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Chapter 21

Introduction to trigonometry
21
...

The theorem of Pythagoras and trigonometric ratios
are used with right-angled triangles only
...

In this chapter, three trigonometric ratios – i
...
sine,
cosine and tangent – are defined and then evaluated
using a calculator
...


21
...


From equation (1):
b = a2 + c2
Transposing√equation (1) for a gives a 2 = b2 − c2 , from
which a = b2 − c 2
2
2
2
Transposing
√equation (1) for c gives c = b − a , from
2
2
which c = b − a
Here are some worked problems to demonstrate the
theorem of Pythagoras
...
In Figure 21
...

In the right-angled triangle ABC shown in Figure 21
...
1
DOI: 10
...
00021-1

a

c 5 3 cm

B

Figure 21
...
e
...


A
c

A

a

C

Hence,
√
25 = ±5 but in a practical example like this an answer
of a = −5 cm has no meaning, so we take only the
positive answer
...


182 Basic Engineering Mathematics
PQR is a 3, 4, 5 triangle
...
e
...

Problem 2
...
3, find the length of EF

From Pythagoras’ theorem,
BC 2 = 12002 + 8802
= 1440000 + 774400 = 2214400
√
BC = 2214400 = 1488 km
...


D
e5 13 cm

f 5 5 cm
E

Now try the following Practice Exercise

F

d

Figure 21
...


Find the length of side x in Figure 21
...


2

41 cm

169 = d 2 + 25

x

d 2 = 169 − 25 = 144
√
d = 144 = 12 cm

Thus,

40 cm

d = EF = 12 cm

i
...


DEF is a 5, 12, 13 triangle, another right-angled
triangle which has integer values for all three sides
...
5

2
...
6(a)
...


Find the length of side x in Figure 21
...


Problem 3
...
One travels due north at an average
speed of 300 km/h and the other due west at an
average speed of 220 km/h
...
7 mm

as shown in Figure 21
...
The distance apart after
4 hours = BC
...
6

E
S

C

Figure 21
...
3 mm
(b)

B

1200 km

880 km

A

4
...
Determine the length of
AC, correct to 2 decimal places
...


A tent peg is 4
...
0 m high
tent
...


In a triangle ABC, ∠B is a right angle,
AB = 6
...
78 cm
...


14
...
8 shows a cross-section of a component that is to be made from a round bar
...

x

90◦ ,

7
...
83 mm and CE = 28
...

Determine the length of DE
...


Show that if a triangle has sides of 8, 15 and
17 cm it is right-angled
...


183

Triangle PQR is isosceles, Q being a right
angle
...
46 cm find (a)
the lengths of sides PQ and QR and (b) the
value of ∠QPR
...
A man cycles 24 km due south and then 20 km
due east
...
Find the distance
between the two men
...
A ladder 3
...
0 m from the
wall
...
8

21
...
9,
opposite side
hypotenuse

sine θ =

‘Sine’ is abbreviated to ‘sin’, thus sin θ =

BC
AC
C

12
...
One
travels due west at 18
...
6 knots
...

13
...
7 shows a bolt rounded off at one
end
...


m

4m

␾7

se

nu

te
po

Hy

Opposite

␪
A

Adjacent

B

Figure 21
...
Remembering these three equations
is very important and the mnemonic ‘SOH CAH TOA’
is one way of remembering them
...
7

184 Basic Engineering Mathematics
SOH indicates sin = opposite ÷ hypotenuse

sin C =

CAH indicates cos = adjacent ÷ hypotenuse
TOA indicates tan = opposite ÷ adjacent

cos C =

Here are some worked problems to help familiarize
ourselves with trigonometric ratios
...
In triangle PQR shown in
Figure 21
...
10

sin θ =

opposite side
PQ
5
=
=
= 0
...
9231
hypotenuse
PR
13

tanθ =

opposite side
PQ
5
=
=
= 0
...
In triangle ABC of Figure 21
...
47
= 0
...
778
4
...
7996
5
...
47
=
= 0
...
62
4
...
7996
5
...
47
=
= 0
...
778
4
...
3314
3
...
If tan B = , determine the value of
15
sin B, cos B, sin A and tan A
A right-angled triangle ABC is shown in Figure 21
...

8
If tan B = , then AC = 8 and BC = 15
...
12

i
...

from which

AB 2 = 82 + 152

AB = 82 + 152 = 17
AC
AB
BC
cos B =
AB
BC
sinA =
AB
BC
tanA =
AC
sin B =

3
...
62 cm

C

Figure 21
...
e
...
472 + 4
...
472 + 4
...
778 cm

from which

AB
AC
BC
AC
AB
BC
BC
AC
AB
AC
BC
AB

By Pythagoras, AB 2 = AC 2 + BC 2

A

B

opposite side
=
hypotenuse
adjacent side
=
hypotenuse
opposite side
=
adjacent side
opposite side
=
hypotenuse
adjacent side
=
hypotenuse
opposite side
=
adjacent side

8
17
15
=
17
15
=
17
15
=
8
=

or 0
...
8824
or 0
...
8750

Problem 7
...
Determine (a) the distance AB and
(b) the gradient of the straight line AB

Introduction to trigonometry
f (x)
8
7
6

f(x)
8
7
B
6

4
3
2

4
3
2

0

A

2

4
(a)

6

B

␪

0

8

C

A

2

4
(b)

6

15
, find sin X and cos X , in frac4
...

5
...
15, find (a) sin α (b) cos θ (c) tan θ
...
13

185

␣

17
␪
15

Figure 21
...
13(a)
...
13(b), the horizontal and vertical
lines AC and BC are constructed
...
211 correct to 3
decimal places
...
e
...
If tan θ = , find sin θ and cos θ in fraction
24
form
...
Point P lies at co-ordinate (−3, 1) and point
Q at (5, −4)
...

(b) the gradient of the straight line PQ
...
4 Evaluating trigonometric ratios
of acute angles
The easiest way to evaluate trigonometric ratios of any
angle is to use a calculator
...

sin 29◦ = 0
...
62◦ = 0
...
3907 cos 83
...
1120

Practice Exercise 83 Trigonometric ratios
(answers on page 349)
1
...

Determine sin Z , cos Z , tan X and cos X
...
In triangle ABC shown in Figure 21
...

B
5
A

3
C

Figure 21
...
If cos A = , find sin A and tan A, in fraction
17
form
...
6745 tan 67
...
4541
43◦
= sin 67
...
◦ = 0
...
466666
...
9725
60
54◦
tan 56◦54 = tan 56
= tan 56
...
5340
60
sin 67◦43 = sin 67

If we know the value of a trigonometric ratio and need to
find the angle we use the inverse function on our calculators
...
5 then
the value of the angle is given by
sin−1 0
...
5)
Similarly, if
cos θ = 0
...
4371 = 64
...
5984 then A = tan−1 3
...
47 ◦
each correct to 2 decimal places
...

