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Chapter 20

Angles and triangles
20
...
1)
...
This chapter involves the measurement of
angles and introduces types of triangle
...
2

c

Angular measurement

h e
g f

R

An angle is the amount of rotation between two straight
lines
...

If a circle is divided into 360 equal parts, then each part
is called 1 degree and is written as 1◦
i
...

or

1 revolution = 360◦
1
1 degree is
th of a revolution
360

Some angles are given special names
...

• An angle equal to 90◦ is called a right angle
...

• Any angle greater than 180◦ and less than 360◦ is
called a reflex angle
...

• If two angles add up to 90◦ they are called complementary angles
...

• Parallel lines are straight lines which are in the same
plane and never meet
...
1
...
1016/B978-1-85617-697-2
...
1

With reference to Figure 20
...
Such pairs of
angles are called vertically opposite angles
...
Such pairs of
angles are called corresponding angles
...
Such pairs of angles are called
alternate angles
...
Such pairs of
angles are called interior angles
...
2
...

i
...


1 degree = 60 minutes

which is written as

1◦ = 60
...
41◦ 29
29◦
= 41
...

1 minute further subdivides into 60 seconds,
1 minute = 60 seconds

i
...


which is written as

1 = 60
...

43◦ 29
+ 27◦ 43
71◦ 12
1◦

(i) 29 + 43 = 72
(ii) Since 60 = 1◦, 72 = 1◦12

(Notice that for minutes, 1 dash is used and for seconds,
2 dashes are used
...


20
...
2 Radians and degrees
One radian is defined as the angle subtended at the centre
of a circle by an arc equal in length to the radius
...
)
With reference to Figure 20
...

(iv) 43◦ + 27◦ + 1◦ (carried) = 71◦
...

This answer can be obtained using the
follows
...

1
...
Press ◦ ’ ’ ’
◦
4
...
Press +
6
...
Press ◦ ’ ’ ’
8
...

10
...

Problem 2
...

(ii) 1◦ or 60 is ‘borrowed’ from the degrees column,
which leaves 83◦ in that column
...

(iv) 83◦ − 56◦ = 27◦, which is placed in the degrees
column
...
2

When s is the whole circumference, i
...
when
s = 2πr,
s
2πr
θ= =
= 2π
r
r
In one revolution, θ = 360◦
...
e
...
30◦
1 rad =
π

Here are some worked examples on angular measurement
...

3
...
Enter 84
2
...
Press ’ ’ ’
5
...

◦
7
...
Enter 39
9
...
Press =
Answer = 27◦ 34

calculator as
Enter 13
Enter 56
Press ◦ ’ ’ ’

Thus, 84◦ 13 − 56◦39 = 27◦ 34
...


Evaluate 19◦ 51 47 + 63◦ 27 34
19◦ 51 47
+ 63◦ 27 34
83◦ 19 21
1◦ 1

Angles and triangles
(i) 47 + 34 = 81
(ii) Since

60

=

1 , 81

= 1 21

(iii) The 21 is placed in the seconds column and 1
is carried in the minutes column
...
Enter 63
4
...
Press =

0
...
753 × 60 = 45
...
18 = 0
...


This answer can be obtained using the calculator as
follows
...
Enter 51
1
...
Press ◦ ’ ’ ’
◦
4
...
Enter 47
6
...
Press +
8
...
Press ◦ ’ ’ ’
◦
10
...
Press ’ ’ ’
12
...
Press ◦ ’ ’ ’
14
...

Problem 4
...
45◦ by calculator
60
27◦
39◦27 = 39
= 39
...

3
...
Enter 39
2
...
Press =
6
...
Press ’ ’ ’
◦
Answer = 39
...
Convert 63◦ 26 51 to degrees in
decimal form, correct to 3 decimal places
51
= 63◦ 26
...
85◦
= 63
...
85 = 63
60
63◦ 26 51 = 63
...

This answer can be obtained using the calculator as
follows
...
Enter 26
6
...
4475◦

Problem 6
...
753◦ to degrees, minutes
and seconds

(v) Since 60 = 1◦, 79 = 1◦19

(vii) 19◦ + 63◦ + 1◦ (carried) = 83◦
...


2
...
Enter 51
8
...
753◦ = 53◦ 45 11
This answer can be obtained using the calculator as
follows
...
Enter 53
...
Press =
Answer = 53◦45 10
...
Press ◦ ’ ’ ’
Now try the following Practice Exercise
Practice Exercise 76 Angular measurement
(answers on page 348)
1
...


Evaluate 76◦ 31 − 48◦37

3
...


Evaluate 41◦ 37 16 + 58◦ 29 36

5
...


