b) What is the matrix of the combined
transformations?
(c) What are the verticles of ∆P′′Q′′R′′?
(d) Find the inverse of the matrix in (b)
and state its eect on ∆P′′Q′′R′′.
10. Table R1.9 is a frequency distribution of
the monthly salaries (in K£) of some
government employees. Use calculation
method to nd the median and state the
modal class.
Salary (K£) 301–305 306–310 311–315 316–320
No of
13 25 43 65
employees
Salary (K£) 321–325 326–330
No of
55 34
employees
Table R1.9
11. Copy and complete Table R 1.10.
x 0° 30° 60° 90° 120° 150° 180°
sin x 0 0.5 0.866
2sin (x + 30°) 1.732 1.732
–
1
x 210° 240° 270° 300° 330° 360°
sin x
–
0.5
–
0.866
2sin (x + 30°)
–
1.732 0
Table R1.10
(a) On the same axes, draw the graphs of
y = sin x and y = 2sin (x + 30°) for
0° ≤ x ≤ 360°.
(b) For each graph, state
(i) the amplitude,
(ii) the period.
(c) Use your graphs to solve the equation
(i) sin x = 2 sin (x + 30°).
(ii) 2sin (x + 30°) =
–
1.4
(d) Fully describe the transformation(s)
that maps sin x onto 2 sin (x + 30°).
12. Construct a parallelogram ABCD in which
AB = 9 cm, AD = 5 cm and ∠BAD = 60°.
Measure and state the length of AC.
On the same diagram construct
(a) the locus of a point P which moves so
that it is equidistant from A and C,
(b) the locus of a point Q which moves so
that ∠BQD = 90°,
(c) the position of a point X which lies
inside the parallelogram so that
AX ≥ XC and ∠BXD ≤ 90°.
72
Geometrical properties of common
solids
In Book 1, we looked at some common solids.
Since solids have length, area and volume (or
since measurements on them can be taken in
three directions which are at right angles to each
other), they are said to be three dimensional.
A vertex, on a solid, is a point where three or
more edges meet while an edge is a line along
which two faces meet.
In many solids, some faces, lines, or lines and
faces are parallel while others are not.
Fig. 5.1 shows a cuboid.
Fig. 5.1
(a) Name lines which are parallel to
(i) BC (ii) EF (iii) BE (iv) ED
(b) Name a face which is parallel to face
(i) ABCD (ii) BCGF (iii) CDHG
(c) Name lines which are parallel to the faces
in (b).
(d) Name lines which are not parallel to and do
not intersect with AD however much they
are extended.
(e) Name all the lines which are perpendicular
to face CDHG.
(f) Name the point of intersection between
faces ADHE, ABCD and DCGH.
(g) Name the line where faces EFGH and
BCGF intersect.
– – – – – – – – – – – – – – –
– – – – – –
– – – – – – – –
C
H
G
D
E
F
A
B
Points, lines and planes
In geometry, a point is said to mark a particular
position, and it therefore has no size. As it has
no length, breadth or thickness, it is said to be
dimensionless.
A line is said to be a set of points: It is straight
and extends indenitely in two directions as in
Fig. 5.2(a). A line segment is a part of a line
with two denite ends [Fig. 5.2(b)] and a
half-line (or ray) is a part of a line with one
denite end and extending indenitely in one
direction [Fig. 5.2(c)].
(a) (b) (c)
Fig. 5.2
A line has length but no breadth or thickness.
It is, therefore, said to be one-dimensional.
A plane is a set of points in a at surface and
extends indenitely in all directions. When
bounded by one straight line, it is said to be a
half-plane. [Fig. 5.3(a)]. When bounded by
any number of lines or curves, it is said to be a
region [Fig. 5.3(b)]. However, a region does
not necessarily have to be bounded all the way
round.
A half-plane A region
(a) (b)
Fig. 5.3
A plane has length and breadth but no thickness,
and it is therefore said to be two dimensional.
A line is
unbounded
A line segment
is bounded at
both ends
A ray is bounded
at one end
• •
•
TRIGONOMETRY (3)
5
73
A solid occupies space. It has length, area and
volume, and has a denite (i.e. xed) shape. It
is therefore said to be three-dimensional.
Note: The word ‘line’ is often loosely used
when referring to a half-line or a line-segment.
Likewise, the word ‘plane’ is also loosely used
when referring to a half-plane or a region.
Skew lines
Two distinct parallel lines have no point of
intersection. However, is it always true that two
lines which do not intersect are parallel?
In point (d) of g 5.1 of the previous section,
you found lines such as BF and CG which do
not intersect with line AD however much they
are produced. Yet, these lines are not parallel to
AD. Such pairs of lines as AD and BF, AD and
CG are said to be skew lines.
Skew lines are lines in space which are not
parallel and do not intersect however much
they are produced.
Determination of a plane
Three points which are in a straight line are
said to be collinear. But if any three points are
non-collinear, they determine a plane.
However, four points are not necessarily
coplanar (i.e. in the same plane), i.e. points A,
B, C and D are not coplanar while points A, B
and C are coplanar (Fig. 5.4). This is the reason
why many tables, desks and chairs are not
stable while a tripod is very stable.
Fig. 5.4
Two parallel lines determine a plane [Fig. 5.5(a)].
Also, two intersecting lines determine a plane
[Fig. 5.5(b)].
(a) (b)
Fig. 5.5
Note:
1. A plane is determined by either
(a) three non-collinear points, or
(b) two parallel or intersecting lines.
2. Three collinear points in space determine an
innite number of planes.
3. A pair of skew lines does not determine a
plane.
4. Two lines in space are either parallel,
intersecting or skew.
Exercise 5.1
In this exercise, restrict your answers to the
points, lines and planes named (not necessarily
drawn) in Fig. 5.6. Fig. 5.6(a) is a model of
a cuboid while Fig. 5.6(b) is a model of a
right square-based pyramid. You may nd
framework models of these useful.
(a)
(b)
Fig. 5.6
1. Name all lines that are skew with
(a) line BG (b) line SQ
2. Name the lines which are parallel to plane
ACGE.
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – –
G
A
E
B
H
C
F
D
P
– – – – – – – – –
– – – – – – –
– – – – – – – – – – – – –
S
O
T
R
Q
B
D
C
A
•
•
•
•
74
3. Which of the following pairs of lines
determine a plane?
(a) AD and FG (b) BD and EF
(c) BD and HF (d) BG and HG
(e) FC and HG (f) PR and OT
(g) ST and PR (h) PT and SP
4. Which of the following sets of points
determine a plane?
(a) A, B, F, E (b) B, C, H, F
(c) A, E, G, C (d) A, F, D, H
(e) B, D, F (f) C, E, H
(g) T, S, R, Q (h) P, O, T
(i) P, S, O, T (j) O, Q, T, R
(k) O, P, R, T (l) S, O, R
5. If a plane cut is made through AD to come
out at F, does it come out through any other
vertex? If so, which one? What solids
result from this cutting?
6. If the top of the square-based pyramid
[Fig. 5.6 (b)] is sawn o, with the cut being
parallel to the base, what is the shape of the
exposed surface? What name is given to
the remaining lower part of the pyramid?
Projections and angles
Projections
Fig. 5.7 shows a pole standing
on horizontal ground. It is kept
vertical by three taut wires
attached to the pole
at S and to the ground
at P, Q and R.
The pole TO is
perpendicular to the
ground. It is said to
meet the ground normally (i.e. perpendicularly).
The wires SP, SQ and SR meet the ground
obliquely (i.e. not perpendicularly).
Suppose that the sun is vertically overhead.
What is the shadow of
(i) point T (ii) wire SP (iii) wire SQ
(iv) wire SR on the ground?
These are the respective projections of
the point T and lines SP, SQ and SR on the
horizontal ground.
The projection of a point onto a plane (or
a line) is the foot of the perpendicular from
the point to the plane (or line). In Fig. 5.8, P
is the projection of point X onto planeABCD
or onto line PQ, drawn on the plane.
If plane ABCD is thought of as being lit from
a perpendicular direction, the shadow that
line QX forms on the plane (i.e. QP) is the
projection of the line onto the plane.
Fig. 5.8
Angle between two lines (including
skew lines)
The lines m and n in Fig. 5.9 are co-planar and
θ
is the angle between them, i.e.
the angle between two intersecting lines is
dened as the acute angle formed at their
point of intersection.
Fig. 5.9
Line n′ is the image of line n under a translation
and is also co-planar with line m. What can you
say about
θ
and the acute angle between m and n′?
– – – – – – – –
– – – – –
light
D
A
Q
P
X
C
B
– – – – – –
– – – – – –
– – – – – –
R
P
T
Q
O
Fig. 5.7
S
m
θ
n
n′
– – – – – – – – –
75
A
– – – – – – – – –
– – – – – – –
– – – – – – – – – – – – –
D
O
V
C
B
Fig. 5.12
Fig. 5.13
G
B
A
C
E
F
H
D
– – – – – – – – – – – – – – – – – –
– – – – – – – –
– – – –
A
C
F
B
D
E
This idea, that angles are unchanged by a
translation, is used to dene the angle between
two skew lines.
If the line a, in Fig. 5.10, is translated so
that its image a′ intersects b, then the angle
between a and b is dened to be the angle
between a′ and b.
Fig. 5.10
For example, the angle between TV and PS, in
Fig. 5.11, is ∠SPR. It is obtained by translating
TV to PR. Alternatively, the angle between
the two lines could be obtained by translating
PS to TW or to UV. State the size of the angle
between UV and PQ and between TU and SR.
Fig. 5.11
Exercise 5.2
1. Referring to the framework
ABCDV, in Fig. 5.12, state
the projection of:
(a) AV onto ABCD
(b) AV onto BDV
(c) BV onto ACV
(d) CV onto AC
2. Referring to the
framework of the cube
ABCDEFGH in
Fig. 5.13, name the
projection of:
(a) AG onto ABCD
(b) AG onto ADHE
(c) FD onto EFGH (d) BH onto DCGH
(e) FH onto ABCD (f) ED onto BCGF
(g) EG onto HG (h) EF onto AB
3. Referring to Fig. 5.12, the angle between
VD and AB is ∠VDC. Name the angle
between:
(a) VC and AB (b) VA and DC
(c) VA and BC (d) VC and AD
4. Referring to Fig. 5.13, state the size of the
angle between:
(a) AB and BE (b) GH and BE
(c) AB and DH (d) HD and FC
(e) EH and BC (f) ED and BG
(g) AH and BG (h) AH and BE
5. Name the lines, in Fig. 5.13, which are skew
to HG and make an angle of 90° with HG.
6. Name the lines, in Fig. 5.13, which are skew
to EG and make an angle of 45° with EG.
7. Fig. 5.14 shows a triangular prism.
Fig. 5.14
(a) Name the lines which are skew to AE.
(b) Name two angles which are equal to
the angle between:
(i) BF and AD (ii) BC and ED
(iii) EF and AC
Angle between a line and a plane
Activity 5.1
Consider Fig. 5.15(a). It shows a piece of paper
on which a point O is marked and ve half-lines
drawn from the point. The paper is placed on
a at surface and a piece of straight sti wire
stood vertically at O.
What angle does the wire make with each of
the half-lines?
b
a′
a
– – – – –
– – – – – – –
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – –
R
T
W
P
V
Q
S
U
76
Answer this by taking your piece of paper and
drawing half-lines and measuring the angles
using a protractor.
Now consider Fig. 5.15(b). The wire is now
placed in a sloping position. What angle does
the wire now make with each of the half-lines?
Again, answer this by placing your own wire
in a similar position and measuring the angles.
(a) (b)
Fig. 5.15
You should observe that:
(a) For any half-line in the plane of the paper
in Fig. 5.15(a), the angle is 90°. We say
that the wire is perpendicular to the plane.
(b) In Fig. 5.15(b), the wire makes dierent
angles with the dierent half-lines.
In the second case, we need to dene which one
is the angle between the wire and the plane.
Thinking of the plane as being lit from a
perpendicular direction, the projection of the
wire onto the plane is OP (Fig. 5.16).
We dene the angle
θ
, between the wire and its
projection onto the paper, as the angle between
the wire and the plane.
Fig. 5.16
Thus:
The angle between a line and a plane is
dened as the angle between the line and its
projection onto the plane.
Angle between two planes
Activity 5.2
Open a book and stand it on a at surface, as
shown in Fig. 5.17. Using a light pencil, mark
the indicated points and draw the lines shown
on the two facing pages.
(You will need to rub
them out when you
are through with this
activity).
The angle between
planes PABQ and
PCDQ is the angle
through which you
would turn PABQ to t onto PCDQ.
(a) Is ∠BQD equal to
(i) ∠APC, (ii) ∠AQC?
(b) What angle do BQ and DQ make with PQ?
(c) How does ∠BSD compare in size with
∠BQD?
(d) RS and TS are both perpendicular to PQ.
How does ∠RST compare in size with
∠APC?
(e) Which of the angles APC, RST and BQD is
equal to the angle between planes PABQ
and PCDQ?
We dene the angle between two planes as
follows:
The angle between two planes is the angle
between any two lines, one in each plane,
which meet on and at right angles to the line
of intersection of the planes.
In Fig. 5.18, lines OX
and OY, drawn at point
O, are perpendicular to
the line of intersection,
PQ, of the planes
ABCD and EFGH.
O O
O
P
θ
Fig. 5.18
Y
A
Q
D
X
C
G
P
E
B
– – – – – – – – –
H
θ
O
– – – – – –
F
– – – – – – – –
– – – – – – – – –
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – –
A
T
R
S
C
P
D
B
Q
Fig. 5.17
77
Fig. 5.20
C
B
G
4 cm
3 cm
12 cm
B
A
G
Hence, the angle between the planes ABCD and
EFGH is ∠XOY = θ.
Exercise 5.3
1. Fig 5.19 represents a
square-based pyramid.
Name the angle between
(a) VQ and PQRS
(b) VR and VQS
(c) VS and VPR
2. Fig. 5.20 shows the framework of a cube
ABCDEFGH. The angle between EC and
plane CDHG is ECH. Name the angle
between
(a) EC and ABCD
(b) FD and BCGF
(c) CF and ABCD
(d) AH and CDHG
(e) EB and BDHF
(f) GB and ACGE
3. Referring to Fig. 5.20, state the size of the
angle between
(a) GF and CDHG (b) EF and ABCD
(c) CH and EFGH (d) BF and ACGE
4. Referring to Fig. 5.20, name the line of
intersection between the planes
(a) ABCD and ADHE
(b) ABFE and BCGF
(c) CDHG and ACGE
(d) ACGE and BDHF
5. Referring to Fig. 5.19, name the line of
intersection and state the size of the angle
between planes
(a) PQRS and QSV (b) PRV and QSV
6. State the size of the angle between each of
the pairs of planes in Question 4.
7. Referring to Fig. 5.19, name the angle
between PQRS and QRV.
Calculating lengths and angles in solids
In three dimensional geometry, unknown
lengths and angles can, in most cases, be
determined by solving right-angled triangles.
The following examples illustrate this. Note
that it is quite helpful to sketch the triangles
separately from the solids.
Example 5.1
ABCDEFGH is a cuboid with dimensions as
shown in Fig. 5.21.
Calculate
(a) the length of AG,
(b) the angle that AG makes with plane BCGF,
(c) the shortest distance between line BF and
plane ACG.
Solution
(a) In Fig. 5.22(a), the diagonal AG and its
projection BG onto the plane BCGF are
drawn in. Fig. 5.22(b) and (c) show the
triangles used to calculate the length of AG.
(a)
(b) (c)
Fig. 5.22
In ∆BCG, ∠BCG = 90°.
∴By Pythagoras’ theorem,
BG
2
= BC
2
+ CG
2
= 4
2
+ 3
2
= 25
∴BG = √25 = 5 cm.
P
– – – – – – – – –
– – – – – – –
– – – – – – – – – – – – –
S
O
V
R
Q
Fig. 5.19
C
H
G
F
E
D
P
A
B
Q
Fig. 5.21
– – – – – – – – – – – – – – – – – – –
C
E
H
A
G
B
D
F
– – – – –
– – – – – – –
4 cm
3 cm
12 cm
– – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
C
E
H
A
G
B
D
F
– – – – –
– – – – – – –
4 cm
3 cm
12 cm
78
In ∆
ABG, ∠
ABG = 90°
∴ By Pythagoras’ theorem,
AG
2
= AB
2
+ BG
2
= 12
2
+ 5
2
= 169
∴ AG = √169 = 13 cm.
We could also use the projection of AG on
ABCD.
(b) The angle that AG makes with plane BCGF
is ∠
AGB since BG is the projection of AG
onto plane BCGF.
In ∆
ABG, tan AGB =
AB
GB
=
12
= 2.4
5
∴ ∠AGB = 67.38° (2 d.p.)
(c) The shortest distance between a line and
a plane is the distance between a point
on the line and its projection onto the
plane.
In Fig. 5.23(a), BP is the shortest distance
between line BF and plane ACG.
(a)
(b)
Fig. 5.23
In Fig. 5.23(b)
AC
2
= AB
2
+ BC
2
= 12
2
+ 4
2
= 160
∴AC = √160 cm
In ∆ABP, sin θ =
BP
=
BP
AB 12
In ∆ABC, sin θ =
BC
=
4
AC
√160
Hence,
BP
=
4
12
√160
∴ BP =
4 × 12
=
48
√160 √16 × 10
=
48
=
12
4√10 √10
=
12√10 (rationalising the
10
denominator)
=
12 × 3.162
10
=
37.944
10
= 3.79 cm (2 d.p).
Example 5.2
A rectangular-based pyramid with vertex V is
such that each of the edges VA, VB, VC, VD is
26 cm long. The dimensions of the base are
AB = CD = 16 cm and AD = BC = 12 cm.
Calculate.
(a) the height VO of the pyramid,
(b) the angle between the edges AD and VC,
(c) the angle between the base and a slant
edge,
(d) the angle between the base and face VBC.
Solution
Fig. 5.24 is an illustration of
the pyramid.
(a) Fig. 5.25 shows the
triangles used to
calculate VO.
In ∆ABC,
AC
2
= AB
2
+ BC
2
(Pythagoras’ theorem)
= 16
2
+ 12
2
= 400
∴AC = √400 = 20 cm
OC = AC = 10 cm.
12 cm
B
A
C
D
P
θ
– – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
C
E
H
A
G
B
D
F
– – – – –
– – – – – – –
4 cm
3 cm
12 cm
– – – – – – – – – – – – – – – – – – – – – – – –
P
A
– – – – – – – – – – –
– – – – – – – –
– – – – – – – – – – – – – –
D
V
C
B
Fig. 5.24
O
– – – – –
26 cm
16 cm
12 cm
M
1
2
–
79
Fig. 5.25
In ∆VOC,
VO
2
= VC
2
– OC
2
(Pythagoras’ theorem)
= 26
2
– 10
2
= 576
∴VO = √576 = 24 cm.
(b) AD and VC are skew lines. We therefore
translate AD to BC to form the required
angle VCB.
Fig. 5.26 shows the triangle
used to calculate ∠VCB.
In ∆VMC, where M is the
midpoint of BC,
cos ∠VCB =
6
= 0.2308
26
∴∠VCB = 76.66°.
(c) Since VO is perpendicular to the base,
VCO is one of the angles between the base
and a slant edge.
In ∆VCO (Fig. 5.25),
tan ∠VCO =
VO
=
24
= 2.4
CO 10
∴∠VCO = 67.38°.
(d) BC is the line of intersection between the
two planes, and M is the mid-point of BC.
VM and OM are lines, in the planes, which
are both perpendicular to BC. Thus, ∠VMO
is the angle between the base and face VBC.
Fig. 5.27 shows the triangle
used to nd ∠VMO.
In ∆VMO,
tan ∠VMO =
24
= 3
8
∴∠VMO = 71.57°.
V
C
O
26 cm
B
16 cm
16 cm
C
A
V
C
M
26 cm
B
– – – – – – – – – – – – – – – –
6 cm
6 cm
Fig. 5.26
M
8 cm
24 cm
V
O
Fig. 5.27
B
P
V
C
M
26 cm
– – – – – – – – – – – – – – – –
26 cm
12 cm
– – – – – – –
Example 5.3
Calculate the angle between the faces VAB and
VBC of the pyramid in Example 5.2.
Solution
VB is the line of intersection
of the two planes.
AP and CP are lines, on
the planes, that are both
perpendicular to VB
(Fig. 5.28). Thus, ∠APC
is the angle between
faces VAB and VBC.
Fig. 5.29 shows the
triangles used to
calculate ∠APC.
In ∆VAB, N is the
mid-point of AB.
Thus,
VN
2
= VB
2
– NB
2
= 26
2
– 8
2
= 612
∴ VN = √612
∴ Area of ∆VAB = 8√612 = × 26 × AP
∴ AP =
2 × 8√612
26
= 15.22 cm.
In ∆VBC, M is the mid-point of BC.
VM
2
= VC
2
– MC
2
= 26
2
– 6
2
= 640
∴ VM = √640
∴ Area of ∆VBC = 6 √640 = × 26 × CP
∴ CP =
2 × 6 × √640
26
= 11.68 cm.
Fig. 5.30 shows the
triangle used in
nding ∠APC.
Since the triangle
is not right-angled,
we nd ∠APC using the cosine rule.
Fig. 5.29
V
B
N
26 cm
A
– – – – – – – – – – – – – – – –
16 cm
26 cm
P
– – – – – – – – –
1
2
–
1
2
–
A
– – – – – – – – – – –
– – – – – – –
– – – – – – – – – – – – – –
D
V
C
B
Fig. 5.28
P
26 cm
16 cm
12 cm
A
20 cm
C
P
11.68 cm
15.22 cm
Fig. 5.30
80
Thus,
cos ∠APC =
PC
2
+ PA
2
– AC
2
2 × PC × PA
=
11.68
2
+ 15.22
2
– 20
2
2 × 11.68 × 15.22
=
136.4 + 231.6 – 400
355.5
=
–
32
=
–
0.090 0
355.5
∴ ∠APC = 95.16°.
Exercise 5.4
1. Fig. 5.31 shows a prism whose cross-section
is a right-angled triangle. Find the angle
between EB and plane ABCD.
Fig. 5.31
2. Fig. 5.32 shows a cuboid. Given that M is
the midpoint of EH, nd the inclination of
BM to ADHE.
Fig. 5.32
3. ABCDEFGH is a cuboid. The base ABCD
is such that AB = DC = 8 cm and
AD = BC = 6 cm. The height of the cuboid
is 4 cm. Calculate the angle between
(a) AG and plane ABCD
(b) EC and plane ADHE
(c) ED and BG
(d) AG and EF
(e) planes EFGH and EBCH.
4. The slant edges VA, VB, VC, VD of a
square-based pyramid are each 20 cm long.
The base is of side 16 cm. Calculate
(a) the height VN of the pyramid,
(b) the angle between a slant edge and the
base,
(c) the angle between a sloping face and
the base.
5. Fig. 5.33 shows a rectangle
ABCD on horizontal ground
with AB = 4 m and BC = 3 m.
AP is a vertical pole to which
three taut wires PB, PC and
PD are attached.
Calculate.
(a) the angle that PC
makes with the
ground,
(b) the angle between
planes PBD and ABCD.
6. The edges of a regular tetrahedron are all
equal in length. Find
(a) the angle between an edge and a face,
(b) the angle between two faces of the
tetrahedron.
(Hint: Let the length of an edge be 2l).
7. VABCD is a square-based pyramid. The
base is of side 12 cm and each slant edge is
18 cm long. Calculate the angle between
the faces VAB and VBC.
8. Referring to Fig. 5.32 in Question 2,
calculate
(a) the length of the diagonal of the
cuboid,
(b) the inclination of plane CDEF to plane
CDHG,
(c) the shortest distance between AE and
plane BDHF.
9 m
B
P
A
D
C
3 m
4 m
Fig. 5.33
– – – – – – – – – – – – – – – – –
– – – – – –
– – – – – – – – – – –
A
12 cm
E
D
B
C
F
16 cm
25°
– – – – – – – – – – –
C
E
H
A
G
B
D
– – – – – – – – – – – –
– – – – – – – –
– – – – – –
3 cm
5 cm
4 cm
F
M
81
The earth
The earth is nearly spherical in shape. For most
purposes, we take the earth to be a perfect sphere
with radius approximately equal to 6 370 km.
The line joining the north (N) and south (S)
poles through the centre of the earth is known as
the earth’s polar axis or Earth’s axis (Fig. 6.1).
The earth continuously revolves about this axis,
making one revolution every 24 hours.
Fig. 6.1
Great circles and small circles
Fig. 6.2 shows a plane that cuts through a solid
sphere.
Fig. 6.2
The intersection of the plane and the sphere is a
surface bounded by a circle.
If the plane cuts the sphere closer to the centre
of the sphere, the intersection of the plane and
the sphere is bounded by a larger circle.
Suppose the plane intersects the sphere through
its centre as in Fig. 6.3.
Fig. 6.3
The intesection in Fig. 6.3 gives one of the
largest possible circles that can be obtained on
the sphere. This is known as a great circle.
The radius of a great circle of a sphere is equal
to the radius of the sphere.
How many great circles are there on a sphere?
The circle of the intersection in Fig. 6.2 is
known as a small circle. The radius of the
small circle is shorter than the radius of the
sphere.
On any sphere, including the earth, we can
identify two points on the surface, N and S,
to represent the north and south poles. There
is only one great circle that can be drawn
horizontally, i.e. perpendicular to the polar axis
(Fig. 6.3). This circle divides the sphere into
two and is known as the equator.
Great circles can also be obtained if the plane
cuts through the axis of the earth, i.e. the plane
contains the axis (Fig. 6.4).
Intersection of
plane and sphere
Plane
•
– – – – – – – – – – – – – – – – – – – – –
axis
– – – – – – – – –
radius
centre
N
S
•
Intersection of
plane and sphere
Plane
S
N
– – – – – – – – – – – – – – – – – – – – –
– – – – – – – – –
•
LATITUDES AND LONGITUDES
6
82
cut
cut
cut
cut
cut
cut
cut
cut
Fig. 6.4
Small and great circles need not always be
horizontal (i.e. circles whose planes meet the
axis at right angles) or vertical (i.e. circles
whose planes contain the axis). However, for
the purposes of this course, we will work with
vertical and horizontal circles only.
Exercise 6.1
1. Name some common solids in everyday life
that roughly represents spheres.
2. Name an example of a great circle on the
surface of the earth.
3. Do all great circles on a given sphere have
the same radius?
4. Name a small circle on the surface of the
earth.
5. Do all small circles on a given sphere have
the same radius?
6. Assuming that the earth is a sphere, decide
which of the following statements are true
and which are false. For those, which are
false, give reasons why they are false.
(a) There is one and only one great circle
on the earth.
(b) The planes of a great circle and a small
circle can be parallel.
(c) If a plane cuts the earth through the
north and south poles, and through the
centre of the earth, the intersection of
the plane and the surface of the earth
denes a great circle.
(d) There is only one great circle through
the poles.
(e) The centre of each great circle on the
earth is the centre of the earth.
(f) The centre of each small circle is the
centre of the earth.
(g) The plane of a small circle contains the
centre of the earth.
(h) The diameter of every great circle on
the earth is equal to the diameter of the
earth.
(i) Only one great circle can be drawn
through any two points on the earth’s
surface.
(j) Only one small circle can be drawn
through any two points on the earth’s
surface.
Latitude and longitude
Activity 6.1
This activity is meant to help us visualise
latitudes and longitudes and clearly distinguish
between them.
1. Using thin cardboard, e.g. manila paper,
cut three discs of the same radius, say 7 cm
(Fig. 6.5).
(a) (b) (c)
Fig. 6.5
2. Cut slots in each disc as shown in Fig. 6.5.
In Fig. 6.5(a) and (b), the short cuts are
each shorter than the radius while the long
ones are equal to the radius. In Fig. 6.5(c),
the two cuts cross at right angles and at
the centre of the circle. Make each of them
longer than the radius.
– – – – – – – –
– – – – – – – – – – – – – – – – – – – – –
Centre of
small circle
Centre of
sphere
N
S
Great circle
Small circle
•
– – – – – –
•
83
3. Slot discs (a) and (b) together and fold each
semi-circular projecting part back onto itself
as in Fig. 6.6.
Fig. 6.6
4. Push the assembly in Fig. 6.6 through
the cross shaped slot in Fig. 6.5(c) and
straighten out the folded projections again
to form the model shown in Fig. 6.7.
Fig. 6.7
The model in Fig. 6.7 represents a simple model
of the skeleton of the earth. Please retain your
model for later use.
1. The north and south poles are the points
where the polar axis cuts the surface of
the earth.
2. The equator is the great circle lying in
the plane which meets the polar axis at
right angles at the centre of the earth. It
is also called a line of latitude.
3. The half of a great circle, which passes
through the poles, is called a meridian
or a line of longitude.
Folded back
projections
Equator, (a
latitude)
North Pole
South Pole
– – – – – –
– – – – – – – – – –
Longitudes
or meridians
Identify these on the model that you made in
Activity 6.1.
The particular meridian, which passes through
Greenwich in London, is called the Greenwich
meridian or the prime meridian. This
meridian is signicant in that it is used as the
standard against which all the other meridians
are xed and referred to, using degrees east
or west of it. Thus, any longitude is described
with reference to Greenwich meridian (G.M.),
so many degrees east or west. Similarly, any
latitude is given, with reference to the equator,
so many degrees north or south. Latitudes and
longitudes are used in the same way x and y are
used in the Cartesian system.
Locating a point on the surface of the
earth
Recall that in order to locate a point on the
Cartesian plane, we need
1. to dene an origin by drawing the x- and
y- axes, and
2. represent the point by an ordered pair (x, y).
Similarly, to locate a point on the surface of the
earth, we need a reference point, and an ordered
pair of values.
Let us use the model that we made in Activity
6.1. Fig. 6.8 is a sketch of the model, with a
point P marked on it.
Fig. 6.8
Colour the edges of one of the meridians to
represent the Greenwich meridian (GM) and
Line of
longitude
30°E
Equator
Coloured
half disk
N
S
Greenwich
meridian
(longitude 0°)
P
30°
– – – – – – – – – – – – – – – – – – – – –
•
•
84
on the surface of the earth, which enable any
point to be dened in terms of two angles.
3. Parallels of latitude are small circles parallel
to the equator, and whose respective centres
lie on the earth’s axis.
4. Latitudes of points on the surface of the
earth range from 0° to 90° north or south
of the equator.
5. Meridians are half great circles passing
through the poles and meeting the parallels
of latitude at right angles.
6. The Greenwich meridian is half a great
circle designated longitude 0°.
7. Longitudes of points on the surface of the
earth range from 0° to 180° east or west of
the prime meridian.
8. In locating points on the surface of the earth,
the Greenwich meridian and the equator
have the same function as the x- and y- axes
in the Cartesian plane.
Note: In some books, a meridian is dened as a
great circle through the poles.
Example 6.1
A is a point on the surface of the earth whose
co-ordinates are (30°S, 27°W).
(a) Find the new position of A after
(i) a rotation of 60°W, followed by a
rotation of 70°N about the centre of
the earth,
(ii) a rotation of 50°E, followed by a
rotation of 50°S about the centre of the
earth,
(iii) a reection in the plane of the equator,
(iv) a reection in the plane of the great
circle passing through the poles and
containing the Greenwich meridian.
(b) If B is another point on the surface of the
earth such that AB is a diameter of a great
circle, nd the coordinates of B.
also colour half the disk containing the edge
(Fig. 6.8). Rotate this half disc to the east with
the earth’s axis xed. In Fig. 6.8, the plane has
turned through 30°. The angle between the new
plane and the plane of the Greenwich meridian
is called the angle of longitude.
The plane of the prime meridian is by convention
given longitude 0°. Thus, point P in g. 6.9
lies on the line of longitude 30°E (east of
Greenwich meridian).
Since there are many other points on the
meridian, 30°E, we need another coordinate in
order to describe the position of P exactly.
Now consider Fig. 6.9.
Fig. 6.9
Suppose OH, a radius of the earth, rotates to the
north, about centre O so that H moves along the
meridian of P towards P. When OH has turned
through say 40°, H coincides with P.
The small circle through P, parallel to the
equator is called a parallel of latitude 40°N or
simply latitude 40°N (north of the equator). All
points on this circle have latitude 40°N. We can
now describe P as the point (40°N, 30°E).
Note that P is the only point with those
coordinates. Also note that we give the latitude
rst and then the longitude.
Points to remember:
1. The equator is the great circle around the
earth designated latitude 0°.
2. Latitudes and longitudes are imaginary lines
Line of
longitude
30°E
Equator
Greenwich
meridian
– – – – – – – – – – – – – – – – – – – – –
N
S
40°
O
H
G
P
C
Parallel
of latitude
40°N
•
•
85
Solution
Fig. 6.10 is a sketch showing the location of A.
Fig. 6.10
(a) (i) From longitude 27°W, a rotation 60°W
about the centre of the earth brings us
to longitude (27 + 60)°W = 87°W.
From latitude 30°S, a rotation of 70°N
brings us to latitude (70 – 30)°N = 40°N.
∴the new position of A is (40°N, 87°W).
(ii) From longitude 27°W, a rotation of
50°E brings us to (50 – 27)°E = 23°E.
From latitude 30°S, a rotation of 50°S
brings us to latitude (30 + 50)°S = 80°S.
∴ the new position of A is (80°S, 23°E).
(iii) From latitude 30°S, reection in the
plane of the equator brings us to 30°N.
From longitude 27°W, reection in the
plane of the equator leaves us on the
same meridian.
∴ the new position of A is (30°N, 27°W).
(iv) From latitude 30°S, reection in GM
leaves us on the same latitude. From
longitude 27°W, reection in GM brings
us to longitude 27°E.
∴the new position of A is (30°S, 27°E).
(b) Fig. 6.11 shows the positions of points A
and B.
Fig. 6.11
PQ is a diameter through the equator.
α = 30° (vertically opposite angles).
Since AB is a diameter of a great circle, B lies
on longitude (180° – 27°)E = 153°E and on
latitude 30°N.
∴ B has coordinates (30°N, 153°E).
Note: A pair of points such as P and Q or A
and B, which are the end points of the diameter
of a circle, are said to be diametrically opposite.
Exercise 6.2
1. Use Fig. 6.12 to answer this question.
Given that ∠POR = 60°, ∠PCQ = 25°, and
NXVS is longitude 50°E
Fig. 6.12
(a) state the latitude of
(i) P (ii) Q (iii) T (iv) R
(b) What is the size of
(i) ∠NOR (ii) ∠NOP (iii) ∠NOS?
(c) state the size of ∠ROT.
.