Problem 8
...
65◦ = 0
...
Press sin

2
...
Press◦ ”’

4
...
Press◦ ”’

6
...
Press =

Answer = 0
...


Problem 9
...
20◦ = 2
...
Evaluate tan 2
...
93 means 2
...

Hence, tan 2
...
2148
It is important to know when to have your calculator on
either degrees mode or radian mode
...

Problem 13
...
4128 in
degrees, correct to 2 decimal places
sin−1 0
...
4128’
...
Press shift

2
...
Press )

5
...
Enter 0
...
380848
...

Hence, sin−1 0
...
38◦ correct to 2 decimal
places
...
Find the acute angle cos−1 0
...
2437 means ‘the angle whose cosine is 0
...

Using a calculator,

1
...
Press shift

2
...
Press )

5
...


Press◦

7
...
Press cos

3
...
Enter 12

6
...
Press =

Answer = 2
...


Problem 10
...
481, correct to 4
significant figures
sin 1
...
481 radians
...
e
...
Therefore
a calculator needs to be on the radian function
...
481 = 0
...
Evaluate cos(3π/5), correct to 4
significant figures
As in Problem 10, 3π/5 is in radians
...
884955
...
3090
Since, from page 166, πradians = 180◦ ,
3
3π/5 rad = × 180◦ = 108◦
...
e
...
Check with your calculator that
cos 108◦ = −0
...
Enter 0
...
894979
...

6
...
93 is displayed
...
2437 = 75
...

Problem 15
...
4523 in
degrees and minutes
tan−1 7
...
4523’
...
Press shift

2
...
Press )

5
...
Enter 7
...
357318
...

6
...
35 is displayed
...
4523 = 82
...


Introduction to trigonometry
Problem 16
...
16,
calculate angle G
E
2
...
71

cm

F

2
...


3
...


4
...
16

With reference to ∠G, the two sides of the triangle given
are the opposite side EF and the hypotenuse EG; hence,
sine is used,
2
...
e
...
26406429
...
71
G = sin−1 0
...

from which,

5
...
6734 in degrees,
correct to 2 decimal places
...


Find the acute angle cos−1 0
...


7
...
4385 in degrees,
correct to 2 decimal places
...


Find the acute angle sin−1 0
...


9
...
8539 in degrees
and minutes
...
311360◦

i
...


∠G = 15
...


Hence,

Problem 17
...
2 tan 49◦26 − 3
...
1 cos 29◦34
By calculator:



26 ◦
= 1
...
9137 and cos 29◦ 34 = 0
...
681 (c) tan 3
...
Find the acute angle tan−1 0
...

11
...
17, determine angle θ, correct to 2 decimal places
...
2 tan 49◦ 26 − 3
...
1 cos29◦ 34
=

(4
...
1681) − (3
...
9137)
(7
...
8698)

4
...
3807 1
...
1756
6
...
2470 = 0
...


␪
9

Figure 21
...
In the triangle shown in Figure 21
...


=

␪

23

Now try the following Practice Exercise
Practice Exercise 84 Evaluating
trigonometric ratios (answers on page 349)
1
...


8

Figure 21
...
Evaluate, correct to 4 decimal places,
4
...
Evaluate, correct to 4 significant figures,
(3 sin 37
...
5 tan 57
...
1 cos12
...
5

Solving right-angled triangles

‘Solving a right-angled triangle’ means ‘finding the
unknown sides and angles’
...
2
=
, from which
AB
AB
6
...
343 mm
AB =
cos 42◦

Pythagoras, AB 2 = AC 2 + BC 2
√
√
from which AB = AC 2 + BC 2 = 5
...
22
√
= 69
...
343 mm
...
Sketch a right-angled triangle ABC
such that B = 90◦ , AB = 5 cm and BC = 12 cm
...
20
...


5 cm

Six pieces of information describe a triangle completely;
i
...
, three sides and three angles
...

Here are some worked problems to demonstrate the
solution of right-angled triangles
...
In the triangle ABC shown in
Figure 21
...
20

By Pythagoras’ theorem,
By definition: sin A =

AC =



52 + 122 = 13

opposite side 12
=
or 0
...
9231
hypotenuse
13

tan A =

opposite side 12
=
or 2
...
In triangle PQR shown in
Figure 21
...
2 mm

P

C

Figure 21
...


388
Q

In triangle ABC,
tan 42◦ =

AC
AC
=
BC
6
...
2 tan 42◦ = 5
...
5 cm

R

Figure 21
...
5

PQ = 7
...
5(0
...
860 cm

Introduction to trigonometry
cos 38◦ =

7
...
5
7
...
518 cm
◦
cos 38
0
...
5)2 + (5
...
59 = (9
...

Problem 21
...
22
A

35 mm

B

37 mm

XZ
, hence XZ = 20
...
0
= 20
...
3953) = 7
...
0 cos 23◦17
20
...
0(0
...
37 mm
Check: using Pythagoras’ theorem,
(18
...
906)2 = 400
...
0)2 ,
cos 23◦17 =

Now try the following Practice Exercise
Practice Exercise 85 Solving right-angled
triangles (answers on page 349)

C

Figure 21
...

sin C =

sin 23◦17 =

189

1
...
24(a) to (f), each correct to 4
significant figures
...
94595, hence
37
C = sin−1 0
...
08◦

708
13
...
08◦ = 18
...
92◦ = 37(0
...
0 mm

sin B =

(a)
15
...
0 mm
...
Solve triangle XYZ given
∠X = 90◦, ∠Y = 23◦17 and YZ = 20
...
Such a sketch is shown in
Figure 21
...

17

...
0 mm
(c)

238179
X

Figure 21
...
24

190 Basic Engineering Mathematics
E

4
...
0

x

D
8
...
0

J

(c)

L

6
...


438

(e)

278

K

M
(d)

7
...
0

538
P

(f)

Figure 21
...
Find the unknown sides and angles in the rightangled triangles shown in Figure 21
...
The
dimensions shown are in centimetres
...
0
5
...
0
(a)

Figure 21
...
25

Introduction to trigonometry

3
...
If the foot of the ladder is
2 m from the wall, calculate the height of the
building
...
Calculate the height of the pylon to the
nearest metre
Figure 21
...

A

4
...
26
...
29

tan 23◦ =

x

AB
AB
=
BC
80

Hence, height of pylon AB = 80 tan 23◦
= 80(0
...
96 m
= 34 m to the nearest metre
...
26

21
...
27, BC represents horizontal ground and
AB a vertical flagpole, the angle of elevation of the top
of the flagpole, A, from the point C is the angle that the
imaginary straight line AC must be raised (or elevated)
from the horizontal CB; i
...
, angle θ
...
A surveyor measures the angle of
elevation of the top of a perpendicular building as
19◦
...
Determine
the height of the building
The building PQ and the angles of elevation are shown
in Figure 21
...