Evaluate 79◦26 19 − 45◦58 56 + 53◦ 21 38

7
...


8
...


9
...
952◦ to degrees and minutes
...
Convert 58
...

Here are some further worked examples on angular
measurement
...
State the general name given to the
following angles: (a) 157◦ (b) 49◦ (c) 90◦ (d) 245◦
(a) Any angle between 90◦ and 180◦ is called an
obtuse angle
...

(b) Any angle between 0◦ and 90◦ is called an acute
angle
...

(c)

An angle equal to 90◦ is called a right angle
...

Thus, 245◦ is a reflex angle
...

48◦ 39

Find the angle complementary to

(b) An angle of 180◦ lies on a straight line
...
3(b),
180◦ = 53◦ + θ + 44◦
from which,

θ = 180◦ − 53◦ − 44◦ = 83◦

Problem 11
...
5

If two angles add up to 90◦ they are called complementary angles
...

74◦ 25

108⬚

Find the angle supplementary to

39⬚
␪

58⬚

If two angles add up to 180◦ they are called supplementary angles
...
5

180◦ − 74◦ 25 = 105◦ 35
Problem 10
...
3

418

␪

␪

538
(a)

448

Problem 12
...
If ∠AOC is 43◦ , find ∠AOD,
∠DOB and ∠BOC

(b)

Figure 20
...

Thus, 360◦ = 58◦ + 108◦ + 64◦ + 39◦ + θ
from which, θ = 360◦ − 58◦ − 108◦ − 64◦ − 39◦ = 91◦

The symbol shown in Figure 20
...


From Figure 20
...

Hence, ∠ AOD = 180◦ − 43◦ = 137◦
...

Hence,
∠ DOB = 43◦ and ∠BOC 137◦

D

A

438

Figure 20
...
3(a),
θ + 41◦ = 90◦
from which,
θ = 90◦ − 41◦ = 49◦

Figure 20
...
Determine angle β in Figure 20
...
7

α = 180◦ − 133◦ = 47◦ (i
...
supplementary angles)
...

Problem 14
...
8
A

23⬚37⬘

F

␪

B
G

E

Although we may be more familiar with degrees, radians
is the SI unit of angular measurement in engineering
(1 radian ≈ 57
...

π
(a) Since 180◦ = π rad then 1◦ =
rad
...
274 rad
...
616666
...
616666
...
616666
...
447 rad
...
Convert 0
...
8

Let a straight line FG be drawn through E such that FG
is parallel to AB and CD
...

∠ECD = ∠FEC (alternate angles between parallel lines
FG and CD), hence ∠FEC = 35◦ 49
...
Determine angles c and d in
Figure 20
...
9

a = b = 46◦ (corresponding angles between parallel
lines)
...

Hence, 46◦ + c + 90◦ = 180◦ , from which, c = 44◦
...

◦
Alternatively, 90 + c = d (vertically opposite angles)
...
Convert the following angles to
radians, correct to 3 decimal places
...
57076
...
743 rad = 0
...


State the general name given to an angle of
197◦
...


State the general name given to an angle of
136◦
...


State the general name given to an angle of
49◦
...


State the general name given to an angle of
90◦
...


Determine the angles complementary to the
following
...


Determine the angles supplementary to
(a) 78◦ (b) 15◦ (c) 169◦41 11

7
...
10
...


With reference to Figure 20
...

y
56
7
8

2
1
3
4

x

Figure 20
...


(b)

vertically opposite angles
supplementary angles
corresponding angles
alternate angles

In Figure 20
...

137⬚29⬘

378

␣
␪
558

␪

16⬚49⬘

508

Figure 20
...
In Figure 20
...


708
␪

298

␪

1508
␪

a

698

608

808

b

(f)

(e)

688

578

␪

Figure 20
...
78

c

␪

11
...
14
...
14
368
␪
(i)

Figure 20
...
Convert 76◦ to radians, correct to 3 decimal
places
...
Convert 34◦ 40 to radians, correct to 3 decimal places
...
Convert 0
...


171

Angles and triangles
20
...

The sum of the three angles of a triangle is equal
to 180◦
...
3
...
e
...
An
example is shown in triangle ABC in Figure 20
...

A right-angled triangle is one which contains a right
angle; i
...
, one in which one of the angles is 90◦
...
15(b)
...
17

548
508

A

c
678

598

B

C

E

F

(a)

(b)
␪

Figure 20
...
e
...
An example is shown in triangle PQR in
Figure 20
...

An equilateral triangle is one in which all the sides and
all the angles are equal; i
...
, each is 60◦
...
16(b)
...
18

With reference to Figure 20
...

(b) Angle θ is called an exterior angle of the triangle
and is equal to the sum of the two opposite interior
angles; i
...
, θ = A + C
...