E quat or
N
S
30°S
A
27°W
– – – – – – – – – – – – – – – – – – – – –
30°
α
P
B
Q
Equator
N
GM
R
– – – – – – – – – – – – – – – – – – – – –
O
Q
P
T
V
C
X
Parallel of
latitude
S
•
•
– – – – – – –
– – – – – – –
– – – – –
– – – –
– – – – – – – –
E quat or
N
S
Lat i t ude
30°S
Gr eenwich
meridian
A
– – – – – – – – – – – – – – – – – – – – –
27°W
86
(d) state the size of
(i) ∠QCX (ii) ∠PCX (iii) ∠ROV
(e) Now state the coordinates of
(i) P (ii) Q (iii) X (iv) R
(v) T (vi) V (vii) N (viii) S
2. A is the point (60°S, 72°W). Write down
the coordinates of the new position of A
after
(a) a rotation of 30°W,
(b) a rotation of 30°W followed by a
rotation of 70°N,
(c) a rotation of 80°E followed by a
rotation of 40°N,
(d) a reection in the plane of the equator,
(e) a reection in the plane of Greenwich
meridian.
3. Three pilots y from Nairobi (1°17′S ,
36°48′E), Mumbai (18°55′N , 72°50′E)
and Cairo (30°l′N , 31°14′E) respectively.
Find their respective positions if each ew
(a) 40°W (b) 80°E (c) 50°N.
4. Imagine that you are standing at the north
pole, and that you are able to dig straight
through the centre of the earth. At what
point would you eventually emerge?
Answer the same question if you were
originally at
(a) (0°, 0°) (b) (30°N , 60°E)
(c) (60°S , 80°W)
5. A pilot takes o from a town P (60°N ,
20°W) and wishes to follow a circular
route so that he nally comes back to P.
Describe any two possible great circle
routes.
Distance along a great circle
The nautical mile (nm)
Let us make a trip around a great circle, say
starting from point A as shown in Fig. 6.13.
In making a round trip from A and back, we
Fig. 6.13
have turned through an angle of 360°.
Since 1° = 60′, we have turned through an angle
equal to 21 600′ (= 360 × 60′).
The distance we have covered in the round trip
is equivalent to the circumference of the earth
along the equator. The distance is
C = 2πR
= 2 × 3.142 × 6 370 km
≈ 40 029 km
For every 1′ angle that we turn through in our
trip, we cover a distance equal to
C
=
40 029
≈ 1.853 km.
21 600 21 600
This distance is equivalent to one nautical mile
(nm).
Thus:
A nautical mile is equivalent to the length of
an arc of a great circle, on the earth, which
subtends an angle of
1′ at the centre
of the earth
(Fig. 6.14).
Fig. 6.14
Some sea and air navigators often use the
nautical mile as the unit of measuring distance.
1 nautical
mile
1′
Great circle
N
S
– – – – – – – – – – – – – – – – – – – – –
A
•
•
O
87
Distance along the equator
In order to nd the distance between two points
along the equator, we rst nd the longitude
dierence between the points. Then we
calculate the length of the arc that subtends
this angle at the centre of the earth. The length
of the arc therefore represents the distance
between the two points.
Note that for any two points on the equator, or
any other circle, there are two possible values
representing this distance, i.e. one along the
major arc and the other along the minor arc.
Usually, we consider the distance along the
minor arc.
Example 6.2
Two towns A and B lie on the equator, such
that their longitudes dier by 20°. Find the
distance between A and B measured in
(a) nautical miles,
(b) in kilometres.
(Take the radius of the earth to be 6370 km.)
Solution
Fig. 6.15 shows possible positions of A and B
on the equator.
Fig. 6.15
(a) A and B lie on the equator.
∴ AB is an arc of a great circle and
∠AOB = 20°.
Since 1° = 60′,
then, 1° is subtended by 60 nm, and
∴ 20° is subtended by 20 × 60 nm.
Thus, arc AB = 1 200 nm long.
(b) The distance between A and B is the length
of the arc AB measured along the equator.
∴ arc AB =
20
× 2πR
360
=
20
× 2 ×
× 6 370
360
= 2 224 km.
Alternative method:
1 nm ≈ 1.853 km
∴ 1200 nm = 1.853 × 1200 km
= 2223.6 km
(Dierence is due to rounding o)
Note that:
The distance in nautical miles along the
equator is ‘longitude dierence × 60 nm’,
i.e. if longitude dierence is
θ
, then
distance =
θ
× 60 nm = 60
θ
nm.
In kilometres, the distance is
longitudedierence
× 2π R =
θ
× 2π R,
360 360
where R is the radius of the earth or
60
θ
× 1.853 km.
Distance along a meridian
In order to nd the distance between two places
on the same meridian we rst nd the latitude
dierence between them, and then proceed as in
Example 6.2.
Example 6.3
Calculate the distance between P(60°N , 30°E)
and Q(20°S , 30°E), giving your answer in
(a) nautical miles
(b) kilometres.
Solution
P and Q lie on the same meridian, 30°E, as
shown in Fig. 6.16.
22
7
––
N
– – – – – – – – – – – – – – – – – – – – –
O
A
B
Longitude
of B
S
Longitude
of A
20°
•
88
22
7
––
Fig. 6.16
(a) The angle subtended by arc PQ is
20° + 60° = 80°.
1° is subtended by 60 nm.
∴ 80° is subtended by 60 × 80 nm.
Thus, length of arc PQ = 4 800 nm.
(b) Length of arc PQ =
80°
× 2 ×
× 6370
360°
= 8 898 km
or 4800 × 1.853 = 8 894 km.
The dierence is due to rounding o.
We should note that:
The length of an arc along a meridian is
‘latitudedierence× 60 nm’.
In kilometres, the length of the arc is
 latitudedierence
× 2πR km,
360
where R is the radius of the earth.
Exercise 6.3
In this exercise, take π = and R = 6 370 km.
1. Calculate the distance, measured along
a meridian, between the north and south
poles. Give your answer in
(a) nautical miles (b) kilometres.
2. Calculate the distance between the following
pairs of points, measured along a great
22
7
––
circle. Give your answers in
(i) nautical miles (ii) kilometres
(a) P(0° , 153°E) and Q(0° , 178°E)
(b) R(50°N , 20°W) and S(80°N , 20°W)
(c) T(70°N , 30°E) and U(20°S , 30°E)
(d) V(58°N , 28°E) and W(10°N , 28°E).
3. Two towns A and B lie on the equator and
their longitudes dier by 34°. Given that A
is on latitude 18°E, state the coordinates of B.
(There are two possible answers.)
4. A pilot takes o from an airport
N(20°S , 36°50′E) and ies due north for
1 380 nm to another town M. Find this
distance in kilometres. What are the
coordinates of M?
5. How far is the tropic of Capricorn from the
north pole?
6. Towns P and Q lie on the same meridian
and are 4 920 nm apart. If P is (24°S , 38°E)
and Q is north of P, nd the coordinates of
Q. Calculate the distance PQ in kilometres.
7. Towns A and B lie on the same meridian,
4 368 nm apart. If A is on latitude 81.4°N,
on what latitude does B lie?
8. The distance between P and Q measured
along a meridian is 1 840 nm. Given that O
is the centre of the earth, calculate ∠POQ.
If P is (40°S , 50°E), nd all the possible
positions of Q.
Radius of, and distance along, a small
circle.
In order to nd distance along a circle, we must
rst know the radius of the circle.
We have already seen that a great circle passes
along the surface of the earth and has its centre
at the centre of the earth. Therefore the radius
of any great circle is equal to the radius of the
earth.
In this section, we consider small circles whose
centres lie on the axis of the earth. In this case,
N
– – – – – – – – – – – – – – – – – – – – –
B
Line of
longitude
30°E
S
Equator
20°
Q
P
60°
O
– – – – – – – – – – – – – – – – – – – – –
•
89
the radius of any small circle must be shorter
than the radius of the earth.
Consider Fig. 6.17.
Fig. 6.17
Both the equator and the parallel of latitude
through P have their centres on the axis NS. P
and Q lie on the surface of the earth such that
Q is on latitude 0° while P is on latitude θ N.
Both points lie on the same meridian.
PX and OQ are parallel (they meet NS at
right angles). OQ represents the radius of the
earth, and PX the radius of the small circle on
latitude θ N.
∠XPO = ∠QOP =
θ
(alternate angles).
Using ∆PXO,
cos
θ
= XP
OP
= XP
R
∴ XP = R cos
θ
∴ r = R cos
θ
(where r denotes the
radius of any small circle).
We note that:
1. Given a parallel of latitude
θ
° N or S, the
radius (r) of the small circle is always
equal to radius of the earth × cos
θ
,
i.e. r = R cos
θ
2. The circumference of a small circle on
θ N or S is equal to 2πr = 2πR cos
θ
.
22
7
––
22
7
––
Example 6.4
A(40°S, 50°W) and B(40°S, 130°E) lie on the
surface of the earth.
(a) Find the radius of the parallel of latitude
on which A and B lie.
(b) Calculate the distance along the small
circle between A and B, giving your answer
in (i) nautical miles and (ii) kilometres.
(c) Find also the distance between A and B
along a great circle, in nautical miles and
in kilometres.
Comment on the answers to parts (b) and (c).
[Take R = 6 370 km, and π = ]
Solution
Fig. 6.18 represents the parallel of latitude on
which points A and B lie.
Fig. 6.18
(a) ∠AXB = 180° (50°W + 130°E = 180°).
X is the centre of the small circle.
∴ r = R cos 40°
= 6 370 cos40° km
= 4879 km (4 s.f.)
(b) Points A and B lie on a diameter of a small
circle.
∴arc AB represents half of a small circle.
Length of arc AB = 180 × 60 × cos 40° nm
= 8 273 nm (4 s.f.)
In kilometres, length of arc AB
=
180
× 2πr, where r = R cos 40°
360
=
180
× 2 × × 6 370 cos 40°
360
= 15 340 km (4 s.f.)
Longitude 50°W
Longitude 130°E
50°
130°
GM
A
B X
N
Equator
(great circle)
– – – – – – – – – – – – – – – – – – – – –
O
Q
P
X
Parallel
of latitude
(small circle)
S
– – – – – – – – –
– – – – – – – – – – –
– – – – – – – – – – –
θ
•
•
90
(c) Since A and B are diametrically opposite,
they lie on the same great circle of
longitude.
Fig. 6.19
∠AOB = 100° (Fig. 6.19).
The distance along a great circle is from A
via S to B.
∴ in nautical miles,
AB = 100 × 60
= 6 000 nm.
In kilometres,
AB =
100
× 2πR
360
=
100
× 2 ×
× 6 370
360
= 11 120 km (4 s.f.).
Comment: The great circle distance is shorter
than the small circle distance.
Note: In Example 6.4:
1. There is another alternative route between A
and B via the north pole. Find its length!
2. The distance between A and B via the south
pole is the shortest.
Shortest distance between two points
The shortest distance between two points
on the surface of the earth is always along
a great circle.
Latitude
40°S
N
S
A
Longitude
50°W
– – – – – – – – – – – – – – – – – – – – –
100°
O
B
Longitude
130°E
40°
40°
– – – – – – – – – – –
– – – – – – – – – – –
– – – – – – –
22
7
––
Example 6.5
Find the shortest distance between points
P(65°N , 25°W) and Q(65°N , 31°E) on the
earth’s surface.
Solution
Fig. 6.20 is a sketch showing the position of the
two points.
Fig. 6.20
The shortest distance between P and Q is the
length of the arc of the great circle through P
and Q, centre O.
Let X be the centre of the circle of latitude 65°N
and M be the the midpoint of chord PQ. Then,
PX = QX
PM = QM
∠PXQ = 25° + 31° = 56°
∴ ∠PXM =
56°
= 28°
2
Using ∆PMX,
sin 28° =
PM
Rcos 65°
∴ PM = Rcos 65° sin 28°
Using ∆PMO,
sin
q
=
PM
=
Rcos 65° sin 28°
R R
= 0.198 4
∴
q
= 11.44°
⇒ 2
q
= 22.88°
65°N
S
N
– – – – – – – – – – – – – – – – – – – – –
X
O
31°E
25°W
Q
P
•
•
– – – – – – – – –
•
M
– – – – – – –
– – – – – –
– – – –
– – – – – –
91
22
7
––
Length of great circle arc PQ is
22.88°
× 2π × 6 370 = 2 543.7
360°
∴ the shortest distance between P and Q is
2 543.7 km
Exercise 6.4
In this exercise, take π = and R = 6 370 km.
1. Use Fig. 6.21 to answer this question.
Fig. 6.21
(a) State the coordinates of A, B, …, H.
(b) Find the angle subtended by each of
the following arcs at the centre of the
appropriate circle.
(i) AB (ii) BC (iii) CD
(iv) AE (v) NB (vi) CS.
(c) Find the radii of the parallels of latitude
(i) 50°N and (ii) 65°S in kilometres.
(d) Find the lengths of the arcs in (b)
above.
2. Find the distance between P(49°S , 9°E) and
Q(49°S , 2°W) in
(a) nautical miles (b) kilometres.
3. P(30°N , 73°E) and Q(30°N , 107°W) are
two towns on the surface of the earth. Find
the distance, in kilometres, between them
(a) along the parallel of latitude,
(b) along a great circle.
Equator
N
S
65°S
– – – – – – – – – – – – – – – – – – – – –
35°W
50°N
X
Y
O
80°E
39°E
25°W
H
G
F
E
D
C
B
A
•
•
•
4. X and Y are two points on the surface of
the earth, on the same parallel of latitude.
X is 90°E and Y is 90°W and the distance
between them, measured along a great
circle, is 4 444 km. Find
(a) their latitude,
(b) the distance between them along the
parallel of latitude.
5. A small circle on a parallel of latitude has a
circumference of 14 940 km. Find the angle
of latitude.
6. Two towns A and B are 1 250 nm apart. If
they lie on the same latitude 45°N and A is
20°E, nd the coordinates of B.
(There are two possible answers.)
7. Four dierent aircrafts y a distance of
1 860 nm due west from their respective
bases A(30°N , 155°W), B(40°S , 175°E),
C(60°S , 135°W) and D(50°S , 40.8°E).
Find the new longitude of each aircraft.
8. Find the shortest distance between town
C(50°S , 19°E) and town D(50°S , 80°E).
Longitude and time
Fig. 6.22
Recall that the earth continually revolves about
its polar axis from west to east. Suppose P and
•
P

•
Q
92
Q are points on the earth’s surface such that P is
to the east of Q (Fig. 6.22). Due to the earth’s
rotation, P will experience daylight before
Q. To nd the timedierence between P and
Q, we use the fact that the earth makes one
complete west to east revolution in 24 hours,
i.e. it completes a 360° turn in 24 hours.
This means that a change in longitude of 1°
corresponds to a change in time of
24
h =
24
× 60 min = 4 min, i.e a change in
360 360
longitude of 1° corresponds to a change in time
of 4 minutes.
Example 6.6
The longitude of Nairobi is approximately
37°E and that of London is 0°. Find the time
dierence between the two cities.
Solution
Nairobi is on a meridian 37° east of Greenwich
meridian.
1° dierence in longitude is equivalent to 4
minutes dierence in time.
∴ 37° dierence corresponds to
37 × 4 min =
37 × 4
hours
60
= 2 h 28 min dierence in time.
i.e. time dierence between Nairobi and London
is 2 h 28 min. Thus, Nairobi experiences
sunrise 2 h 28 min before London.
The result of Example 6.6 means that when it
is, say, 6.00 a.m in London, it would be 8.28
a.m in Nairobi. This is known as the local time
in Nairobi, i.e. time based on the longitude
dierence with the Greenwich Meridian. Thus,
places, which are on the same meridian, have
the same local time.
However, places that have nearly the same
local times, i.e. that lie in a specic range of
longitudes, are usually designated the same
average times, known as Standard Times.
Thus, countries of the East African region and
those in the same longitude band have the same
Standard Time. Thus, when it is 6.00 a.m. in
London, it is 9.00 a,m. East African Standard
Time, which is based on longitude 45°E.
Exercise 6.5
1. Two towns lie on the equator, 630 nm apart.
Find the time dierence between them.
2. Towns A and B are such that their local
times dier by 2 hours. If A is on longitude
20°W, nd two possible positions for B.
3. Given that the local time at A(0° , 25°W) is
11:05 a.m, nd the local time at B(0° , 38°E).
4. Find the dierence between the local times
of each of the following pairs of towns.
(a) P lies on 113.93°E and Q lies on
48.73°E,
(b) A lies on 81.6°E and B lies on 67.2°W,
(c) R lies on 12° W and S lies on 76.5°W.
5. The longitudes of A and B dier by 78.5°.
(a) Find the dierence between their local
times.
(b) If A has coordinates (35°N , 0°E), nd
the coordinates of B.
Speed in knots and kilometres per
hour
On any journey, the speed varies. Thus, we
talk about average speed for any given journey
within a given time interval. We already know
that the average speed for a journey is given as
Distance travelled
in the appropriate unit.
Time taken
For example, if the distance is measured in
kilometres and the time in hours, the speed
is stated as kilometres per hour (km/h).
Similarly, if the distance is given in nautical
miles (nm) and time in hours, the speed is
nautical miles per hour or nm/h. A speed of 1
nautical mile per hour is dened as 1 knot.
93
Example 6.7
A motor boat travels from a harbour A, 20°S
due north to another harbour B, 13°N.
(a) Calculate the distance between A and B,
giving your answers in
(i) nautical miles (ii) kilometres.
(b) Given that the boat took a total of 90
sailing hours, nd the speed of the vessel in
(i) knots (ii) kilometres per hour.
Solution
(a) Harbours A and B lie on the same meridian
(Fig. 6.23).
Fig. 6.23
Since A is 20°S and B 13°N, the latitude
dierence is 20° + 13° = 33°.
(i) 33° on a great circle represents
33 × 60 nm = 1 980 nm.
(ii) 1 nm ≈ 1.853 km
∴1 980 nm = 1 980 × 1.853
= 3 668.94 km
≈ 3 669 km.
Alternatively,
Length of arc AB =
33
× 2π × 6 370
360
= 3 668.86 km.
≈ 3 669 km
(b) (i) Speed =
Distance
=
1 980
Time 90
=
1 980
= 22 knots.
90
N
S
– – – – – – – – – – – – – – – – – – – – –
O
B (13°N)
A (20°S)
33°
•
(ii) Speed =
Distance
Time
=
3 669
90
= 40.76
≈ 40.77 km/h.
Exercise 6.6
1. A shing boat sails at an average speed of
6 knots.
(a) Express this speed in kilometres per hour.
(b) What distance does the boat cover in
40 minutes?
2. Coastal towns A and B lie on the equator
such that A is 10°W and B is 53°E. If a
boat travels at 8 knots, how long would it
take to travel from A to B?
3. Find the distance between ports
P(40°N , 30°W) and Q(20°S , 30°W). How
long will a ship, travelling at an average
speed of 24 knots, take to travel from P to Q?
4. An aircraft ies from A(40°N , 8°W) to
B(5°S , 8°W) at an average speed of 675
km/h. How long does the ight take?
Express the given speed in knots.
5. P and Q are two towns 2 400 nm apart.
(a) Given that P is (0° , 20°W), and that Q
is West of P, nd the position of Q.
(b) From Q, a ship sails due north to R, a
distance of 3 335 km. Find the location
of R.
(c) If the average speed between Q and R
is 25 knots, how long does the journey
take?
6. An aircraft ew due east from
A(20°S , 60°W). It landed at port B, whose
longitude is 10°W. From port B, the aircraft
ew due south to C, a distance of 3 240 nm.
(a) State the respective positions of B and C.
(b) If the total ying time was 30 hours, nd
the average speed of the aircraft in knots.
.
94
1
3
–
Inequalities
In Book 2, we learnt how to solve linear
inequalities in one and two unknowns. We also
learnt how to form inequalities from inequality
graphs and from given situations. Recall the
following important facts about operations on
inequalities:
1. An inequality still holds if:
(a) The same number is added to or
subtracted from both sides.
(b) Both sides are multiplied or divided
by the same positive number.
2. Multiplying or dividing both sides by
the same negative number reverses the
direction of the inequality.
Thus in general:
Given that a, b and c are numbers and that
a ≤ b, then,
a + c ≤ b + c and a – c ≤ b – c,
ac ≤ bc if c > 0, i.e. c is positive, but
ac ≥ bc if c < 0, i.e. c is negative (the
inequality sign changes).
a
≤
b
if c > 0, but
c c
a
≥
b
if c < 0 (the sign of the inequality
c
c
changes).
Note that the above results are also true when
a ≥ b, a < b or a > b.
Analytical solution of linear inequalities
Example 7.1
Solve the inequality
2 – 6x
≤ 3 – x.
3
Solution
Multiplying both sides by 3,
2 – 6x
≤ 3 – x
3
becomes
2 – 6x ≤ 9 – 3x
2 – 3x ≤ 9 (adding 3x to both sides)
–
3x ≤ 7 (subtracting 2 from both sides)
x ≥
–
2 (dividing both sides by –3).
Representing simultaneous linear
inequalities on a number line
Solutions of linear inequalities involving one
variable can be illustrated on a number line
as in Example 7.2 (Remember we did similar
work in Book 2).
Example 7.2
Solve the simultaneous linear inequalities
5 + 3x < 23 and
25 < 6x + 1
and illustrate your solution on a number line.
State the combined solution.
Solution
We solve each inequality separately as follows:
5 + 3x < 23 25 < 6x + 1
3x < 18 and 24 < 6x
x < 6 4 < x, i.e. x > 4
Illustrate the two solutions on the same number
line (Fig. 7.1).
Fig. 7.1
The shaded portion of the number line
represents the solution set.
∴ 4 < x < 6.
0 1 2 3 4 5 6 7
° °
x < 6
x > 4
LINEAR PROGRAMMING
7
95
1
3
–
Exercise 7.1
1. Solve the following inequalities.
(a) 3 x – 5 < 4 (b) 8 – x > 2
(c) 2(x – 3) ≤ 4(x – 1)
(d) x ≤ 3x – 1 ≤ x + 3
(e) 2(x – 3) ≤ 4 ≤ 4(x – 1)
2. Solve the following simultaneous
inequalities.
(a) 2 x + 7 ≤ 29 , 5x – 1 > 29
(b) a – 9 < 48 , 2a – 46 ≥ 48
(c) 7 p + 8 < 36 , 34 – 8p < 10
(d)
–
1 + 2x ≥
–
6 , 2x – 1 < 7
3. Solve the following inequalities.
(a)
3x + 7
≥
1 + 2x
(b)
2
–
1
> 1
5 6 m 2m
(c)
2(3 – x)
≤
3(2 – x)
3 2
(d)
(1 – x)
–
(2 – x)
≤ 1
3 2
4. Solve the following inequalities.
(a)
1
–
2
< 3
(b)
P
+
P – 1
≤ P
x – 1 3 3
4
5. Solve the following inequalities.
(a) 5 + x ≤
2x (b)
2 –
y
≥
3y + 1
3
3 4
(c)
3 –
x + 1
≤
x – 1
+ 1
2 4
(d) b
+
b
< 1 –
b + 1
3 2 4
Graphical solution of linear inequalities
It is important to recall that in order to represent
an inequality graphically, we must rst identify
the boundary line. The line is drawn solid if
the boundary is included in the required region,
or broken if the boundary is not included.
In order to identify the required region, we
pick any point, not on the line, and substitute
its coordinates in the given inequality to test
whether it satises the inequality or not. We
then shade the unwanted region. If the required
region is an intersection of two or more regions,
the individual regions are illustrated, one at a
time, but on the same graph.
Example 7.3
Represent, on the same graph, the solution of
the three simultaneous inequalities x < 7,
y < 5 and 8x + 6y ≥ 48.
Solution
We are looking for the intersection of the three
regions: x < 7, y < 5 and 8x + 6y ≥ 48.
The boundary lines are x = 7, y = 5 and
8x + 6y = 48 respectively.
Thus, we draw the line
(i) x = 7 (broken) and shade the region x > 7,
(ii) y = 5 (broken) and shade the region y > 5,
(iii) 8x + 6y = 48 (solid) and shade the region
8x + 6y < 48 (Fig. 7.2).
Substituting (0 , 0) in 8x + 6y ≥ 48 gives 0 ≥ 48,
which is false. Therefore, the point (0, 0) does
not belong to the region we want so we shade
the region that contains (0 , 0). For y < 5, it is
obvious that any number below 5 satises the
inequality, therefore we shade the region above
the line y = 5. Similarly, for x < 7, we shade
the region to the right of x = 7.
Fig. 7.2
y
x
8
6
4
2
0
2 4 6 8 10
– – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – –
y = 5
8x + 6y = 48
x = 7
96
4
5
–
(a) x < 4, 3y – x ≤ 6, 3y + 2x > 6
(b) 0 ≤ x ≤ 4, 0 ≤ 3y – x ≤ 6
(c) x ≥ 0, 5x + 6y ≤ 60, 9x + 6y ≤ 72,
x + y ≥ 8
4. Find the points with integral coordinates
which satisfy the simultaneous inequalities
x ≤ 4, 3y ≤ x + 6, 2x + 3y > 6.
5. R represents the region in a Cartesian plane
whose points satisfy the inequalities
0 ≤ x < 5, 0 ≤ 3y + x ≤ 9. Solve the
inequalities graphically.
6. On the same graph, show the region Q that
is satised by the inequalities x ≥ 0, y ≥ 0,
x + y ≤ 12, x + 2y ≤ 16 and y ≥
–
x + 4.
Forming linear inequalities
Given a narrative, and required to form
inequalities from it, our rst task is to dene
the variables that we intend to use. Since our
method is graphical, we can only deal with
problems that can be fully dened in terms of
two variables. Once we explain the meaning
of the letters we introduce, then we write down
as many inequalities as we can from the given
information.
Inequalities can also be formed from a given
graph(s). In order to form such inequalities, we
1. nd the equation of the boundary line(s),
2. test for the required region, as in Example
7.2, and then state the appropriate inequality.
Example 7.5
A games master wishes to buy new sports shoes
for his students. He has sh 12 000 to spend.
In his town, only two shops, A and B, stock the
kind of shoes he wants. At shop A, they cost
sh 500 a pair and in shop B they cost sh 600 a
pair. In shop A, only six pairs are remaining.
Write down as many inequalities as possible
from the given information.
Example 7.4
Use graphical method to solve, simultaneously,
the inequalities
x ≥ 0, y ≥ 0, x + y ≤ 4, x + 2y < 6.
Solution
The required boundary lines are x = 0 , y = 0,
x + y = 4, x + 2y = 6
On the same axes, we draw the boundary lines
and shade the unwanted regions, one at a time
(Fig. 7.3).
Fig. 7.3
The unshaded region represents the solution set.
Exercise 7.2
1. Draw separate diagrams to show the regions
representing the following inequalities.
(a) 5 x + 4y < 60 (b) 3 x – y > –6
(c) 8 x + 3y ≥ 24
2. A region R is given by the inequalities
x ≤ 6, y ≤ 6, x + y < 9 and 6x + 5y ≥ 30.
Represent this region graphically and list all
the points in the region which have integral
coordinates.
3. Show the regions and list the points whose
coordinates are integers and which satisfy
the following simultaneous inequalities.
x + y = 4
x + 2y = 6
1 2 3 4 5 6
0
5
4
3
2
1
x = 0
y = 0
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
x
y
97
∴ with reference to l
2
, the region satises the
inequality x ≤ 6.
Line l
3
passes through points (0, 8) and (7, 0)
∴ y-intercept is 8, and gradient of line is
–
8
.
7
∴ equation of l
3
is: y =
–
8
x + 8.
7
With reference to l
3
, region R satises the
inequality y ≤
–
x + 8.
Similarly,
y-intercept for l
4
is 2 and gradient is
–
.
∴ equation of l
4
is y =
–
x + 2
and it satises the equality y >
–
x + 2.
Thus, region R is the intersection of the four
inequalities, i.e. y < x + 2, x ≤ 6, y ≤
–
x + 8,
and y >
–
x + 2.
1 2 3 4 5 6 7
y
10
9
8
7
6
5
4
3
2
1
0
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
R
l
3
l
2
l
1
Fig. 7.4
l
4
x
Solution
Though it is reasonable to argue that he should
buy the six pairs from shop A and the rest from
shop B, he does not have to.
Let x be the number of pairs of shoes bought
from shop A and y be the number of pairs of
shoes bought from shop B.
The cost of the shoes is 500x + 600y
∴ 500x + 600y ≤ 12 000
x ≤ 6 (since there are only 6 pairs of shoes
in shop A)
x ≥ 0 and y ≥ 0 (since x and y cannot be
negative)
∴ the inequalities are
x ≥ 0, y ≥ 0, x ≤ 6, 5x + 6y ≤ 120.
Example 7.6
Use Fig. 7.4 to write down the inequalities that
satisfy x and y in the unshaded region R.
Solution
Fig. 7.4 represents an unshaded region R,
enclosed by four lines, l
1
, l
2
, l
3
and l
4
. These are
the boundary lines for R.
Now, we nd the equations of the four lines, and
dene the region with reference to each line.
Thus, l
1
passes through (0 , 2) and (3 , 6).
∴ y-intercept is 2 and
gradient is
4
.
3
∴ equation of l
1
is y =
4
x + 2
3
Now, choosing one point, say (2 , 2), we test for
the region at (2 , 2), y =
4
x + 2 ⇒ 2 <
8
+ 2
.
3 3
∴ with reference to l
1
, the region satises the
inequality y <
4
x + 2 (since l
1
is a broken line).
3
l
2
has the equation x = 6.
For all the points in the unshaded region, the
value of x is less than or equal to 6.
1
3
–
8
7
–
1
3
–
1
3
–
4
3
–
1
3
–
8
7
–
y =
3
7
x + 8
y =
2
3
x + 2
y =
1
3
x + 2
98
(a)
(b)
(c)
Fig. 7.5
x
y
8
6
4
2
0
1 2 3 4 5
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
Exercise 7.3
In each of Questions 1 to 5, explain the
meaning of the letters you introduce, and then
write down as many inequalities as you can
from the given information.
1. Jane went shopping, with sh 540, to buy
some Christmas cards for her friends. She
found two types of cards, one costing
sh 60.00 each and the other costing sh 90.00
each. She decided to buy some of each type,
but not less than four cards altogether.
2. A small aircraft company is required
to transport 600 people and 45 tonnes
of baggage. Two types of aircraft are
available. Type A carries 60 people and
6 tonnes of baggage. Type B carries 70
people and 3 tonnes of baggage. Only eight
aircrafts of type A and seven of type B are
available for use.
3. A milk transporter distributes 900 crates
of milk per day, using trucks and vans. A
truck carries 150 crates while a van carries
60 crates. The cost of each trip by a truck
is sh 500 and that of a van is sh 400. He
has sh 4 400 available to use for transport.
It is advisable that he uses both modes of
transport.
4. Amin went shopping, with sh 72 only,
to buy some reworks. He bought two
dierent types, A and B, at sh 6.00 and
sh 9.00 each respectively. For every one of
type B, he bought at least two of type A.
5. Joan wishes to buy x kg of beans and y kg
of maize. The cost of beans is sh 40 per
kg while that of maize is sh 25 per kg. She
wishes to buy more maize than beans, and
she has only sh 400 to spend.
6. Use Fig. 7.5 to write down all the
inequalities that satisfy x and y in the
unshaded region in each case.
x
8
6
4
2
0
2 4 6 8 10
y
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – –
x
y
8
6
4
2
0
2 4 6
8 10
99
Fig. 7.6
If we draw line 3x + 4y = k, where k is a
constant, say k = 12 or 24, etc., we notice that
the value of k gets larger as the line moves
further away from the origin. Thus, to nd the
maximum value of 3x + 4y, we draw a line
3x + 4y = k on our graph and using a ruler and
a set square we translate the line as far away
from the origin as the conditions will allow. The
furthest point in the region which falls on the
line gives the values of x and y which make
3x + 4y a maximum.
In our example, the furthest point is (4 , 6), at
the intersection of the lines x + y = 10 and
x + 2y = 16.
∴ the maximum value of 3x + 4y is
3 × 4 + 4 × 6
= 12 + 24
= 36
Note that if we needed the minimum value, we
would translate the line towards the origin. We
would then pick the point within the region that
is nearest to the origin.
A line such as 3x + 4y = k in Example 7.7 is
known as the search line.
Exercise 7.4
1. A region is dened by the inequalities
x ≥ 0, y ≥ 0, 3 x + 2y ≤ 12 and 7x + 3y ≤ 21.
Show the region on a graph and nd, within
x
y
10
8
6
4
2
4 8 12 16 20
0
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – –
3x +
4
y = 12
x + 2y = 16
x + y = 10
y = 0
7. A poultry farmer plans to keep some layers
and broilers. Layers cost sh 65 per day old
chick while broilers cost sh 45 per day old
chick. He cannot aord to spend more than
sh 40 000. He nds it uneconomical to
keep less than 400 birds. He wishes to keep
more broilers than layers and not less than
300 broilers.
Write down as many inequalities as possible
to describe the given situation.
8. The sum of three consecutive integers
is less than 999. If the smallest of these
integers is n, what are the other two integers
in terms of n?
Write down an inequality involving n.
9. The sum of the lengths of any two sides of
a triangle is greater than the length of the
third side. The lengths of the sides of such
a triangle are 5 cm, 9 cm and x cm. Write
down as many inequalities in x, as possible.
Maximising or minimising a function
Sometimes, when we are solving problems
graphically, we may be required to nd the
maximum or minimum value of a function in
the region obtained.
Example 7.7
A region, A, given by the inequalities x ≥ 0,
y ≥ 0, x + y ≤10 and x + 2y ≤ 16 represents
the solution set of a certain problem. Find the
maximum value of the function 3x + 4y in this
region.
Solution
Fig. 7.6 illustrates the region that satises all
the given inequalities.
We could nd the maximum value by listing the
coordinates of all the points in the region and
substituting them in the function 3x + 4y. But
this is not practical since the number of points
in the region is innite. Therefore, we use the
method described below.
100
the region,
(a) the maximum value of x + y,
(b) the maximum value of 5x + y.
2. Solve the simultaneous inequalities x ≥ 0,
y ≥ 0, 2x + 2y ≥ 11, 2x + y ≥ 8 and
2x + 5y ≥ 18 and nd, within the solution
set, the minimum value of
(a) 3 x + 2y, (b) 2x + 3y, (c) x + 3y.
3. Find the minimum value of
(a) y – x (b) 6x + 5y (c) 4x + 6y
in the region dened by y ≤ 2x, x ≤ 6,
y ≥ 2 and 2x + 3y ≤ 30.