P

A

h
478

␪

C

B

Q

R

198

S

120

x

Figure 21
...
30
P

␾

Hence,
Q

h
x + 120
h = tan 19◦(x + 120)

In triangle PQS, tan 19◦ =
R

i
...
h = 0
...
28

If, in Figure 21
...
e
...
(Note, ∠PRQ is
also φ − alternate angles between parallel lines
...
An electricity pylon stands on
horizontal ground
...
e
...
0724x

(1)

In triangle PQR, tan 47◦ =
Hence,

Equating equations (1) and (2) gives
0
...
0724x
0
...
3443)(120) = 1
...
3443)(120) = (1
...
3443)x
41
...
7281x
41
...
74 m
0
...
0724x = 1
...
74) = 60
...

Problem 25
...
Find the distance of the ship
from the base of the cliff at this instant
...
Determine the speed of the ship in
km/h
Figure 21
...
Since the angle
of depression is initially 30◦, ∠ACB = 30◦ (alternate
angles between parallel lines)
...
31

75
AB
=
hence,
BC
BC
75
BC =
tan 30◦
= 129
...
A vertical tower stands on level ground
...
Find the
height of the tower
...
If the angle of elevation of the top of a vertical
30 m high aerial is 32◦ , how far is it to the
aerial?
3
...
0 m high
the angle of depression of a boat is 19◦ 50
...

4
...
0 m high the
angles of depression of two buoys lying due
west of the cliff are 23◦ and 15◦, respectively
...
9 + x

5
...
He moves 50 m nearer
to the flagpole and measures the angle of elevation as 26◦22
...

6
...
At a point 200 m from the building
the angles of elevation of the top and bottom of the pole are 32◦ and 30◦ respectively
...

7
...
If the lighthouse is
25
...


Thus, the ship sails 76
...
e
...
From a window 4
...
Determine,
correct to the nearest centimetre, the width of
the road and the height of the building
...
16
=
m/s
Hence, speed of ship =
time
60
76
...
57 km/h
...
The elevation of a tower from two points, one
due west of the tower and the other due east
of it are 20◦ and 24◦ , respectively, and the two
points of observation are 300 m apart
...


Hence,
from which

129
...
06 m
tan 20◦

x = 206
...
9 = 76
...
The marks available are shown in brackets at
the end of each question
...


Determine 38◦48 + 17◦ 23
...


Determine 47◦43 12 − 58◦ 35 53 + 26◦17 29
...
683◦

(2)

3
...


Convert 77◦ 42 34 to degrees correct to 3 decimal
places
...


Determine angle θ in Figure RT8
...


to degrees and minutes
...


In Figure RT8
...

(3)
A

B

(2)

(3)

C
928

538

E

478

␪

1248

D

298

Figure RT8
...


Find angle J in Figure RT8
...


Figure RT8
...


(2)
H

Determine angle θ in the triangle in Figure RT8
...

(2)

G

288

␪

J

Figure RT8
...
Determine angle θ in Figure RT8
...


␪

508

Figure RT8
...


(3)

328

Determine angle θ in the triangle in Figure RT8
...

(2)
438
␪
␪

688

Figure RT8
...
6

11
...

(2)

194 Basic Engineering Mathematics
K

12
...
7, determine angles x, y and z
...
60

y

598

378
z
J

x

5
...
10

Figure RT8
...
In Figure RT8
...

c

e

b

a

(5)

d

1258

608

16
...
Ship X
travels due west at 30 km/h and ship Y travels due
north
...

Calculate the velocity of ship Y
...
If sin A = , find tan A in fraction form
...
Evaluate 5 tan 62 11 correct to 3 significant
figures
...
Determine the acute angle cos 0
...

(2)
20
...
11, find angle P
in decimal form, correct to 2 decimal places
...
2

29
...
8

14
...
9, determine the length of AC
...
11

B

21
...
27◦ − 5 cos 7
...
81◦
...
In triangle ABC in Figure RT8
...

(4)

10 m

A

D

A
3m
C
8m

408 359

E

Figure RT8
...
In triangle J K L in Figure RT8
...

(b) sin L and tan K , each correct to 3 decimal
places
...
12

23
...
How far is point P
from the base of the pylon, correct to the nearest
metre?
(3)

List of formulae
Laws of indices:

Areas of plane figures:

a m × a n = a m+n
a m/n =

√
n m
a

am
an

= a m−n (a m )n = a mn

a −n =

Quadratic formula:
If ax 2 + bx + c = 0

1
an

Area = l × b

(i) Rectangle

a0 = 1
b

√
−b ± b2 − 4ac
x=
2a

then

Equation of a straight line:

l

(ii) Parallelogram Area = b × h

y = mx + c

Definition of a logarithm:
If y = a x

then

h

x = loga y

Laws of logarithms:
log(A × B) = log A + log B
 
A
= log A − log B
log
B

b

(iii) Trapezium

log An = n × log A

1
Area = (a + b)h
2
a

Exponential series:
ex = 1 + x +

x2 x3
+
+···
2! 3!

h

(valid for all values of x)
b

Theorem of Pythagoras:
b 2 = a 2 + c2

(iv) Triangle

Area =

1
×b×h
2

A

c

B

b
h

a

C
b

List of formulae
Area = πr 2 Circumference = 2πr

(v) Circle

(iii) Pyramid
If area of base = A and
perpendicular height = h then:

r

s

␪

Volume =

1
× A×h
3

r

2π radians = 360 degrees

Radian measure:

h

For a sector of circle:
θ◦
(2πr) = rθ
360

(θ in rad)

1
θ◦
(πr 2 ) = r 2 θ
360
2

(θ in rad)

s=

arc length,

shaded area =

Equation of a circle, centre at origin, radius r:
x 2 + y2 = r 2
Equation of a circle, centre at (a, b), radius r:

Total surface area = sum of areas of triangles
forming sides + area of base
(iv) Cone
1
Volume = πr2 h
3
Curved Surface area = πrl

(x − a)2 + (y − b)2 = r 2

Total Surface area = πrl + πr2

Volumes and surface areas of regular
solids:
l

(i) Rectangular prism (or cuboid)

h

Volume = l × b × h
Surface area = 2(bh + hl + lb)

r

(v) Sphere

l

h

4
Volume = πr3
3
Surface area = 4πr2

b

(ii) Cylinder
Volume = πr2 h
Total surface area = 2πrh + 2πr2
r

r

h

337

338 Basic Engineering Mathematics
Areas of irregular figures by approximate
methods:
Trapezoidal rule


 
width of 1 first + last
Area ≈
interval
2 ordinate

+ sum of remaining ordinates
Mid-ordinate rule
Area ≈ (width of interval)(sum of mid-ordinates)
Simpson’s rule