8

A
1318

608

278
Q

(a)

R

B

608
(b)

C

Figure 20
...
An example is shown in triangle
EFG in Figure 20
...

A scalene triangle is one with unequal angles and
therefore unequal sides
...
17(b)
...
19

c

a

B

172 Basic Engineering Mathematics
A right-angled triangle ABC is shown in Figure 20
...

The point of intersection of two lines is called a vertex
(plural vertices); the three vertices of the triangle are
labelled as A, B and C, respectively
...
The side opposite the right angle is given
the special name of the hypotenuse
...
19, is always the longest side of
a right-angled triangle
...

With reference to angle A, BC is the opposite side and
AC is the adjacent side
...
So, in the triangle ABC, length AB = c, length
BC = a and length AC = b
...

∠ is the symbol used for ‘angle’
...
Another way of indicating
an angle is to use all three letters
...
e
...
Similarly, ∠BAC means ∠A and ∠ACB means
∠C
...

Problem 18
...
20

Name the types of triangle shown in

(d) Obtuse-angled scalene triangle (since one of the
angles lies between 90◦ and 180◦ )
...


Problem 19
...
21, with reference to angle θ, which side
is the adjacent?
C

A

␪
B

Figure 20
...
With reference to angle θ, the opposite side
is BC
...

Problem 20
...
22, determine angle θ

2
...
1

2

398

2
...
5
␪

2
...
5
(e)

Figure 20
...


(b) Acute-angled scalene triangle (since all the
angles are less than 90◦)
...


Figure 20
...

The triangle is right-angled
...

Figure 20
...

Determine the value of θ and α in

Angles and triangles

173

Problem 23
...
25

A
62⬚
D

B

C

e

␪
15⬚
E
␣

a
558 b
628
d c

Figure 20
...
23

180◦

In triangle ABC, ∠A + ∠B + ∠C =
(the angles in
a triangle add up to 180◦)
...
Thus,
∠DCE = 28◦ (vertically opposite angles)
...

Hence, ∠θ = 28◦ + 15◦ = 43◦
...


a = 62◦ and c = 55◦ (alternate angles between parallel
lines)
...

b = d = 63◦ (alternate angles between parallel lines)
...

Check: e = a = 62◦ (corresponding angles between
parallel lines)
...
ABC is an isosceles triangle in
which the unequal angle BAC is 56◦
...
24
...
Also, calculate ∠DBC
A

Practice Exercise 78
page 348)
1
...
26
...
24

Since triangle ABC is isosceles, two sides – i
...
AB and
AC – are equal and two angles – i
...
∠ABC and ∠ACB –
are equal
...

Hence, ∠ABC + ∠ACB = 180◦ − 56◦ = 124◦
...

2
An angle of 180◦ lies on a straight line; hence,
∠ABC + ∠DBC = 180◦ from which,
∠ DBC = 180◦ − ∠ABC = 180◦ − 62◦ = 118◦
...
e
...


97⬚

45⬚
45⬚

(d)

(c)

53⬚
60⬚
5

5
37⬚

(e)

Figure 20
...


Find the angles a to f in Figure 20
...


O
658

328

578
N
838

114°

b

a
(a)

M

c

␪

(b)
P

1058
d

Figure 20
...


f

Determine ∠φ and ∠x in Figure 20
...


1058

e

E

(c)

588

Figure 20
...


B

In the triangle DEF of Figure 20
...
30

7
...
31(a) and (b), find angles w, x, y
and z
...
28

4
...
28, determine
angle D
...


MNO is an isosceles triangle in which
the unequal angle is 65◦ as shown in
Figure 20
...
Calculate angle θ
...
31

8
...
32(a) and (b)
...
32

9
...
33
...


Problem 24
...
35 are congruent and name their
sequence

f

1258 e

(b) two sides of one are equal to two sides of the other
and the angles included by these sides are equal
(SAS),

h

C

i

G
E

B

D
L

k 998

J

H
A

Figure 20
...
Triangle ABC has a right angle at B and
∠BAC is 34◦
...
If the
bisectors of ∠ABC and ∠ACD meet at E,
determine ∠BEC
...
If in Figure 20
...

A

C

978

K
(b)
F

T
S
V

M

U

E
C

B

O

N

D

Q

P

(c)

R

W

X
(d)

A
(e)

Figure 20
...
e
...

(b) Congruent GIH, JLK (side, angle, side; i
...
, SAS)
...
34

(c) Congruent MNO, RQP (right angle, hypotenuse,
side; i
...
, RHS)
...
4

Congruent triangles

Two triangles are said to be congruent if they are equal
in all respects; i
...
, three angles and three sides in one
triangle are equal to three angles and three sides in the
other triangle
...
It is not indicated that
any side coincides
...
e
...