4. 500 men and 42 tonnes of equipment are to
be transported to a new camp. There are two
types of trucks that can be used. Truck type
A carries 50 men and 5 tonnes of equipment,
while truck type B carries 40 men and 3
tonnes of equipment. Given that x trucks
of type A and y trucks of type B were used,
write down, from this information, as many
inequalities as you can and represent them
on a graph.
Find the minimum value, in the region, of
1050x + 900y.
5. A famine relief food agent has to transport
900 bags of maize and beans from the city to
one of the distributing centres. He intends
to use trucks which can carry 150 bags at a
time and vans which can carry 60 bags at a
time.
(a) If the number of lorries used is l, and
the number of vans v, write down
an inequality in terms of l and v to
represent the given situation.
(b) The cost of running a truck for the
journey is sh 5 000, and that of running
a van is sh 4 000. If the agent has a
maximum of sh 44 000 to spend, write
down another inequality in terms of l
and v.
(c) Represent the inequalities in (a) and
(b) graphically, and clearly label the
region which must be satised by the
inequalities in l and v.
(d) (i) Use your graph to list all the
possible combinations of vehicles
that could be used.
(ii) Identify the combination that
would keep the cost at a minimum,
and hence state the minimum cost.
(iii) Find the combination that would
be most expensive.
Optimisation
We are now in a position to look at some
problems which may arise from real life
situations and which can be solved by means of
graphs of linear inequalities. The whole process
of nding the possible solutions to a given
problem is called linear programming.
By drawing appropriate graphs, we rst nd a
region containing the points which represent
possible solutions. When the solution set has
been found, the next task is to decide which
element of the solution set best meets the
requirements of the problem. This process is
called optimisation. Usually, the best solution
is the one that will make, say, the prot as large
as possible, the cost as little as possible, the
time taken for a process as short as possible,
and so on.
In order to minimise or maximise a value, we
use a method similar to the one we used in the
previous section. This involves drawing the
graph of the function we wish to maximise
or minimise. This function is known as the
objective function and is usually of the form
C = ax + by, where a, b and C are constants.
The following example illustrates this method.
Example 7.8
A transport company has two types of lorries,
9 of type A and 5 of type B. There are 11 drivers
available. The company has been contracted
101
to transport at least 3 600 bags of coee from a
certain co-operative store to the Coee Board
of Kenya stores in Nairobi. Type A lorries
can each make 4 trips and carry 90 bags per
trip. Type B lorries make 3 trips per day and
carry 150 bags each per trip. It costs sh 5
000 per day to run a type A lorry and sh 8 000
per day to run a type B lorry. How should the
contractor organise the use of his lorries so as to
(a) run the lorries at a minimum cost?
(b) carry the maximum number of bags each
day?
(c) use the minimum number of drivers?
Solution
Let x be the number of type A lorries used per
day and y be the number of type B lorries used
per day. Due to the limitation on the number of
lorries and the number of drivers available,
x ≤ 9, y ≤ 5 and x + y ≤ 11
are some of the required inequalities.
In one day, a type A lorry can carry 90 × 4 bags,
while a type B can carry 150 × 3 bags.
A minimum of 3 600 bags are to be transported,
using x type A lorries and y type B lorries.
∴ 360x + 450y ≥ 3 600 which can be written
as 4x + 5y ≥ 40
∴ all the possible inequalities from the given
information are
x ≤ 9, y ≤ 5, x + y ≤ 11 and 4x + 5y ≥ 40
Now, we graph the inequalities to nd the
solution set of the problem (Fig. 7.7).
From our graph, it is clear that the points
(4, 5), (5, 4), (5, 5), (6, 4), (6, 5), (7, 3), (7, 4),
(8, 2), (8, 3), (9, 1) and (9, 2) represent possible
solutions.
Using these possible solutions, we must now
nd the solution that suits the transporter’s
requirements best.
1. The expression that gives the total cost is
5 000x + 8 000y. In order to minimise the
cost, we use the objective function
Fig. 7.7
C = 5 000x + 8 000y,
where C represents the total cost.
By choosing an appropriate value of C, we
get the search line. For example, if
C = 40 000, we get the search line
5000x + 8000y = 40 000, i.e. 5x + 8y = 40.
By translating this line towards the solution
region, we nd that the solution point which
gives the minimum cost is (9, 1).
∴The contractor should use 9 lorries
of type A and 1 of type B so as to incur a
minimum cost of sh (5 000 × 9 + 1 × 8 000)
= sh 53 000.
2. In order to transport the maximum number
of bags, our objective function is 4x + 5y.
This function gives the best solution at
(6 , 5), i.e. use 6 type A lorries and 5 type B
lorries to carry a maximum of 1 290 bags
per day.
3. In order to use the minimum number of
drivers, the objective function is x + y.
The best solution is 9 drivers which is given
by the point (5 , 4).
It is important to note that the maximum or
minimum usually occurs at one of the corners
of the region. Thus, instead of using a search
x
y
12
10
8
6
4
2
0
2 4 6 8 10 12
x + y = 1 1
– – – – – – – – – – – – – – – – – – – – – – – – – – –
5x + 8y = 40
4x + 5y = 40
x = 9
102
line, one could get the optimum solution by
testing the coordinates of the corner points.
Re-do Example 7.8 using this technique.
Exercise 7.5
In this exercise, the solution set will consist of
only positive numbers or zero.
1. At a charity show, 600 tickets are available
at two dierent prices. Type A is oered
at sh 300 each and type B at sh 500 each.
The organiser wishes to raise a minimum
of sh 150 000. He also wishes to sell at
least twice as many of the dearer tickets
as the cheaper ones. Write down as many
inequalities as you can from the given
information.
(a) Represent all the inequalities
graphically.
(b) Find the maximum possible collection.
(c) If the organiser has to cover his
expenses of sh 12 000, nd the
minimum possible amount that goes
to charity.
2. Mr. Koech has an 18 ha piece of land. He
wishes to plant part of the land with beans
and the other part with potatoes. The total
cost per ha for beans is sh 1 000, and the
cost per ha for potatoes is sh 800. He has
to hire 2 men per ha for beans, and 1 man
per ha for potatoes. He cannot hire more
than 18 men. For the whole project he
has a maximum of sh 12 000 available. If
he plants x ha with beans and y ha with
potatoes, write down all the inequalities that
must be satised by x and y. Hence, nd his
maximum prot if prot per ha of beans is
sh 2 400 and sh 1 600 per ha of potatoes.
3. A factory is in the process of installing two
new types of machines. For machine type
A, the oor space available is 500 m
2
,
labour needed per machine is 9 men, and
the output per week is 300 units. For type B,
oor space available is 600 m
2
, labour per
machine is 6 men, and the output per week
is 200 units. The factory has 4 500 m
2
oor
space available and only 54 skilled workers
who can work on the new machines.
(a) If the manager buys x machines of type
A and y machines of type B, write down
the inequalities satised by x and y.
(b) Represent the inequalities graphically,
and identify clearly the possible
solutions open to the manager.
(c) How many machines of each type
should he buy in order to
(i) achieve maximum output?
(ii) utilise all his skilled workers?
4. A factory makes tables and chairs. One
table takes 1 hour of machine time and
3 hours of craftman’s time. A chair takes
2 hours of machine time and 1 hour of
craftman’s time. In a day, the factory has a
maximum of 28 machine hours and 24 hours
of craftman’s time. The factory makes
x tables and y chairs on a particular day,
(a) Write down all the inequalities satised
by x and y.
(b) Represent the inequalities graphically.
(c) How many tables and chairs must be
made in a day if the factory is to work
at full capacity?
(d) If a chair gives a prot of sh 500 while
a table gives a pot of sh 800, nd the
maximum prot to the company in a
day if it produces
(i) chairs only,
(ii) tables only,
(iii) an equal number of tables and
chairs,
(iv) at full capacity.
5. Lily is a fresh fruit juice supplier. She is
particular about avour and colour density.
She uses x mangoes and y oranges. For a
strong good avour, she ensures that
5x + 2y is at least 80. For an attractive
colour, 3x must be less than 2y.
103
(a) Write down all the possible inequalities
which satisfy the given conditions,
assuming that both types of fruits must
be used.
(b) Illustrate your inequalities graphically.
(c) Given that one mango costs sh 20
while an orange costs sh 5, state the
objective function, and hence nd
the cheapest combination of fruits for
making the juice.
6. In a musical concert, a local musician
was to sing x classicals and y raps. Each
classical takes 3 minutes while each rap
takes 4 minutes. Allowing for applause and
change over, the musician is expected to
perform for a maximum of 36 minutes. His
manager advises him to sing more classicals
than raps. His fans demand that he sings
more than 3 classicals and at least 2 raps.
(a) Write down all the inequalities that
satisfy the given conditions in x and y.
(b) Represent the inequalities in part (a)
graphically.
(c) Use your graph to write down all the
possible combinations of songs.
7. A science laboratory is to be improved
by adding some extra science kits. There
are two types available. Each type A kit
occupies 3 m
2
of oor space while type B
occupies 0.5 m
2
of oor space.
The available oor area does not exceed
40 m
2
. Each type A kit costs sh 12 000
and each of type B costs sh 30 000. The
purchasing ocer can only approve a
maximum of sh 270 000 for this purpose.
(a) If the school plans to buy x kits of type
A and y of type B, form as many
inequalities in x and y as possible to
represent this information.
(b) Use graphical method to nd the
greatest number of kits of each type
that the school may be able to buy.
8. A transport company uses two types of
trucks, P and Q. Type P carries 200 bags
of maize while type Q carries 300 bags of
maize per trip. There are more than 12 000
bags to be moved, and the trucks are to
make no more than 60 trips. Type Q trucks
are to make at most twice the number of
trips made by type P trucks.
(a) If x represents the number of trips made
by type P trucks and y the number of
trips made by type Q trucks, write
down all the inequalities representing
this information.
(b) The transporter makes a prot of
sh 1 000 per trip on type P truck and
sh 2 000 per trip on type Q truck. Write
down an objective function for the prot.
State the number of trips he should
make in order to maximise his prot.
104
1
4
–
3
2
–
Revision exercise 2.1
1. With reference to Fig. R2.1, state
(a) all the lines that are skew with line BC,
(b) all the lines that are parallel to plane
ADHE,
(c) which of the following sets of lines or
points determine a plane:
(i) A, H, G, B (ii) AD and FG
(iii) FB and HD (iv) A, H, G, C
(d) the projection of line (i) EG onto plane
ABCD, (ii) FG onto line BC.
Fig. R2.1
2. Solve the following inequalities and
illustrate each solution set on a number line.
(a) 3(x – 2) + 4 > –2 – (3 + 2x)
(b) (x + 2) + (x + 3) < (x + 4)
(c) ≤ +
3. A and B lie on a parallel of latitude 7 788
km apart. Their longitudes are 10°E and
140°W respectively. Calculate the latitude
on which they lie.
4. The length of the edge of a cube is 10
cm. Calculate the angle between its main
diagonal and the base.
5. Find the greatest integral value of y which
satises 2 – y > y + 3.
6. A ship sails 250 nm along the parallel of
latitude 60°N. Calculate its change in
longitude.
7. A point A lies on the latitude 50°N and
longitude 10°W, B lies on latitude 50°N
and 85°W and C lies on 35°S and 10°W.
Calculate
(a) the distance between A and C along a
great circle,
(b) the radius of the parallel of latitude
50°N,
(c) the distance between A and B along a
parallel of latitude,
(d) the distance of B from the equator,
(e) the length of the chord AB,
(f) the angle subtended by AB at the centre
of the earth, hence the length of the arc
AB along a great circle.
8. Given that p, q and r are numbers such that
2 ≤ p ≤ 3, 4 ≤ q ≤ 5 and 6 ≤ r ≤ 7, nd
(a) the maximum value of
r – p
,
q
(b) the minimum value of q + r – p.
9. Fig. R2.2 represents a wedge in which the
base and back are rectangular. Find the
angle indicated
θ
.
Fig. R2.2
10. A plane ies from Alexandria (30°N, 30°E)
straight west to New Orleans (30°N, 90°W),
then ies south to Guatemala (15°N, 90°W).
REVISION EXERCISES 2 (CHAPTERS 5–7)
– – – – – – – – – – –
– – – – – – – – – – – – – – – –
A
H
G
C
B
D
F
E
– – – – – – – – – –
1
2
–
1
4
–
1
3x
––
1
2x
––
1
4
–
– – – – – – – – – – – – – – – – – – – – – – –
– – – – – –
– – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – –
12 cm
4 cm
3 cm
θ
5-7
105
Revision exercise 2.2
1. Fig. R2.5 shows a regular octahedron in
which each edge is 4 cm long.
(a) Name two squares which are congruent
to ABCD.
(b) Find the length of EF.
(c) Find the volume of the octahedron.
Fig. R2.5
2. By shading the unwanted region, indicate
the region satised by the inequalities y > 0,
0 ≤ x ≤ 3 and x + y < 4.
3. An isosceles triangle ABC has base
AB = 6 cm and the two equal sides each
9 cm long. It rests with AB on a horizontal
surface, with its plane inclined at 30° to the
horizontal surface. How high is C above
the surface?
4. Form linear inequalities to describe each of
the following conditions:
(a) The perimeter of a rectangular plot
of land must be at least 80 m and
the length must not exceed twice the
breadth.
(b) The sum of the ages of two sisters is
not more than 20 years and one of them
is at least 3 years older than the other.
5. A and B are points on the same latitude
55°N. Their longitudes are 95°W and 35°E
– – – – – – – – – –
– – – – – – – – – – – – – –
– – – – – – – – – – – –
– – – – – – – –
N
B
C
G
H
A
D
E
F
M
•
•
Taking the radius of the earth to be 6 370
km, calculate,
(a) the total distance covered by the plane,
(b) the local time at which the plane reaches
New Orleans if it leaves Alexandria
at 8.00 a.m and moves at an average
speed of 1 000 km/h.
11. The unshaded region in Fig. R2.3 represents
the solution set of three simultaneous
inequalities. Find the inequalities.
Fig. R2.3
12. ABCDEFGH, in Fig. R2.4 is a frustum of a
pyramid whose base is a square of side 12
cm. M and N are the centres of ABCD and
EFGH respectively. Given that MN = 5 cm,
calculate,
(a) NT, where T is the midpoint of FG.
(b) the inclination of the face BCGF to the
horizontal.
Fig. R2.4.
– – – – – – – – – – – – – – – – – – – – – – – – – – – – –
y
x
(2 , 4)
(
–
5 , 3)
O
A
B
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – –
– – – – – – – – –
– – – – – – –
F
E
D
C
A
B
106
10. (a) Calculate the dierence in local time
between (40°20′N , 74°W) and
(22°N , 72°15′E).
(b) The local times at points P and Q dier
by 2 hours. If P lies on longitude
23°E, nd two possible longitudes of Q.
11. Fig. R2.7 shows a right pyramid on a
rectangular base ABCD, where AB = 14
cm and BC = 10 cm. Its slant edge is 24 cm
long. Calculate
(a) the height of the pyramid,
(b) the angle between faces ADV and DCV,
(c) the angle between VA and AD.
Fig. R2.7
12. A classroom globe is in the form of a sphere
of radius 28 cm. Calculate, on this globe,
(a) the length of the circle of latitude 65°N.
(b) the shortest distance of a point, on this
circle, from the south pole.
Revision exercise 2.3
1. In Fig. R2.8, PQRS is a vertical wall
running in the south-north direction.
PQ = 16 m, PS = 5 m, QR = 13 m and point
T is 12 m due east of P. Calculate
(a) TR (b) ∠QTR
(c) the bearing of T from Q.
2
3
–
1
4
–
1
3
–
respectively. Calculate
(a) the radius of the latitude,
(b) the length of line segment AB,
(c) the angle subtended by AB at the centre
of the earth,
(d) the length of the arc AB.
6. A ship sails from (40°S , 10.5°W), a distance
of 1 200 km along the parallel of latitude.
Find the change in longitude. Find the
coordinates of two possible nishing points.
7. Illustrate the region that satises the
inequalities, y < 9 – x, 2x + y < 24, and
x + y > 12. Find, within the region,
(a) the maximum value of
(i) x + 2y (ii) 3x + y
(b) the minimum value of x + 2y.
8. Fig. R2.6 shows a door opened at an angle
of 60°. Calculate
(a) DE (b) ∠DCE (c) ∠DBE
Fig. R2.6
9. Draw a graph to illustrate the region
satisfying the inequalities y ≥ 0, y + 2x ≥ 8,
4y + 3x < 24 and 3y ≤ 2x.
(a) List the coordinates of all the points
that give integral solutions.
(b) Find the minimum value of y – x in
this region.
(c) Find the maximum value of 2y + 3x in
this region.
C
E
D
A
B
F
1 m
2 m
– – – – – – – –
– – – – – – – – – –
V
B
A
C
D
24 cm
10 cm
– – – – – – – – – – – – – –
14 cm
107
6. A computer teacher has sh 2 500 to spend
on some basic calculators and some packets
of diskettes. A packet of diskettes costs
sh 250 and a calculator sh 100. She intends
to buy at least ve of each and spend over
sh 500 more on diskettes than on calculators.
If she buys x calculators and y packets of
diskettes,
(a) write down all the possible inequalities
representing this information,
(b) nd all the possible combinations of the
number of calculators and the number of
packets of diskettes that she can buy, and
(c) nd the greatest number of packets of
diskettes that she can buy.
7. Fig. R2.9 shows a cage in which base
ABCD and roof PQRS are both rectangular.
AP, BQ, CR and DS are perpendicular to
the base. Calculate
(a) QR (b) ∠QRC
(c) the angle between planes ABCD and
PQRS,
(d) the inclination of PR to the horizontal.
Fig. R2.9
8. Fig. R2.10 shows a right pyramid with
a triangular base. The base ABC is an
equilateral triangle of side 8 cm. The edges
VA, VB and VC are 12 cm each. Find
(a) the angle between VA and the base,
(b) the angle between face VBC and the
base,
P
12 m
T
R
Q
S
13 m
16 m
5 m
N
1
3
–
Fig. R2.8
2. Show, on the xy-plane, the region
representing the inequalities
4y + 3x ≥ 18, y < 2 x + 2, 5 y + 2 x < 32, y ≥ 0.
Maximize x + y in this region.
3. A ship sails from P(0°, 70°W) due north
to Q(25°N, 70°W), then due east to
R(25°N, 12°E) and nally a further
1 800 nm due S. Calculate
(a) the total distance travelled in nm,
(b) its average speed in knots if the total
time taken for the journey is 300 hours,
(c) the nal position of the ship.
4. A spherical globe of radius 55 cm is marked
with lines of latitude and longitude. An
ant resting at a point (60°N, 20°E) starts
moving south, following a line of longitude,
at a steady speed of 1 cm/s and stops after
30 seconds. Calculate
(a) the total distance the ant moves,
(b) the nal position of the ant,
(c) the angle between the nal position and
the initial position.
5. A pilot ies from P(40°S, 148°E) to
Q(40°N, 148°E) at an average speed of
576 km/h. Calculate his time of ight in
hours.
If from P the pilot ies due east at the same
speed, nd his new position in 2 hours.
– – – – – – – – – – – – – – –
– – – – – – – – – –
– – – – – – – – – – – – – – –
P
60 cm
B
C
D
A
Q
R
S
75 cm
50 cm
80 cm
108
(c) the height of the pyramid.
Fig. R2.10
9. ABCD is a rectangular base of a pyramid
with vertex V. The edges VA, VB, VC, and
VD are each 25 cm long. AB = CD = 14 cm
and AD = BC = 10 cm. Calculate
(a) the height of the pyramid,
(b) the angle between the base and an edge,
(c) the angle between the base and plane
VBC,
(d) the angle between the base and plane
VCD.
10. A farmer plans to grow two types of crops,
X and Y. He has identied up to 70 ha of
land for this purpose. He has 240 man-days
of labour available and he can spend up to a
maximum of sh 36 000. The requirements
for the crops are as follows.
X Y
Minimum number of
hectares to be sown 10 20
Man-days per ha 2 4
Cost per ha in sh 600 400
If x and y represent the number of hectares
to be used for crops X and Y respectively,
write down, in their simplest form, the
inequalities which x and y must satisfy.
On a graph paper, using a suitable scale,
show the region within which the point
(x , y) must lie if the inequalities are to be
satised simultaneously.
11. An aircraft leaves A(60°N, 13°W) at 1300
h, and arrives at B(60°N, 47°E) at 1700 h.
(a) Calculate the average speed of the
aircraft in knots.
(b) Town C(60°Ν, 133°W) has an airstrip.
Two helicopters leave B at the same
time. One moves due west to C, while
the other one moves due north to C.
If the two aircrafts are moving at
600 knots, nd the time taken by each
to reach C.
(c) The local time at a town D(23°N, 5°W)
is 1000 h. What is the local time at B?
12. A town council plans to build a car park for
x cars and y buses. Cars are allowed 10 m
2
of ground space, and buses 20 m
2
. There is
only 500 m
2
available. The park can hold a
maximum of 40 vehicles at a time, and only
a maximum of 15 buses at a time.
(a) Write down three inequalities other
than x ≥ 0 and y ≥ 0.
(b) On a graph paper, show the region that
satises the given conditions.
(c) If the parking charges are sh 50 for cars
and sh 200 for buses per day, nd the
number of vehicles of each type that
should be parked to obtain maximum
income.
(d) Calculate the maximum income.
– – – – – – – – – – – –
– – – – – – – – – – – –
– – – – – – – – – – – –
C
V
B
A
109
Gradient, average and instantaneous
rate of change.
Point of interest
Sir Isaac Newton (1642–1727), an English
mathematician, is taken by many people
to be one of the greatest intellectuals of
all time. He and Gottfried Wilheim von
Leibniz (1646–1716), a German philosopher/
logician, independently invented the branch
of mathematics called calculus. Calculus is
the field of mathematics which deals with
dierentiation and integration of functions and
related concepts and applications.
By the age of 23, Isaac Newton had made three
major discoveries: the nature of colours, the
calculus and the law of gravitation. It was by
using his version of the calculus that he gave
the rst satisfactory explanation of the motion
of the sun, the moon and the stars. However,
Newton was extremely sensitive to criticism,
and so was very secretive. All the same, he was
eventually persuaded to publish his discoveries
in 1687.
In addition to discovering calculus, Leibniz
also invented many of its symbols. He also
invented a mechanical multiplication machine.
In Book 2, we learnt that if a line passes
through the points (x
1
, y
1
) and (x
2
, y
2
), then its
gradient is given by:
Gradient =
y
2
– y
1
=
y
1
– y
2
x
2
– x
1
x
1
– x
2
In Book 3, we saw that the gradient of a curve
changes continuously. Thus, the gradient of
a curve at a particular point could only be
approximated. We approximated it by the
gradient of the tangent to the curve at that point.
Gradient is a measure of the rate of change of
one variable with respect to another, e.g. change
of distance with time, change of velocity with
time, change of pressure with height above sea
level, etc.
In Fig. 8.1, AB is a line that cuts the given
curve at two points. Such a line is known as a
secant to the curve. The gradient of secant AB
gives the average rate of change (i.e. average
gradient) of y with respect to x.
The gradient of tangent PT gives the rate of
change of y with respect to x at the point P, i.e.
an instantaneous rate of change.
Fig. 8.1
Exercise 8.1
1. Given that s = 4 + 6t – t
2
, where s represents
distance in metres and t represents time in
seconds, is an equation of motion, make a
table of values for t = 0 to t = 6 and draw
the graph of this motion.
2. What does the gradient of the curve from
Question 1 represent?
3. Using your graph from Question 1, nd the
average speed between t = 1 and t = 4.
4. Use your graph from Question 1 to nd the
speed when
(a) t = 2 (b) t = 3
y
x
T
B
P
A
•
•
•
DIFFERENTIATION
8
110
With Q at (3 , 9), the gradient of
PQ =
9 – 1
= 4.
3 – 1
If Q moves to point Q
1
(2 , 4), the gradient of
PQ
1
=
4 – 1
= 3.
2 – 1
Table 8.1 shows the gradient of PQ for dierent
positions of Q.
Position of Q Change Change Gradient
in x in y
(3 , 9) 2 8 4
(2 , 4) 1 3 3
(1.5 , 2.25) 0.5 1.25 2.5
(1.2 , 1.44) 0.2 0.44 2.2
(1.1 , 1.21) 0.1 0.21 2.1
(1.01 , 1.020 1) 0.01 0.0201 2.01
(1.001 , 1.002 001) 0.001 0.002 001 2.001
Table 8.1
From Table 8.1, it is apparent that as the change
in x becomes smaller, the gradient of PQ
becomes very nearly equal to 2.
If h is a small change in x, then Q is
[1 + h , (1 + h)
2
] and the change in y is
(1 + h)
2
– 1 = (1 + 2h + h
2
) – 1 = 2h + h
2
.
The gradient of PQ now is
2h + h
2
.
h
As Q approaches P, the small change h
approaches zero (i.e. h 0).
Since tangent PT is the limiting position of
secant PQ as Q approaches P, the gradient of PT
is the limit of
2h + h
2
as h
0.
h
This is usually expressed as:
Gradient of PT = lim
h 0
2h + h
2
and
h
simplies to:
Gradient of PT = lim
h 0
(2 + h) = 2.
Hence, the gradient of the curve y = x
2
at (1 , 1)
is 2.
5. Describe how the gradient of the curve
from Question 1 is behaving
(a) between t = 0 and t = 3,
(b) at t = 3,
(c) between t = 3 and t = 6.
Gradient of a curve at a point
We have already seen that the gradient of a
curve at a point is given by the gradient of the
tangent to the curve at that point.
Consider the gradient of the curve y = x
2
at the
point (1 , 1).
In Fig. 8.2, P(1 , 1) is the point on the curve.
Q(3 , 9) is another point of the curve.
Fig. 8.2
Consider secant PQ to be a straight rod hinged
at P. It is rotated clockwise about P to take up
successive positions PQ
1
, PQ
2
, …
Notice that the points at which it meets the
curve i.e. Q
1
, Q
2
… are successively nearer to
the xed point P. The closer this point gets to
P, the closer the gradient of the secant is to the
gradient of the tangent PT.
If we use to denote ‘approaches’or ‘tends
to’, we can then write:
As Q P along the curve, the gradient
of PQ the gradient of PT.
The tangent PT is called the limit of secant
PQ, and the gradient of the curve at P is the
limit of the gradient of PQ.
0
x
Q
1
Q(3 , 9)
•
•
•
•
•
•
y = x
2
Q
3
Q
2
P(1 , 1)
y
T
111
Gradient of PQ = (x + h )
2
– x
2
h
= x
2
+ 2hx + h
2
– x
2
h
= 2hx + h
2
h
As Q P , h 0 and gradient of the curve at P
= Lim
h 0
2hx + h
2
h
= Lim
h 0
(2x + h)
= 2x
i.e. the gradient of the curve y = x
2
at the point
(x , y) is 2x.
The expression 2x, obtained in Example 8.2,
is called the gradient function of the curve
y = x
2
.
The gradient function can be used to nd the
gradient of the curve at any point. For example,
the gradient of the curve y = x
2
at:
1. (1 , 1) is 2 × 1 = 2 (as in Table 8.1),
2. (2 , 4) is 2 × 2 = 4 (as in Example 8.1),
3. (3 , 9) is 2 × 3 = 6.
What is the gradient of the curve y = x
2
when
1. x = 1.5, 2. x = –2, 3. x = 0, and
4. y = 25?
Note: The method of nding the gradient
function using small changes is referred to as
workingfromrstprinciples.
Example 8.3
For the curve y = x
3
, nd
(a) the gradient function,
(b) the gradient at the point (2, 8).
Solution
(a) Fig. 8.4 is a sketch of the curve y = x
3
.
P and Q are points on the curve with
coordinates (x , x
3
) and [x + h , (x + h)
3
]
respectively.
Example 8.1
Deduce the gradient of the curve y = x
2
at (2 , 4)
by taking the x-coordinate at Q as (2 + h).
Solution
In this case, the tangent to the curve is drawn at
P(2 , 4).
The x-coordinate of Q is (2 + h). Hence, its
y-coordinate is (2 + h)
2
= 4 + 4h + h
2
.
Thus, the change in y-coordinate is
(4 + 4h + h
2
) – 4 = 4h + h
2
∴ Gradient of PQ = Lim
h 0
4h + h
2
h
= Lim
h 0
(4 + h)
= 4.
i.e. the gradient of the curve y = x
2
at (2 , 4) is 4.
Gradient function
The method used to nd the gradient of
the curve y = x
2
at a particular point can be
extended to nd the gradient at any point.
Example 8.2
Find the gradient of y = x
2
at the point (x , y).
Solution
(x, y) = (x , x
2
) is a general point on the curve
y = x
2
.
If h is a small change in x, we have another
point on the curve with coordinates
[x + h , (x + h)
2
] as in Fig. 8.3.
Fig. 8.3
0
x
P(x , x
2
)
y
y = x
2
Q[x + h , (x + h)
2
]
– – –
– – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – –
h
(x + h)
2
– x
2
•
•
112
Fig. 8.4
Gradient of PQ =
(x + h)
3
– x
3
h
=
x
3
+ 3x
2
h + 3xh
2
+ h
3
– x
3
h
=
3x
2
h + 3xh
2
+ h
3
h
= 3x
2
+ 3xh + h
2
As Q P , h 0 and
gradient of the curve at P
= Lim
h 0
(3x
2
+ 3xh + h
2
)
= 3x
2
i.e. the gradient of the curve y = x
3
at the
point (x , y) is 3x
2
.
(b) At (2 , 8), the gradient of the curve y = x
3
is
3 × 2
2
= 12.
Exercise 8.2
1. Find the average gradient of the curve y = x
4
between x = 1 and x = 4.
2. P(1 , 1) is a point on the curve y = x
3
. Q is a
movable point on the same curve. Copy and
complete Table 8.2 for dierent positions of
Q. Hence deduce the gradient of y = x
3
at
P(1 , 1).
Position of Q Change Change Gradient
in x in y
(2 , 8) 1 7 7
(1.5 , 3.375) 0.5 2.375 6.75
(1.1 , 1.331) 0.1 0.331
(1.01 , 1.030 301)
(1.001 , 1.003 003 001)
Table 8.2
3. Using h as a small change in x, nd the
gradient of the curve y = x
4
at x = 2.
4. Find the gradient function of y = c, where
c is a constant.
5. Use h, a small change in x, to nd the
gradient functions of the following.
(a) y = x
4
(b) y = x
5
.
6. Find the gradient function of each of
the following and, in each case, nd the
gradient of the curve when x = 2.
(a) y = x
6
(b) y = x
7
Gradient function of a polynomial
A polynomial in x is an expression of the form
a
0
x
n
+ a
1
x
n – 1
+ … + a
n – 1
x + a
n
,
where n is a positive integer and a
0
, a
1
, …, a
n
are constants.
The method of Examples 8.2 and 8.3 can be
used to nd the gradient function of such an
expression.
Example 8.4
Find the gradient function of y = ax
3
, where a
is a constant.
Solution
P(x, y) = P(x, ax
3
) is a general point on the
curve y = ax
3
(Fig. 8.5).
Q[x + h, a(x + h)
3
] is another general point on
the curve, where h is a small increase in x.
Gradient of PQ =
a(x + h)
3
– ax
3
h
x
P(x , x
3
)
y
y = x
3
Q[x + h , (x + h)
3
]
– – – –
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – ––
h
(x + h)
3
– x
3
0
16
12
8
4
1 2
•
•
113
1
2
–
1
2
–
3
2
–
Fig. 8.5
=
ax
3
+ 3ax
2
h + 3axh
2
+ ah
3
– ax
3
h
=
3ax
2
h + 3axh
2
+ ah
3
h
= 3ax
2
+ 3axh + ah
2
.
As Q P , h 0 and gradient of the curve at P
= Lim
h 0
(3ax
2
+ 3axh + ah
2
)
= 3ax
2
.
Use a similar approach, to that of Example 8.4,
to show that the gradient function of y = ax
4
is 4ax
3
.
Example 8.5
Find the gradient function of y = x
2
+ 4x – 1 and
nd the gradient of this curve at the point (1 , 4).
Solution
Fig. 8.6 is a sketch of the curve y = x
2
+ 4x – 1.
Points P and Q, whose coordinates are
[x , (x
2
+ 4x – 1)] and
[x + h, (x + h)
2
+ 4(x + h) – 1] respectively
are two points on the curve.
PR = h, and
RQ = [(x + h)
2
+ 4(x + h)–1] – (x
2
+ 4x – 1)
= (x
2
+ 2xh + h
2
+ 4x + 4h – 1) – (x
2
+ 4x – 1)
= 2xh + 4h + h
2
Hence, gradient of PQ
=
RQ
=
2xh + 4h + h
2
PR h
= 2x + 4 + h.
Fig. 8.6
∴ Gradient of the curve at P
= Lim
h 0
(2x + 4 + h)
= 2x + 4
i.e. the gradient function of the curve is 2x + 4.
At the point (1 , 4),
gradient = 2 × 1 + 4 = 6.
Using a similar approach to that of Example
8.5, conrm the following:
Polynomial Gradient function
2x
4
8x
3
4x
2
+ 3x –1 8x + 3
x
3
– 5x 3 x
2
– 5
Note:
1. The gradient function of a polynomial is the
same as the sum of the gradient functions of
the individual terms of the polynomial.
2. Tondthegradientfunctionofatermofa
polynomial in x, reduce the power of x by 1
 andmultiplythecoecientofxbythe
original power.
Exercise 8.3
1. Calculate the gradients of the curve y = 8x
2
at the points where x =
–
1 , , , 3.
2. Find the gradient function of the curve
y =
x
2
– 3x. Hence nd the gradients of
the curve when x =
–
3 , 0 , 3 , 6.
x
P(x , ax
3
)
y
y = ax
3
Q[x + h , a(x + h)
3
]
– – –
– – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – –
h
a(x + h)
3
– ax
3
0
•
•
x
P
y
Q
– – – – – – –
– – – – – – – – – – – – – – – – –
0
R
–
1
y = x
2
+ 4x – 1
•
•
114
5
4
–
1
2
–
2
3
–
1
10
––
1
4
–
y = x
2
. PR = δx, RQ = δy, ON = x + δx and
NQ = y + δy. Hence, the coordinates of Q are
(x + δx, y + δy).
Fig. 8.7
Gradient of PQ =
RQ
=
δy
.
PR δx
This is a measure of the average rate of change
of y with respect to x between P and Q.
At P, y = x
2
……… ……… (i)
and at Q, y + δy = (x + δx)
2
……… (ii)
Subtracting (i) from (ii) gives
δ y = (x + δx)
2
– x
2
= x
2
+ 2xδx + (δx)
2
– x
2
i.e. δy = 2x × δx + (δx)
2
.
Dividing through by δx gives
δ y
= 2x + δx.