1 width of
first + last
Area ≈
ordinate
3 interval




sum of even
sum of remaining
+4
+2
ordinates
odd ordinates

For a general sinusoidal function y = A sin (ωt ± α),
then
A = amplitude
ω = angular velocity = 2π f rad/s
ω
= frequency, f hertz
2π
2π
= periodic time T seconds
ω
α = angle of lead or lag (compared with
y = A sin ωt )

Cartesian and polar co-ordinates:
If co-ordinate (x, y) = (r, θ) then

y
r = x 2 + y 2 and θ = tan−1
x
If co-ordinate (r, θ) = (x, y) then
x = r cosθ and y = r sin θ

Mean or average value of a waveform:
area under curve
length of base
sum of mid-ordinates
=
number of mid-ordinates

mean value, y =

Triangle formulae:
Sine rule:
Cosine rule:

b
c
a
=
=
sin A sin B
sin C
a 2 = b2 + c2 − 2bc cos A

B

b

a

If a = first term and d = common difference, then the
arithmetic progression is: a, a + d, a + 2d,
...

The n’th term is: arn−1
Sum of n terms, Sn =

A
c

Arithmetic progression:

a (1 − r n )
a (r n − 1)
or
(1 − r )
(r − 1)

If − 1 < r < 1, S∞ =

a
(1 − r )

C

Area of any triangle
1
= × base × perpendicular height
2
1
1
1
= ab sin C or
ac sin B or
bc sin A
2
2
2

a +b+c
= [s (s − a) (s − b) (s − c)] where s =
2

Statistics:
Discrete data:

#
mean, x¯ =

x

n
)
(
 #

(x − x¯ )2

standard deviation, σ =
n

List of formulae
Grouped data:

#
fx
mean, x¯ = #
f

)
 #$
%(

f (x − x)
¯ 2

#
standard deviation, σ =
f

Standard integrals
y
axn
cos ax

Standard derivatives
sin ax

y or f(x)

dy
= or f (x)
dx

axn

anxn−1

sin ax

a cos ax

cos ax

−a sin ax

eax

aeax

ln ax

1
x

eax
1
x

&

a

y dx
x n+1
+ c (except when n = −1)
n +1

1
sin ax + c
a
1
− cos ax + c
a
1 ax
e +c
a
ln x + c

339

Answers

Answers to practice exercises
Chapter 1

Chapter 2

Exercise 1 (page 2)
1
...

7
...

13
...

17
...
16 m
3
...
£565
6
...
−36 121
9
...
1487
12
...
−70872
15
...
25 cm
d = 64 mm, A = 136 mm, B = 10 mm

1
...

4
...

7
...


5
...

13
...


Exercise 2 (page 5)
(a) 468 (b) 868
2
...

(a) 259 (b) 56
8
...


2
...

6
...

10
...


(a) 12 (b) 360
(a) 90 (b) 2700
(a) 3 (b) 180
(a) 15 (b) 6300
(a) 14 (b) 53 900

Exercise 4 (page 8)
2
...
68

3
...
5

DOI: 10
...
00040-5

3
...

10
1
14
...
1
21
6
...


16
...


2
6
8
2
...

35
9
11
3
1
7
...

5
13
5
3
2
12
...
3
2
5
12
4
3
1
17
...
13
4
9
(a) £60, P£36, Q£16

7
...

15
...


22
9
8
25
3
16
4
27
17
60

4
...

12
...

20
...
5
9
...
−33
10
...
1
2
12
14
...


71
8
11
15
3
16
51
8
52
17
20

19
...

15
5
...
4
20
...
2880 litres

Exercise 7 (page 14)
1
18
7
6
...
2
1
...
22
11
...

(a) 4 (b) 24
(a) 10 (b) 350
(a) 2 (b) 210
(a) 5 (b) 210
(a) 14 (b) 420 420

2
...

3
...

7
...


1
7
4
11
17
30
43
77
9
1
40

1
...
4

1
9
19
7
...
2
20
2
...
1
8
...
7
3

4
...

15
5
...
0
...
14
...
1
...


13
20

7
40
21
1
141
(b)
(c)
(d)
(e)
25
80
500
11
3
7
...
10
25
200
11
1
7
(b) 4
(c) 14
(d) 15
40
8
20
2
...
4
40
41
10
...
(a)

11
...
625

9
250

3
...


6
125

9
...

4
...

10
...


2
...
11
...
185
...
8307
5
...
1581
6
...
571
5
...
1
...
0
...
068
11
...
5 ×10
12
...
5 ×103
−6
−3
4
...
202
...
18
...
6
...
0
...
11
...
0
...
14
...
1
...
2
...
65
...
0
...
329
...
18
...
43
...
72
...
12
...
−124
...
4
...
0
...
−

9
10

4
...
732
8
...
0
...
0
...
0
...
−0
...
0
...
0
...
5
...
2
...
0
...
0
...
0
...
998 2
...
544
3
...
02 4
...
42
456
...
434
...
626
...
1591
...
444 10
...
62963
11
...
563 12
...
455
13
...
8
...
(a) 24
...
812
(a) 0
...
0064 17
...
4˙ (b) 62
...
4
...
0
...
3
...
13
...
50
...
53
...
36
...
12
...
0
...
46
...
1
...
2
...
2
...
30
...
0
...
219
...
5
...
5
...
52
...
0
...
25
...
591
...
69
...
17
...

4
...

10
...
11927
6
...
0944
10
...
325

2
...
30
...
84
...
10
...
2
...


Exercise 10 (page 19)

1
...

9
...

16
...


2
...
0
...
137
...
19
...
515
...
15
...
52
...
0
...
80
...
295
...
59 cm2
159 m/s
0
...

5
...

11
...
78 mm
0
...
8 m2
281
...

6
...

12
...
5
5
...
5
2
...

4
...

10
...


£589
...
508
...
V = 2
...
5
5
...
81 A 6
...
79 s
E = 3
...
I = 12
...
s = 17
...
184 cm2 11
...
327
(a) 12
...
p
...

(c) 13
...
15 h

342 Basic Engineering Mathematics
Exercise 26 (page 43)

Chapter 5

1
...
£66 3
...
450 g 5
...
56 kg
6
...
00025 (b) 48 MPa 7
...
76 litre

Exercise 21 (page 34)
1
...
32%
2
...
4% 3
...
7% 4
...
4%
5
...
5%
6
...
20
7
...
0125 8
...
6875
9
...
462% 10
...
2% (b) 79
...
(b), (d), (c), (a) 12
...