176 Basic Engineering Mathematics
Problem 25
...
36, triangle PQR is
isosceles with Z , the mid-point of PQ
...
Determine
the values of angles RPZ and RXZ

L

F

A
E

G

K

M

I

O

H
C
N

R

J

D

P

(a)

678

X

(b)

(c)

Y

288
P

B

V

288
Z

U

Q
R

Q

Figure 20
...

∠RXZ = ∠QPR + 28◦ and ∠RYZ = ∠RQP + 28◦ (exterior angles of a triangle equal the sum of the two interior
opposite angles)
...


2
...

Hence, XZ = YZ
...

∠QRZ = 67◦ and thus ∠PRQ = 67◦ + 67◦ = 134◦
...

2
∠ RXZ = 23◦ + 28◦ = 51◦ (external angle of a triangle equals the sum of the two interior opposite
angles)
...
37

∠PXZ = 180◦ − ∠RXZ and ∠QYZ = 180◦ − ∠RYZ
...


Triangles PRZ and QRZ are congruent since
PR = RQ, ∠RPZ = ∠RQZ and PZ = ZQ (SAS)
...


Z

T
(d)

20
...
Show that triangles AEB and
CDB are congruent
...
With
reference to Figure 20
...
e
...


State which of the pairs of triangles in
Figure 20
...


588
a

C

Q

658 588
p
R

Figure 20
...

side a

In Figure 20
...
82
p
=
=
12
...
44 10
...
82
= 4
...
44
10
...
e
...
0 cm

D
f 5 5
...
42 cm

C

p q
=
z
y

p
6
...
97 10
...
82
p = 12
...
32 cm
10
...
39

In triangle ABC, 50◦ + 70◦ + ∠C = 180◦ , from which
∠C = 60◦
...

Hence, triangles ABC and DEF are similar, since their
angles are the same
...
0
a
=
i
...

=
d
f
4
...
0
Hence, side, a =

Hence,

A

12
...
42) = 10
...

5
...
42

z5

q 5 6
...
63

y51

R

cm

X

Figure 20
...

In triangle XYZ, ∠X = 180◦ − 90◦ − 55◦ = 35◦
...
The triangles may he redrawn as
shown in Figure 20
...


x 5 7
...
82 cm

Figure 20
...
97

cm

Also,
358

R

Z

Since BD is parallel to AE then ∠CBD = ∠CAE and
∠CDB = ∠CEA (corresponding angles between parallel lines)
...

Since the angles in triangle CBD are the same as in
triangle CAE, the triangles are similar
...
e
...
44 cm

12

Q

9

12


...
e
...
In Figure 20
...
In Figure 20
...
63 cm

X

9
CD
=
, from which
6+9
12
 
9
= 7
...
A rectangular shed 2 m wide and
3 m high stands against a perpendicular building of
height 5
...
A ladder is used to gain access to the
roof of the building
...
58 mm
25
...
43, where AF
is the minimum length of the ladder
...
Hence, triangles
BAD and EDF are similar since their angles are the
same
...
36 mm

Figure 20
...


PQR is an equilateral triangle of side 4 cm
...
If PS is 9 cm, find the length of ST
...
Find the
length of PX
...


In Figure 20
...


AB = AC − BC = AC − DE = 5
...
5 m
AB
BD
2
...
e
...
4 m = minimum distance
Hence, EF = 2
2
...


4
...
5 m

Shed
D

F

E

C

Figure 20
...

Figure 20
...
46, AF = 8 m, AB = 5 m and
BC = 3 m
...

C

B

D

Now try the following Practice Exercise
Practice Exercise 80 Similar triangles
(answers on page 349)
1
...
44, find the lengths x and y
...
46

E

F

Angles and triangles
20
...
48:

To construct any triangle, the following drawing instruments are needed:

A

(a) ruler and/or straight edge

b 5 3 cm

(b) compass
608

(c) protractor

a 5 6 cm

B

(d) pencil
...
48

Here are some worked problems to demonstrate triangle
construction
...

Problem 30
...

(iii) From C measure a length of 3 cm and label A
...


D
G

C

F

Problem 32
...
47

With reference to Figure 20
...

(ii) Set compass to 5 cm and with centre at A describe
arc DE
...

(iv) The intersection of the two curves at C is the vertex of the required triangle
...

It may be proved by measurement that the ratio of the
angles of a triangle is not equal to the ratio of the sides
(i
...
, in this problem, the angle opposite the 3 cm side is
not equal to half the angle opposite the 6 cm side)
...
Construct a triangle ABC such that
a = 6 cm, b = 3 cm and ∠C = 60◦

708
Q

448
5 cm

R

Figure 20
...
49:
(i) Draw a straight line 5 cm long and label it QR
...
Draw QQ
...
Draw RR
...