δx
This is the gradient of secant PQ.
As δx approaches zero, δy also approaches zero
and Q approaches P (or secant PQ virtually
coincides with the tangent at P).
Hence, gradient of the curve at P
= Lim
δ x 0
δy
δx
= Lim
δ x 0
(2x + δx)
= 2x.
The expression Lim
δ x 0
δ y
is usually
δ x
abbreviated as
dy
.
dx
3. For each of the following, nd the gradient
function and the gradient of the tangent at
the given point.
(a) y = 3x
2
+ 2, (3 , 29).
(b) y = 5x
3
– 2x
2
– x, (2 , 30).
(c) y = 3x
4
– x, point where x =
–
1.
(d) y = 4x
5
– x, point where x = .
4. At what point is the gradient of the curve
(a) y = x
2
+ 4x equal to 6?
(b) y = 5x
2
– 3x + 2 equal to 17?
5. Find the x-coordinate of the point where the
gradient of the curve y = x
2
– 3x is equal
to zero.
6. From the work done so far, deduce a rule
for nding the gradient function of the
general polynomial
a
0
x
n
+ a
1
x
n – 1
+ … + a
n – 1
x + a
n
.
Hence, nd the gradient function of each of
the following curves.
(a) y = 5x
3
– 6x
2
+ 4x
(b) y = x
15
+ x
4
.
Delta (δ)notationandthedierential
coecient
We have so far used h to represent a small
change in x. However, any other letter could be
used instead of h. The Greek letter “delta” (i.e.
∆ or δ) is also commonly used to represent a
small change. A small change in x is denoted as
δx (read ‘delta x’). Likewise, a small change in
y is denoted as δy.
Note that δx is a single quantity and not the
product of δ and x. Similarly, δy is a single
quantity.
Example 8.6
Using δx in place of h, nd the gradient
function of the curve y = x
2
.
Solution
In Fig. 8.7, P and Q are points on the curve
O
x
•
P(x , y)
y
y = x
2
Q(x + δ x , y + δ y)
– – – – – –
– – – – – – – – – – – – – – – – –
R
•
δ y
δ x
N
M
– – – – – –
•
115
Thus, if
δy
= 2x + δx,
δx
then Lim
δx 0
δy
= 2x
δx
⇒
dy
= 2x (gradient function of y = x
2
).
dx
dy
(read ‘dee y dee x’) is called the dierential
dx
coecientofy with represent to x.
The process of nding the dierential
coecient of an expression is called
dierentiation, and the gradient function thus
obtained is often called the derived function or
derivative of the expression.
The process of dierentiation is denoted by
d
.
Thus
d
(x
2
) =
d(x
2
)
= 2x.
dx
dx
We say that y or x
2
has been dierentiatedwith
respect to x.
Example 8.7
Working with small change δx, nd the derived
function for the expression x
3
– 3x
2
.
Solution
Let y = x
3
– 3x
2
………………………………… (i)
Let δx and δy be small changes in x and y
respectively. Then,
y + δy = (x + δx)
3
– 3(x + δx)
2
……… (ii)
Subtracting (i) from (ii) gives:
δy = [(x + δx)
3
– 3(x + δx)
2
] – [x
3
– 3x
2
]
= [x
3
+ 3x
2
δx + 3x(δx)
2
+ (δx)
3
–
(3x
2
+ 6xδx + 3(δx)
2
)] – [x
3
– 3x
2
]
= 3x
2
δx + 3x(δx)
2
– 6xδx + (δx)
3
– 3(δx)
2
.
⇒
δ y
= 3x
2
– 6x + 3xδx + (δx)
2
– 3δx.
δx
Lim
δ x 0
δy
= Lim
δx 0
(3x
2
– 6x + 3xδx +
δx
(δx)
2
– 3δx)
= 3x
2
– 6x
(since 3xδx 0, (δx)
2
0
and 3δx 0).
Hence
dy
= 3x
2
– 6x .
dx
Example 8.8
Dierentiate (a)
1
(b)
1
with respect to x.
x
x
2
Solution
(a) Let y =
1
.
x
Working with small changes,
y + δy =
1
.
x + δx
Subtracting y from y + δy gives
δy =
1
–
1
x + δx x
=
x – x – δx
=
–
δx
.
x(x + δx) x(x + δx)
⇒
δ y
=
–
1
.
δx x(x + δx)
∴
dy
= Lim
δx → 0
δy
dx
δx
=
–
1
.
x
2
(b) Let y =
1
.
x
2
Working with small changes
y + δy =
1
.
(x
+
δx)
2
Subtracting y from both sides gives
δy =
1
–
1
(x + δx)
2
x
2
=
x
2
– (x + δx)
2
x
2
(x + δx)
2
=
x
2
– x
2
– 2xδx – (δx)
2
x
2
(x + δx)
2
=
–
2xδx + (δx)
2
.
x
2
(x + δx)
2
⇒
δ y
=
–
2x + δx
.
δx x
2
(x + δx)
2
∴
dy
= Lim
δx 0
δy
dx δx
=
–
2x
=
–
2
.
x
4
x
3
dx
116
(
)
4
Dierentiationbyrule
From what we have done so far in this chapter,
it should be clear that dierentiation can be
done by rule, without going through the process
of nding limits. For example:
If y =
x
2
,
dy
= 2x ;
dx
If y =
x
3
,
dy
= 3x
2
;
dx
If y = 5x
4
,
dy
= 20x
3
;
dx
If y = 3x
3
+ 4x
2
– 5x,
dy
= 9x
2
+ 8x – 5 ; etc.
dx
We can now write down the following rules of
dierentiation.
If y = x
n
,
dy
=
nx
n – 1
;
dx
If y = ax
n
,
dy
=
anx
n – 1
;
dx
If y = c, i.e. y = cx° (where c is a constant),
then
dy
=
0 × cx
–1
= 0.
dx
If y = a
0
x
n
+ a
1
x
n – 1
+ ... + a
n – 1
x + a
n
,
dy = a
0
nx
n – 1
+ a
1
(n – 1)x
n – 2
+ … + a
n – 1
.
From the examples done so far, it seems
like these rules hold for all values of n, even
negative ones. For instance, in Example 8.8,
where x has negative indices, we have seen that:
For y =
1
= x
–1
,
dy
=
–
1
=
–
x
–2
, and
x
dx
x
2
for y =
1
=
x
–2
,
dy
=
–
2
=
–
2x
–3
, which
x
2
dx
x
3
agree with the rules.
As a matter of fact, these rules are true for all
values of n (both positive and negative, as well
as fractional values), but the proof is beyond the
scope of this book.
1
2
–
1
3
–
Example 8.9
Dierentiate the following expressions with
respect to x.
(a) x
2
(x + 2) (b)
x
3
– 3x
2
x – 3
Solution
Note that in order to be able to dierentiate the
given expressions, we need to rst rewrite them
in a form for which the rules of dierentiation
apply.
(a) Let y = x
2
(x + 2).
⇒ y = x
3
+ 2x
2
∴
dy
= 3x
2
+ 4x.
dx
(b) Let y =
x
3
– 3x
2
x – 3
⇒ y =
x
2
(x – 3)
x – 3
i.e. y = x
2
∴
dy
= 2x, except for x = 3 (where y
dx
is undened!)
Exercise 8.4
1. Dierentiate the following expressions
with respect to x.
(a) 9x
3
(b)
–
5x
4
(c)
–
4x
5
+ x
3
(d)
8
x
(e)
–
2
(f)
5
(g)
5x – 4
x
2
x
4
(h) 4x
2
+ 3x – 4 (i) 5x
3
– 3x
2
+ 4x
(j) x –
1
(k)
x
2
+ 2x +
3
x
x
3
(l)
1
+
1
–
1
(m) x
– x
3x
2
2x
3
5x
4
(n)
3 –
1
2
(o) √x + √x
x
2. Find
dy
given that:
dx
(a) y = x(x – 1)
(b) y = (x
2
– 1)(x + 2)
1
2
–
117
(c) y = (√x + 3)(√x – 3)
(d) y =
x
3
+ x
2
– x – 1
x
(e) y =
(x
2
+ 2)(x
2
+ x)
x
(f) y =
x
3
(x
2
– 1)
(g)
y =
x
2
– 4x + 4
x – 1 x – 2
(h) y = x
-2
(x – x
4
) (i)
y = √x(x
– x
–
).
3. Find the derivatives of the following
relations and hence nd the gradients of
their curves at the given points.
(a) y = 3x – x
2
, (–2 , 2)
(b) y = x
2
+ 3x – 1, (
–
2 ,
–
3)
(c) y = x
3
+ 3x
2
– 9x + 3, (1 ,
–
2).
4. At what point(s) is the gradient of the curve
(a) y = x
2
+ 3x – 1 equal to 3?
(b) y = x
3
+ 2x + 1 equal to 5?
(c) y = x
3
+ 3x
2
– 9x + 3 equal to 15?
Equations of tangent and normal to a
curve
We have seen that the gradient of a curve at a
point is the same as the gradient of the tangent
to the curve at that point. This fact is useful in
nding the equation of the tangent and normal
to a curve at a specied point. The normal is
the line that is perpendicular to the curve at
the given point.
Example 8.10
Find the equations of the tangent and the
normal to the curve y = x
2
+ 4x – 3 at the
point (1 , 2).
Solution
y = x
2
+ 4x – 3
⇒
dy
= 2x + 4 (the gradient function)
dx
At the point (1 , 2), x = 1
⇒ the gradient of the curve at this point is
2 × 1 + 4 = 6.
Hence, the gradient of the tangent to the curve
at (1 , 2) is 6, i.e. the same as the gradient of
the curve.
The equation of the tangent is
y – 2
= 6
x – 1
⇔ y – 2 = 6(x – 1)
i.e. y = 6x – 4
The normal to the curve at (1, 2) is
perpendicular to the tangent at the same point.
Thus, if the gradient of the normal is m, then
m × 6 =
–
1
⇔ m =
–
1
6
Hence, the equation of the normal is
y – 2
=
–
1
x – 1 6
⇔ 6(y – 2) =
–
1(x – 1)
⇔ 6y – 12 =
–
x + 1
i.e. y =
13
–
x
6
6
Exercise 8.5
1. Find the equations of the tangents to the
following curves at the indicated points.
(a) y = 3x
2
– 12x, (2 , 3)
(b) y = x
2
– 4x + 3, (3 , 1)
(c) y = 1 + 2x + x
3
, (1 , 4)
(d) y = x
2
+ 3x – 1, point where x =
–
2
2. Find the gradient and the equations of the
tangent and the normal to the curve
y = 5x
3
– 7x
2
+ 3x + 2 at the point (1 , 3).
3. Find the equations of the tangent and the
normal to the curve x
2
= 4y at the point
(6 , 9).
4. Determine the coordinates of the points on
the curve y =
8
at which the gradient is
–
2.
x
1
2
–
1
2
–
118
Find also the equations of the tangents at
these points.
5. What is the gradient of the curve y = x –
at the point where x =
–
1? Find the
equations of the tangent and the normal to
the curve at this point.
Stationary points
Consider the curve y = x
2
+ 3.
Its gradient function is
dy
= 2x.
dx
Table 8.3 gives the gradients of the curve at
dierent points.
Point where
–
3
–
2
–
1 0 1 2 3
x is
Gradient
dy
–
6
–
4
–
2 0 2 4 6
dx
Table 8.3
Fig. 8.8 is a sketch of the curve y = x
2
+ 3.
Fig. 8.8
Between points A and B in Fig. 8.8, the gradient
is negative, but steadily increases to zero at B.
Between points B and C, the gradient is
positive, increasing steadily as we move
towards C.
A point such as B is called a turning point of
the curve. The gradient at the turning point of
any curve is always zero. Since the curve
y = x
2
+ 3 has its least value at B, point B is
said to be a minimum point. Hence B is a
minimum turning point.
Now consider the curve y =
–
x
2
+ 3, whose
sketch is shown in Fig. 8.9.
Fig. 8.9
We see that the gradient of the curve y =
–
x
2
+ 3
is positive between D and E, but it decreases
steadily to zero at E.
Between E and F, the gradient is negative,
decreasing further as we move towards F.
Since the gradient at E is zero, point E is a
turning point.
The value of y =
–
x
2
+ 3 is greatest at E.
Hence E is said to be a maximum point or a
maximum turning point.
We notice that on moving along a curve from
left to right:
1. When a point on the curve is a maximum
point, the value of y is greatest at this
point, and the gradient of the curve is zero
at the point [Fig. 8.10 (a)]. The gradient
changes from positive to negative.
2. When a point on the curve is a minimum
point, the value of y is least at that point,
and the gradient of the curve is zero at the
point [Fig. 8.10 (b)]. The gradient changes
from negative to positive.
Fig. 8.10
1
x
3
–
+
–
–
+
+
E
D
dy
is –ve
dx
dy
= 0
dx
dy
is +ve
dx
3
– – – – – – – – – – – – – – – – – – – –
x
y
0
F
(a)
–
A
x
y
–
C
–
B
+
+
+
– – – – – – – – – –
M
Greatest value of y is MB
0
x
y
–
–
–
+
+
+
D
E
F
– – –
N
Least value of y is NE
(b)
0
+
–
B
C
3
– – – – – – – – – – – – – – – – – – – –
x
y
0
+
+
y = x
2
+3
–
dy
is +ve
dx
dy
is –ve
dx
–
dy
= 0
dx
A
119
+
0
–
–
0
+
+
0
+
–
0
–
Some curves have sections like those shown in
Fig. 8.11.
Fig. 8.11
The gradient is zero at both G and H. On both
sides of G, the gradient is positive and on both
sides of H, it is negative.
A point such as G or H is called a stationary
pointofinexion (sometimes spelt as
‘inection’) or a horizontalpointofinexion.
G is said to be an increasingpointofinexion
while H is a decreasingpointofinexion.
We note that on moving along a curve from left
to right:
1. When a point on the curve is an increasing
pointofinexion, the gradient remains
positive on either side of the point.
2. When a point on the curve is a decreasing
pointofinexion, the gradient remains
negative on either side of the point.
3. At a point of inexion, the direction of
curvature of the curve changes.
Note:
1. A stationary point is a point on a curve
at which the gradient of the curve is zero.
Thus, maximum points, minimum points and
points of inexion described above are all
stationary points.
2. A turning point is a point on a curve at
which the sign of the gradient of the curve
changes. Thus, maximum and minimum
points are turning points, but points of
inexion are not.
Table 8.4 summarises how to identify the
type of stationary point. The symbols – and +
are used to indicate whether the values of the
gradients on either side of the stationary points
are negative or positive.
Gradient Gradient at Gradient Diagrammatic Type of
to the stationary to the representation stationary
left point right point
+ 0 – maximum
– 0 + minimum
increasing
+ 0 + point of
inexion
decreasing
– 0 – point of
inexion
Table 8.4
Example 8.11
Identify the stationary points on the curve
y = x
3
– 2x
2
+ 3x.
For each point, determine whether it is a
maximum point, a minimum point or a point of
inexion.
Solution
y = x
3
– 2x
2
+ 3x
⇒
dy
= x
2
– 4x + 3.
dx
At stationary points,
dy
= 0
dx
i.e. x
2
– 4x + 3 = 0
⇔ (x – 1)(x – 3) = 0
⇔ x = 1 and x = 3.
when x = 1, y = (1)
3
– 2(1)
2
+ 3(1) =
when x = 3, y = (3)
3
– 2(3)
2
+ 3(3) = 0
∴ the stationary points are (1 , ) and (3 , 0).
A little to the left of x = 1, say at x = 0,
dy
= (0)
2
– 4(0) + 3 = 3, which is positive.
dx
A little to the right of x = 1, say at x = 2,
dy
= (2)
2
– 4(2) + 3 =
–
1, which is negative.
dx
1
3
–
1
3
–
4
3
–
(b)
x
y
–
0
– – – – – – – – –
H
–
–
–
dy
= 0
dx
•
dy
= 0
dx
(a)
x
y
+
+
+
– – – – – – – – – –
0
G
+
•
1
3
–
4
3
–
1
3
–
120
4
3
–
Thus, (1 , ) is a maximum point.
A little to the left of x = 3, say at x = 2,
dy
=
–
1 (found above), which is negative.
dx
A little to the right of x = 3, say at x = 4,
dy
= (4)
2
– 4(4) + 3 = 3, which is positive.
dx
Thus, (3, 0) is a minimum point.
Note:
The value of y at the maximum or minimum
point is not necessarily the greatest or least
value of the curve. In Example 8.11, for
instance, the value of y at the maximum point
(1, ) is . But the curve has other greater
values at other points. For example, when x = 6,
y = (6)
3
– 2(6)
2
+ 3(6) = 18, which is much
larger than .
Likewise, the curve has other values smaller
than 0, the value of y at the minimum point
(3 , 0).
From this observation, we see that a maximum
or a minimum point is actually a local
maximum or a local minimum point. At such
a point, the curve has a greater or smaller value
than at points immediately on either side of it,
and not necessarily the greatest or least value.
Exercise 8.6
1. Find the maximum or minimum value of y
on the curve y = 4x – x
2
.
2. Find the local maximum and minimum
values of y on the curve
y = 2x
3
– 3x
2
– 12x + 8.
3. Identify the stationary points on the curve
y = x
3
– 3x + 2 and, for each, determine
whether it is a maximum point, a minimum
point or a point of inexion.
4. Identify the stationary points, and determine
the nature of each, on the curve
y = x
3
– 2x
2
+ 3.
5. Identify the stationary points, and determine
their nature, on the curve y = x
3
– 3x
2
+ 3x + 4.
6. Investigate the stationary points, if any, on
the curve y = –x
4
+ 4x
3
+ 1.
Curve sketching
In order to draw a curve, we need to nd the
coordinates of a number of points from the
given equation and plot the points. In this way,
an accurate drawing is made of a part of the
curve, but some characteristic features of the
curve may be missed out because they may
occur beyond the range of the points plotted.
When a sketch is required, we are to show all
the characteristic features of the curve, and an
accurate drawing is not necessary. What is of
major importance is the information concerning
the stationary points and the points at which the
curve crosses the axes.
Example 8.12
Find the stationary points of the curve
y = x
3
+ x
2
– x. Hence, sketch the curve.
Solution
y = x
3
+ x
2
– x
⇒
dy
= 3x
2
+ 2x – 1
dx
At stationary points,
dy
= 0
dx
i.e. 3x
2
+ 2x – 1 = 0
(x + 1)(3x – 1) = 0
∴ x
=
–
1 and x =
when x =
–
1, y = (
–
1)
3
+ (
–
1)
2
– (
–
1) = 1
when x =
, y = ( )
3
+ ( )
2
– =
Thus, the stationary points are (
–
1 , 1) and
( , ).
Table 8.5 shows the signs of the gradients
a little to the left and right of each of the
stationary points.
4
3
–
4
3
–
1
3
–
4
3
–
1
3
–
1
3
–
1
3
–
–
5
27
––
1
3
–
1
3
–
1
3
–
–
5
27
––
1
3
–
121
x
–
2
–
1 0 0 1
dy
at x 7 0
–
1
–
1 0 4
dx
Diagramatic
representation
Table 8.5
Thus, (
–
1 , 1) is a maximum point and ( , ) is
a minimum point, so that the maximum value of
y is 1 and the minimum value is
.
The main features of the curve are now known.
But as an additional aid, we nd the points at
which the curve cuts the y- and x- axes.
If x = 0, y = 0
3
+ 0
2
– 0 = 0
Also, when y = 0, x ≈ 0.62 or
–
1.62.
The curve may now be sketched, and it is as in
Fig. 8.12.
Notice that we do not need graph paper for a
sketch.
Fig. 8.12
Exercise 8.7
In each of the following cases, nd the
stationary points and determine their nature.
Hence, sketch the curve.
1. y = x
2
– 3x + 2 2. y =
–
3x
2
– 4x + 4
3. y = x(4 – x) 4. y = 4 – x
3
5. y = x
3
– 3x
2
+ 3x – 1
6. y = 2 + 3x – 3x
2
+ x
3
7. y = 2x
3
– 3x
2
– 12x + 5
8. y =
–
x
3
+ 3x
2
– 3x + 2
9. y = x
3
–
x
2
– 2x –
10. y = x
4
– 8x
2
1
3
–
1
3
–
1
3
–
1
3
–
dv
dt
3
2
–
1
3
–
Applicationofdierentiation
Velocity and acceleration (km)
Suppose that a body moves along a straight
line and that its distance from the initial point
is s metres after t seconds while its velocity is
v m/s.
The value
ds
gives the gradient of the
dt
distance-time graph of the body while gives
the gradient of the velocity-time graph.
As we saw in Book 1, under linear motion,
these gradients respectively dene the velocity
and acceleration of the moving body.
Thus:
The velocity, v, in metres per second (m/s)
of a body that moves in a straight line is the
rate of change of distance, s, with respect to
time, t, given by
v = .
The acceleration, a, in metres per second per
second (m/s
2
) of the body is the rate of change
of velocity, v, with respect to time, t, given
by a =
dv
.
dt
Example 8.13
A particle moves in a straight line in such a way
that its distance s metres from a xed point O,
after t seconds, is given by s =
t
3
–
t
2
+ 2t.
(a) Find the position of the particle when it is
momentarily at rest.
(b) Find the acceleration of the particle at this
time and interpret the result.
(c) Find the maximum or minimum velocity of
the particle and interpret the result.
Solution
s =
t
3
–
t
2
+ 2t
⇒
ds
= t
2
– 3t + 2
dt
ds
dt
–
5
27
––
1
3
–
–
5
27
––
(
–
1 , 1)
x
y
(
–
1.62 , 0)
(0.62 , 0)
(
,
)
1
3
–
5
27
––
–
(
y
= x
3
+
x
2
– x
)
3
2
–
1
3
–
122
3
2
–
3
2
–
3
2
–
1
4
–
3
2
–
1
3
–
i.e. velocity, v = t
2
– 3t + 2
⇒
dv
= 2t – 3
dt
i.e. acceleration, a = 2t – 3.
(a) When the particle is momentarily at rest, its
velocity is zero,
i.e. t
2
– 3t + 2 = 0
or (t – 1)( t – 2) = 0
⇒ t = 1 and t = 2
∴the particle is at rest when t = 1 s
and when t = 2 s.
Its positions at these times are
(1)
3
– (1)
2
+ 2(1) = m from O
when t = 1 s and
(2)
3
–
(2)
2
+ 2(2) =
=
m from O
when t = 2 s.
(b) The accelerations of the particle when it is
momentarily at rest are as follows:
When t = 1 s,
a = 2 × 1 – 3 =
–
1 m/s
2
i.e. the particle is being retarded (slowing
down).
When t = 2 s,
a = 2 × 2 – 3 = 1 m/s
2
i.e. the particle is being accelerated
(speeding up).
(c) The maximum or minimum velocity occurs
when acceleration is zero, i.e. when
dv
= 2t – 3 = 0
dt
⇒ t =
s.
A little to the left of t =
, say at t = 1,
dv
= 2 × 1 – 3 = –1 (negative).
dt
A little to the right of t =
, say at t = 2,
dv
= 2 × 2 – 3 = 1 (positive).
dt
Thus, the velocity when t =
s is a
minimum.
Hence, the minimum velocity is
(
)
2
– 3(
) + 2 =
–
m/s.
This means that the particle is moving
towards the initial point O.
Exercise 8.8
1. A particle moves along a straight line in
such a way that after t seconds, the distance
s metres that it has moved is given by
s = t
2
+ 3t + 4. Find its
(a) velocity when
(i) t = 0 s (ii) t = 2 s
(b) acceleration
2. The velocity of a moving particle is given
by v = 6t – t
2
.
(a) Find the maximum velocity of the
particle.
(b) Find the time when the velocity of the
particle is zero.
(c) State the sign, and meaning, of the
velocity of the particle when t > 6.
3. A body moves in a straight line such that its
distance s metres after t seconds is given by
s = 3t + 5t
2
–
t
3
. Find
(a) the acceleration of the body at
(i) t = 1 s, (ii) t = 3 s,
(b) the maximum velocity of the body.
4. The motion of a particle is such that
s = 2t
3
– t
2
– 4t, where s is the distance it has
moved after t seconds.
(a) Find its position when it is momentarily
at rest. What does the result mean?
(b) Find its acceleration at this point.
(c) What is its minimum velocity?
5. A particle moves in a straight line such that
its distance s m from a xed point O at time
t s is given by s =
–
2t
3
+ 9t
2
.
(a) Find the velocity and acceleration of
3
2
–
1
3
–
5
6
–
3
2
–
1
3
–
4
6
–
3
2
–
2
3
–
3
2
–
123
the particle when t = 3 s and interpret
the results.
(b) What is the distance of the particle from
O when t = 4 s? Show that the particle
is, at this time, moving towards O.
6. A particle moves along a straight line in
such a way that its distance s m from an
initial point at time t is given by
s = 9t – 6t
2
+ t
3
. Determine the
(a) times when the particle is stationary,
(b) distance of the particle from the initial
point at these times,
(c) acceleration of the particle at these
times,
(d) velocity of the particle when its
acceleration is zero.
7. The motion of a particle, in a straight line, is
such that its velocity v m/s at time t seconds
is given by v = –5t
2
+ 12t. Find the
(a) value of t when the acceleration of the
particle is zero,
(b) maximum velocity of the particle.
8. A stone is thrown vertically upwards into
the air. Its height, h metres, above the
ground is given by h = 12t – t
2
, where t is
time, in seconds.
(a) What is its initial velocity?
(b) After how many seconds does it reach
its maximum height?
(c) What is its greatest height?
(d) What is the value of t when the stone
comes back to the ground?
Maxima and minima
The theory of turning points can be applied to
practical problems in which the maximum or
minimum value of a quantity is required. Such
problems require that we set up an equation
in the form required for the application of
dierentiation. The procedure is:
1. Write an expression for the required
quantity.
2. Use the given conditions to rewrite it in
terms of a single variable.
3. Find the turning point(s) and their type(s).
It is often obvious from the problem itself
whether a maximum or minimum value
has been obtained.
Example 8.14
I wish to use 28 m length of fencing wire to
make a rectangular enclosure for my goats.
What is the maximum area that the enclosure
can have?
Solution
Let the lengths of the sides of the enclosure be
x and y.
Then, perimeter = 2x + 2y = 28
⇒ y = 14 – x
Hence, area A = x(14 – x)
= 14x – x
2
.
⇒
dA
= 14 – 2x.
dx
When A is a maximum,
dA
= 0,
dx
i.e. 14 – 2x = 0
⇒ x = 7.
When x = 7, y = 14 – 7 = 7.
Hence, maximum area is 7
2
m
2
= 49 m
2
.
Exercise 8.9
1. A games committee wishes to use 120 m
of rope to form three sides of a rectangular
spectators’ stand, the fourth side being an
existing wall. Find the maximum possible
area of the stand.
2. The sum of two numbers is 60. Find their
values if their product is to be a maximum.
28 – x
2
= 14 – x
x
124
3. A cylindrical solid is to have a volume of
128π cm
3
. If the total surface area of the
solid is to be minimum, what should the
radius be?
4. A rectangular block has a square
cross-section. The sum of a side of the
cross-section and the length of the block is
30 cm. Determine the maximum volume of
the block.
5. A rectangular sheet of metal has sides 16 cm
and 6 cm. A square of side x cm is cut from
each corner of the sheet and the remaining
piece is folded to make an open box.
(a) Find an expression, in terms of x, for
the volume V of the box.
(b) Find the value of x for which the
volume of the box is a maximum and
calculate it.
125
Fig. 9.1
How does your answer compare with the known
area of Kenya?
Solution
(a) Make a tracing of the outline of the map
of Kenya (Fig. 9.1) on a centimetre square
graph paper, as in Fig. 9.2.
Fig. 9.2
(b) By ticking, count the whole squares and
note their number down. (23).
(c) Coun all partially enclosed squares and
take them as half squares
(d) The sum of (b) and (c) gives an estimate of
the aea in square centimetres
Total number of ticks is 23.
Total number of crosses is 16.
∴ Approximate area in square cemtimetres
is 39.
Approximation of areas of irregular
shapes
In our earlier work on area, we learnt how to
calculate areas of geometrical gures such as
squares, trapezia, circles and other polygons.
However in practical problems, we sometimes
need to estimate areas of irregular shapes. It
becomes necessary to use approximation methods
to estimate such areas. This section provides a
practical approach of nding approximate areas
of irregular shapes and gures.
The simplest method of estimating area is that
of counting squares.
Area by counting squares
This technique is best illustrated by use of an
example.
Example 9.1
Fig. 9.1 shows a map of Kenya drawn to a scale
of 1:12 000 000. Use the method of counting
squares to estimate the area of Kenya in km
2
.
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×























AREA APPROXIMATION
9
126
1
2
–
1 cm on the map represents 12 000 000 cm
∴ 1 cm
2
represents 12 000 000 × 12 000 000 cm
2
= 120 × 120 km
2
= 14 400 km
2
∴ 39.5 cm
2
represents 14 400 × 39.5 km
2
= 568 600 km
2
The area of Kenya is stated as being 582 600 km
2
We have estimated it to be 561 600 km
2
.
∴ our approximation is correct to 1 s.f.
Note: The accuracy of this answer can be
improved by
1. (a) using smaller squares, e.g. cm squares,
or
(b) matching parts of squares that in one’s
judgment appear to make a complete
square when combined.
2. An alternative way to estimate the area is to
consider all fractional squares, divide their
number by two, add to the number of whole
squares and proceed as before.
Example 9.2
Draw the graph of y = 2x
2
for 0 ≤ x ≤ 5.
Estimate the area bounded by the curve, the
x-axis and the line x = 5.
Solution
Using Table 9.1 and a suitable scale, e.g. 1 cm
to represent 10 units on the vertical axis and
1 cm to represent 1 unit on the horizontal axis,
we draw the graph of y = 2x
2
(Fig. 9.3).
x 0 1 2 3 4 5
y 0 2 8 18 32 50
Table 9.1
The shaded region in Fig. 9.3 represents the
required area.
We count the number of whole squares in the
region: they are 4. The fractions of squares
which are half or more are 4.
∴ total number of squares is 4 + 4 = 8
According to our scale 1 cm
2
represents an area
0
1 2 3 4 5
x
y
50
40
30
20
10
y = 2x
2
Fig. 9.3
of 10 × 1 = 10 square units.
∴ 8 squares = 8 × 10
= 80 square units.
Area by comparing masses
For a small region, area can also be estimated
by rst tracing the region on cardboard, cutting
it out, and then comparing the mass of the
cut-out with that of a 1 cm
2
cut-out of the same
cardboard.
ACTIVITY 9.1
Fig. 9.4 represents the area of a region bounded
by two perpendicular lines and a curve.
Fig. 9.4
1. Estimate the area in Fig. 9.4, in square
centimetres, by counting squares.
2. Copy Fig. 9.4 accurately, onto hard
cardboard.
3. Cut out the tracing along the boundaries and
nd its mass in grams.
127
x
8
0
642
40
y
30
20
10
Verocity (m/s)
Time (s)
10
4. From the same cardboard cut out a square
piece of side 1 cm and nd its mass in
grams.
5. By comparing the two masses estimate the
area of the region in square centimetres.
6. How does your answer to (1) above compare
with that in (5) above?
Which of the two methods described above
would you prefer to use in estimating a given
area? Give reasons for your answer.
Exercise 9.1
1. Draw the graph of y = x
3
for 0 ≤ x ≤ 3. Use
your graph to estimate the area enclosed by
the curve, the x-axis and the line x = 4.
2. Draw a circle of radius 4 cm on graph paper.
(a) Estimate its area by counting the squares.
(b) Use your answer to part (a) to estimate a
value for π correct to 2 decimal places.
How accurate was your answer to part (a)
above?
3. Letters x and y represent variables whose
corresponding values are given in Table 9.2.
x 0 1 2 3 4 5 6 7 8 9 10
y 3 4 4.25 4.5 3.75 2.5 1 1 1.5 1.75 0
Table 9.2
Plot y against x, and join consecutive points
by line segments. By the method of counting
squares, estimate the area bounded by the
coordinate axes and the graph.
4. Fig. 9.5 represents the velocity-time graph
of a particle.
(a) For how many seconds was the
particle in motion?
(b) Estimate the distance covered by the
particle.
5. A car starts from rest and accelerates as
shown in Table 9.3.
Fig. 9.5
Time (s) 0 1 2 3 4 5
Speed (m/s) 0 8 21 43 58 66
Time(s) 6 7 8 9 10
Speed (m/s) 72 76 79 81 82
Table 9.3
(a) Represent this motion graphically.
(b) Estimate the average speed for the
whole 10 seconds.
Area by triangulation
The method of counting squares to estimate
area can be cumbersome at times. A more
practical method is to subdivide the region into
shapes whose areas can be easily calculated.
Such shapes include squares, rectangles,
etc. This method is known as triangulation.
Example 9.3 illustrates this method.
Example 9.3
Use triangulation to estimate the area of the
shaded region in Fig. 9.6(a), part of a quarter
circle of radius 8 units, by subdividing the
region into suitable shapes.
(a) (b)
Fig. 9.6
x
y
6
8
8
0
x
y
6
8
8
0
C
D
B
A
128
Solution
We can subdivide the region into two parts,
rectangle OABC and the irregular shape BCD
[Fig. 9.6(b)].
Shaded area = area of rectangle OABC +
area of BCD.
Area of BCD can be approximated by the area
of ∆BCD.
Since Fig. 9.6(a) represents part of a circle
whose equation is x
2
+ y
2
= 8
2
,
at x = 6, y
2
+ x
2
= 64
⇒ y
2
+ 6
2
= 64
⇒ y
2
= 64 – 36 = 28
∴ y = √28
i.e. AB = 2√7
= 5.29
and CD = 8 – 5.29 = 2.71
∴ Area of rectangle OABC = 6 × 5.29
= 31.74
Area of ∆BCD = × 6 × 2.71
= 8.13
∴ the shaded area ≈ 31.74 + 8.13
= 39.87 square units.
Note that by estimating BCD using the area of
∆BCD, we have an under-approximation of the
area.
Exercise 9.2
In this exercise, use triangulation method of
approximating area.
1. Draw the graph of y
2
= x for 0 ≤ x ≤ 4,
joining the points with a smooth curve.
Estimate the area of the region bounded by
the y-axis, the line y = 2 and the curve.
2. Estimate the area enclosed by a circle,
centre (0 , 0), radius 3 cm.
3. Draw the graph of y = sin(x + 30°) for
0
c
≤ x ≤ π
c
. Estimate the area of the region
between the curve and the x-axis.
4. Given that y =
–
x
2
+ 10, draw the graph of y
against x for 0 ≤ x ≤ 4. Use your graph to
estimate the area bounded by the curve, the
two axes, and the line x = 4.