14
...
A = , B = 50%, C = 0
...
30,
2
17
3
F = , G = 0
...
85, J =
10
20

Exercise 27 (page 45)
1
...
170 fr
3
...
8 mm
4
...
(a) 159
...
5 gallons
6
...
4 MPa
7
...
2 mm 8
...

3
...

7
...

14
...
8 kg 2
...
72 m
(a) 496
...
657 g
(a) 14%
(b) 15
...
49% 11
...
2%
2
...
5
...
73 s 4
...
36% 6
...
76 g
9
...
17% 13
...
3
...
25%
37
...
7%

1
...
5 weeks
2
...
(a) 9
...
12 (c) 0
...
50 minutes
5
...
375 m2 (c) 24 × 103 Pa

Chapter 7
Exercise 29 (page 48)

Exercise 23 (page 38)
1
...

9
...

14
...

16
...
5%
2
...
£310
4
...
£20 000 7
...
45 8
...
25
£39
...
£917
...
£185 000 12
...
2%
A 0
...
9 kg, C 0
...
3 t
20 000 kg (or 20 tonnes)
13
...
5 mm 17
...
27
6
...
128
7
...
100 000
8
...
24
10
...
96
9
...
1 6
...
16 4
...
01 10
...
76
7
...
1000 9
...
36 13
...
34 15
...
25
1
1
1
17
...
49 19
...
5
20
...
128

2
...
36 : 1

2
...
5 : 1 or 7 : 2 3
...
96 cm, 240 cm 5
...
£3680, £1840, £920 7
...
£2172

Exercise 31 (page 52)
1
...
9

Exercise 25 (page 42)
1
...
76 ml
3
...
12
...
14
...
25 000 kg

147
148
17
13
...


2
...
±3
10
...
64

19
56

32
25
1
7
...


11
...
4

1
2

4
...
±
13
1
12
...

3
...

7
...

11
...

15
...

19
...

23
...

27
...

4
...

8
...

12
...

16
...

20
...

24
...

28
...

y
20
...
(x − y)(a + b)
1
...

5
...

9
...

13
...


2x(y − 4z)
2x(1 + 2y)
4x(1 + 2x)
x(1 + 3x + 5x 2 )
r(s + p + t )


2 p q 2 2 p2 − 5q
2x y(y + 3x + 4x 2 )
7y(4 + y + 2x)
2r
18
...

t
21
...
(a − 2b)(2x + 3y)

2
...

6
...

10
...

14
...


1
...
2

6
...
2

11
...
2

4
...
6
1
8
...
6
18
...
−10
17
...
0
14
...
12y 2 − 3y
4
...
1

7
...
6a 2 + 5a −
11
...
9x 2 +

1
...
−2

3
...
15

7
...
5

11
...
2

16
...
6

21
...
−3

1
4
1
3

10
...
10a 2 − 3a + 2

15
...
5

1
2

1
3

4
...
12

9
...
13

13
...
−11

15
...
3

19
...
10

23
...
±4

Exercise 44 (page 79)
1
...

5
...


10−7
2
...
3
...
8  (b) 30 
digital camera battery £9, camcorder battery £14
800 
7
...
12 cm, 240 cm2
4
...
30 kg

2
...
004
5
...
12 m, 8 m

3
...
£312, £240
9
...
5 N

Chapter 12
Exercise 46 (page 84)
1
...
3

Exercise 43 (page 76)

Exercise 41 (page 72)
1
...
4b − 15b2
3
5
...
2

c
p
V
5
...
r =
2π
3
...
a =
t
1
6
...
x =
m
2
...
R =
I
5
13
...
T =

10
...
C =

ω ωL −

12
...
f =

13
...
λ =

R2


5

345

+ , 63
...
r =
or 1 −
S
S
2
...
f =

AL
3F − AL
or f = F −
3
3

4
...
t =

R − R0
R0 α

6
...
b =
9
...
R =

t 2g
4π 2


360 A
12
...
080
14
...
L =

11
...
a =
m −n
3(x + y)
3
...
b = √
1 − a2
a( p2 − q 2 )
7
...
t2 = t1 +
11
...
725



√
v 2 − 2as

M
+ r4
π
mrCR
4
...
r =
x+y



2
...
965
10
...
03L
8
...
5
p = 2, q = −1
x = 3, y = 2
a = 5, b = 2
s = 2, t = 3
m = 2
...
5
x = 2, y = 5

2
...

6
...

10
...

14
...


x = 3, y = 4
x = 4, y = 1
x = 1, y = 2
a = 2, b = 3
x = 1, y = 1
x = 3, y = −2
a = 6, b = −1
c = 2, d = −3

Exercise 50 (page 94)

13
...

3
...

7
...

11
...

15
...

3
...

7
...

4
...

8
...
30, b = 0
...
x = , y =
2
4
1
1
3
...
c = 3, d = 4
3
7
...
a = , b = −
3
2
4
...
r = 3, s =
2
8
...

3
...

6
...


a = 0
...
5, c = 3
α = 0
...
56 
a = 4, b = 10

2
...
47, I2 = 4
...
£15 500, £12 800
7
...
40
9
...
5, F2 = −4
...
x = 2, y = 1, z = 3
3
...
x = 2, y = −2, z = 2
4
...

7
...

10
...


x = 2, y = 4, z = 5 6
...
x = −4, y = 3, z = 2
x = 1
...
5, z = 4
...

4
...

8
...


1
...
0
...
905 A
0
...
38 m
1
...
835 m or 18
...

5
...

9
...
84 cm
0
...
78 cm
7m

Chapter 14
Exercise 58 (page 110)

Exercise 54 (page 104)
1
...
−1
...
5
7
...

13
...


4
−2 or −3
4 or −3
2

19
...
4 or −8
4
5
...
−5
11
...
2 or 7
17
...
−1
...
x = 1, y = 3 and x = −3, y = 7

6
...
x = , y = − and −1 , y = −4
5
5
3
3

9
...

15
...


23
...
5

1
4
25
...

5

1
1
26
...

or −
3
2

1
2 or −1
−4
3 or −3

1
8
1
24
...

8
5
30
...


22
...
x 2 − 4x + 3 = 0
33
...
x 2 − 36 = 0

3
...
3
7
...
−2
9
...

2
3
1
11
...
10 000 13
...
9 15
...
0
...

18
...
4

or −2
3
2

32
...
4x 2 − 8x − 5 = 0
36
...
7x − 1
...
4

3
...
−3

5
...
−3
...
268
3
...
468 or −1
...
2
...
307

1
8

3
...
−3
...
637
4
...
290 or 0
...
−2
...
351

1
...
log 12

2
...
log 500

3
...
log 100

4
...
log 6

9
...
log 1 = 0 11
...
log 243 or log 35 or 5 log3
13
...
log 64 or log26 or 6 log2

Exercise 56 (page 107)
1
...