Problem 33
...
5 cm and
∠X = 90◦

180 Basic Engineering Mathematics
V
Z
P U

centred at Y and set to 6
...


S

(iv) The intersection of the arc UV with XC produced,
forms the vertex Z of the required triangle
...


C
R
B

A

Q
X

A9

Y

Figure 20
...
50:
(i) Draw a straight line 5 cm long and label it XY
...
With compass centred at X make an arc at A and A
...
) With compass centred at A
draw the arc PQ
...
Join the intersection of the arcs, C to X , and a right angle to
XY is produced at X
...
)
(iii) The hypotenuse is always opposite the right angle
...
Using a compass

Now try the following Practice Exercise
Practice Exercise 81 Construction of
triangles (answers on page 349)
In the following, construct the triangles ABC for
the given sides/angles
...


a = 8 cm, b = 6 cm and c = 5 cm
...


a = 40 mm, b = 60 mm and C = 60◦
...


a = 6 cm, C = 45◦ and B = 75◦
...


c = 4 cm, A = 130◦ and C = 15◦
...


a = 90 mm, B = 90◦ , hypotenuse = 105mm
...

The n’th term is: a + (n − 1)d
n
Sum of n terms, Sn = [2a + (n − 1)d]
2

Geometric progression:
If a = first term and r = common ratio, then the geometric progression is: a, ar, ar2 ,
...

4
...

10
...

16
...


Exercise 5 (page 11)

19 kg
2
...
479 mm
−66
5
...
−225
−2136
8
...
£10 7701
−4
11
...
5914
189 g
14
...
$15 333
89
...

3
...

5
...

9
...

9
...

17
...

(a) 8613 kg (b) 584 kg
(a) 351 mm (b) 924 mm
(a) 10 304 (b) −4433
6
...

(a) 8067 (b) 3347
10
...

4
...

8
...


(a) 48 m (b) 89 m
(a) 1648 (b) 1060
18 kg

1
...
14
7
...
88
8
...
1016/B978-1-85617-697-2
...


3 4 1 3 5
, , , ,
7 9 2 5 8
9
10
...
1
15
16
18
...


6
...

21
...
2
3
...
11
8
...

13
...

18
...

11
...

19
...

8
...

16
...


4
...
2

5
...
5

5
12
1
9
...

23
4
...
15

3
28
8
10
...


15
...
400 litres
22
...

15

1
...
59
6
...
−1

2
5

Exercise 6 (page 13)

11
...
7

(a) £1827 (b) £4158

Exercise 3 (page 6)
1
...

5
...

9
...
2

11
...
4
20
17
12
...
−

3
...
2

1
6

3
4
1
9
...
4

13
20
1
10
...


Answers to practice exercises
Exercise 14 (page 25)

Chapter 3

1
...
571
5
...
96
8
...
0871

Exercise 8 (page 17)
1
...
23
8
...


13
20
21
6
...
(a) 1
50
5
...
0
...


4
...
6

7
16

(e) 16

17
80

1
...

7
...

13
...
182
2
...
122
3
...
82
0
...
0
...
2
...
273
8
...
256
9
...
30366
6
1
...
3
...
37
...
2 × 10
14
...
767 ×10
15
...
32 ×106

12
...
6875 13
...
21875 14
...
1875

Exercise 16 (page 27)
1
...
4667

Exercise 9 (page 18)
1
...
18
5
...
297

2
...
785
3
...
38
6
...
000528

2
...
3
6
...
3

4
...
27

3
...
54
7
...
52 mm

4
...
83

13
14

3
...
458

6
...
7083

7
...
2
...
3

1
3

10
...
0776

1
...
9205
5
...
4424
9
...
6992

2
...
7314
6
...
0321
10
...
8452

3
...
9042
7
...
4232

4
...
2719
8
...
1502

Exercise 18 (page 28)

4
...
47
...
385
...
582
...
9 6
...
82
7
...
1
8
...
6
0
...
0
...
1
...
53
...
84 14
...
69
15
...
81 (b) 24
...
00639 (b) 0
...
(a) 8
...
6˙
2400

1
...
995
5
...
6977
9
...
520

Exercise 12 (page 23)
3
...
62
7
...
330

4
...
832
8
...
45

Exercise 13 (page 24)
1
...
25
2
...
0361 3
...
923 4
...
296 × 10−3
5
...
4430 6
...
197 7
...
96 8
...
0549
9
...
26 10
...
832 × 10−6

2
...
782
6
...
92
10
...
3770

3
...
72
7
...
0

4
...
42
8
...
90

Exercise 19 (page 29)
1
...