5. Table 9.4 shows the corresponding values of
two variables x and y.
x 1 2 3 4 5 6
y 9.0 5.0 3.7 3.0 2.6 2.3
Table 9.4
Using a scale of 1 cm to 1 unit on both axes,
draw the graph of y against x. Use your
graph to estimate the area between the curve,
the x-axis, x = 1 and x = 6.
Finding area under a curve using
trapezia
You should be familiar with the formula for
nding the area of a trapezium. In this section,
we extend the knowledge of the use of this
formula. We focus on such areas as can be
represented on the Cartesian plane.
The area in Fig. 9.6(a) could also be
approximated using a trapezium as in Fig. 9.7.
Fig. 9.7
Area of the trapezium = × 6(8 + 5.29)
= 13.29 × 3
= 39.87 square units
Our approximation could be improved if we
divided the region into smaller trapezia so as
to minimise the amount of area that is cut out.
The process of using many trapezia and then
adding the results is called the trapezium rule
of estimating area.
C
B
A
O
8
5.29
6
1
2
–
1
2
–
129
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
Thus,
Area of trapezium A = × 1(4.4 + 4.8) = 4.6
Area of trapezium B = × 1(4.8 + 4) = 4.4
Area of trapezium C = × 1(4 + 3.5) = 3.75,
etc.
∴ total area of the 9 trapezia is
× 1(4.4 + 4.8) + × 1(4.8 + 4) + × 1(4 + 3.5)
+ × 1(3.5 + 3.3) + × 1(3.3 + 3.6) + ×
1(3.6 + 4) + × 1(4 + 4) + × 1(4 + 2.5) +
× 1(2.5 + 1.5)
= × 1(4.4 + 4.8 + 4.8 + 4 + 4 + 3.5 + 3.5 +
3.3 + 3.3 + 3.6 + 3.6 + 4 + 4 + 4 + 4 +
2.5 + 2.5 + 1.5)
[(
× 1
)
is a common multiplier
]
= × 1(4.4 + 2 × 4.8 + 2 × 4 + 2 × 3.5 + 2 ×
3.3 + 2 × 3.6 + 2 × 4 + 2 × 4 + 2 × 2.5 +
1.5)
=
{
4.4 + 1.5 + 2(4.8 + 4 + 3.5 + 3.3 + 3.6 +
4 + 4 + 2.5)
}
=
{
5.9 + 2(29.7)
}
=
{
5.9 + 59.4
}
= × 65.3
= 32.65 square units.
Note: In the process of calculating areas of the
series of trapezia, all ordinates except the rst
and the last, appear in pairs.
Exercise 9.3
Estimate the areas in Exercise 9.2 using the
trapezium rule with not less than ve strips in
each case.
Deriving the trapezium rule
Fig. 9.9 represents the graph of y against x for a
certain relation between x and y.
We need to estimate the area between the curve,
the x-axis and the lines x = a and x = b.
Example 9.4
Table 9.5 shows corresponding values of x
and y. Draw a graph using these values and
estimate the area under the curve between the
y-axis and the line x = 9.
x 0 1 2 3 4 5 6 7 8 9
y 4.4 4.8 4 3.5 3.3 3.6 4 4 2.5 1.5
Table 9.5
Solution
Fig. 9.8 shows the graph of y against x drawn
using the values in Table 9.5.
Fig. 9.8
We require to estimate the area enclosed by the
x-axis, y-axis, the curve and the line x = 9.
The base of the area along the x-axis is 9 units
long, i.e. from x = 0 to x = 9. It is convenient
to divide this interval into 9 equals parts, each
of width 1 unit. Thus, along the x-axis, we have
10 ordinates whose lengths can be read from
the graph, i.e. at zero ordinate, length is 4.4;
at ordinate 1, length is 4.8; etc.
Now, through the ordinates, we divide the area
into 9 strips and join consecutive tops of the
ordinates in order to form the trapezia.
The total area of these trapezia will give an
approximation to the required area.
x
y
5
4
3
1
A B C D E F G H
I
2 4 6 108
0
2
130
y
x
h
h
hh
1
n
n – 1
2
………
y
0
y
n
y
n
– 1
y
2
y
1
y
n
– 2
b
a
1
2
–
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. 9.9
The base of the region is (b – a) units long. If we
divide the interval into n equal parts, each part
will be
(b – a)
units wide. Let the width of each
n
part be h units. We now complete the trapezia
each of which is h units wide, and whose
ordinates are denoted by
y
0
, y
1
, y
2
, …, y
n
– 1
, y
n
.
The total area of these n trapezia will give us an
approximation to the required area.
Now,
the area of the rst trapezium = h(y
0
+ y
1
)
the area of the second trapezium = h(y
1
+ y
2
)
the area of the third trapezium = h(y
2
+ y
3
),
and so on, and
the area of the n
th
trapezium = h(y
n
– 1
+ y
n
).
∴ The required area is
h(y
0
+ y
1
) + h(y
1
+ y
2
) + h(y
2
+ y
3
) + …
+ h(y
n
– 2
+ y
n
– 1
) + h(y
n
– 1
+ y
n
)
= h(y
0
+ y
1
+ y
1
+ y
2
+ y
2
+ … + y
n
– 2
+ y
n
– 1
+
y
n
– 1
+ y
n
) (factoring out h)
= h(y
0
+ 2y
1
+ 2y
2
+ … + 2y
n
– 2
+ 2y
n
– 1
+ y
n
)
= h{y
0
+ y
n
+ 2(y
1
+ y
2
+ … + y
n
– 2
+ y
n
– 1
)}
Therefore, the required area
= h
{
y
0
+ y
n
+ 2(y
1
+ y
2
+ … + y
n
– 2
+ y
n
– 1
)
}
This formula is called the Trapezium rule, also
known as the Trapezoidal rule.
Note:
1. It is possible to use this formula without
drawing an accurate graph. The ordinates
can be calculated using the function relating
x and y (or any other variables).
2. The smaller the value of h, the more the
number of trapezia and the better the
approximation.
3. It is easier to remember the formula as:
Approximate area
= h (sum of ‘ends’ + 2 sum of
‘middles’).
Using the trapezium rule
Example 9.5
Use the trapezium rule with seven ordinates to
estimate the area between the curve y = , lines
x = 1 and x = 3 and the x-axis.
Solution
We need to show the required area in a sketch
diagram (Fig. 9.10).
Fig. 9.10
y
2
x
1
0
1
3
2
y =
1
x
1
2
–
1
2
–
1
x
–
131
2
6
–
1
3
–
1
1
–
1
3
–
2
3
–
1
2
–
1
3
–
1
3
–
2
3
–
2
3
–
1
3
–
1
3
–
3
4
–
3
5
–
1
2
–
3
7
–
1
2
–
1
3
–
1
2
–
3
8
–
1
6
–
1
3
–
1
3
–
1
6
–
1
3
–
1
6
–
1
6
–
Since the base is (3 – 1) units, each trapezium
will be = units wide. (With 7 ordinates, we
make 6 trapezia)
Thus, the ordinates are
y
0
at x = 1 ⇒ y
0
= = 1
y
1
at x = 1 ⇒ y
1
=
1
=
3
1 4
y
2
at x = 1 ⇒ y
2
=
1
=
3
1 5
y
3
at x = 2 ⇒ y
3
=
y
4
at x = 2 ⇒ y
4
=
1
=
3
2 7
y
5
at x = 2 ⇒ y
5
=
1
=
3
2 8
y
6
at x = 3 ⇒ y
6
= .
By the trapezium rule,
A = h
{
y
0
+ y
6
+ 2(y
1
+ y
2
+ y
3
+ y
4
+ y
5
)
}
= ×
{
1 + + 2
(
+ + + +
)}
=
{
1 + 2(0.75 + 0.6 + 0.5 + 0.428 6 +
0.375)
}
=
{
1 + 2(2.653 6)
}
=
{
1 + 5.307 2
}
=
× 6.640 5
= 1.106 75 ≈ 1.107 square units.
Exercise 9.4
1. Table 9.6 shows the corresponding values of
x and y for a certain curve
x 1 1.2 1.4 1.6 1.8 2
y 6.5 6.2 5.2 4.3 4.0 2.6
Table 9.6
Use trapezium rule to estimate the area
between the curve, the x-axis and the lines
x = 1 and x = 2. (You need not sketch the
curve).
2
3
–
1
3
–
2. Sketch the curve y
2
+ x
2
= 16.
Using ve ordinates, estimate the area
between the curve and the positive
directions of the Cartesian axes.
Hence state the area of a circle centre (0 , 0)
radius 4 units.
3. Sketch the graph of 4y
2
+ 9x
2
= 36. Use 6
trapezia to estimate the area enclosed by the
curve.
4. The velocity of a particle at time t seconds
is given by the formula v = 6t – 3t
2
. Use the
trapezium rule with 9 ordinates to estimate
the distance covered by the particle during
the rst 2 seconds of motion.
5. Draw the graph of y = 10sin θ for
π
c
≤ θ ≤ 2π
c
.
Using the trapezium rule with seven
ordinates, estimate the area between the
curve, and the x-axis, leaving your answer
in terms of π.
Area under a curve using rectangles
In this method of approximating area, the
basic gure used is a rectangle. Just as in the
trapezium rule, this method uses a series of
strips, usually of uniform width. Rather than
use the end ordinates of the strips, this method
uses the mid-ordinate of each strip and so it is
referred to as the mid-ordinate rule.
Fig. 9.11 is a sketch showing the area enclosed
by the curve y =
, the x-axis and the lines
x = 1 and x = 3.
Fig. 9.11
1
x
y
2
x
1
0
1
3
2
y =
1
x
132
In order to estimate this area by the mid-ordinate
rule, we must divide it into strips of equal width
as for the trapezium rule. If we divide the area
into 4 strips, each strip will be
2
units wide.
4
Thus we divide the area into 4 strips of width
0.5 units each (Fig. 9.12).
Fig. 9.12
Now consider strip ABIJ.
At A, x = 1,
At B, x = 1.5,
and AB = 0.5.
Now, locate a point P on the x-axis such that
AP = PB.
Thus, at P, x = 1 +
0.5
= 1.25.
2
Through P draw a line parallel to AJ to meet the
curve at Q (Fig. 9.12). Through Q, draw a line
parallel to AB to meet AJ at N and BI produced
at M.
Now, ABMN is a rectangle of length PQ =
1
=
4
and breadth AB = 0.5.
1.25
5
∴ Area of ABMN = AB × PQ (since BM = PQ)
= 0.5 × 0.8
= 0.4 square unit.
Area ABMN is an approximation of the area of
the strip ABIJ.
PQ is called the mid-ordinate of the strip. The
mid-ordinates of the other strips are calculated
y
2
x
1
0
1
3
2
A
– – – – – – – –
B C D E
P
– – – – –
I
M
N
F
G
H
J
( )
– –
Q 1.25 ,
1
1.25
y =
1
x
y
x
b
0
( )
in the same way, i.e.
mid-ordinate of BCHI =
1
=
4
,
1.5 + 0.25 7
mid-ordinate of CDGH =
1
=
4
and
2 + 0.25 9
mid-ordinate of DEFG =
1
=
4
2.75 11
∴ Total area
=
1
×
4
+
1
×
4
+
1
×
4
+
1
×
4
2 5 2
7 2 9 2 11
=
1 4
+
4
+
4
+
4
2 5 7 9 11
=
1
(0.8 + 0.571 4 + 0.444 4 + 0.363 6)
2
=
1
× 2.179 4
2
= 1.089 7 square units.
Note: To get the total area, each mid-ordinate
has been multiplied by the width of the
strip and then the results added.
Deriving the mid-ordinate rule
Now consider Fig. 9.13.
Fig. 9.13 (a) shows the area enclosed between
the coordinate axes, line x = b and a curve of y
against x.
(a)
133
(b)
Fig. 9.13
In order to use the mid-ordinate rule, we divide
the area into n strips each of width h units
[Fig. 9.13(b)], where h =
b
.
n
Let the mid-ordinates be y
1
, y
2
, … y
n
.
∴ area of 1
st
strip = h × y
1
,
area of 2
nd
strip = h × y
2
, and so on.
∴ Total area = h × y
1
+ h × y
2
+ h × y
3
+ … +
h × y
n
= h(y
1
+ y
2
+ y
3
+ … + y
n
)
The required area is:
A = h(y
1
+ y
2
+ y
3
+ … + y
n
)
This formula is called the mid-ordinate rule.
Using the mid-ordinate rule
Example 9.6
Sketch the graph of y =
12
for x > 0. Use the
x
mid-ordinate rule with 8 ordinates to estimate
the area between the curve, the x-axis and lines
x = 1 and x = 8.
Solution
Fig. 9.14 (a) shows the required area.
Now, we divide the area into 7 strips of width 1
unit each [Fig. 9.14(b)].
(a)
(b)
Fig. 9.14
Let the mid-ordinates be y
1
, y
2
, … y
7
and mid-
points of the strips be x
1
, x
2
, ...., x
7
.
At x
1
= 1.5, y
1
=
12
= 8
1.5
At x
2
= 2.5, y
2
=
12
= 4.8
2.5
At x
3
= 3.5, y
3
=
12
= 3.429
3.5
At x
4
= 4.5, y
4
=
12
= 2.667
4.5
At x
5
= 5.5, y
5
=
12
= 2.182
5.5
At x
6
= 6.5, y
6
=
12
= 1.846
6.5
At x
7
= 7.5, y
7
=
12
= 1.6
7.5
y
x
8
1
0
y =
12
x
y
x
8
1
0
1 2 3 4 5 6 7
y =
12
x
– – – – – – – – – – – – – – – – – – – – – – –
– – – – – –
– – – – –
y
n
y
2
y
x
……………
y
1
b
h
134
∴ Area = 1× 8 + 1 × 4.8 + 1 × 3.429 + 1 ×
2.667 + 1 × 2.182 + 1 × 1.846 +
1 × 1.6
= 8 + 4.8 + 3.429 + 2.667 + 2.182 +
1.846 + 1.6
= 24.524 square units.
Execise 9.5
1. Draw the graph of y
2
= 4x. Use the
mid-ordinate rule with ve ordinates to
estimate the area between the curve, the
axes and the line x = 5.
2. Draw the graph of y = 2sin
θ
for 0
c
≤
θ
≤ 2π
c
.
Use the mid-ordinate rule with six strips to
estimate the area between the curve and the
θ-axis.
3. Draw the graphs of y = x
2
and y = x + 2 on
the same axes. State the coordinates of
the points where the two graphs intersect.
Use the mid-ordinate rule with six strips to
estimate the area between the curve, the
x-axis, line x = –1 and line x = 2. Hence
estimate the area enclosed by the curve and
the line y = x + 2. Estimate the area using
the trapezium rule.
4. A particle moves in a straight line such that
its velocity, v (m/s) in time, t (sec) is given
by v = 6t – t
2
. Draw the velocity-time graph
for 0 ≤ t ≤ 8.
(a) Use 4 strips and the mid-ordinate rule
to estimate the area between the curve,
the t-axis and the line t = 8.
(b) Hence estimate the average speed for
the whole 8 seconds.
5. Draw the graphs of y = –x
2
+ 8 and y = x
2
on
the same axes. State the coordinates of the
intersections of the two graphs. Use 4 strips
and the mid-ordinate rule to estimate the
area between the two curves.
6. Draw the graph of y = 3x(4 – x) for values
of x from
–
2 to 6. Use the mid-ordinate
rule with six ordinates to estimate the area
bounded by the curve, the x-axis and the
lines x =
–
1 and x = 5.
135
x
y
c
2
0
y
= 3
x
+
c
y
= 3
x
y
= 3
x
+ 2
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
–
Dierentiation
In Chapter 8, we learnt how to obtain the
gradient function from a given relation. For
example, if y = 3x,
dy
= 3.
dx
What is the gradient function if (a) y = 3x + 2,
(b) y = 3x + c (where c is a constant)?
We notice that the gradient functions are the
same, and equal to 3, for all the three cases
above.
The three equations above represent parallel
straight lines of gradient 3, as shown in Fig. 10.1.
Fig. 10.1
Reversedierentiation
Suppose we are just given the gradient function
as
dy
= .
dx
The constant gradient indicates a straight line
y = x, i.e. a straight line of gradient passing
through the origin.
This is so because dierentiating y = x gives
dy
= . However, as seen above, there are other
dx
possibilities such as y = x + 5, y = x + ,
y = x – 1, etc.
Any straight line of gradient may be written
as y = x + c, where c is a constant. This is the
general equation which gives
dy
=
.
Hence,
dx
dy
= is the gradient function for the equation
dx
y = x + c, i.e. all straight lines of gradient .
Write down the equation of the line whose
gradient function is:
(a) 5 (b)
–
3 (c)
–
In each case, write down the general form of the
equation of the line.
Quite often, the gradient function is not a
constant. For example, if
dy
= 4x, y = 2x
2
is a
dx
curve with this gradient function. The curves
y = 2x
2
+ 3, y = 2x
2
– 5, etc. also have the same
gradient function.
Thus, y = 2x
2
+ c, where c is a constant, is the
general equation of all curves with gradient
function
dy
= 4x. The equation y = 2x
2
+ c
dx
represents family of curves ‘parallel’ to y = 2x
2
,
as shown in Fig. 10.2.
Fig. 10.2
We notice, from Fig. 10.2, that for any given
value of x, such as x = a, the gradients of the
2
3
–
y =
2
x
2
– 5
–
5
0
c
3
y =
2
x
2
+ 3
y =
2
x
2
y =
2
x
2
+
c
x
y
– – – – – – – – – – – – – – – – – – – – – – – –
a
INTEGRATION
10
136
1
4
–
1
4
–
curves are the same. For instance, since
dy
= 4x,
dx
then if a = 1, the gradient is 4 for each curve.
The process of nding an expression for y
in terms of x, given the gradient function is
known as integration.
Integration is the reverseofdierentiation.
The function y = 2x
2
+ c is called the integral
of 4x with respect to x and c is called the
arbitrary constant of integration.
Equations such as
dy
=
and
dy
= 4x are
dx dx
called dierentialequations and y = x + c
and y = 2 x
2
+ c are the general solutions of
the respective dierential equations.
Now nd y in terms of x if
dy
is
dx
(a) x
3
(b) x
4
(c) 4x
3
(d)
–
x
2
.
Integration by rule
Recall that when a power of x is dierentiated,
the power of the gradient function is one
degree less,
i.e. if y = x
n
, then
dy
= nx
n
– 1
.
dx
Hence, in integration (i.e. reverse dierentiation),
we have to start by adding 1 to the power of the
given gradient function.
Example 10.1
Find y given that
dy
= x
5
.
dx
Solution
Adding 1 to the power gives x
6
.
However, dierentiating x
6
gives 6x
5
, which
indicates that we divide x
6
by 6 (the new
power).
Thus,
x
6
is the required expression.
6
Adding an arbitrary constant gives the general
solution as y =
x
6
+ c.
6
Thus, the method is y =
x
5 + 1
+ c .
5 + 1
It appears that the method of Example 10.1 can
be put down as a general rule for integration.
Did you use the same method to nd y given
that
dy
is (a) x
3
, (b) x
4
, (c) 4x
3
, (d)
–
x
2
?
dx
In each of the cases that we have so far
considered, the power of x is a positive integer.
If the power is zero, i.e. if
dy
= k
dx
(since kx
0
= k × 1), then y =
kx
0 + 1
+ c
0 + 1
i.e. y = kx + c,
so that the method still holds when the power is
zero.
Suppose the power of x is negative, e.g. if
dy
=
1
= x
–4
,
dx x
4
then y =
x
–
4 + 1
+ c
–
4 + 1
⇒ y =
x
–
3
+ c
–
3
⇒ y =
–
1
+ c.
3x
3
Dierentiating y =
–
1
+ c gives
dy
=
1
,
3x
3
dx x
4
thus conrming that the method still works when
the power of x is negative.
There is one exception though. This is when the
power is
–
1, i.e.
dy
=
1
= x
–1
.
dx
x
We see that y =
x
–
1 + 1
+ c
–
1 + 1
y =
x
0
+ c.
0
The method does not work in this case, since
division by zero is undened.
The integral of
1
is a special case which is
x
beyond the scope of this book.
137
( )( )
( )
( )
9. Given that
dy
=
–
1
(x
2
– 2x), nd y.
dx x
10. Find the general equation of a curve whose
gradient function is
3
–
2
+ 1.
x
3
x
2
Finding the arbitrary constant
Often, questions will give enough information
to enable us to calculate the arbitrary constant.
Examples 10.3 and 10.4 illustrate how to nd
the constant.
Example 10.3
The gradient at any point on a curve is given by
3x – 2. Given that the curve passes through
(
–
2 , 1), nd the equation of the curve.
Solution
The gradient is given by
dy
.
dx
Thus,
dy
= 3x – 2
dx
⇒ y =
3x
2
– 2x + c.
2
But this equation is satised by the coordinates
of the point (
–
2 , 1). Thus, we substitute
–
2 for x
and 1 for y to get
1 =
3(
–
2)
2
– 2(
–
2) + c
2
⇒ 1 = 6 + 4 + c
⇒ 1 = 10 + c
⇒ c =
–
9
Thus, the equation of the curve is
y =
3x
2
– 2x – 9.
2
Example 10.4
Find s in terms of t given that
ds
= 3t
2
+ 4t
dt
and that s = 1 when t = 0.
Solution
ds
= 3t
2
+ 4t
dt
We can now state the rule thus:
If
dy
= ax
n
, then y =
ax
n + 1
+ c
for all
dx n + 1
integral values of n except n =
–
1.
In words, the rule is ‘add 1 to the power,
divide by the new power and add a constant
of integration’.
Example 10.2
Integrate the expression 5x
2
– 4x + 5.
Solution
dy
= 5x
2
– 4x + 5
dx
So, y =
5
× x
3
–
4
× x
2
+ 5x + c
3
2
=
5x
3
– 2x
2
+ 5x + c.
3
Note that just like in dierentiation, the
expression is integrated term by term.
Exercise 10.1
Integrate the expressions in Questions 1 to 8
with respect to x.
1. (a) 5 x (b)
–
x (c) 3x
2
2. (a) 5 x – 2 (b) x
2
+ 5 (c) 3x
2
+ x
3. (a) ax + b (b) ax
2
+ bx (c) x(x – 3)
4. (a)
1
(b)
2
(c)
–
3x
–3
3x x
2
5. (a) 4x
3
+ 3x
2
– 2x (b)
x
4
–
x
2
–
x
+
1
2
3
2
4
(c) x
3
– x
2
+ x – 1
6. (a) (x – 1)(x + 1) (b) (x + 2)
2
(c) (x + 2)(x + 3)
7. (a) x
–3
+ x
–2
– 1 (b)
1
– x
1
+ x
x
x
(c)
2
+ 1 1 –
2
x
2
x
2
8. (a)
x
3
– x
2
(b)
x
3
– 3x – 1
(c)
kx
2
– bx
3
x x
3
x
5
138
1
2
–
1
2
–
⇒ s =
3t
3
+
4t
2
+ c
3 2
i.e. s = t
3
+ 2t
2
+ c.
When t = 0, s = 1 (given).
Thus, 1 = 0 + 0 + c
⇒ c = 1
∴s = t
3
+ 2t
2
+ 1.
Exercise 10.2
1. Find y in terms of x given that
dy
= 4 – 6x
dx
and that y = 1 when x = 1.
2. A line passes through (4 ,
–
3) and its
gradient is
–
2. Use integration to nd its
equation.
3. The gradient of a curve at any point is given
by 2x
2
– 4. Given that the curve passes
through (1 , 3), nd its equation.
4. A curve passes through the origin and its
gradient function is 3x
2
+ 6x – 2. Find its
equation.
5. At any point (x , y), a curve has gradient
1 – 3x
2
. Given that the curve passes through
(
–
1 , 5), nd its equation.
6. The point (1 , 4) lies on a curve whose
gradient at any point is 2x + 4. Find the
coordinates of the point at which the
gradient is zero.
7. If v and t are two variables such that
dv
= 4t – 3, and v = 5 when t = 2, nd v in
dt
terms of t.
8. Two variables s and t are such that
ds
= 12t
2
+ 6t. If s =
–
1, when t =
–
1, nd s
dt
in terms of t.
Area under a curve
In Chapter 9, we approximated the area under
a curve by the sum of the areas of trapezia
(Fig. 9.9).
We stated the trapezium rule as:
Area ≈
h
{
(y
0
+ y
n
) + 2(y
1
+ y
2
+ … + y
n
– 1
)
}
2
where h represents the uniform width of the
approximately trapezoidal strips of area and y
0
,
y
1
, …, y
n
are the ordinates.
An easy way to remember the formula is
Area ≈
h
{
sum of ‘ends’ + 2sum of ‘middles’
}
2
Example 10.5 reminds us how to nd the area
using the trapezium rule.
Example 10.5
Estimate the area under the curve y = 2x
2
+ 3
between x = 0 and x = 3 using trapezia of width
1 unit.
Solution
The working may be laid out as in Table 10.1.
y = x
2
x
ends middles
0 3
1 5
2 11
3 21
Total 24 16
Table 10.1
Area under the curve =
h{
sum of ‘ends’ + 2sum
of ‘middles’
}
=
{
24 + 2(16)
}
= 28 square units.
The estimate of the area may be improved
by using more trapezia, i.e. by reducing the
trapezium width.
139
1
2
–
2
3
–
Now estimate the area in E xample 10.5 using
trapezia of width unit. You should get 27.5
square units, which is a better estimate.
Rather than use trapezia when estimating the
area under a curve, we could use rectangles.
Again, the accuracy of the estimate will
be improved by increasing the number of
rectangles, i.e. decreasing the width of the
rectangles.
The method of using rectangles leads us to a
way of nding the exact area under a curve.
Consider Fig. 10.3, which shows the area
enclosed by part of the curve y = 2x
2
+ 3, the
axes and line x = 3.
This is area PQRO.
Fig. 10.3
S is the point (x , y) on the curve, SV its
y-coordinate.
Let area PSVO be A.
Now, if we move point S along the curve, A
increases or decreases as x increases or decreases,
i.e. the size of A depends on the value of x.
Our aim is to nd an expression for A in terms
of x.
Point T, with coordinates (x +
δ
x, y +
δ
y), is a
point close to S and TU is its y-coordinate.
If we move to the right from SV to TU, we
increase x by
δ
x and the resulting increase in A
is the shaded area STUV, denoted as
δ
A, i.e.
δ
A
is an increase in A corresponding to the increase
δ
x in x.
The shaded area
δ
A can be approximated by
either the area of rectangle SWUV (which is
y
δ
x) or the area of rectangle XTUV [which is
(y +
δ
y)
δ
x].
Thus, y
δ
x <
δ
A < (y +
δ
y)
δ
x.
Dividing by δx, which is positive, we get
y <
δ
A
< y +
δ
y
δ
x
As
δ
x becomes smaller (i.e. approaches zero),
δ
y
also approaches zero, and (y +
δ
y) approaches y,
i.e. as
δ
x 0,
δ
y 0 and (y +
δ
y) y.
Thus,
δ
A
lies between y and a value which we
δ
x
can make as close to y as we please, by making
δx suciently small.
∴ Lim
δ x
0
δ
A
= y
δ
x
Writing Lim
δ x→0
δ
A
as
d A
, we have
δ
x dx
dA
= y
dx
i.e.
d A
= 2x
2
+ 3.
dx
By integration, A = x
3
+ 3x + c.
If we move the right hand boundary of the area
A from SV to PO, we reduce the area to zero.
Thus, when x = 0, A = 0.
Substituting these values in the last equation,
we get c = 0.
∴ A = x
3
+ 3x.
We have now found an expression for A in
terms of x.
If we now move the right hand boundary of A
from SV to QR, we increase x to 3.
When x = 3, A = × 3
3
+ 3 × 3
= 18 + 9 = 27 square units.
Compare this result with what you got using the
trapezium rule (E xample 10.5).
2
3
–
– – – –
y
= 2
x
2
+ 3
S(x
, y)
δy
δx
δA
A
V
P
U
R
X T
Q
W
3
O
3
– – – – –
y
x
y
2
3
–
140
y
= 2
x
2
+ 3
S
T
P
R
Q
4
O
x
y
M
2
1
N
3
2
3
–
2
3
–
2
3
–
2
3
–
1
3
–
2
3
–
1
3
–
1
3
–
1
2
–
1
2
–
Example 10.6
Find the area enclosed by the x-axis, the curve
y = 2x
2
+ 3 and the lines x = 2 and x = 4.
Solution
The required area is shown in Fig. 10.4 as
SQRT.
Fig 10.4
This area may be found as the dierence
between the areas PQRO and PSTO.
Using A as before,
dA
= y = 2x
2
+ 3
dx
∴ A = x
3
+ 3x
(we have already shown that the constant of
integration is zero).
When x = 4,
A =
× 4
3
+ 3 × 4
=
128
+
36
3
3
=
164
= 54
square units (area PQRO).
3
When x = 2
A = × 2
3
+ 3 × 2
=
16
+
18
3 3
=
34
= 11 square units (area PSTO).
3
∴ area SQRT = 54 – 11
= 43 square units.
Exercise 10.3
1. Using the trapezium rule with strip width
of unit, nd an approximation of the area
under the curve y = 2x
2
+ 3 between x = 2
and x = 4. Find the percentage error in this
approximation compared with the exact area
obtained in Example 10.6.
2. Using integration, nd the area bounded by
the line y = 2x + 5, the y-axis and the line
x = 4.
3. Find the exact area enclosed by the x-axis,
the curve y = 2x
2
+ 3 and the lines:
(a) the y-axis and x = 6,
(b) x = 1 and x = 2.
4. Find the area enclosed by the curve y = x
3
,
the y–axis and the line x = 10.
5. What is the area bounded by the curve
y = 3x
2
– 4x + 2, the x-axis and the lines
x = 2 and x = 5?
Deniteandindeniteintegrals
In the previous section, we used the symbol A
to denote an area having the y-axis as its left-
hand boundary. Suppose that instead we had
dened a similar area A′ having line x = 1 as its
left-hand boundary (Fig. 10.5).
Fig. 10.5
y
= 2x
2
+ 3
S
T
P
R
Q
4
O
x
y
3
2
141
2
3
–
2
3
–
2
3
–
2
3
–
( ) ( )
1
4
–
1
2
–
Reasoning as before, we should get the result
dA = y = 2x
2
+ 3
dx
A = x
3
+ 3x + k (where k is the
constant of integration)
But A = 0 when x = 1.
Substituting these values in the last equation,
we get
k =
–
11
3
A = x
3
+ 3x –
11
.
3
Example 10.7
Using the left-hand boundary as the line x = 2,
redo Example 10.6.
Solution
The required area is shown in Fig. 10.5 as
SQRT, and the left-hand boundary from which A
is measured to the right is shown as MN.
The required area SQRT is found as area
MQRN – area MSTN.
Thus,when x = 4
A = 4
3
+ 3 4 –
11
3
=
164
–
11
(area MQRN),
3 3
and when x = 2,
A = 2
3
+ 3 2 –
11
3
=
34
–
11
(area MSTN).
3
3
area SQRT =
164
–
11
–
34
–
11
3 3 3 3
=
164
–
34
3
3
=
130
= 43
square units
(same result
3
as before).
In Examples 10.6 and 10.7, we have determined
the constant of integration as zero (using A)
and as
–
11
(using A). It is clear from Example
3
10.7 that the constant drops out on subtraction.
We could thus have measured A from any
convenient left-hand boundary and found area
SQRT by subtraction, without evaluating the
constant of integration.
Example 10.8
Find the area bounded by the x-axis, x = 2,
x = 4 and the curve y = x
3
+ x.
Solution
The required area is shaded in Fig. 10.6.
Fig 10.6
dA
= y = x
3
+ x
dx
A = x
4
+ x
2
+ c
When x = 4
A = 4
4
+ 4
2
+ c
= 72 + c.
When x = 2,
A = 2
4
+ 2
2
+ c
= 6 + c
the required area = (72 + c) – (6 + c)
= 72 + c – 6 – c
= 66 square units.
Alternative layout
The following square bracket notation is
an instruction to substitute and subtract. It
shortens the working.
dA
= x
3
+ x
dx
A = x
4
+ x
2
+ c
1
3
–
1
4
–
1
2
–
x
y
– – – – – – – – – – – – – – – – –
– – – –
4
2
0
y = x
3
+ x
1
4
–
1
2
–
1
4
–
1
2
–
142
2
1
4
–
1
2
–
1
4
–
1
2
–
1
4
–
1
2
–
1
4
–
1
2
–
1
4
–
1
2
–
The required area, between lines x = 2 and
x = 4, is
[
x
4
+ x
2
+ c
]
4
= ( × 4
4
+ × 4
2
+ c) –
( × 2
4
+ × 2
2
+ c)
= (72 + c) – (6 + c)
= 66 square units
The value obtained in Example 10.8 is called
the area under the curve y = x
3
+ x from
x = 2 to x = 4.
2 and 4 are called, respectively, the lower and
upper limits of integration.
The integral x
4
+ x
2
+ c, containing the
arbitrary constant of integration, is called an
indeniteintegral.
When the limits are given, and the integral
may be evaluated, e.g.
[
x
4
+ x
2
+ c
]
4
, it is
called a deniteintegral.
Note: Since the constant of integration drops
out in a denite integral, it is not necessary to
evaluate it or to write it inside the bracket.
Exercise 10.4
1. Evaluate the following
(a)
[
4x + c
]
5
(b)
[
2t
2
+ 3t
]
8
(c)
[
x
2
– 3x
]
0
(d)
[
3x
2
– x + c
]
3
2. Find the area under the curve y = 3x + 4
from x = 0 to x = 8, by integration. Check
by another method.
3. Find the area under the curve y = 2x
2
from
x = 1 to x = 3.
4. Find the area under the curve y = 3x
2
– 2
from x = 1 to x = 3.
5. (a) Using the trapezium rule, with trapezia
of width 1 unit, estimate the area under
the curve y = x
2
– 2x + 1 from x = 0
to x = 6.
(b) Use integration to nd the exact area
under the curve in (a) between the
given limits.
(c) Calculate the percentage error in the
estimate obtained in (a).
6. Find the area under the curve
y = –x
2
+ 3x + 18. (Hint: Y ou need to sketch
the curve in order to get the limits.
The integral notation
We have already used integration to nd the
area under a curve.
The word integration implies the putting
together of parts to make a whole. This
fundamental aspect of the process is brought
out in the following alternative approach to the
area under a curve.
Suppose that we wish to nd the area under the
curve in Fig. 10.7 from x = a to x = b.