5
...

9
...

13
...
637 or −3
...
781 or 0
...
608 or −1
...
851 or −2
...
481 or −1
...
167
4
...
438

2
...

6
...

10
...


0
...
792
0
...
693
1
...
232
2
...
086
4
...
676
7
...
641

15
...
5
16
...
5
19
...
x = 2

17
...
5 18
...
a = 6 22
...
1
...
3
...
0
...
6
...
2
...
3
...
2
...
−0
...
316
...

2
...

4
...


(a)
0
...
0988
(a) 4
...
04106
2
...


Exercise 67 (page 134)

0
...
064037
2
...
07482
120
...
446
8
...
08286

Exercise 63 (page 120)
1
...
0601

2
...
389 (b) 0
...

x − 2x 4
3
1
4
...
(a) Horizontal axis: 1 cm = 4 V (or 1 cm = 5 V),
vertical axis: 1 cm = 10 
(b) Horizontal axis: 1 cm = 5 m, vertical axis:
1 cm = 0
...
2 mm
2
...
5 (d) 5
3
...
5
4
...
1 (b) −1
...
The 1010 rev/min reading should be 1070 rev/min;
(a) 1000 rev/min (b) 167 V

1 − 2x 2 −

Exercise 64 (page 122)
1
...
95, 2
...
(a) 28 cm3 (b) 116 min

2
...
65, −1
...
(a) 70◦C (b) 5 minutes

Exercise 65 (page 124)
1
...

3
...

11
...

17
...
55547 (b) 0
...
8941
(a) 2
...
33154 (c) 0
...
4904 4
...
5822 5
...
197
6
...
2
0
...
11
...
1
...
1
...
962 12
...
4
147
...
4
...
3
...
e
...
500 19
...
Missing values: −0
...
25, 0
...
25, 2
...
(a) 4, −2 (b) −1, 0 (c) −3, −4 (d) 0, 4
1
1
1 1
(b) 3, −2 (c) ,
2
2
24 2
4
...
(a) 2,

1
2
2
1
2
5
...
(a) (b) −4 (c) −1
5
6
7
...
(2, 1)

9
...
5, 6)

10
...
(a) 89 cm (b) 11 N (c) 2
...
4 W + 48
12
...
15 W + 3
...
a = −20, b = 412

Exercise 69 (page 144)
1
...
(a) 850 rev/min (b) 77
...
(a) 150◦ C (b) 100
...
(a) 0
...
25L + 12
2
...
21 kPa

3
...
32 volts (b) 71
...
(a) 1
...
293 m 5
...
45 s
6
...
37 N
7
...
04 A (b) 1
...
2
...
£2424

347

9
...
07 A (b) 0
...
5 N (e) 592 N (f) 212 N
4
...
003, 8
...
(a) 22
...
43 s (c) v = 0
...
5
6
...
9L − 0
...
(a) 1
...
89% (c) F = −0
...
21
8
...
00022 (c) 28
...
a = 0
...
3 kPa, 275
...
(a)

Chapter 18

9
...
6, 13
...
6, 0
...
6 or 0
...
x = −1
...
5 (a) −30 (b) 2
...
50
(c) 2
...
8

Exercise 74 (page 161)

Exercise 70 (page 149)
1
...
(a) y (b)

√

x (c) b (d) a

y
1
(c) f (d) e 4
...
(a) (b) 2 (c) a (d) b
x
x
6
...
5, b = 0
...
78 mm2 7
...
15

1
...
5, y = −5
...
(a) x = −1
...
5 (b) x = −1
...
24
(c) x = −1
...
0

3
...
(a) 950 (b) 317 kN
9
...
4, b = 8
...
4 (ii) 11
...
x = −2
...
5 or 1
...
x = −2, 1 or 3, Minimum at (2
...
1),
Maximum at (−0
...
2)
3
...
x = −2
...
4 or 2
...
x = 0
...
5
6
...
3, 1
...
8
7
...
5

Exercise 71 (page 154)
1
...

3
...

5
...

7
...


(a) lg y (b) x (c) lg a (d) lg b
(a) lg y (b) lg x (c) L (d) lg k
(a) ln y (b) x (c) n (d) ln m
I = 0
...
75 candelas
a = 3
...
5
a = 5
...
6, 38
...
0
R0 = 26
...
42
8
...
08e0
...
4 N, μ = 0
...
0 N, 1
...
5, y = 1
...
3, y = −1
...
4, b = 1
...
82◦ 27
2
...
51◦11 4
...
15 44 17 6
...
72
...
27
...
37◦ 57
10
...
reflex 2
...
acute 4
...
(a) 21◦ (b) 62◦ 23 (c) 48◦56 17

Chapter 19

1
...

5
...


Chapter 20

2
...
x = −1, y = 2
6
...
(a) Minimum (0, 0) (b) Minimum (0, −1)
(c) Maximum (0, 3) (d) Maximum (0, −1)
2
...
4 or 0
...
−3
...
9
4
...
1 or 4
...
−1
...
2
6
...
5 or −2, Minimum at (−1
...
1)
7
...
7 or 1
...
(a) ±1
...
3

6
...
(a) 60◦ (b) 110◦ (c) 75◦ (d) 143◦ (e) 140◦
(f ) 20◦ (g) 129
...
Transversal (a) 1 & 3, 2 & 4, 5 & 7, 6 & 8
(b) 1 & 2, 2 & 3, 3 & 4, 4 & 1, 5 & 6, 6 & 7,
7 & 8, 8 & 5, 3 & 8, 1 & 6, 4 & 7 or 2 & 5
(c) 1 & 5, 2 & 6, 4 & 8, 3 & 7 (d) 3 & 5 or 2 & 8
9
...
a = 69◦ , b = 21◦ , c = 82◦ 11
...
1
...
0
...
40◦55

Exercise 78 (page 173)
1
...
a = 40◦ , b = 82◦, c = 66◦,
d = 75◦, e = 30◦ , f = 75◦
3
...
52◦
5
...
5◦
6
...
40◦, 70◦, 70◦, 125◦, isosceles
8
...
a = 103◦, b = 55◦ , c = 77◦ , d = 125◦,
e = 55◦, f = 22◦, g = 103◦, h = 77◦ ,
i = 103◦, j = 77◦, k = 81◦
10
...
A = 37◦, B = 60◦ , E = 83◦

349

4
3
4
3
2
...
sin A = , tan A =
17
15
112
15
, cos X =
4
...
(a)
(b)
(c)
17
17
15
7
24
6
...
(a) 9
...
625

Exercise 79 (page 176)
1
...
proof

Exercise 84 (page 187)
1
...