7
...


Chapter 4

2
...
1
...
12
...


Exercise 17 (page 27)

Exercise 11 (page 20)

1
...
797
5
...
42
9
...
59

1
21
9
...
567
5
...

5
...

13
...

18
...
40
3
...
13459
4
...
9
6
...
4481 7
...
36 × 10−6
9
...
625 × 10−9
10
...
70

Exercise 15 (page 25)

3
125

15
...
28125

1
...
3
5
...
3

341

A = 66
...
144 J
14 230 kg/ m3

2
...

8
...


C = 52
...
407 A
628
...
1 m/s

3
...

9
...


R = 37
...
02 mm
224
...
526 

Exercise 20 (page 30)
1
...

7
...

12
...
27
2
...
1 W
3
...
61 V
F = 854
...
I = 3
...
T = 14
...
96 J
8
...
77 A 9
...
25 m
A = 7
...
V = 7
...
53 h (b) 1 h 40 min, 33 m
...
h
...
02 h (d) 13
...

3
...

6
...

12
...

16
...

20
...


Exercise 35 (page 62)

m3

cubic metres,
2
...
metres per second, m/s
kilogram per cubic metre, kg/m3
joule
7
...
watt
radian or degree
10
...
mass
electrical resistance 13
...
electric current
inductance
17
...
pressure
angular velocity
21
...
m, ×10−3
−12
6
×10
24
...

3
...

7
...


Exercise 33 (page 56)
1
...
39 × 10 (b) 2
...
9762 × 102

4
...

11
...


3
...
401 × 10−1 (b) 1
...
23 × 10−3

15
...
(a) 5 × 10−1 (b) 1
...
306 × 102 (d) 3
...

4
...

8
...


1
...
(a) 2
...
317 × 104 (c) 2
...
(a) 1
...
004 × 10 (c) 1
...
8a 2
3
...
5
6
...

9
...
3x
14
...
3 p + 2q 17
...
z 8
4
...
a 8
5
...
(a) 1
...
1 × 105

7
...
x −3 or

9
...
(a) 2
...
4 × 10−1
(c) 3
...
11 × 10−1 MeV
(e) 9
...
241×10−2 m3 mol−1

Exercise 34 (page 58)
1
...

5
...

9
...

13
...

17
...


60 kPa
50 MV
100 kW
1
...
50 m
13
...

4
...

8
...

12
...

16
...

20
...
15 mW
55 nF
0
...
5 mV
15 pF
46
...
025 MHz
0
...
0346 kg
4 × 103

6q 2
1
−
2
5x z 2

2ab
2a 2 + 2b2

Exercise 37 (page 66)

6
...
7 (c) 54100 (d) 7
7
...
0389 (b) 0
...
008
1
...


2 × 102

343

1
x6
1
13
...
p6 q 7r 5
10
...
n 3
6
...
t 8

9
...
c14

1
15
...
s −9
x 12
y
18
...
x 5 y 4 z 3 , 13 20
...
b−12 or

Exercise 38 (page 67)
1
...
a −4 b5 c11
1
7
...


1+a
b

3
...


p2 q
q−p

6
...


a 11/6 b1/3 c−3/2

√
6

z 13

√
6 11 √
3
a
b
or √
c3

344 Basic Engineering Mathematics
Chapter 10

Chapter 11

Exercise 39 (page 69)
1
...

5
...

9
...

13
...

17
...

21
...

25
...


x 2 + 5x

+6
+9
4x 2 + 22x + 30
a 2 + 2ab + b2
a 2 − 2ac + c2
4x 2 − 24x + 36
64x 2 + 64x + 16
3ab − 6a 2
2a 2 − 3ab − 5b2
7x − y − 4z
x 2 − 4x y + 4y 2
0
4ab − 8a 2
2 + 5b2
4x 2 + 12x

Exercise 42 (page 75)

2
...

6
...

10
...

14
...

18
...

22
...

26
...


2x 2 + 9x

+4
− 12
2 pqr + p2 q 2 + r 2
x 2 + 12x + 36
25x 2 + 30x + 9
4x 2 − 9
r 2 s 2 + 2rst + t 2
2x 2 − 2x y
13 p − 7q
4a 2 − 25b2
9a 2 − 6ab + b2
4−a
3x y + 9x 2 y − 15x 2
11q − 2 p
2 j2 +2 j

Exercise 40 (page 71)
2(x + 2)
p(b + 2c)
4d(d − 3 f 5)
2q(q + 4n)
bc(a + b2 )
3x y(x y 3 − 5y + 6)
7ab(3ab − 4)


2x y x − 2y 2 + 4x 2 y 3
3x
17
...
(a + b)(y + 1)
22
...