Fig. 10.7
The area of the shaded strip in Fig. 10.7 forms
part of the required area. The area of the strip
may be approximated by that of a rectangle
of height y (corresponding to some number x
between a and b) and width
δ
x. Thus, the area
of the strip is approximately y
δ
x.
The sum of the areas of all such strips, into
which the required area may be subdivided,
2
1
2
1
–
3
P(x , y)
b
O
x
y
a
Q(x +
δ
x , y +
δ
y)
y
– – – – – – – –
•
x
δ
x
– – – – – – – – – – – – – – – – – – – – – – – – – – –
•
•
•
143
Fig. 10.8
The shaded element of area is
y
δ
x = (6x – x
2
)
δ
x
∴ the required area =
∫
(6x – x
2
)dx
=
[
3x
2
– x
3
]
= (27 – 9) – (3 – )
= 15 square units.
The following example extends our work to
cover negative area (i.e. area under the x-axis).
Example 10.10
Evaluate the area enclosed by the curve y = x
3
,
the x-axis and the lines x =
–
2 and x = 2.
Solution
A sketch of the curve is shown in Fig. 10.9, with
the required area shaded.
Fig. 10.9
x = a
x = b
x = a
x = b
x = a
x = b
x = a
x = b
x = a
x = b
a
b
a
b
0
x
y
1
y = 6x – x
2
y
– – – – – – – – – – – – –
δ
x
– – – – – – – – – – – – – – – – – – –
3
1
3
1
3
–
gives the area under the curve between x = a
and x = b.
Thus, area under the curve between x = a and
x = b is
A ≈ ∑ y
δ
x , which may also be written as
A ≈
S
yδx, where ∑ or
S
denotes ‘sum of’.
As the width
δ
x of the strips becomes smaller,
i.e. as we increase the number of rectangles, the
sum approaches the exact area under the curve,
i.e. A = Lim
δ
x → 0
∑
y
δ
x
or A = Lim
δ
x → 0
S
y
δ
x
This can also be written as
A =
∫
ydx, or simply A =
∫
ydx
In our previous work, we calculated area under
a curve using integration. Thus:
The symbol
∫
(…)dx denotes ‘integration with
respect to x’.
The symbol
∫
, which is an elongated
S
,
for ‘sum’, is a reminder that integration is
essentially summation.
∫
ydx is a denite integral while
∫
ydx is an
indenite integral.
We shall, henceforth, think of every area under
a curve as a summation.
Example 10.9
Find the area under the curve y = 6x – x
2
from
x = 1 to x = 3.
Solution
Fig. 10.8 shows the required area.
1
3
y
x
y = x
3
0
2
–
2
1
3
–
1
3
–
144
2
1.5
1
0
8
y = 4x
2
– 12x + 9
y
x
3y = 16x + 24
–
2
–
2
[ ]
2
–
2
–
2
[ ]
0
[ ]
2
0
0
0
[ ]
1.5
0
1.5
Working as before, we get:
Required area =
∫
x
3
dx
=
x
4
2
4
= 16 – 16
4 4
= 0 square units.
But we can clearly see from the graph that the
area is not zero.
Let us now calculate the two shaded portions of
area separately.
Area below the x-axis
=
∫
x
3
dx
=
x
4
0
4
= 0 –
16
4
=
–
4
Thus, when the area under a curve is below the
x-axis, it bears a negative sign. Thus the area
is 4 square units.
Area above the x-axis
=
∫
x
3
dx
=
x
4
2
4
=
16
+ 0
4
= 4
Thus, the two portions of area are numerically
equal but opposite in signs. That is why they
cancel out giving an area equal to zero.
The correct area is 4 + 4 = 8 square units.
Note that the cancellation of areas does not
always lead to zero. It leads to zero only if the
ordinates are equal and opposite in sign. If the
ordinates are not equal and opposite in sign, one
portion of the area will be larger than the other,
and so the cancellation will be partial.
Example 10.10 shows that:
If the area under a curve consists of two
portions, one below the x-axis and the other
one above it, then we should evaluate the
two areas separately and add their absolute
values.
The example also illustrates the advisability of
making a rough sketch if in doubt as to the shape
of a curve and the position of the required area.
Example 10.11
Given that a curve has equation
y = 4x
2
– 12x + 9, use integration to calculate
the area between the curve and the line
3y = 16x + 24.
Solution
Fig. 10.10 is a sketch of the two graphs, on the
same axes.
Fig. 10.10
The graph of y = 4x
2
– 12x + 9 meets the line at
points (0 , 8) and (1.5 , 0).
Area between the curve, x = 0 and y = 0 is
represented by
∫
(4x
2
– 12x + 9)dx =
4
x
3
–
12
x
2
+ 9x
3 2
=
4
×
27
– 6 ×
3
2
+ 9 ×
3
3 8 2
2
2
=
9
–
27
+
27
2 2 2
= 4.5
145
1
2
–
1
2
–
1
2
1
2
2
1
2
7. Find the area enclosed between the curve
y = 2x – x
2
and the line x + y = 2.
Application of integration in
kinematics
Kinematics is the study of the motion of a
particle or a body without reference to the
forces acting on the particle or body. In this
section, we shall use integration when looking
at the motion of various bodies.
In Chapter 8, we saw that given an expression
for the distance moved in given direction by a
body, in terms of t, we could nd the velocity
by dierentiating the expression. Likewise,
dierentiating the expression for velocity gives
acceleration.
The reverse process, i.e. integration, enables us
to obtain an expression for velocity from that of
acceleration and an expression for the distance
from that of velocity.
Thus:
Given that s represents displacement, v
velocity and a acceleration at any time t, then
v =
∫
a dt and s =
∫
v dt.
Remember that s and v must be expressed
in terms of t for them to be integrable with
respect to t.
Example 10.12
A body moves in a straight line with acceleration
(5 – 12t) m/s
2
, t seconds after the start. Given
that the body started with a velocity of 3 m/s,
nd
(a) its velocity and displacement in terms of t,
(b) how far the body is from its starting point
after 2 seconds and its velocity then.
Interpret these results.
0
( )
Area between line 3y = 16x + 24, the x-axis and
the y-axis is
× base × height = × 1.5 × 8
= 6
∴ shaded area, i.e. the required area
= 6 – 4.5 = 1.5 square units.
Exercise 10.5
1. Find the following indenite integrals.
(a)
∫
(2x + 3)dx (b)
∫
(3x
2
– 2x)dx
(c)
∫
x(16 – x)dx (d)
∫
10 –
1
dx
x
2
2. Evaluate the following denite integrals.
(a)
∫
(4x – x
2
)dx (b)
∫
(5x
2
– 4x)dx
(c)
∫
1
dx
(d)
∫
x
2
(x – 2)dx
x
2
3. In each of the following cases, nd the area
enclosed by the x–axis, the curve and the
straight lines.
(a) y = 2x
3
, x = 1, x = 3
(b) y = x
2
+ 3, x =
–
2, x = 5
(c) y = x
3
, x =
–
2, x = 0
(d) y =
3
, x = 1, x = 4
x
2
4. Sketch each of the following curves and
nd the area under it between the given
ordinates.
(a) y = x
2
– 4; x = 0 and x = 4
(b) y = x
3
; x =
–
1 and x = 2
(c) y = x
2
– 4x; x =
–
2 and x = 4
(d) y = x
2
– 5x + 6; x = 1 and x = 4
5. P and Q are the two points at which the
curve y = 6 – x – x
2
cuts the x–axis. Find the
area enclosed by the x–axis and the portion
of the curve which lies between P and Q.
6. Use integration to calculate the area
between the curve y = x
2
– 4x + 5 and the
line y = x + 1.
146
( )
3
3
4
Solution
Velocity, v =
ds
= 3t
2
dt
∴s =
∫
3t
2
dt
Required is the distance moved between t = 3
and t = 4.
Thus, required distance =
∫
3t
2
dt
= [t
3
]
4
= 4
3
– 3
3
= 37 m.
Example 10.14
A stone is thrown vertically upwards with
a velocity of 15 m/s. Assuming that the
acceleration due to gravity is 10 m/s
2
, nd
expressions for its velocity and displacement t
seconds later.
After how many seconds does it attain maximum
height? Find this maximum height.
Solution
The stone is projected upwards while the
acceleration due to gravity is downwards.
Taking upwards to be positive, then
acceleration, a =
dv
=
–
10
dt
∴ velocity, v =
∫
(
–
10)dt
=
–
10t + c
1
But when t = 0, v = 15
∴15 =
–
10(0) + c
1
⇒c
1
= 15
Hence, v =
–
10t + 15.
Now, velocity, v =
ds
=
–
10t + 15
dt
∴ displacement, s =
∫
(
–
10t + 15)dt
=
–
5t
2
+ 15t + c
2
But when t = 0, s = 0
⇒ c
2
= 0
Hence, s =
–
5t
2
+ 15t.
At maximum height, velocity is zero (since the
Solution
(a) Acceleration, a =
dv
= 5 – 12t
dt
v =
∫
(5 – 12t)dt
= 5t – 6t
2
+ c
1
when t = 0, v = 3 m/s (given in question).
Substituting these values gives
3 = 5(0) – 6(0)
2
+ c
1
∴ c
1
= 3
Thus, velocity, v = (5t – 6t
2
+ 3) m/s.
(b) Velocity, v =
ds
= 5t – 6t
2
+ 3
dt
∴ s =
∫
(5t – 6t
2
+ 3)dt
=
5
t
2
– 2t
3
+ 3t + c
2
2
But s = 0 when t = 0
∴ c
2
= 0
Thus, displacement, s =
5
t
2
– 2t
3
+ 3t m
2
After 2 s, i.e. when t = 2
s =
5
(2
2
) – 2(2)
3
+ 3(2)
2
= 10 – 16 + 6
= 0 m.
This result means that the body is back to its
starting point.
At t = 2
v = 5(2) – 6(2)
2
+ 3
= 10 – 24 + 3
=
–
11 m/s.
This result means that the body is moving at
11 m/s, in the opposite (note negative sign)
direction from the initial one.
Example 10.13
A particle moves in a straight line with velocity
(3t
2
) m/s, t seconds after the start. Find the
distance moved in the 4th second.
147
stone must be starting to fall downwards)
i.e. v =
–
10t + 15 = 0
⇒ t = 1.5 s
i.e. the stone attains maximum height after 1.5 s.
Maximum height =
–
5(1.5)
2
+ 15(1.5)
= 11.25 m.
Exercise 10.6
1. A particle moves in a straight line with a
constant acceleration of 5 m/s
2
. If its velocity
after t seconds is v m/s, nd v in terms of t,
given that velocity after 2 seconds is 14 m/s.
2. A particle moves in a straight line in such a
way that its velocity after t seconds is
(2t + 7) m/s. Find the distance moved in
the rst 3 seconds.
3. A stone is thrown vertically upwards from
the top of a cli at 20 m/s. If the acceleration
due to gravity is 9.8 m/s
2
, nd expressions
for its velocity and position t seconds later.
4. A ball is thrown vertically upwards from
ground level with an initial velocity of
30 m/s. Assuming that the acceleration due
to gravity is 10 m/s
2
, determine
(a) expressions for its velocity and height
in terms of t.
(b) its velocity after (i) 2 seconds
(ii) 5 seconds
(c) how long the ball takes to attain
maximum height and the maximum
height attained.
(d) the length of time that the ball takes to
come back to the ground.
5. Find the displacement (s) in terms of time
(t) if
(a)
ds
= 5 and s = 4 when t = 0,
dt
(b)
ds
= 2t – 3 and s = 6 when t = 2,
dt
(c) v = t
2
– 3t – 2 and s = 10 when t = 6,
(d) v = 5t(t – 3) + 2 and s = 0 when t = 0.
6. In each of the following cases, nd the
distance moved in the given interval,
assuming that distance is in metres and time
is in seconds.
(a)
ds
= 3t + 4; t = 0 to t = 2
dt
(b) v = 2t
3
– 3; t = 1 to t = 2
(c) v = (2t –1)(2t + 1); t = 1 to t = 3
(d) v = 3t
2
–
1
; t = 1 to t = 3
t
2
7. A particle starts from rest, moving in a
straight line, and attains an acceleration of
(24t – 12t
2
) m/s
2
after t seconds.
(a) How long does it take to return to its
starting point and what is its velocity
then?
(b) What is its maximum displacement
during this interval?
(c) What is its maximum velocity during
this interval?
8. A particle moves along a straight line with
an acceleration a. At time t = 0, its velocity
is u. If its velocity after t seconds is v, nd
(a) v in terms of u, and t,
(b) the distance s covered, after t seconds,
in terms of u, a and t.
148
1
3
–
( )
1
2
–
1
2
–
6. Find the following integrals:
(a)
∫
3x
2
– 6x + 5)dx (b)
∫
–
x
2
+ 3x – 4)dx
(c)
2
–
4
+ 3x – 1 dx
x
2
x
–
3
(d)
x – 1
x
2
+
1
dx
x
3
7. On the same axis, draw the graphs of y = x
and y = 5x – x
2
. Use the trapezium rule to
estimate the area between
(a) the x-axis, y = x and x = 4.5.
(b) the x-axis, y = 5x – x
2
and x = 4.5.
Hence estimate the area between y = x
and y = 5x – x
2
.
8. The gradient of a curve is given by 2x – 5.
If the curve passes through the point (3 , 2),
nd its equation.
9. Dierentiate the following expressions
with respect to x.
(a)
x
5
–
1
+
3
5
x
5
4
(b)
(x
2
– x)(x
2
+ x)
+
x
2
– 3x – 4
x x + 1
10. (a) Draw the graph of y
2
= x for 0 ≤ x ≤ 4.
(b) Use the trapezium rule, with 9
ordinates, to estimate the area of the
region bounded by the curve and the
line x = 4.
11. Solve the dierential equation x
2
dy
= x
4
+ 1,
given that y = 2 when x = 1.
dx
12. Use 6 ordinates and
(a) the trapezium rule,
(b) the mid-ordinate rule to estimate the
area bounded by the y-axis, the x-axis,
x = 5 and the curve y = 3x
2
+ 2.
Revision exercise 3.1
1. Working with small change δx, nd the
derived function
dy
for the expression
dx
2x
3
– x
2
.
2. The distance of a particle from a point P,
moving in a straight line, is given by
s = t
3
– 2t
2
+ 3t, where t is time in seconds.
Find
(a) the values of t when the velocity, v, is
zero.
(b) the value of t when the acceleration, a,
is zero.
3. Copy and complete Table R3.1 for
y = 5x
2
– 2x + 2.
x 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0
y 18 28.25 74 142.25 233
Table R3.1
Use your table to estimate the area bounded
by the curve, the x-axis, x = 2 and x = 7
using
(a) the trapezoidal rule with 11 ordinates,
(b) the mid-ordinate rule with 6 ordinates.
4. Integrate the following with respect to x.
(a)
2
– √2
2
+ √2
x
2
x
2
(b)
–
1
(x
3
– 3x) +
4x
3
– x
x
x
5. Find the point(s) at which the gradient of
the curve
(a) y = x
5
– 4 is equal to 80
(b) y = x
3
–
x
2
– 2x is equal to zero.
2
( )
∫
( )
( ) ( )
∫
1
2
–
REVISION EXERCISES 3 (CHAPTERS 8–10)
8-10
149
Revision exercise 3.2
1. Find the equations of the tangent and
normal to the curve y = 2x
2
+ 1 at the point
where x = 2.
2. Use the trapezium rule, with strips of 1 unit
width to estimate the area bounded by the
curve y =
1
, the lines x = 0 and x = 5.
x +1
3. Evaluate (3x
2
– 11 x + 6)dx.
4. A stone is thrown vertically upwards. Its
height, h metres, at time t seconds is given
by h = 20t – 2t
2
. Calculate the maximum
height attained by the stone.
5. For the curve y = 3 – 6x + x
2
, nd
(a) the point on the curve where the
tangent has gradient 2,
(b) the equation of the tangent at the point
in (a),
(c) the equation of the normal at the point
in (a).
6. Sketch the curve y = x
2
– 4. Use the mid-
ordinate rule and 6 ordinates to estimate the
area enclosed between the curve, the line
x = 4 and the x-axis.
7. Sketch the curve y = (x + 2)(x + 1)(x – 1)
and hence nd the exact area which is
completely enclosed by the curve and the
x-axis.
8. Evaluate the following integrals:
(a)
∫
( x
3
– x
2
)dx (b)
∫
(3x
2
+ 1)dx
(c)
∫
(x – 2)(x – 1)(x + 1)dx
(d)
∫
(2s
2
– 3s + 4)ds
9. Identify the stationary points on the curve
y = (x + 3)
2
(x + 1) and distinguish between
them.
10. Find the turning points of the curve whose
equation is y = x
3
+ 5x
2
+ 3x – 9, and
distinguish between them. Hence sketch the
curve for values of x such that
–
5 ≤ x ≤ 2.
11. (a) Using (i) the trapezium rule, and
(ii) mid-ordinate rule, with strip width
of 1 unit, nd two approximations to
the area enclosed by the curve
y = 16 – x
2
, the x-axis and the ordinates
x =
–
2 and x = 2.
(b) Find the exact area under the curve in
(a) and hence nd the percentage error
in each of the approximations.
12. Sketch the curve y = x(5 – x) and the
line y = x. Use the trapezium rule and 6
ordinates to estimate the area between the
two graphs.
Revision exercise 3.3
1. The equation of a curve is y = 3x
2
– 4x – 6.
Find the equation of the normal to the
curve at the point (3 ,
–
2).
2. A particle is projected vertically up in the
air. Its height, in m, above the ground after
t seconds is given by h = 16t – 4t
2
.
(a) Determine its velocity at t = 1 s and
at t = 3 s and interpret your results.
(b) Find the maximum height that the
particle achieves.
(c) Find the average velocity of the
particle over the period t = 1.5 s to
t = 3.5 s.
3. Sketch the curve y = 10 – x
2
. Hence, use
integration method to estimate the area
between the curve and the x-axis.
4. A particle starts from rest and moves with
an acceleration, a m/s
2
, given by a = t – 6,
where t represents time in seconds. Given
∫
1
2
1
2
–
0
4
–
2
2
0
3
1
2
–
1
–
1
150
2
3
–
that its initial velocity is 2 m/s, nd
expressions for its velocity, v, and
displacement, s, in terms of t.
5. With six ordinates, estimate the area
between the curve y = 3x
2
+ 1, the x-axis
and the ordinates x = 1 and x = 3 using
(a) the trapezium rule,
(b) the mid-ordinate rule.
6. Fig. R3.1 shows a sketch of the curve
y = x
2
and the line y = 3x. Calculate the
area enclosed between the curve and line
(shaded) as a dierence of two areas.
Fig. R3.1
7. Two variables, x and y, are related such that
y = 3x
2
+ 2x.
(a) Use dierentiation to identify the
coordinates of the turning point.
(b) State the coordinates of the points
where the curve crosses the x-axis.
(c) Sketch the curve.
(d) Calculate the exact area of the region
between the curve and the x-axis and
the lines x =
–
and x = 2.
8. Find the gradient of the curve
y = (x
2
– 9)(x + 3) at the points where it
cuts or touches the x-axis. Also, nd its
turning points and hence sketch it.
9. The acceleration, a m/s
2
, of a particle
is given by a = 2t – 3 where t is time in
seconds. When t = 0, the displacement
s = 10 m and the velocity v = 24 m/s. Find
(a) its velocity at t = 5 s,
(b) the distance it travels in the rst 4 s.
10. The velocity of a particle moving in a
straight line is given by v = 4 + 6t – t
2
m/s.
Use the trapezium rule and six ordinates
to estimate the distance covered by the
particle in the rst 5 seconds. Calculate its
average speed.
11. A particle moves in a straight line in such
a way that its distance, s metres, after t
seconds is given by the equation
s =
t
3
– 3t
2
+ 5t. Find the times when
3
(a) the particle is stationary,
(b) its velocity is 5 m/s,
(c) its acceleration is 10 m/s
2
.
12. Given that y =
–
x
2
+ 9, draw the graph of y
against x for 0 ≤ x ≤ 4. Use mid-ordinate
rule with, eight strips, to estimate the area
bounded by the curve, the axes, and the
line x = 4.
y
x
y
= 3
x
y
=
x
2
151
In this chapter, you are provided with three sets
of model examination papers to help you in
revision and examination practise. The rst set
is fully solved and a marking scheme provided
to show you how marks are scored.
When an examination paper is marked, three
types of marks are normally awarded:
1. Method marks, denoted M.
2. Accuracy marks, denoted A, dependent on
the method mark, M.
3. Accuracy marks not dependent on M mark
are called bonus marks, denoted B. B marks
are sometimes given other names such as L
(for correct line), S (for correct scale), etc.
The method marks, M, are given for knowing
and carrying out a correct method for solving
a section of the problem. Accuracy marks
are then awarded for carrying out the method
correctly.
Sometimes, the work involved in a part of a
question is too little for it to merit more than
one mark. In such cases, where an M mark
would not be particularly appropriate, we use
a B mark. For example, if you were asked to
state the range of the set of values:
35, 25, 82, 71, 60,
the answer, range = 82 – 25 = 57 could attract
B1; any other answer would attract B0 mark
(0 mark), even if 82 – 25 is correctly shown. It
is the answer (57) that earns the mark.
Example
Solve the equation 5sin
2
x + 2cos x = 2,
0° < x < 180°.
Solution
5sin
2
x + 2cos x = 2
5(1 – cos
2
x) + 2cos x = 2
Using the substitution sin
2
x = 1 – cos
2
x could
attract a B1 mark.
5 – 5cos
2
x + 2cos x = 2
5cos
2
x – 2cos x – 3 = 0
Rearranging into a proper quadratic form could
attract an M1 mark.
(5 cos x + 3)(cos x – 1) = 0
Factorisation could attract an M1 mark.
x = 90° or x = 126.9°
Each correct () value of x could attract an A1
mark.
Note:
For each set of model papers, Paper 1 mainly
covers topics from Book 1 and 2 while Paper
2 mainly covers topics from Book 3 and 4, but
knowledge and skills acquired in Book 1 and 2
are assumed.
MODEL EXAMINATION PAPERS
11
152
Model examination papers: Set 1 (fully solved)
PAPER 1
Time: 2 hours
SECTION I (50 marks)
Answer all the questions in this section.
1. Use mathematical tables to evaluate
0.3
+ √0.498 (3 marks)
0.035 1
2. Express y in terms of x given that
10x
2
– 9xy + 2y
2
= 0 (3 marks)
3. A circle of radius 4.9 cm ts exactly inside
a square. Find the area of the space between
the circle and the square. (2 marks)
4. Solve the simultaneous equations
xy = 4
x + y = 5 (4 marks)
5. Table MEPS 1.1 shows the number of
goals scored in 40 soccer matches during a
certain season.
No. of goals 0 1 2 3 4 5 6 7
No. of matches 3 9 6 8 5 5 2 2
Table MEPS 1.1
Calculate the mean number of goals
scored per match. (3 marks)
6. Find the length of AC in Fig. MEP 1.1.
(3 marks)
Fig. MEP 1.1
D
30°
B
A
C
5 cm
3 cm
1
3
–
1
2
–
7. A right pyramid is cut along a plane
parallel to the base such that the height
of the resulting frustrum is that of the
original pyramid. Express the volume of
the resulting smaller pyramid as a fraction
of the volume of the initial pyramid.
(3 marks)
8. I owe a certain American nancier
US$ 10 000. He has given me the option
of paying through my account in Kenya
or through my account in the UK. Find
which option is cheaper, and by how much
(in shillings), given that the prevailing
exchange rates are:
1 US dollar = 81.338 9 Kenya shillings.
1 Sterling pound = 151.015 1 Kenya
shillings.
1 US Dollar = 0.538 8 sterling pounds.
(3 marks)
9. A spherical ball is deated so that its
volume decreases in the ratio 8:27. Find
the ratio in which the radius decreases.
(2 marks)
10. Solve the equation
64
x
– 121 = 7 – 4
3
x
. (3 marks)
11. Simplify
2x
2
+ x – 6
+
1
.
(4 marks)
x
2
– 4 x – 2
12. A line passes through the point (
–
1 , 2) and
has gradient
–
. Write down its equation in
the form ax + by = c. (3 marks)
13. A surveyor nds that she needs 28 beacons
placed 40 m apart when she surveys a length
of road. If she were to place the beacons
30 m apart, how many would she need ?
(2 marks)
1
2
–
153
were measured and recorded as in Table
MEPS 1.2.
Height
0 ≤ x < 5 5 ≤ x < 15 15 ≤ x < 25 25 ≤ x < 45 45 ≤ x < 75
x cm
No. of
7 46 72 64 11
seedlings
Table MEPS 1.2
(a) Estimate the mean height (5 marks)
(b) Using a scale of 1 cm to represent
10 units along the horizontal axis, and a
scale of 1 cm to represent 1 unit along
the vertical axis, draw a histogram to
represent the distribution. (5 marks)
19. Atieno and Muthoni invested in a commuter
taxi (i.e. matatu) business, buying a
minibus whose carrying capacity was 26
passengers, 25 of whom would be paying.
They put the minibus on a route connecting
two towns, A and B, where the fare was
sh 120 one way (even if one alighted along
the way). Every day, the minibus made
three round trips between the two towns.
On each day, fuel consumption was
sh 2 500, and the driver and conductor
were paid sh.450 and sh 250 respectively.
A further sh 3 500 was set aside, daily, for
maintenance, insurance and loan repayment.
(a) On a day when the minibus was full on
every trip, how much was the
(i) amount of the day’s collection,
(2 marks)
(ii) net prot? (2 marks)
(b) The agreement between Atieno and
Muthoni was that they would be sharing
each day’s prots in the ratio 3 : 4.
Calculate how much each got on a day
when the minibus was 75% full, on
average, per round trip. (6 marks)
20. A triangle has vertices A(1 , 2), B(7 , 2),
and C(5 , 4).
(a) Draw triangle ABC on the Cartesian
plane. (1 mark)
6 cm
9 cm
O
P
Q
– – – – – – – – – – – – – –
6 m
2 m
–
–
14. A test tube is made up of a hemispherical
bottom and a cylindrical stem, both of
internal radius 0.7 cm. Calculate the
capacity of the test tube, given that its
height is 12 cm. (4 marks).
15. Last year, Nafula was four times as old
as her son, Wekesa. In three years time
from now, the sum of their ages will be 53.
Determine their present ages. (4 marks)
16. The cross section of a house is in the shape
of a pentagon, as shown in Fig. MEPS 1.2.
Determine the area of the cross-section.
(4 marks)
Fig. MEPS 1.2
SECTION II (50 marks)
Answer any ve questions in this section.
17. (a) In Fig. MEPS 1.3, O is the centre of a
circle whose radius is 6 cm and PQ is
9 cm.
Fig. MEPS 1.3
Calculate the area of the major segment.
(7 marks)
(b) Find the area of a triangle ABC with
sides 7 cm, 9 cm and 11 cm long.
(3 marks)
18. The heights of tree seedlings in a nursery
154
(b) Construct the image ∆A′B′C′ of ∆ABC
under a rotation of 90° clockwise
about the origin. (2 marks)
(c) Draw ∆A′′B′′C′′, the image of
∆A′B′C′ under reection in the line
y = x. State the coordinates of A′′, B′′
and C′′. (3 marks)
(d) ∆A′′′B′′′C′′′ is the image of ∆A′′B′′C′′
under reection in the line y = 0. Draw
∆A′′′B′′′C′′′ and state the coordinates
of its vertices. (2 marks)
(e) Describe a single transformation that
maps ∆A′′′B′′′C′′′ onto ∆ABC.
(2 marks)
21. Two airports A and B are such that B is 500
km due east of A. Two planes (call them
planes A and B) lift o from the airports at
exactly the same time. Plane A ies on a
bearing of 030° at 360 km/h and plane B
ies on a bearing of 315° at 240 km/h, and
then they land after ying for 90 min.
(a) Using a scale of 1 cm to represent 100
km, make a scale drawing to show the
planes’ nal positions. (6 marks)
(b) Use the scale drawing to nd the
distance between the two planes after
the ight. (2 marks)
(c) Determine the bearing, after the ight, of
(i) plane B from plane A. (1 mark)
(ii) plane A from plane B. (1 mark)
22. The container of a petrol tanker is a cylinder
with its axis horizontal. Its internal length
is 7 m and its internal diameter is 2.1 m.
Fig. MEPS 1.4 shows the vertical section
of the container when the depth of the
petrol it contains is 0.525 m, AB being the
horizontal level of the petrol.
Fig. MEPS 1.4
Calculate
(a) ∠AOB, where O is the centre of the
circular section. (3 marks)
(b) the area of sector AOB. (2 marks)
(c) the shaded area in Fig. MEP1.4.
(3 marks)
(d) the mass of the petrol in the tanker,
given that one cubic metre of petrol
has a mass of 700 kg. (2 marks)
23. Three men, Kimani, Said and Sang decided
to start a business. The initial capital
that they needed was sh 3 000 000, of
which they were able to raise 40%, by
making contributions in the ratio 3 : 3 : 2
respectively. They approached a nancier
who agreed to provide the remaining
amount of capital, which was to be paid
back within one year, with an interest of
20% and in the same ratio of 3 : 3 : 2. The
three men were to share the proceeds of the
business in the ratio of contribution, i.e.
3 : 3 : 2. During the year, the business
realised sh 3 200 000.
(a) How much of the initial amount did
Sang raise? (2 marks)
(b) How much did Kimani pay to the
nancier at the end of the year?
(4 marks)
(c) After paying the nancier at the end of
the year, how much was Said left with?
(4 marks)
B
A
O
•
0.525 m
155
PAPER 2
Time: 2 hours
√
O
30°
C
A
B
x
3
x
1 1
( )
0 1
3 1
( )
2 1
3 4
( )
24. A car manufacturing company exports
one third of its cars. Three fths of the
remainder are sold locally to dealers and
the rest to individuals.
(a) Calculate the percentage of cars that
are sold to each category of people.
(5 marks)
(b) Construct a pie chart to represent the
above information and state the angle
of each of the sectors. (5 marks)
SECTION I (50 marks)
Answer all the questions in this section.
1. Make s the subject of the formula
a =
s
2
+ q
(3 marks)
p
2
2. In Fig. MEPS 1.5, O is the centre of the circle.
AB//OC and ∠ABO = 30°. Find ∠ACO.
(3 marks)
Fig. MEPS 1.5
3. Find the values of x for which the matrix
does not have an inverse. (3 marks)
4. Given that
2x – y
=
4,
nd
y
. (2 marks)
x – 2y 3 x
5. Solve for x in
log
10
5 – 2 + log
10
(2x + 10) = log
10
(x – 4).
(3 marks)
6. The transformation represented by the
matrix M = maps a triangle ABC
onto another triangle Α′Β′C′ of area
36 cm
2
. Find the area of ∆ABC. (3 marks)
7. The mathematics club of a certain school
consists of 7 boys and 5 girls. Three
members are to be chosen at random to be
ocials of the club. Use a tree diagram to
nd the probability that
(a) all the ocials are boys.
(b) one ocial is a girl. (4 marks)
8. Fig. MEPS 1.6 shows a line segment AB.
Draw the locus of a point P such that
∠APB = 90°. (2 marks)
A B
Fig. MEPS 1.6
9. A triangle ABC is such that ∠ACB = 30°
and BC = 5 cm. It has an area of 10 cm
2
.
Find AC. (3 marks)
10. Given that matrix A = , nd matrix B
such that A
2
= A + B. (4 marks)
11. Fig MEPS 1.7 shows a circle, centre O and
radius 3.5 cm. ∠POQ = 144°. Find the
shaded area. (3 marks)
Fig MEPS 1.7
12. Use binomial expansion to nd the value of
1.002
5
correct to 4 d.p. (3 marks)
1
2
–
144°
Q
P
O
3.5 cm
156
13. Find the x-coordinate of the point at which
the gradient of the curve y = x
2
– 3x + 3
is
–
7. (2 marks)
14. The mass of a solid metallic cylinder varies
jointly as its height and the square of its
radius. A cylinder with radius 8 cm and
height 25 cm has a mass of 9.6 kg. Find
the mass of a similar cylinder, made of the
same metal, having radius 6 cm and height
35 cm. (4 marks)
15. In Fig MEPS 1.8, OA = a, OB = b, AP = PQ
and AQ = QB. Express OP in terms of a
and b in the simplest form. (3 marks).
Fig MEPS 1.8
16. A biscuit factory has a machine which
produces 18 000 biscuits per hour. The rate
of production decreases by 20% every hour.
(a) Calculate the number of biscuits
produced in the rst two hours.
(2 marks)
(b) If the machine has to be shut down after
running for four hours, how many
biscuits will it have produced?(3 marks)
SECTION II (50 marks)
Answer any ve questions in this section.
17. Table MEPS 1.3 shows the distribution of
masses of a sample of pupils in a certain
academy.
Mass 20–24 25–29 30–34 35–39 40–44
(kg)
No. of 1 5 9 11 20
pupils
Mass
45–49 50–54 55–59 60–64 65–69
(kg)
No. of
20 19 8 4 3
pupils
Table MEPS 1.3
Using 47 as a working mean, calculate
(a) the mean mass. (5 marks)
(b) the standard deviation of the masses.
(5 marks)
18. The roof of a warehouse is in the shape of a
triangular prism, as shown in Fig. MEPS 1.9.
Fig. MEPS 1.9
Calculate
(a) the angle between faces RSTU and
PQRS. (3 marks)
(b) the space occupied by the roof.
(3 marks)
(c) the angle between plane QTR and
PQRS. (4 marks)
19. A body moves in a straight line in such a
way that at any time, t seconds, its distance,
s metres from the starting point, is given by
s = 8t – t
2
.
(a) How fast is the body moving at
(i) t = 1 second?
(ii) t = 3 seconds? (3 marks)
(b) What is the maximum displacement
that the body achieves? (4 marks)
(c) Find the acceleration and comment
on it. (1 mark)
(d) After how long will the body be back at
the starting point? (2
marks)
– – – – – – – –
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – –
24 m
10 m
8 m
12 m
Q
T
P
U
R
S
Q
P
O
B
a
b
A
157
20. At the beginning of a certain year, Muasya
deposited sh 100 000 in an investment
account, which earned compound interest
at 15% per annum. At the beginning
of each subsequent year, he deposited
a further sh 5 000 in the same account.
Determine
(a) how much money he had in the account
after 5 years. (7 marks)
(b) the percentage interest earned over the
ve years. (3 marks)
21. Use a ruler and a pair of compasses only
for all constructions in this question.
(a) Draw a line segmnt AB, 8 cm long. On
the line, construct ∆ABC such that
∠BAC = 75° and line AC = 7 cm.