5
...

13
...
7550 2
...
846
3
...
52
(a) 0
...
1010 (c) 0
...
33◦
6
...
25◦
7
...
78◦
8
...
41 54 11
...
05
12
...
3586 14
...
803

Exercise 80 (page 178)

Exercise 85 (page 189)

1
...
54 mm, y = 4
...
9 cm, 7
...
(a) 2
...
3 m

1
...
22 (b) 5
...
87 (d) 8
...
595 (f ) 5
...
(a) AC = 5
...
04◦ , ∠C = 30
...
928 cm, ∠D = 30◦, ∠F = 60◦
(c) ∠J = 62◦, HJ = 5
...
59 cm
(d) ∠L = 63◦ , LM = 6
...
37 cm
(e) ∠N = 26◦, ON = 9
...
201 cm
(f ) ∠S = 49◦, RS = 4
...
625 cm

Exercise 81 (page 180)
1–5
...


3
...
54 m

4
...
40 mm

Chapter 21
Exercise 82 (page 182)

Exercise 86 (page 192)

1
...

7
...

11
...


1
...
15 m
2
...
249
...
110
...
53
...
9
...
107
...
9
...
56 m
9
...
24 m
3
...
54 mm
20
...
7
...
11
...
11 mm 8
...
20 cm each (b) 45◦
10
...
81 km
3
...
132
...
94 mm
14
...
sin Z = , cos Z = , tan X = , cos X =
41
41
9
41

1
...
78◦ and 137
...
53◦ and 351
...
(a) 29
...
92◦ (b) 123
...
14◦
3
...
21◦ and 224
...
12◦ and 293
...
t = 122◦7 and 237◦53
5
...
θ = 39◦44 and 219◦44

Exercise 88 (page 202)
1
...
180◦
3
...
120◦
◦
◦
◦
5
...
2, 144
7
...
5, 720◦
7
9
...
6, 360◦ 11
...
5 ms
2
13
...
100 μs or 0
...
625 Hz 16
...
leading

Exercise 91 (page 209)
1
...
2 cm, Q = 47
...
65◦,
area = 77
...
p = 6
...
83◦, R = 44
...
938 m2
3
...
33◦, Y = 52
...
05◦,
area = 27
...
Z = 29
...
50◦ , Z = 96
...
(a) 40 (b) 25 Hz (c) 0
...
29 rad (or 16
...

3
...

7
...
(a) 122
...
80◦ , 40
...
54◦
(a) 11
...
55◦
4
...
4 m
BF = 3
...
0 m 6
...
35 m, 5
...
48 A, 14
...
(a) 75 cm (b) 6
...
157 s
(d) 0
...
94◦ ) lagging 75 sin 40t
3
...
01 s or 10 ms
(d) 0
...
61◦) lagging 300 sin 200πt
4
...
43) volts

π

A or
5
...
524)A
6
...
2 sin(100πt + 0
...
(a) 5A, 50 Hz, 20 ms, 24
...
093A (c) 4
...
375 ms (e) 3
...
C = 83◦ , a = 14
...
9 mm,
area = 189 mm2
2
...
568 cm, a = 7
...
65 cm2
3
...
0 cm,
area = 134 cm2
4
...
08 mm,
area = 185
...
J = 44◦29 , L = 99◦31 , l = 5
...
132 cm2 , or, J = 135◦ 31 , L = 8◦29 ,
l = 0
...
917 cm2
6
...
2 mm,
area = 820
...
19 mm, area = 174
...
80
...
38◦, 40
...
(a) 15
...
07◦
3
...
25 cm, 126
...
19
...
36
...
x = 69
...
130◦ 8
...
66 mm

Chapter 24
Exercise 94 (page 215)
1
...

3
...

5
...

7
...


(5
...
04◦) or (5
...
03 rad)
(6
...
82◦) or (6
...
36 rad)
(4
...
57◦) or (4
...
03 rad)
(6
...
58◦) or (6
...
54 rad)
(7
...
20◦) or (7
...
55 rad)
(4
...
31◦) or (4
...
12 rad)
(5
...
04◦) or (5
...
74 rad)
(15
...
75◦) or (15
...
37 rad)

Exercise 95 (page 217)
(1
...
830)
2
...
917, 3
...
362, 4
...
(−2
...
154)
(−9
...
400)
6
...
615, −3
...
750, −1
...
(4
...
233)
(a) 40∠18◦, 40∠90◦, 40∠162◦, 40∠234◦, 40∠306◦
(b) (38
...
36), (0, 40), (−38
...
36),
(−23
...
36), (23
...
36)
10
...
0 mm
1
...

5
...

9
...
p = 105◦ , q = 35◦
3
...
r = 142◦, s = 95◦

Exercise 97 (page 225)
1
...
91 cm
2
...
7 cm2 3
...
27
...
18 cm
6
...
(a) 29 cm2 (b) 650 mm2
8
...
3
...
6750 mm
11
...
30 cm2
12
...

4
...

9
...

13
...

18
...


113 cm2
2
...
1790 mm2
2
2
802 mm
5
...
1269 m2
2
1548 m
8
...
0 cm (b) 783
...
46 m 10
...
80 cm, 74
...
86 mm (b) 197
...
26
...
67 cm, 54
...
82
...
748
(a) 0
...
2 m2
17
...
47 m2
2
(a) 396 mm (b) 42
...
701
...
74 mm

Exercise 104 (page 237)
1
...
Centre at (3, −2), radius 4
3
...
Circle, centre (0, 0), radius 6

Chapter 27
Exercise 98 (page 226)
1
...

3
...

7
...
27 cm2 (b) 706
...
(a) 20
...
41 mm
(a) 53
...
9 mm2
6
...
89 m

Exercise 99 (page 228)
1
...
1624 mm2 3
...
918 ha (b) 456 m

Exercise 105 (page 243)
1
...

5
...

10
...

14
...

17
...


1
...
5 cm3
3
...
15 cm3 , 135 g
7
...
(a) 35
...
3 cm2
1
...
37
...
63 cm
13
...
709 cm, 153
...
99 cm
16
...
099 m2
8
...
22 m
18
...
5 min
4 cm
20
...
08 m3

Exercise 100 (page 229)
1
...
80 m2

3
...
14 ha

Chapter 26
Exercise 101 (page 231)
1
...
24 cm 2
...
5 mm 3
...
629 cm 4
...
68 cm
5
...
73 cm 6
...
97
...
201
...
0 cm2 2
...
68 cm3 , 25
...
113
...
1 cm2 4
...
131 cm 5
...
2681 mm3 7
...
083 cm3
(b) 20 106 mm2 or 201
...
8
...
(a) 512 × 106 km2 (b) 1
...
664

Exercise 102 (page 232)

Exercise 107 (page 251)

5π
5π
π
(b)
(c)
6
12
4
2
...
838 (b) 1
...
054
3
...
(a) 0◦ 43 (b) 154◦8 (c) 414◦ 53

1
...