3
...

7
...

11
...

15
...
0 19
...
( p + q)(x + y)
23
...

4
...

8
...

12
...

16
...
1

2
...
1

7
...
2
16
...
−4

3
...

2
13
...
−3

12
...
6

9
...
−2

2
...
4 + 3a
6
...
10y 2 − 3y +
9
...


1
− x − x2
5

1
7

1
4

8
...
5

2
...
−4

6
...
−4

8
...
−10

12
...
9

17
...
±12

22
...
−15t
12
...
2
...
2

5
...
2

10
...
3

14
...
−6

18
...
4

20
...
±3

24
...

4
...

6
...
8 m/s2
3
...
472
(a) 1
...
30 m/s2

Exercise 45 (page 80)
1
...
45◦ C
7
...
0
...
50
8
...
30
6
...
3
...
d = c − e − a − b

1
− 4x
3

10
...
2x + 8x 2
3
...

− 4x
2

5
...
R =
I
c
7
...
v =

y
7
v −u
4
...
y = (t − x)
3
y−c
8
...
x =

346 Basic Engineering Mathematics
5
...

9
...

11
...
x = 1, y = 6, z = 7
x = 5, y = 4, z = 2 8
...
5, y = 2
...
5
i1 = −5, i2 = −4, i3 = 2
F1 = 2, F2 = −3 F3 = 4

Exercise 57 (page 109)
1
...

6
...

10
...
191 s 2
...
345 A or 0
...
619 m or 19
...
066 m
1
...
165 m
12 ohms, 28 ohms

3
...

7
...


7
...
0133
86
...
4 or −4
4
...
5 or 1
...

10
...

16
...
−2 or −

2
3

2
...
0 or −
3
8
...
−3 or −7
14
...
−3
20
...
5

1
...
2 or −2

2
1
2
1
2
...

12
...

18
...
−1 or 1
...

or −
2
5
2
or −3
28
...
1 or −
3
7
4
1
29
...

2
3
27
...

4
21
...
4 or −7

31
...
x 2 + 5x + 4 = 0
35
...
2 or −6

or −
or

or −

Chapter 15
Exercise 59 (page 112)

1
3

1
3
1
1
6
...
2
8
...
1 10
...
2
12
...
100 000 14
...

32
1
16
...
01 17
...
e3
16
1
...
x 2 + 3x − 10 = 0
34
...
x 2 − 1
...
68 = 0

2
...
3

4
...


Exercise 60 (page 115)

Exercise 55 (page 106)
1
...
732 or −0
...
1
...
135
5
...
443 or 0
...
x = 0, y = 4 and x = 3, y = 1

2
...
137 or 0
...
1
...
310
6
...
851 or 0
...
log 6
5
...
log 15
6
...
log 2
7
...
log 3
8
...
log 10 10
...
log 2
12
...
log 16 or log24 or 4 log2
14
...

3
...

7
...

11
...


0
...
137
2
...
719
3
...
108
0
...
351
1
...
081
4 or 2
...
562 or 0
...

4
...

8
...

12
...
296 or −0
...
443 or −1
...
434 or 0
...
086 or −0
...
176 or −1
...
141 or −3
...
0
...
1
...
b = 2 20
...
x = 2
...
t = 8
21
...
x = 5

Exercise 61 (page 116)
1
...
690 2
...
170 3
...
2696 4
...
058 5
...
251
6
...
959 7
...
542 8
...
3272 9
...
2

Answers to practice exercises
Chapter 16

Chapter 17

Exercise 62 (page 118)
1
...

3
...

5
...
1653
(a)
5
...
55848
(a) 48
...
739

(b)
(b)
(b)
(b)
6
...
4584
0
...
40444
4
...
7 m

(c)
(c)
(c)
(c)

22030
40
...
05124
−0
...
2
...
(a) 7
...
7408
8 3
3
...
2x 1/2 + 2x 5/2 + x 9/2 + x 13/2
3
1 17/2
1
+ x
+ x 21/2
12
60

1
...
1 V
(c) Horizontal axis: 1 cm = 10 N, vertical axis:
1 cm = 0
...
(a) −1 (b) −8 (c) −1
...
14
...
(a) −1
...
4
5
...
3
...
05
3
...
1
...
30
4
...

2
...

7
...

14
...