(6 marks)
(b) Construct the circumcircle of ∆ABC
and measure its radius. (4 marks)
22. A tailoring business makes two types of
garments, A and B. Garment A requires
3 metres of material while garment B
requires 2 metres of material. The
business uses not more than 600 metres of
material daily in making both garments.
It must make not more than 100 garments
of type A and not less than 80 of type B
each day.
(a) Write down four inequalities from this
information. (4 marks)
(b) Graph these inequalities. (3 marks)
(c) If the business makes a prot of sh 80
on garment A and a prot of sh 60
on garment B, how many garments of
each type must it make in order to
maximise its total prot? (Assume that
all the garments made are sold the
same day.) (3 marks)
23. (a) In Fig. MEPS 1.10, TP and TQ are
tangents to the circle and PR is parallel
to TQ. Find the angles marked a, b, c,
and d. (6 marks)
Fig. MEPS 1.10
(b) In F ig. MEPS 1.11, O is the centre of
the circle and ∠OAD = 25°. Calculate
∠BCD. (4 marks)
Fig. MEPS 1.11
24. (a) A shear parallel to the x-axis maps
point (1 , 2) onto point (7 , 2). T is the
transformation equivalent to this shear
followed by reection in the line y = x.
Find the matrix which denes T
.
(5 marks)
(b) A transformation P maps points (1 , 3)
and (
–
2 ,
–
3) onto points (2 , 4) and
(
–
3 ,
–
11) respectively. Find the matrix
of the transformation. (5 marks)
1
2
–
D
25°
A
C
•
B
O
T
64°
R
Q
P
b
d
a
c
158
Model examination papers: Set 2
PAPER 1
Time: 2 hours
SECTION I (50 marks)
Answer all questions in this section.
1. Use logarithms to evaluate
16.49
2
× √0.631 8
. (4 marks)
327.5
2. Factorise completely, the expression
4x
2
– y
2
+ 3(2x + y). (2 marks)
3. Fig. MEPS 2.1 represents a circle, centre O.
Given that ∠POR = 108°, PQ = QR, nd
(a) the value of y. (1 mark)
(b) ∠OPQ. (2 marks)
Fig. MEPS 2.1
4. Solve the following compound inequality
and illustrate your solution on a number line.
3x + 4 < 25 ≤ 6x – 7 (3 marks)
5. A rectangular water tank 2 m long, 1.5 m
wide and 1.8 m high is half full of water. A
metal sphere of radius 30 cm is put into the
water and is completely submerged. Find
the new water level. (4 marks)
6. Casual workers in a certain company
demanding a rise of 8% in wages were
awarded 3 % rise. As a result, each man’s
weekly wages rose to Ksh 1 345.50. How
much would the weekly wages have been if
the 8% rise had been granted. (3 marks)
1
2
–
7. Find the equation of a line, through the
point (2 , 1), perpendicular to the line
x + 2y =
–
3. (3 marks)
8. Fig. MEPS 2.2 shows two triangles such
that ∠AQP = ∠ABC, AP = 2 cm,
AQ = 3 cm and QC = 1 cm. Find the
length of PB. (3 marks)
Fig. MEPS 2.2
9. Find the value of m and n given that
m – 2 = . (2 marks)
10. A piece of wire 40 cm long is bent to form
a right-angled triangle whose hypotenuse is
17 cm long. Find the length of each of the
other two sides. (4 marks)
11. Two tractors A and B travel between
two towns 120 km apart. Tractor A has
an average speed of 10 km/h more than
tractor B and takes 1 hour less than tractor
B. Find the respective speeds of the two
tractors. (4 marks)
12. Water ows at the rate of 10 m/min through
a cylindrical pipe 10 mm in radius into a
conical container of base radius 10 cm and
height 12 cm. How long does it take to ll
the container? (4 marks)
13. In Fig. MEPS 2.3, ABCD is a cyclic
quadrilateral and ADE is a straight line.
1
2
–
B
A
C
Q
P
( )
–
1
1
( )
3
n
( )
2
1
P
108°
R
O
Q
y
=
=
1
2
–
159
D
A
B
C
3x
y
2x
E
80°
Given that ∠CDE = 80°, ∠ABC = 2x,
∠BCD = 3 x and ∠DAB = y, nd the values
of x and y. (3 marks)
Fig. MEPS 2.3
14. A boy who is 156 cm tall stands on a level
ground 100 m away from the foot of a tree.
He observes the angle of elevation of the
top of the tree to be 20.4°. Calculate the
approximate height of the tree in metres.
(3 marks)
15. A map is drawn using a scale of 1 : 200 000.
Find the actual area reprsented on the map
by a rectangle 2 cm by 2.5 cm, giving your
answer in km
2
. (2 marks)
16. Use tables of reciprocals to nd the value
of x given that
1
=
1
–
1
. (3 marks)
x 0.27
4.6
SECTION II (50 marks)
Answer any ve questions in this section.
17. (a) Town B is 30 km from town A on a
bearing of 042°. Town C is 22 km
from A on a bearing of 350°. By means
of a scale drawing, nd the distance
and bearing of C from B. (4 marks)
(b) A section of a river with straight banks
runs due south. A man standing on
the edge facing the river observes the
bearing of a tree on the opposite bank
of the river to be 300°. After walking
150 m along the river due north, he
observes the bearing of the same tree
to be 220°. Use a scale drawing to nd
the width of the river. (6 marks)
18. A and B are two stations on a double
railway line. At 1400 hours, two trains
leave station A both travelling to station
B. One train does an average of 80 km/h
and arrives 50 minutes late. The other train
does an average of 100 km/h and arrives
20 minutes late. The scheduled time of
arrival of the two trains is the same.
(a) Using a scale of 1 cm to represent
10 km on the vertical axis, and 6 cm to
represent 1 hour on the horizontal axis,
draw (on the same axes) the graphs of
motion of the two trains and use them
to determine
(i) the distance between A and B.
(5 marks)
(ii) the scheduled time of arrival
of the two trains. (2 marks)
(b) At what speed should the trains have
travelled to arrive on schedule?
(3 marks)
19. The length and breadth of a rectangle
are given as (6x – 1) and (x – 2) cm
respectively. If the length and breadth are
each increased by 4 cm, the new area is
three times that of the original rectangle.
(a) Form an equation in x and solve it.
(4 marks)
(b) Find the dimensions of the original
rectangle. (2 marks)
(c) Express the increase in area as a
percentage of the original area.
(4 marks)
20. Using a ruler and pair of compasses only,
construct a parallelogram ABCD whose
diagonals AC and BD intersect at a point O,
given that AC = 7.2 cm, BD = 5.6 cm and
∠AOB = 45°. Measure the length of AB.
Construct the circumcircle of ∆BOC and
mark a point E on AB produced such
160
=
=
6 cm
C
18 cm
24 cm
A
F
G
E
B
D
that ∠BEC = 45°. Calculate the area of
quadrilateral BECO. (10 marks)
21. Fig. MEPS 2.4 represents a cross-section of
a concrete prism whose length is 3 m.
Fig. MEPS 2.4
Given that AB = FE, and AGF is a semicircle,
(a) calculate the total surface area of the
prism, (5 marks)
(b) determine the volume of the concrete
used to make the prism, (3 marks)
(c) given that 1 cm
3
of concrete has a mass of
6.8 g, nd the mass of the prism in kg.
(2 marks)
22. Fig. MEPS 2.5 represents a pyramid with a
rectangular base ABCD in which AB = 3 cm
and BC = 5 cm. It is also given that ∆OAB
is equilateral, ∆OCD is isosceles with
OC = OD and ∆BOC is right-angled at O.
X and Y are the midpoints of AB and CD
respectively.
Fig. MEPS 2.5
(a) Draw a fullsize net of the pyramid,
showing the rectangular face surrounded
by the triangular faces. (5 marks)
(b) On a separate diagram, construct
∆OXY accurately and measure ∠XOY.
(5 marks)
23. A rhombus has vertices at A(
–
1 , 1),
B(0 , 8), C(5 , 3) and D(x , y). T is the
intersection of the diagonals of the rhombus.
(a) Find the coordinates of D and T.
(2 marks)
(b) Given that ∠CBT = a, express ∠BAD
in terms of a. (2 marks)
(c) Calculate the lengths of diagonals AC
and BD. (3 marks)
(d) Calculate the area of the rhombus.
(3 marks)
24. ∆ABC has vertices A(
–
2, 6), B(2 , 3) and
C(
–
2 , 3). ∆A′B′C′ is the image of ∆ABC
under reection in the line x =
–
3.
Under a translation vector , ∆A′B′C′
is mapped onto ∆A″B″C″, and ∆A″B″C″
is mapped onto ∆A′′′B′′′C′′′ under another
transformation.
(a) Given that A′′′(6 ,
–
6), B′′′(2 ,
–
3) and
C′′′(6 ,
–
3), determine and describe
fully the transformation that maps
∆A′′B′′C′′ onto ∆A′′′B′′′C′′′. (4 marks)
(b) Describe fully the transformation that
would map ∆ABC onto ∆A′′′B′′′C′′′.
(3 marks)
(c) Given also that A′′′′B′′′′C′′′′ is a triangle
whose vertices are A′′′′(2 ,
–
6),
B′′′′(
–
2 ,
–
3) and C′′′′(2 ,
–
3), describe
the transformation that maps ∆ABC
onto ∆A′′′′B′′′′C′′′′. (3 marks)
( )
10
2
5 cm
O
– – – – – – – – – – – – – – –
– – – – – –
3 cm
C
D
B
A
=
=
– – – – – – – – – – – –
–
=
–
161
PAPER 2
Time: 2 hours
1
2
–
SECTION I (50 marks)
Answer all the questions in this section.
1. Solve the equation 6
2x + 1
= 2
3x + 1
. (4 marks)
2. Find the area bounded by the curve y = 4x
2
,
the lines x = 1 and x = 4, and the x-axis.
(2 marks)
3. AB and XY are two intersecting chords of
a circle. They meet at R such that
AR = 4 cm, XR = 5 cm and RY = 3 cm.
Calculate the length of AB. (2 marks)
4. P and Q are two towns. Their local times
dier by 2 hours. If the longitude of P is
13°E, nd two possible longitudes of Q.
(3 marks)
5. Simplify the following expressions.
(a)
1 – sin
2
x
(1 mark)
cos x
(b)
1
– tan
2
x
(2 marks)
cos
2
x
6. 3 x
2
– 4x + 1 is the gradient function of a
curve that passes through the point (1 ,
–
2).
Find the equation of the curve. (3 marks)
7. Given that n =
p
, make h the subject
√h
2
+ k
2
of the formula. (3 marks)
8. In Fig. MEPS 2.6, TA and TB are tangents to
the circle at A and B. A D is parallel to BC
and ∠CAD = 51°. Calculate the value of x.
(3 marks)
Fig. MEPS 2.6
1
2
–
C
51°
A
D
B
x
T
9. Solve the equation sin ( x) = 0.5 for
0° ≤ x ≤ 360°. (2 marks)
10. The transformations X and Y have the
matrices and , respectively.
Find the image A′B′ of AB where, A is
(2 , 2) and B is (5 , 4), under the combined
transformation YX. (4 marks)
11. A car was valued at sh 3 000 000 in
January 2000. Each year, its value
decreased by 12% of its value at the
beginning of the year. Find the value of the
car in January 2004, giving your answer
correct to 4 s.f. (3 marks)
12. A bag contains 3 blue marbles, 4 white
marbles and 3 red marbles. Two marbles
are drawn from the bag, one at a time
without replacement. Find the probability
that one of them is red and the other is
white. (3 marks)
13. Two lines, x + 2y =
–
1 and 2x + 3y = 3,
intersect at a point R. Find the equation
of a circle, centre R, radius 5, giving your
answer in the form x
2
+ y
2
+ gx + fy + c = 0,
where g, f and c are constants. (4 marks)
14. Two grades of tea, costing sh 100 and
sh 150 per kg respectively are mixed in the
ratio 3 : 5 by mass. The mixture is then
sold at sh 160 per kg. Find the percentage
prot on the cost price. (4 marks)
15. The rst and thirteenth terms of an AP are
7 and 1 respectively. Calculate the number
of terms which have a sum of zero.
(5 marks)
16. Find the coordinates of the point on the
curve y = x
2
– x + 5 at which the gradient is 2.
(2 marks)
( )
2 1
1 1
( )
1
–
1
–
2 1
1
2
–
162
SECTION II (50 marks)
Answer any ve questions from this section.
17. P(30°N, 20°W), Q(30°N, 40°E),
R(60°N, a°E) and S(b°N, c°W) are four
points on the surface of the earth. R is due
North of Q, and S is due West of R and due
North of P.
(a) State the values of a, b and c. (3 marks)
(b) Given that all distances are measured
along parallels of latitude or along
meridians, and in nautical miles, nd
the distance of R from P using two
alternative routes, one via Q and the
other via S. (3 marks)
(c) Two pilots start ying from P to R,
one along the route PQR at 400 knots,
and the other along PSR at 300 knots.
Which one reaches R earlier and by
how long? (2 marks)
(d) Find the position of the second pilot by
the time the rst one reaches R.
(2 marks)
18. Table M EPS 2.1 shows the marks scored by
a group of 55 candidates in a mathematics
quiz.
Mark 5–9 10–14 15–19 20–24 25–29
f 1 5 8 11 12
Mark 30–34 35–39 40–44 45–49
f 9 5 3 1
Table MEPS 2.1
(a) Draw a cumulative frequency graph to
represent this data. (3 marks)
(b) Use your graph to estimate
(i) the median mark (2 marks)
(ii) the interquartile range (3 marks)
(iii) the 70th percentile (2 marks)
19. Triangle ABC has coordinates A(0 ,
–
1),
B(4 , 3) and C(2 , 2).
(a) Find the image ∆A′B′C′ of ∆ABC
under a translation
. (2 marks)
(b) Given that ∆A′′B′′C′′ is the image
of ∆A′B′C′ under an enlargement
scale factor 3, centre (0 , 0), nd the
coordinates of A′′, B′′ and C′′.
(3 marks)
(c) Given that the area of ∆ABC is 6 square
units, state the area of A′′B′′C′′.
(2 marks)
(d) State the coordinates of the image of
∆A′′B′′C′′ under a reection in the
x-axis. (3 marks)
20. The values of x and y in Table MEPS 2.2
are such that y varies inversely as the
square of x, and x > 0.
x 2 4 8 12 16
y 240 9.6
Table MEPS 2.2
(a) Find the constant of proportionality
between y and x
2
. (2 marks)
(b) Copy and complete the table. (2 marks)
(c) Plot the graph of y against x
2
and use
it to nd the value of y when x = 3 and
when x = 15, correct to 1 d.p. (6 marks)
21. Two circles, centres A and B, have radii
4 cm and 8 cm respectively. If the two
circles share a common chord 6 cm long,
calculate
(a) the area of the common region.
(5 marks)
(b) the perimeter of the gure. (5 marks)
22. (a) Expand and simplify
(2t – 1)(t – 1)(t – 2). (2 marks)
(b) A particle moves along a straight line
such that its distance s m from a xed
point A after t seconds is given by
s = 2t
3
– 7t
2
+ 7t –2. Find
(i) the times when the particle is at
point A. (2 marks)
(ii) the velocity and acceleration of the
particle at t = 0 and t = 2.
(4 marks)
( )
1
–
2
163
(iii) the times at which the particle is
instantaneously at rest. (2 marks)
23. (a) Draw the graph of y = 2x
2
+ 3x + 1
for
–
4 ≤ x ≤ 3. (6 marks)
(b) Use your graph to solve the equation
(i) 2x
2
+ 4x – 3 = 0. (2 marks)
(ii) x
2
– x – 4.5 = 0. (2 marks)
24. PQRSTUVW is a prism such that PQRS,
TUVW, QRVU and PSWT are rectangles
and PQUT and SRVW are trapezia. Its
measurements in cm are as given in
Fig. MEPS 2.7.
Calculate
(a) the length of PV. (4 marks)
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – –
– – – – – – –
P
U
Q
T
R
S
W V
3.5 cm
12 cm
8 cm
6 cm
Fig. MEPS 2.7
(b) the angle between the planes PQRS
and UQRV. (2 marks)
(c) the angle between planes PQRS and
PQVW. (3 marks)
(d) the volume of the prism. (2 marks)
164
Model examination papers: Set 3
PAPER 1
Time: 2 hours
1
2
–
SECTION I (50 marks)
Answer all the questions in this section.
1. Use logarithms to evaluate
68.53
. (4 marks)
(13.8 × 0.07421)
2. Solve the equation 8
x
+ 2
3x
+ 3 = 35.
(3 marks)
3. The unshaded region in Fig. MEPS 3.1 is
bounded by the lines y = 4, x =
–
2 and
y = x + 3.
Fig. MEPS 3.1
State the inequalities that satisfy the region.
(2 marks)
4. The area of ∆ABC, Fig. MEPS 3.2, is
given as 36 cm
2
.
Fig. MEPS 3.2
–
4
4
3
2
1
0
–
3
–
2
–
1
y = 4
y
=
x + 3
x =
–
2
x
y
Given that all the dimensions are in
centimetres, calculate the value of x.
(3 marks)
5. Solve the equation
x
–
2x – 3
=
x – 5
. (3 marks)
6 5 3
6. A returning U.S. tourist changed sh 14
400 into U.S. dollars on a day when the
exchange rate was 1 U.S. dollar =
sh. 73. 455 6. How many dollars did he
receive if the bank charged him a 10%
commission? (3 marks)
7. A line is drawn through point (3 , 5)
perpendicular to line 2y + x = 3. Find the
equation of the line. (3 marks)
8. ABCD is a trapezium in which AB is
parallel to DC and DA is perpendicular to
AB. If AB = 10 cm, AC = 15 cm and
∠ACD = 28°, calculate the area of the
trapezium. (4 marks)
9. Three points A(2 , 3), B(2k , 5) and
C(
–
3 , 6) lie on a straight line. Find the
value of k. (3 marks)
10. Joan has some money in two denominations
only: twenty-shilling coins and one
hundred-shilling notes. She has four times
as many twenty-shilling coins as one
hundred-shilling notes. Altogether she has
sh. 2 160. How many one hundred-shilling
notes does she have? (3 marks)
11. Solve the equation
log 5 + log(2x + 10) – 2 = log(x – 4).
(3 marks)
A
C
B
x
5
2x
1
2
–
165
( )
–
3
2
( )
2
3
( )
9
7
12. Given that x = , y = and z = ,
nd the constants a and b so that
ax + by = z. (2 marks)
13. Expand and simplify the expression
(2x – 1)(2x + 1) – (7x – 4) and hence
factorise it. (4 marks)
14. A map has a scale of 1:25 000. On this
map, a square piece of land is represented
by an area of 2 cm
2
. Calculate the actual
area of the plot in hectares. (3 marks)
15. A lampshade is in the shape of an open
frustum of a cone. Its top and bottom
diameters are 10 cm and 20 cm
respectively, and its height is 12 cm. Find,
in terms of π, the area of the material used
to make the lampshade. (4 marks)
16. Triangle ABC in Fig. MEPS 3.3 is
right-angled at A.
Fig. MEPS 3.3
Calculate sin
θ
. Hence nd
θ
. (3 marks)
SECTION II (50 marks)
Answer any ve questions in this section.
17. A metal water pipe has an internal radius
of 10 cm. The pipe is made of metal of
uniform thickness of 1 cm.
(a) Find the volume, in cubic centimetres,
of metal used to make a portion of the
pipe, 5 m long.
(4 marks)
(b) Given that the metal used to make the
pipe has a density of 8.5 g/cm
3
, nd the
mass of the piece of pipe in kilograms.
(3 marks)
(c) Water runs through the pipe at a rate of
30 cm per second. Find the time taken
to ll a water tank of capacity 50 000
litres. (3 marks)
18. A dam containing 4 158 m
3
of water is to
be drained. A pump is connected to a pipe
of radius 3.5 cm and the pump operates for
8 hours per day. Water ows through the
pipe at the rate of 1.5 m per second.
(a) Find the number of days it takes to
drain the dam. (6 marks)
(b) The cost of hiring the pump, excluding
the cost of the fuel, is sh 500 per day.
The pump consumes 10 litres of diesel
per day, and the cost of diesel is
sh 79.99 per litre. Calculate the total
cost of draining the dam. (4 marks)
19. According to new company rules, a tourist
van driver is required to cover the distance
between two towns A and B in x hours. If
he travels from A at 90 km/h, he will reach
B ten minutes too early. If he travels at
75 km/h he will arrive at B six minutes
later than the scheduled time.
(a) Find the distance between A and B and
the
recommended time of arrival.
(3 marks)
(b) If he leaves A at 11.45 a.m, at what time
will he be expected at B? (2 marks)
(c) The driver started from A and arrived
at B on time. At town C, two-thirds of
the way from A, he got a puncture and
spent ve minutes changing the wheel.
If his average speed between A and C
was 80 km/h, nd his average speed for
the remaining part of the journey.
(5 marks)
20. Three solids, a cylinder, a sphere and a
cone, are such that their radii are equal. It
is also given that their surface areas are
equal. If the volume of the sphere is
A
B
C
θ
9 cm
25 cm
166
904.9 cm
3
, nd the volume, to 4 s.f., of:
(a) the cylinder, (6 marks)
(b) the cone. (4 marks)
21. Kamau and Mumo decided to start a
business together. Kamau contributed
sh 500 000 and Mumo sh 640 000. They
agreed that in any one year, 15% of the
prot shall be divided equally between
them, and 20% shall be used to meet the
cost of running the business the following
year. The rest of the prot shall be shared
in the ratio of their contribution. In the rst
year, the business made a prot of
sh 432 000.
(a) How much did they set aside for the
running of the business? (2 marks)
(b) What was Mumo’s share of the prot at
the end of the rst year? (4 marks)
(c) Kamau decided to keep a deposit for a
house which was going for sh 1 500 000.
What percentage of the cost of the
house did he pay? (4 marks)
22. Table MEPS 3.1 shows marks scored by
100 pupils in a mathematics examination.
(a) Determine the value of x. (1 mark)
(b) Calculate the mean score. (3 marks)
(c) Estimate the median score. (3 marks)
(d) Represent the data in a histogram.
(3 marks)
Marks 20–29 30–39 40–49 50–59 60–69
Number of 4 6 10 x 14
pupils
Marks 70–79 80–89 90–99
Number of 24 14 6
pupils
Table MEPS 3.1
23. (a) A small eld was surveyed and the
measurements recorded in a surveyor’s
eld book as in Table MEPS 3.2.
100 0 F
E 30 65
50 40 D
C 20 30
20 25 B
A 0 0
Table MEPS 3.2
(i) Using a scale of 1cm to 10m, make
an accurate drawing of the map of
the eld. (4 marks)
(ii) Find the area of the eld. (3 marks)
(b) Assuming that the baseline used in (a)
runs in a northerly direction, give the
position of D, relative to A, using
bearing and distance. (3 marks)
24. A pilot set out to y from town A to town D
through towns B and C. The distance from
A to B is 750 km on a bearing of 050°.
C is 500 km on a bearing of 340° from B,
while D is on a bearing of 250°, a distance
of 1625 km from C.
(a) Use a suitable scale to draw a diagram
representing the route of the ight.
(4 marks)
(b) On the return journey, the pilot ew
directly from D to A. What was his
direction of ight? (3 marks)
(c) How long did the ight in (b) take,
given that the pilot travelled at a speed
of 650 km/h? Give your answer to
the nearest second. (3 marks)
167
PAPER 2
Time 2 hours
1
2
–
B
O
E
D
C
•
9. Rationalise the denominator and simplify
√3 + 2√5 (2 marks)
√5 – √3
10. In Fig. MEPS 3.5, CB is a tangent to the
circle, DEB is a straight line, and
DE : EB = 1 : 3. Find BD and EB, given
that CB = 8√3 cm. (3 marks)
Fig. MEPS 3.5
11. On 1st January 2006, I invested a sum of
money in a xed deposit account at 10%
interest compounded semi–annually. By
31st December 2007, I expected my
account to have a sum of sh 644 204. How
much money did I invest? (3 marks)
12. Three men working for 8 hours a day take
6 days to dig a trench 9 m long. How long
would ve men working 4 hours a day take
to dig a trench 45 m long? (3 marks)
13. The resistance of an electrical conductor
is partly constant and partly varies as the
temperature. When the temperature is
20°C, the resistance is 55 ohms, and when
the temperature is 28°C, the resistance is
58 ohms. Find the relation between the
temperature and the resistance. (4 marks)
14. Solve the following simultaneous equations.
x + y = 0
x
2
+ y
2
– xy = 12 (3 marks)
15. Find the coordinates of the turning points of
the curve y = x
3
+ 2x
2
– 12x. (3 marks)
SECTION I (50 minutes)
Answer all the questions in this section
1. Make n the subject of the formula
P =
–
3
m – n
.
(3 marks)
1 + mn
2. Find the standard deviation of the following
values, giving your answer correct to 3 s.f.
4, 7, 6, 4, 8, 6, 9, 11, 10, 11.
(3 marks)
3. Use binomial expansion to determine the
value of (1.99)
5
correct to 4 d.p. (4 marks)
4. Use the trapezium rule, with strips of unit
width, to estimate the area bounded by the
curve y =
1
, x = 0 and x = 5. (3 marks)
x + 1
5. Two points P and Q are 7.5 cm apart.
Determine, by construction, the locus of R
such that PR = 2.5 cm and ∠PRQ = 90°.
(3 marks)
6. The rst term of an increasing AP is 3.
The third, the sixth and the tenth terms
of the A.P. form a G.P. Find the common
dierence of the AP. (3 marks)
7. Find the x-coordinates of the points on the
curve y = 3x
3
+ 18x
2
+2x +3 at which the
gradient is 2. (3 marks)
8. ABCDEF represents a prism with a
triangular cross–section as shown in
Fig. MEPS 3.4. AB = 6 cm, BC = 9 cm,
EB = 12 cm and ∠ABC = 50°. Calculate
the volume of the prism. (3 marks)
Fig. MEPS 3.4
(
)
√
12 cm
– – – – – – – – – – –
– – – – – – – – – – – – – – –
6 cm
F
C
E
9 cm
– – – – – – – – – – – – – – – – – – – –
D
50°
A
B
168
(i) a slant face and the base. (2 marks)
(ii) a slant edge and the base. (2 marks)
(c) the angle between the planes ABE
and DCE. (3 marks)
Fig. MEPS 3.6
20. Using the same axes, a horizontal scale of
1 cm to represent 30°, and a vertical scale
of 5 cm to represent 1 unit, draw the graphs
of y = sin (x – 15°) and y = 2 cos x for
0° ≤ x ≤ 360°. (5 marks)
Use your graphs to solve the equations
(a) sin (x – 15°) = 0.5 (1 mark)
(b) 2 cos x – 0.5 = 1 (1 mark)
(c) sin (x – 15) – 2cos x = 0 (1 mark)
21. A bus company runs a eet of two types
of buses operating between Nairobi and
Mombasa. Type A bus has capacity to
take 52 passengers and 200 kg of luggage.
Type B carries 32 passengers and 300 kg of
luggage. On a certain day, there were 500
passengers with 3 500 kg of luggage to be
transported. The company could only use a
maximum of 15 buses altogether.
(a) If the company uses x buses of type A
and y buses of type B, write down all
the inequalities satised by the given
conditions. (4 marks)
(b) Represent the inequalities graphically
and use your graph to determine the
smallest number of buses that could be
used. (4 marks)
(c) If the cost of running one bus of type A
is sh 7 200 and that of running one bus
of type B is sh 6 000, nd the minimum
cost of running the buses. (2 marks)
16. The oor of a rectangular room measures
4.8 m by 3.2 m. Estimate the percentage
error in
(a) the perimeter of the oor, (2 marks)
(b) the oor area of the room. (2 marks)
SECTION II (50 marks)
Answer any ve questions in this section.
17. Two towns, A and B, lie on the same
parallel of latitude, 61.5°N.
(a) Find the shorter distance between the
towns and the north pole. (2 marks)
(b) If the longitudes of A and B are 42°W
and 29°E respectively, nd the shortest
distance in km between them.
(3 marks)
(c) If C is another town due south of A
and 960 km away from A, nd the
coordinates of C. (5 marks)
18. Table MEPS 3.3 shows the distribution of
marks scored by a group of 200 candidates
in a mathematics national examination.
Marks 1–10 11–20 21–30 31–40 41–50
No. of
4 10 16 38 48
candidates
Marks 51–60 61–70 71–80 81–90 91–100
No. of
36 20 12 10 6
candidates
Table MEPS 3.3
(a) Use a working mean of 45.5 to
calculate the standard deviation of the
distribution. (6 marks)
(b) Estimate the median mark. (4 marks)
19. ABCDE is a right pyramid on a horizontal
square base of side 10 cm (Fig. MEPS 3.6).
The slant edges are all 8 cm long.
Calculate
(a) the height of the pyramid. (3 marks)
(b) the angle between
– – – – – – – – – – –
– – – – – – – – –
– – – – – – – – –
– – – – – – –
E
A
C
B
D
•
O
8 cm
10 cm
10 cm
169
22. A teacher in a public secondary school
earns sh 19 620 per month. He is paid a
house allowance of sh 20 000 per month,
a medical allowance of sh 2 476 and a
commuter allowance of sh 318. In addition
to income tax, he is deducted sh 392.40
towards a Widows’ and Children’s Pension
Scheme (WCPS).
Calculate the teacher’s net salary after tax,
given that he gets a monthly tax relief of
sh 1 162 per month.
Use the rates of tax given in Table MEPS 3.4.
(10 marks)
Monthly taxable Rate of tax %
income, sh p.m.
0–10 164 10
10 165–19 740 15
19 741–29 316 20
29 317–38 892 25
Over 38 892 30
Table MEPS 3.4
23. AOB is a triangle such that OA = a, OB = b
and ∠AOB is obtuse. P is a point on AB
such that AP = 3PB and Q is the midpoint
of OA.
(a) Express
(i) AB,
(ii) BQ in terms of a and b. (3 marks)
(b) If X is a point on BQ such that
BX = hBQ, express OX in terms of a, b
and h. (3 marks)
(c) Given further that OX = kOP, nd the
values of h and k. (4 marks)
24. Jean is a college student and she stays
alone. She usually sets the alarm clock to
wake her up in the morning. The probability
that she remembers to set the alarm before
going to sleep is . If she does not set the
alarm, she never wakes up before 7.30 a.m.
If she sets the alarm for 7.00 am, the
probability that it wakes her up is only 0.8.
Whenever she wakes up at 7.00 am, she
is never late for class, but if she wakes up
at 7.30 am, the probability that she will be
late for class is 0.8.
Calculate the probability that
(a) Jean wakes up at 7.00 am. (3 marks)
(b) she forgets to set the alarm the night
before but manages to reach college on
time. (2 marks)
(c) she sets the alarm, but it fails to wake
her up and yet she reaches college
punctually. (3 marks)
(d) she is late for college. (2 marks)
1
3
–
170
Marking scheme
B 1 for reciprocal of
0.0351 from tables.
B 1 for square root
from tables.
B 1 for ans.
3
M 1 for splitting middle
term.
A 1 for correct factors.
Equating each factor, in
turn, to zero.
A 1 for both ans.
3
SECTION I (50 marks)
1.
0.3
+ √0.498 = 0.3 × 28.49 + √0.498
0.0351
= 8.547 + 0.7057
= 9.2527
2. 10 x
2
– 9xy + 2y
2
= 0
Factorising LHS:
10x
2
– 5xy – 4xy + 2y
2
= 0
⇒ 5x(2x – y) – 2y(2x – y) = 0
⇒ (2 x – y)(5x – 2y) = 0
Equating each factor, in turn, to zero:
2x – y = 0 and (5x – 2y) = 0
⇒ y = 2x and y =
5
x.
2
3. Sketch (Fig. SMEPS 1.1):
Fig. SMEPS 1.1
4. xy = 4
x + y = 5
y = 5 – x
x(5 – x) = 4
⇒ 5x – x
2
– 4 = 0
or x
2
– 5x + 4 = 0
M 1 for numerical
expression of
required area.
A 1 for ans.
2
Shaded is the required area, given by
(area of square) – (area of circle)
i.e required area = 9.8
2
– × 4.9 × 4.9
= 96.04 – 75.46
= 20.58 cm
2
Solutions to model examination papers — Set 1
Marking scheme
PAPER 1
Time: 2 hours
Writing y in terms of x
using 2nd eqn.
Substituting into 1st eqn.
and simplifying.
B 1 for quadratic
equation.
1
2
–
4.9 cm
•
22
7
––
171
Marking scheme
M 1 for fx column
M 1
A 1
3
M 1
M 1
A 1
3
Factorising LHS: x
2
– x – 4x + 4 = 0
⇒ x(x – 1) – 4(x – 1) = 0
(x – 1) (x – 4) = 0
x = 1 or x = 4
y = 4 or y = 1
Alternatively:
y =
4
x
x +
4
= 5
x
x
2
+ 4 = 5x
x
2
– 5x + 4 = 0
The rest of the working follows the steps in the main scheme.
5. The working may be laid out as in Table SMEPS 1.1.
No of goals, x No of matches, f fx
0 3 0
1 9 9
2 6 12
3 8 24
4 5 20
5 5 25
6 2 12
7 2 14
∑f = 40 ∑fx = 116
Table SMEPS 1.1
Mean no. of goals =
116
40
= 2.9
6.
AD
= tan 30°
BD
∴ AD = BD tan 30° = 3 tan 30°
= 3 × 0.577 3 = 1.731 9
DC = √BC
2
– BD
2
(Pythagoras’ theorem)
= √5
2
– 3
2
= 4 cm
∴ AC = 1.731 9 + 4 = 5.731 9
M 1 for process of
factorising.
A 1 for both ans.
A 1 for both ans.
4
172
Marking scheme
7. Fig. SMEPS 1.2 is a sketch of the pyramid.
Let the height of the original pyramid be h.
Then, the height of the resulting smaller
pyramid is h – h = h.
Linear scale factor (l.s.f)
=
height of smaller pyramid
height of original pyramid
h
=
h
=
Volume scale factor (v.s.f) = (l.s.f)
3
=
(
)
3
Thus, volume of smaller pyramid = of volume of the initial pyramid.
8. Through account in Kenya, I pay
sh 10 000 × 81.339 = sh 813 389
Through account in Britain, I pay
sh 10 000 × 0.538 8 × 151.015 0 = sh 813 669
Thus, it is cheaper to pay through the account in Kenya by
sh (813 669 – 813 389)
= sh 280
9. Linear scale factor = √(volume scale factor)
∴radius decreases in the ratio
√8 : √27
= 2 : 3
10. 64
x
– 121 = 7 – 4
3x
64
x
+ 4
3x
= 128
4
3x
+ 4
3x
= 2 × 4
3
⇒ 2 × 4
3x
= 2 × 4
3
Since the bases are the same;
3x = 3
x = 1
1
3
–
2
3
–
2
3
–
2
3
–
B 1
3
B 1
M 1 for multiplying rate
by sh 10 000.