4
...


1
...
90 cm2
(a) 56
...
82 cm2
3
...
57 kg
5
...
4 m2
29
...
5 cm3 (b) 84
...
4 cm3 (b) 32
...
0 cm3

352 Basic Engineering Mathematics

7
...

11
...


(b) 146 cm2 (vi) (a) 86
...
9 cm (b) 38
...
125 cm3
3
2
10
...
5 m
10
...
220
...
1 cm3 , 1027 cm2
2
(a) 1458 litres (b) 9
...
45

147 cm3 , 164 cm2
10480 m3 , 1852 m2
10
...
14 m

14
...
72◦ to the 5 N force
29
...
04◦ to the horizontal
9
...
70◦
9
...
89 m/s at 159
...
62 N at 26
...
07 knots, E 9
...

3
...

7
...

7
...

10
...


2
...
1707 cm2
6
...
(a) 54
...
16◦ (b) 45
...
66◦
2
...
71 m/s at 121
...
55 m/s at 8
...
83
...
6◦ to the vertical
2
...
22
...
78◦ N

Exercise 109 (page 256)
1
...
137
...
4
...
54
...
63
...
4
...
143 m2

Exercise 111 (page 260)
1
...
59 m3

2
...
20
...
(a) 2 A (b) 50 V (c) 2
...
0
...
1 A
5
...
13 cm2 , 368
...

3
...

7
...


i − j − 4k
−i + 7j − k
−3i + 27j − 8k
i + 7
...
6i + 4
...
9k

2
...

6
...

10
...
5j − 10k
2i + 40j − 43k

Chapter 30
Exercise 118 (page 279)

2
...
5 V (b) 3 A
4
...
83 V (b) 0

1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395)

Chapter 29
Exercise 119 (page 281)
Exercise 113 (page 266)
1
...

2
...
scalar
4
...
scalar
6
...
vector 8
...
vector

Exercise 114 (page 273)
1
...

3
...

5
...
35 N at 18
...
62◦ to the 12 m/s velocity
16
...
57◦ to the 13 N force
28
...
30◦ to the 18 N force
32
...
80◦ to the 30 m displacement

1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395)

Exercise 120 (page 283)
1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395) 4
...
11 sin(ωt + 0
...
8
...
173)

Answers to practice exercises
Exercise 121 (page 284)
11
...
324)
2
...
73 sin(ωt − 0
...
79 sin(ωt − 0
...
695 sin(ωt + 0
...
38 sin(ωt + 1
...
3 sin(314
...
233) V (b) 50 Hz
(a) 10
...
3t + 0
...
(a) 79
...
352)V (b) 150 Hz
(c) 6
...

3
...

5
...

7
...
(a) continuous (b) continuous (c) discrete
(d) continuous
2
...
If one symbol is used to represent 10 vehicles, working correct to the nearest 5 vehicles, gives 3
...
5,
6, 7, 5 and 4 symbols respectively
...
If one symbol represents 200 components, working
correct to the nearest 100 components gives: Mon 8,
Tues 11, Wed 9, Thurs 12 and Fri 6
...

3
...

4
...

5
...

6
...

7
...

P increases by 20% at the expense of Q and R
...
Four rectangles of equal height, subdivided as follows: week 1: 18%, 7%, 35%, 12%, 28%; week 2:
20%, 8%, 32%, 13%, 27%; week 3: 22%, 10%, 29%,
14%, 25%; week 4: 20%, 9%, 27%, 19%, 25%
...

9
...
5◦ , 22
...
5◦, 167
...


353

10
...

11
...
(a) £16 450 (b) 138

Exercise 124 (page 297)
1
...
3–39
...
5–39
...
7–39
...
9–40
...
1–40
...
3–40
...
5–40
...
7–40
...

2
...
35, 39
...
75, 39
...
and
heights of 1, 5, 9, 17,
...
There is no unique solution, but one solution is:
20
...
9 3; 21
...
4 10; 21
...
9 11;
22
...
4 13; 22
...
9 9; 23
...
4 2
...
There is no unique solution, but one solution is:
1–10 3; 11–19 7; 20–22 12; 23–25 11;
26–28 10; 29–38 5; 39–48 2
...
20
...
45 13; 21
...
45 37; 22
...
45 48
6
...
5, 15, 21, 24, 27, 33
...
5
...
3, 0
...
67, 2
...
5 and 0
...

7
...
95 2), (11
...
95 19), (12
...
95
42), (13
...
A graph of cumulative frequency against upper class
boundary having co-ordinates given in the answer
to problem 7
...
(a) There is no unique solution, but one solution is:
2
...
09 3; 2
...
14 10; 2
...
19 11;
2
...
24 13; 2
...
29 9; 2
...
34 2
...
07, 2
...
and heights of 3, 10,
...
095 3; 2
...
195 24; 2
...
295 46; 2
...


(d) A graph of cumulative frequency against upper
class boundary having the co-ordinates given
in part (c)
...
Mean 7
...
Mean 27
...
−2542 A/s
2
...
16 cd/V (b) 312
...
(a) −1000 V/s (b) −367
...
−1
...
(a)
15
2
3
4
...
(a)

5
...


7
...

9
...
(a) 8 x + 8 x3 +
x +c
5

7x 2
+c
2
3
(b) t 8 + c
8
5 4
x +c
(b)
24
3
(b) 2t − t 4 + c
4
(b)

(b) 4θ + 2θ 2 +

5
(a) θ 2 − 2θ + θ 3 + c
2
3
2
3
(b) x 4 − x 3 + x 2 − 2x + c
4
3
2
4
1
(a) − + c
(b) − 3 + c
3x
4x
4√ 5
1√
4 9
(a)
x +c
(b)
x +c
5
9
15 √
10
5
(b)
x +c
(a) √ + c
7
t

7
(b) − cos 3θ + c
3
1
(b) 18 sin x + c
3
−2
(b)
+c
15e5x

13
...
(a) 4x + c

3
sin 2x + c
2
1
11
...
(a) e + c
8
10
...
(a) 1
...
5

2
...
5

3
...
333

4
...
75 (b) 0
...
(a) 10
...
1667

6
...
(a) 1 (b) 4
...
(a) 0
...
638

9
...
09 (b) 2
...
(a) 0
...
099

Exercise 141 (page 334)
1
...
37
...
1
...
proof
6
...
5
9
...
67

3
...
1
10
...
29

---