A 1 for answer.
3
M 1 for multiplying rate
by sh 10 000.
3
3 3
A 1
2
M 1
B 1 for re-arranging.
M 1 for expressing as
powers of 4.
A 1
3
B 1 for height of
smaller pyramid.
2
3
–
8
27
––
Fig. SMEPS 1.2
– – – – – – – – – – – – – – –
– – – – – – – – – – – – – –
– – – – – – – – – –
– – – – – – – – – – – – –
– – – – – – – – – –
– – – – – – – – – – – – –
– – – – – – –
– – – – – – – – – – – – –
– – – – – – – – – – – – –
h
h
h
2
3
–
1
3
–
•
•
173
Marking scheme
B 1 for identifying the
factors of x
2
– 4.
A 1
4
M 1 for expressing as a
single fraction.
M 1 for attempt to
factorise the
numerator.
B 1 for c.
B 1 for eqn. in form
y = mx + c.
B 1
3
M 1
A 1
3
M 1
1
2
–
11.
2x
2
+ x – 6
+
1
x
2
– 4 x – 2
=
2x
2
+ x – 6
+
1
(x + 2)(x – 2) x – 2
=
(2x
2
+ x – 6) + (x + 2)
(x + 2)(x – 2)
=
2x
2
+ 2x – 4
(x + 2)(x – 2)
=
2x
2
+ 4x – 2x – 4
(x + 2)(x – 2)
=
2x(x +2) – 2(x + 2)
(x + 2)(x – 2)
=
(x + 2)(2x – 2)
(x + 2)(x – 2)
=
2x – 2
x – 2
12. Since gradient =
–
,
y =
–
x + c.
Using point (
–
1 , 2), 2 =
–
× (
–
1) + c
c = 1
∴ Equation is y =
–
x +
⇒ 2y + x = 3
⇒ x + 2y = 3
Alternatively:
Using points (x , y) and (
–
1 , 2) on the line;
Gradient =
y – 2
=
–
x + 1
⇒ 2(y – 2) =
–
1(x + 1)
⇒ 2y – 4 =
–
x – 1
⇒ x + 2y = 3
1
2
–
1
2
–
1
2
–
3
2
–
1
2
–
1
2
–
174
Marking scheme
13. Number of gaps between beacons = 28 – 1 = 27
∴ New number of beacons =
40 × 27
+ 1
30
= 37
[Note that the distance marked (40 × 27 = 1 080 m) remains unchanged.
Hence this is the distance divided into the 30 m gaps.]
14. Volume of sphere = πr
3
.
Thus, capacity of hemispherical bottom
= × × × 0.7
3
= 0.718 7 cm
3
Volume of cylinder = πr
2
h.
Height of cylindrical part = 12 – 0.7 = 11.3 cm
Thus, capacity of cylindrical part
= × 0.7
2
× 11.3
= 17.402 cm
3
Total capacity = 0.718 7 + 17.402
= 18.120 7 cm
3
.
15. Let Wekesa’s present age be x.
Last year, Wekesa’s age was (x – 1), and Nafula’s age was 4(x – 1).
In 3 years time, their respective ages will be
(x – 1) + 4 and 4(x – 1) + 4.
∴[(x – 1) + 4] + [4(x – 1) + 4] = 53
⇒ x + 3 + 3x = 53
⇒ 5x + 3 = 53
⇒ x = 10
Thus, Wekesa’s present age is 10 years,
and Nafula’s age present age is 37 years.
16. Extract the triangular section and calculate
the length of the base. (Fig. SMEPS 1.3).
tan 45° =
x
2
∴ x = 2 m
⇒ length of base = 4 m
Note: To get value of x, we could also use the properties of an isosceles
triangle.
M 1
A 1
2
4
3
–
M 1 for expression
of hemispherical
volume.
A 1
4
M 1 for expression of
cylindrical volume.
A 1 for statement of
both ages.
4
M 1 for expression of
total age after
(1 + 3) years and
equating to 53.
B 1 for expression of
Nafula’s age.
A 1
B 1 for length of base.
2 m
x
Fig. SMEPS 1.3
– – – – – – – – – –
45°
B 1
4
3
–
1
2
–
22
7
––
22
7
––
175
Marking scheme
M 1
M 1
A 1
4
M 1
M 1
M 1
M 1
A 1 for value of θ.
A 1
B 1
Area of triangular section
= × 4 × 2
= 4 m
2
Area of rectangular section
= 4 × (6 – 2)
= 16 m
2
∴Total area = 4 + 16 = 20 m
2
.
SECTION II (50 marks)
17. (a) Let ∠POQ =
θ
Then, sin
θ
=
4.5
6
= 0.75
∴
θ
= 48.59°
∴
θ
= 97.18°
Reex angle POQ = 360° – 97.18°
= 262.82°
Area of major sector =
262.82
× 3.14 × 6 × 6
360
= 82.53
Area of ∆POQ = × 6 × 6 sin 97.18°
≈ 17.86 cm
2
∴ Area of major segment
= Area of major sector + Area of ∆POQ
≈ 100.4 cm
2
Alternative method:
— Calculate area of minor sector with angle 97.18°
— Calculate area of ∆POQ
— Hence nd area of the missing segment
— Find area of whole circle
— Subtract area of missing segment to obtain area of major
segment.
(b) Semi-perimeter, s = (7 + 9 + 11)
= 13.5
M 1
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
176
Marking scheme
Area = √s(s – a)(s – b)(s – c)
= √13.5(13.5 – 7)(13.5 – 9)(13.5 – 11)
= √13.5 × 6.5 × 4.5 × 2.5
= √987.187 5
≈ 31.42 cm
2
18. (a) Class Mid-value Freq. xf Height
x f h = f ÷ c
0 ≤ x < 5 2.5 7 17.5 1.4
5 ≤ x < 15 10 46 460 4.6
15 ≤ x < 25 20 72 1 440 7.2
25 ≤ x < 45 35 64 2 240 3.2
45 ≤ x < 75 60 11 660 0.4
Table SMEPS 1.2
∑ xf = 4 817.5
Mean height =
4 817.5
200
≈ 24.09 cm.
(b)
Fig. SMEPS 1.4
M 1
A 1
10
M 1 for x column
M 1 for xf column
M 1
A 1
M 1 for calculation of
bar heights
(h = f ÷ c, where
c = class width)
S 1 use of scale
B 2 for complete 
histogram
(Allow B 1 for 3
bars )
B 1, B 1 for axes
labels
10
0
10 20 30 40 50 60 70 80
y
x
7
6
5
4
3
2
1
5 15 25 45 75
7.2
4.6
1.4
0.4
3.2
Seedling height, x (cm)
Frequency density = f
c
177
Marking scheme
M 1
A 1
B 1
M 1
A 1
A 1
M 1
A 1
10
M 1
A 1
B 1
B 1 construction
shown
B 1 all vertices
B 1
B 1 all coordinates 
B 1 all vertices
B 1 all vertices
B 1 all coordinates 
19. (a) (i) Collection for 1 round trip
= sh 120 × 25 × 2
= sh 6 000
∴ Total day’s collection
= sh 6 000 × 3
= sh 18 000
(ii) Day’s expenditure
= sh (2 500 + 450 + 250 + 3 500)
= sh 6 700
∴ Day’s net prot
= sh (18 000 – 6 700)
= sh 11 300
(b) The day’s collections were
75% of sh 18 000
= sh
75
× 18 000
100
= sh 13 500
Day’s net prot = sh (13 500 – 6 700)
= sh 6 800
Atieno’s share = sh
3
× 6800
7
= sh 2 914.30
Muthoni’s share = sh
4
× 6800
7
= (6 800 – 2 914.30)
= sh 3 885.70
20. See Fig. SMEPS 1.5.
(a) ∆ABC
(b) Image ∆A′B′C′
(c) Line y = x drawn
Image ∆A′′B′′C′′
A′′(
–
1 , 2), B′′(
–
7 , 2), C′′(
–
5 , 4)
(d) Image ∆A′′′B′′′C′′′
A′′′(
–
1 ,
–
2), B′′′(
–
7 ,
–
2), C′′′(
–
5 ,
–
4)
178
Marking scheme
Fig. SMEPS 1.5
(e) 180° clockwise rotation
about (0 , 0).
21. (a) In 90 min (1 h),
plane A ies 360 × 1 = 540 km
and plane B ies 240 × 1 = 360 km
Fig. SMEPS 1.6 is the required scale drawing.
(b) The distance between the planes after 90 min is A′B′.
From the scale drawing, A′B′ = 2.15 cm.
∴the distance between the planes is
2.15 × 100 = 215 km.
B 1 angle & direction
B 1 rotation about (0 , 0)
10
B 1
B 1
B 1 for measurement
of A′B′ in cm.
B 1 for conversion
1
2
–
1
2
–
1
2
–
1
0
–
1
–
5
– – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
1
–
7
2
–
6
3
–
5
4
–
4
5
–
3
6
–
2
7
–
1
x
y
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
C
A′
A
B
B′
C′
B′′
y = x
A′′
C′′
– – – – – – – – – –
– – – – – – – – – –
90°
A′′′
C′′′
B′′′
4
3
2
–
2
–
3
–
4
–
6
–
7
179
Marking scheme
7°
7°
315°
030°
N
B
B′
A
A′
N
N
N
500 km
– – – – – – – – – – –
B 1 for 030° bearing.
B 1 for AA′.
B 1 for 315° bearing.
B 1 for BB′.
B 1
B 1
10
M 1
M 1
A 1
A 1
B 1
Fig. SMEPS 1.6
(c) (i) The bearing of plane B from plane A after 90 min is
180° + 7° = 187°
(obtained by drawing a N-line at A′ and measuring the angle).
(ii) The bearing of plane A from plane B after 90 min is 007°
(obtained by drawing a N-line at B and measuring the angle).
22. Since the diameter of the container is 2.1 m, then r = 1.05 m.
Hence OC, in Fig. SMEPS 1.7, is 1.05 – 0.525 = 0.525 m.
(a) Let ∠AOB = 2
θ
cos
θ
=
0. 525
=
1.05
∴
θ
= 60°
and ∠AOB = 120°
(b) Area of sector AOB
=
120
× × 1.05 × 1.05
360
= 1.155 m
2
(Note that is a more appropriate
value of π than 3.14)
(c) Area of ∆AOB = × 1.05 × 1.05 × sin 120°
= 0.477 4 m
2
∴Area of shaded part = 1.155 – 0.477 4
= 0.677 6 m
2
(d) Volume of petrol is taken = shaded area × length of tank
= 0.677 6 × 7 = 4.7432 m
3
Fig. SMEPS 1.7
M 1
M 1
A 1
22
7
––
1
2
–
22
7
––
1
2
–
1.05 m
0.525 m
0.525 m
O
θ
B
A
C
180
Marking scheme
M 1
A 1
10
2
8
–
M 1
A 1
M 1
A 1
M 1
A 1
M 1
M 1
A 1
A 1
10
B 1
B 2 for pie chart with
labelling. (Allow
B1 if not labelled)
10
B 1
M 1 for any two
expressions.
A 2 for all three 
%ages. Give A 1
if one is wrong.
}
}
M 1 as above.
A 2 as above.
= sh 2 160 000
∴ Kimani paid × 2 160 000
= sh 810 000
(c) At the end of the year, Said got
× 3 200 000 = sh 1 200 000
He paid to the nancier
×
120
(60% of 3 000 000)
100
= sh 810 000
∴ His balance was 1 200 000 – 810 000 = sh 390 000
24. (a) Let the total number of the cars be x.
No. of cars for export = x
No. of cars for local dealers = × x = x
No of cars for individuals = x – ( x + x) = x
%age of cars for export =
x
× 100 = 33.33%
x
%age of cars for local dealers =
x
× 100 = 40%
x
%age of cars for individuals =
x
× 100 = 26.67%
x
(b) Pie chart angles (Fig. SMEPS 1.8)
Export:
x
× 360° = 120°
x
Local dealers
x
× 360° = 144°
x
Individuals
x
× 360° = 96°
x
3
8
–
3
8
–
3
8
–
∴mass of petrol = density × volume = 700 × 0.677 6 × 7
= 3 320.24 kg.
23. (a) Total number of shares = 3 + 3 +2 = 8
Sang raised ×
40
× 3 000 000
100
= sh 300 000
(b) The nancier was to be paid
120
× (60% of 3 000 000) =
120
× 1 800 000
100 100
1
3
–
2
5
–
4
5
–
Fig. SMEPS 1.8
Individuals
Local
dealers
Export
1
3
–
2
5
–
4
15
––
1
3
–
3
5
–
2
5
–
2
3
–
1
3
–
4
15
––
2
5
–
181
Marking scheme
Marking scheme
SECTION I (50 marks)
1. a =
s
2
+ q
p
2
a
2
=
s
2
+ q
p
2
a
2
p
2
= s
2
+ q
s
2
= a
2
p
2
– q
s = ±√a
2
p
2
– q
2. ∠ABO = ∠BAO = 30° (base angles of isosceles ∆ABO)
∠AOB = 180° – (30° + 30°) = 120° (angles of a triangle)
∠BOC = ∠ABO = 30° (alternate angles)
∠AOC = ∠AOB + ∠BOC = 150°
Hence ∠ACO = (180° – 150°) = 15° (base angle of isosceles ∆AOC)
3. The matrix has no inverse if its determinant is zero.
i.e x
3
× 1 – 1 × x = 0
x
3
– x = 0
x(x
2
– 1) = 0
x(x – 1)(x + 1) = 0
The values of x for this are x =
–
1, x = 0 or x = 1.
4. Given:
2x – y
=
4
x – 2y
3
3(2x – y) = 4(x – 2y)
6x – 3y = 4x – 8y
8y – 3y = 4x – 6x (Putting like terms together.)
5y =
–
2x
y
=
–
2
x 5
B 1 for squaring both
sides to remove sq.
root.
M 1 for multiplying
both sides by p
2
.
A 1 C.A.O
3
√
B 1
B 1
3
B 1
B 1 for nding det and
equating to zero.
M 1 for factorisation.
A 1 for all three ans.
3
x
3
x
1 1
( )
M 1 for cross-
multiplication.
A 1
2
1
2
–
PAPER 2
Time: 2 hours
1
2
–
182
Marking scheme
B 1 for writing 2 in log
form.
M 1
A 1
3
( )
5. log
10
5 – 2 + log
10
(2x + 10) = log
10
(x – 4)
log
10
5 – log
10
100 + log
10
(2x + 10) = log
10
(x – 4)
log
10
5(2x + 10)
= log
10
(x – 4)
(Writing LHS as a
100 single log.)
5(2x + 10)
= x – 4
100
10x + 50 = 100x – 400
⇒ 90x = 450
⇒ x = 5
6. M =
Det M = 0 × 1 – 3 × 1 =
–
3
Area of ∆A′B′C′ = |
–
3| × area of ∆ABC
∴area of ∆ABC = × area of ∆ A′B′C′
= × 36
= 12 cm
2
7. (a) The tree diagram is as in Fig. SMEPS 1.9.
B — Boy
G — Girl
Fig. SMEPS 1.9
P(all ocials are boys) = × × = .
( )
0 1
3 1
1
3
–
1
3
–
B 1
M 1
A 1
3
B 2 for tree and all
numerical values 
B 1 for ≥ 5 correct.
B 1
B
G
B
G
B
G
B
G
B
G
B
G
6
11
––
7
11
––
4
11
––
5
12
––
7
12
––
5
10
––
5
10
––
5
11
––
4
10
––
6
10
––
4
10
––
6
10
––
3
10
––
7
10
––
B
G
7
12
––
7
44
––
5
10
––
6
11
––
183
Marking scheme
B 1
4
B 1 for correctly
locating the centre.
B 1 for complete locus
only.
2
M 1 for numerical
expression of area.
M 1 for making AC the
subject.
A 1
3
(b) From the tree diagram in (a);
P(one ocial is a girl)
= P(BBG or BGB or GBB)
= × × + × × + × ×
= 3 × = .
8.
Comment
∠APB = 90°.
It is known that the angle subtended by a diameter of a circle anywhere
on the circumference is a right angle.
∴AB must be a diameter of a circle.
To draw the circle, we locate the centre half-way between A and B.
Fig. SMEPS 1.10 shows the locus.
Fig. SMEPS 1.10
9. Fig. SMEPS 1.11 is a sketch of the triangle.
Area of the triangle = 10 cm
2
∴10 = × 5 × AC sin 30°
i.e. 10 = × 5 × AC ×
AC =
2 × 10
5 ×
= 8 cm.
7
12
––
7
12
––
5
10
––
6
11
––
5
11
––
7
11
––
5
12
––
6
10
––
6
10
––
7
44
––
21
44
––
– – – – – – – – – – – – – – – –
A B
Locus of P
•
1
2
–
1
2
–
1
2
–
1
2
–
Fig. SMEPS 1.11
A
C
B
30°
5 cm
184
Marking scheme
10. A
2
= =
Let B =
Then A + B = + =
Now =
∴a = 5, b = 5, c = 15 and d = 15
∴ B =
11. Area of sector POQ
=
144°
× × 3.5 × 3.5 = 15.4 cm
2
360°
Area of ∆POQ
= × 3.5 × 3.5 × sin 144°
= × 3.5 × 3.5 × 0.587 8
= 3.60 cm
2
Area of shaded segment = 15.4 – 3.6 = 11.8 cm
2
12. By using Pascals triangle,
(a + b)
5
= a
5
+ 5a
4
b + 10a
3
b
2
+ 10a
2
b
3
+ 5ab
4
+ b
5
Now, 1.002
5
= (1 + 0.002)
5
Thus, 1.002
5
= 1
5
+ 5 × 1
4
(0.002) + 10 × 1
3
(0.002)
2
+ ...
= 1 + 0.01 + 0.000 04 + ...
= 1.010 0 (4 d.p)
13. y = x
2
– 3x + 3
⇒
dy
= 2x – 3
dx
Since
dy
= gradient,
dx
⇒ 2 x – 3 =
–
7
x =
–
2
( )
2 1
3 4
( )
7 6
18 19
( )
2 1
3 4
( )
a b
c d
( )
2 1
3 4
( )
a b
c d
( )
2 + a 1 + b
3 + c 4 + d
( )
7 6
18 19
( )
2 + a 1 + b
3 + c 4 + d
( )
5 5
15 15
B 1 for A
2
.
M 1
B 1 for matrix B.
4
A 1
3
M 1 for numerical
expression of area
of sector POQ.
M 1 for numerical
expression of area
of ∆POQ.
A 1
3
B 1
M 1 for substitution in
1st 3 terms only.
B 1 for 
dierentiation.
B 1
2
A 1 for all values.
1
2
–
22
7
––
1
2
–
185
Marking scheme
B 1
B 1 only for simplest
form.
3
B 1 for correct AB.
M 1 for substitution.
A 1
( )
M 1 for substitution.
A 1
4
A 1
M 1 for substitution.
1
2
–
1
2
–
1
4
–
1
4
–
1
4
–
1
4
–
3
4
–
1
5
–
14. From the fact that m α hr
2
, where m = mass, h = height and r = radius,
we get:
m = khr
2
, where k is a constant of proportionality.
∴9.6 = k × 25 × 8
2
⇒ k =
9.6
25 × 8 × 8
= 0.006
For the other cylinder,
m = 0.006 × 35 × 6
2
= 7.56 kg.
15. Since AQ = QB, then AQ = AB
Since AP = PQ, then AP = AQ = AB
∴OP = OA + AB
But AB = AO + OB = b – a
Thus, OP = a + (b – a)
= a + b
16.
Comment
The production pattern takes the form of a GP with
1st term, a = 18 000 and
common ratio, r =
80
=
.
100
Required is the sum of a GP, given by
S
n
= a
1 – r
n
, where n = number of terms.
1 – r
Note that |r| < 1, hence the choice of this formula.
(a) No. of biscuits produced in the rst two hours is
S
2
= 18 000
=
18 000 ×
= 32 400
9
25
––
4
5
–
4
5
–
( )
1 – ( )
2
1 –
4
5
–
186
Marking scheme
Alternatively:
No. of biscuits produced in the rst hour = 18 000
No. of biscuits in the second hour =
80
× 18 000
100
∴ No of biscuits produced in the rst two hours
= 18 000 +
80
× 18 000
100
= 32 400
(b) No. of biscuits produced in four hours is
S
4
=
18 000
[
1 –
( )
4
]
1 –
=
18 000
(
1 –
)
=
18 000 × 369
× 5
625
= 53 136
Note that the step-by-step method could be used for this part also.
SECTION II (50 marks)
17. Using the method of the working mean,
mean, x = A +
∑ f(x – A)
, where A is the working mean, and
∑f
s.d =
∑ fd
2
– ∑ fd
2
, where d = x – A.
∑ f ∑ f
The working may be laid out as in Table SMEPS 1.3.
A = 47
Class x f d = x – A f(x – A) (x – A)
2
f(x – A)
2
20 – 24 22 1
–
25
–
25 625 625
25 – 29 27 5
–
20
–
100 400 2 000
30 – 34 32 9
–
15
–
135 225 2 025
35 – 39 37 11
–
10
–
110 100 1 100
40 – 44 42 20
–
5
–
100 25 500
45 – 49 47 20 0 0 0 0
50 – 54 52 19 5 95 25 475
55 – 59 57 8 10 80 100 800
60 – 64 62 4 15 60 225 900
65 – 69 67 3 20 60 400 1 200
∑ f = 100
∑f(x – A) ∑ f(x – A)
2
=
–
175 = 9 625
Table SMEPS 1.3
M 1
A 1
A 1
5
M 1 for manipulation.
M 1 for substitution.
–
√
B 1 for class
mid-values.
B 1 for (x – A) values.
B 1 for attempt to get
f(x – A).
B 1 for ∑f(x – A).
B 1 for attempt to get
f(x – A)
2
.
B 1 for ∑f(x – A)
2
.
( )
4
5
–
4
5
–
256
625
–––
1
5
–
187
Marking scheme
(a) From the 3rd and 5th columns;
mean mass, x = 47 +
–
175
100
= 45.25 kg
(b) From the 3rd and 7th columns;
s.d =
9 625
–
–
175
2
100 100
= √69.25 – 3.0625
= √93.19
= 9.653 kg
One may also calculate the s.d. as
s.d =
∑ f(x – A)
2
– (x – A)
2
∑ f
! Caution: Though correct, the formula s.d =
∑ fx
2
– x
2
may
∑ f
not be used in this question because it ignores the use
of the working mean in the calculation.
18. (a) Fig. SMEPS 1.12 is a
sketch of the roof:
The required angle
is ∠QRU =
θ
.
Now, given ∆ABC
with sides a, b and c as shown in
Fig. SMEPS 1.13, then
a
2
= b
2
+ c
2
– 2bc cos A
or cos A =
b
2
+ c
2
– a
2
2bc
This is the Cosine rule.
Thus, cos
θ
=
10
2
+ 12
2
– 8
2
2 × 10 × 12
=
180
= 0.75
240
∴
θ
= cos
–
1
(0.75) = 41.41° (or 41° 25′)
(b) Cross-section area of the roof
= × 10 × 12 sin 41.41°
= × 10 × 12 × 0.661 4
M 1 for numerical
expression of cos
θ
.
M 1 for substitution.
A 1
M 1 for substitution.
A 1
10
A 1
M 1 for numerical
expression of
cross-section area.
B 1 for identifying
∠QRU.
–
√
√
( )
√
1
2
–
1
2
–
C
a
B
A
b
c
Fig. SMEPS 1.13
– – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – –
Fig. SMEPS 1.12
R
T
U
Q
8 m
S
P
12 m
24 m
10 m
θ
188
Marking scheme
A 1
10
M 1 for expression for
the volume.
A 1
M 1 for expression of
tan α
B 1 for identifying the
angle α.
A 1
10
B 1 for expression of
v
B 1
M 1
A 1
M 1
B 1
A 1
∴Space occupied by the roof (i.e volume)
= × 10 × 12 sin 41.41° × 24
= 60 × 0.661 4 × 24
= 952.4 m
3
.
(c) The required angle is ∠TBA = α (Fig. SMEPS 1.14)
Now, TA = UB = 10 sin
θ
since sin
θ
=
UB
UR
Also, tan α =
TA
AB
Thus, tan α =
10 sin 41.41°
24
= 0.275 6
∴ α = tan
–
1
(0.275 6)
= 15.41° (or 15° 25′)
19. (a) Velocity is given by v =
ds .
dt
Since s = 8t – t
2
,
v =
ds
= 8 – 2t
dt
(i) At t = 1, v = 8 – 2 = 6 m/s
(ii) At t = 3, v = 8 – 6 = 2 m/s
(b) At maximum displacement,
ds
= 0
dt
Thus 8 – 2t = 0
⇒ t = 4 s
s = 8t – t
2
∴maximum displacement
= 8 × 4 – 4
2
= 32 – 16
= 16 m.
(c) Acceleration
dv
=
–
2 m/s
2
; it is constant throughout.
dt
(d) At the starting point, displacement is zero.
∴ 8t – t
2
= 0
⇒ t(8 – t) = 0
⇒ t = 0 or t = 8
i.e. the body will be back at starting point after 8 seconds.
A 1
M 1
B 1
1
2
–
( )
– – – – – – – –
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – –
Fig. SMEPS 1.14
R
T
U
Q
α
S
P
B
A
24 m
189
Marking scheme
M 1 for 115% initial
investment.
A 1
A 1
10
M 1 for 115% adjusted
investment
M 1 for expression for
%age interest.
B 1 for ∠BAX = 60°.
B 2 for either ∠XAZ
bisected or
∠BAY = 90°.
B 1 for ∠XAC = 15°.
B 1 for location of C.
B 1 for completion of
∆ABC.
B 2 for locating 
centre of ∆ABC.
B 1 for circle.
B 1 for radius stated.
10
20. (a) Let the amounts at the end of the 1st yr; 2nd yr, etc. be A
1
, A
2
, …
respectively.
Each year, the amount increases by 15% to
115
(i.e. 1.15) of what
it was at the beginning of the year. Thus,
100
A
1
= 100 000 × 1.15
= sh 115 000
A
2
= (115 000 + 5 000) × 1.15
= sh 120 000 × 1.15 = sh 138 000
A
3
= (138 000 + 5 000) × 1.15
= sh 143 000 × 1.15 = sh 164 450
A
4
= (164 450 + 5 000) × 1.15
= sh 169 450 × 1.15 = sh 194 867.50
A
5
= (194 867.50 + 5 000) × 1.15
= sh 199 867.50 × 1.15
= sh 229 847.63
A
5
is the amount of money after 5 years.
Note: Figures must be consistently rounded o to 2 d.p where
necessary.
(b) Total investment = sh (100 000 + 4 × 5 000)
= sh 120 000
Total amount after ve years
= sh 229 847.63
Percentage interest =
229 847.63 – 120 000
× 100%
120 000
=
109 847.63
× 100%
120 000
= 91.54 %
21. (a) The construction is as in
Fig. SMEPS 1.15.
(b) Radius = 4.8 cm.
Fig. SMEPS 1.15
A 1
B 1
M 1 for exp. of A
3
.
A 2 for all three ans.
(Give A 1 for any
two ).
30°
15°
60°
8 cm
A
7 cm
C
X
Y
B
Z
}
190
Marking scheme
B 1
B 1 for line and
shading.
B 1 for lines.
B 1 for region.
R
}
}
B 2 for all  (Give B 1
for any two ).
240
220
200
180
160
140
120
100
80
60
40
20
0
20 40 60 80 100 120 140 160 180 200
y
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
x = 100
y = 80
3
x
+ 2
y
= 600
1
2
–
x
22. (a) Let the number of garments of type A made be x and those of
type B be y.
From material usage, we get 3x + 2 y ≤ 600
From numbers that must be made daily, we have
x ≤ 100
y ≥ 80
x ≥ 0
(b) The lines drawn are
3x + 2 y = 600
x = 100
y
= 80
x
= 0
See Fig. SMEPS 1.16.
Fig. SMEPS 1.16
1
2
–
1
2
–
191
Marking scheme
}
B 1 for statement of
objective function.
B 2 for stated nos. of
garments.
10
M 1
B 1
B 1
A 1
B 1
A 1
M 1
A 1
B 1
B 2
10
(c) The objective function is
P = 80x + 60y, where P = total prot.
Either draw a searchline by choosing an appropriate value of P,
e.g. 12000 ⇒ 80x + 60y = 12 000
or inspect for maximum prot at the three vertices of the region
that are furthest from the origin.
For maximum prot, make
100 garments of type A
and 120 garments of type B
23. (a) TP = TQ (both are tangents)
∴a =
(180° – 64°) (base angles of ∆TPQ)
= × 116°
= 58°
PR//TQ (given)
∴b = a = 58° (alternate angles)
∠PRQ = 58° (angle in alternate segment to a)
∴c = 180° – (b + ∠PRQ)
= 180° – (58° + 58°)
= 64°
b is in alternate segment to d
∴d = b = 58°.
Alternatively
d and ∠PRQ are alternate angles
∴d = ∠PRQ = 58°
or
a + c + d = 180° (angles on a straight line)
∴ d = 180° – (58° + 64°)
= 58°
(b) ∠ADC = 90° (angle on a diameter)
∴∠ACD = 180° – (90° + 25°)
= 65°
AOC is a diameter; so AC⊥BD, AC bisects BD as well as ∠BCD.
Hence ∠BCD = 2 × 65°
= 130°
M 1
B 1 for use of
searchline or
inspection.
}
1
2
–
1
2
–
192
Marking scheme
M 1 for premultiplying
object matrice with
operator matrix and
equating to image
matrix.
M 1 for multiplying
matrices in correct
order.
A 1
M 1
B 1
24. (a) represents a shear parallel to the x-axis (the invariant line).
Since (1 , 2) → (7 , 2), we have
=
=
⇒ k = 3
Thus, denes the shear S.
The matrix which denes reection in the line y = x is R = .
Hence the matrix that denes the shear followed by the reection is
T = RS =
=
(b) Let P =
.
Then =
Thus
=
.
So a + 3b = 2
and 2a + 3b = 3
which when solved simultaneously give a = 1, b = .
Also c + 3d = 4
and 2c + 3d = 11
which when solved simultaneously give c = 7, d =
–
1.
Hence, P =
.
M 1 for attempt to
solve simultaneous
equations.
A 1 for all values a, b,
c, d.
B 1 for stating the matrix
provided M marks
scored.
10
M 1
A 1
1 k
0 1
( )
1 k
0 1
( )( )
1
2
( )
1 + 2k
2
( )
7
2
( )
0 1
1 0
( )
0 1
1 0
( )
1 3
0 1
( )
0 1
1 3
( )
a b
c d
( )
a b
c d
( )
1
–
2
3
–
3
2
–
3
4
–
11
( )
a + 3b
–
2a – 3b
c + 3b
–
2c – 3d
( )
2
–
3
4
–
11
( )
( )
1 3
0 1
( )
1
7
–
1
1
3
–
1
3
–
193
Formulae
1. Plane shapes
Shape Perimeter Area
Square (side s) 4 s s
2
Rectangle (length l, breadth b) 2(l + b) lb
Triangle (base b, height h) Sum of sides bh
Parallelogram (base b, height h) Sum of sides bh
Trapezium (height h, parallels a and b) Sum of sides (a + b)h
Circle (radius r) 2πr πr
2
Sector of a circle (radius r, angle
θ
) 2r + 2πr πr
2
2. Solid shapes
Shape Surface area Volume
Cube (edge s) 6 s
2
s
3
Cuboid (length l, breadth b, height h) 2(lb + bh + lh)
lbh
Prism Sum of areas of faces Ah
Cylinder (radius r, height h) 2 πrh + 2πr
2
π
r
2
h
Cone (radius r, slant height l, height h) πrl + πr
2
πr
2
h
Sphere (radius r) 4 πr
2
πr
3
Pyramid (base area A, height h) Sum of areas of faces Ah
MATHEMATICAL FORMULAE AND TABLES
1
2
–
1
2
–
1
3
–
θ
360°
––––
4
3
–
1
3
–
θ
360°
––––
194
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) Sum of an arithmetic progression
The sum to the nth term 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
=
a(1 – r
n
)
if r < 1, where a is the rst
1 – r
term and r is the common ratio
or
a(r
n
– 1)
if r > 1.
r – 1
8. Statistics
(a) Mean
Mean is given by: x =
∑
fx
N
For grouped data:
If N = ∑
f
and d =
x – a
where a is an assumed
c
or working mean and c is the class interval
or other appropriate factor, then
x = a + c
∑
fd
.
d
(b) Median
Given L = lower class boundary of the
median class,
N = total frequency of the
distribution,
(∑
f )
L
= total of all the frequencies
below the median class,
A
c
C
B
a
b
3. Right-angled triangles
In a triangle ABC right-angled at C,
c
2
= a
2
+ b
2
(Pythagoras’ theorem)
tan A =
a
tan B =
b
b a
sin A =
a
sin B =
b
c c
cos A =
b
cos B =
a
c c
tan A =
sin A
and tan B =
sin B
cos A cos B
sin A = cos B sin B = cos A
4. Any triangle ABC
In any triangle ABC, with the usual notation
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
Area = √{s(s – a)(s – b)(s – c)},
where s = ( a + b + c)
or Area = bc sin A = ac sin B = ab sin C
c
B
a
A
b
C
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
–
b±√b
2
– 4ac
2a
195
f
m
= frequency of the median
class, and
c = width of the median class,
then;
median = L +
– (∑
f )
L
c
f
m
(c) Variance
The variance is given by:
v =
∑
f(x – x)
2
or v =
∑
fx
2
– x
2
N
N
N + 1
2
––––
–
√
–
{ }
(
)
or v = c
2
∑
fd
2
–
∑
fd
2
N
N
(d) Standard deviation
The standard deviation is given by:
s = √ variance =
∑
f(x – x
2
)
∑f
–
196
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 2 3 4 5 6 7 8 9
197
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 2 3 4 5 6 7 8 9
198
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
199
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
200
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
201
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
202
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 0.0 875 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 0.1 405 1423 1441 1459 1477 1495 1512 1530 1548 1566 3 6 9 12 15
9 0.1 584 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
203
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
204
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
– – – – – – – – –
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
205
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
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
206
LOGARITHMS 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
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
207
LOGARITHMS 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
–
–
–
–
–
– – – – – – – – –
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
208
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
– – – – – – – – –
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
209
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
210
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
211
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
212
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
213
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
214
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
215
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
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
216
1
x
RECIPROCALS 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
217
RECIPROCALS 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
218