(a) 0
...
91374 (c) 8
...
2293 (b) −0
...
13087
−0
...
−0
...
2
...
816
...
8274 8
...
02
9
...
522 10
...
485
1
...
3
13
...
9 15
...
901 16
...
095
a
t = eb+a ln D = eb ea ln D = eb eln D i
...
t = eb D a
 
U2
18
...
W = PV ln
U1

Exercise 68 (page 140)
1
...
75, 0
...
75, 2
...
75;
1
Gradient =
2
2
...
(a) 6, −3 (b) −2, 4 (c) 3, 0 (d) 0, 7
3
...
(a) 2, − (b) − , −1 (c) , 2 (d) 10, −4
2
3
3
18
3
3
5
6
...
(a) and (c), (b) and (e)
8
...
(1
...
(1, 2)

11
...
4 (d) l = 2
...
P = 0
...
5

13
...
(a) 40◦ C (b) 128 
2
...
5 V

Exercise 66 (page 127)
1
...
5◦C

3
...
25 (b) 12 (c) F = 0
...
99
...
(a) 29
...
31 × 10−6 s
4
...
993 m (b) 2
...
(a) 50◦ C (b) 55
...
30
...
(a) 3
...
46 s
8
...
45 mol/cm3
10
...
(a) 7
...
966 s

(d) 89
...
−0
...
73
5
...
5 m/s (b) 6
...
7t + 15
...
m = 26
...
63
7
...
31 t (b) 22
...
09 W + 2
...
(a) 96 × 109 Pa (b) 0
...
8 × 106 Pa

348 Basic Engineering Mathematics
1
1
(b) 6 (c) E = L + 6 (d) 12 N (e) 65 N
5
5
10
...
85, b = 12, 254
...
5 kPa, 280 K
9
...
(−2
...
2), (0
...
8); x = −2
...
6
10
...
2 or 2
...
75 and −1
...
3 or −0
...
(a) y (b) x 2 (c) c (d) d

2
...
(a) (b) x (c) b (d) c
x
x
1
y
5
...
a = 1
...
4, 11
...
y = 2x 2 + 7, 5
...
x = 4, y = 8 and x = −0
...
5
2
...
5 or 3
...
24 or 3
...
5 or 3
...
(a) y (b)

8
...
a = 0
...
6 (i) 94
...
2

Exercise 75 (page 162)
1
...
0, −0
...
5
2
...
1, −4
...
8, 8
...
x = 1
4
...
0, 0
...
6
5
...
7 or 2
...
x = −2
...
0 or 1
...
x = −1
...

2
...

4
...

6
...

9
...
0012 V2 , 6
...
0, b = 0
...
7, b = 2
...
53, 3
...
0, c = 1
...
y = 0
...
24x
T0 = 35
...
27, 65
...
28 radians

Exercise 72 (page 156)
x = 2, y = 4
x = 3
...
5
x = 2
...
2
a = 0
...
6

Exercise 76 (page 167)
1
...
27◦54
3
...
100◦6 52
◦


◦


5
...
86 49 1 7
...
55◦ 8
...
754◦
9
...
58◦22 52

Exercise 77 (page 169)
1
...
obtuse 3
...
right angle
5
...

3
...

7
...
x = 1, y = 1
4
...
x = −2, y = −3

Exercise 73 (page 160)
1
...
−0
...
6
3
...
9 or 6
...
−1
...
1
5
...
8 or 2
...
x = −1
...
75, −0
...
x = −0
...
6
8
...
63 (b) 1 or −0
...
(a) 102◦ (b) 165◦ (c) 10◦ 18 49
7
...
3◦ (h) 79◦ (i) 54◦
8
...
59◦ 20
10
...
51◦
12
...
326 rad 13
...
605 rad 14
...
(a) acute-angled scalene triangle
(b) isosceles triangle (c) right-angled triangle
(d) obtuse-angled scalene triangle
(e) equilateral triangle (f ) right-angled triangle

Answers to practice exercises
2
...
DF, DE
4
...
122
...
φ = 51◦, x = 161◦
7
...
a = 18◦ 50 , b = 71◦10 , c = 68◦ , d = 90◦,
e = 22◦ , f = 49◦, g = 41◦
9
...
17◦

11
...
sin A = , cos A = , tan A = , sin B = ,
5
5
4
5
3
4
cos B = , tan B =
5
3
8
8
3
...
sin X =
113
113
15
15
8
5
...
(a) sin θ =
(b) cos θ =
25
25
7
...
434 (b) −0
...
(a) congruent BAC, DAC (SAS)
(b) congruent FGE, JHI (SSS)
(c) not necessarily congruent
(d) congruent QRT, SRT (RHS)
(e) congruent UVW, XZY (ASA)
2
...

4
...

9
...


2
...
4
...
36
...
8660 (b) −0
...
5865
42
...
15
...
73
...
7◦56
◦

◦

◦
31 22 10
...
29
...
20◦21
0
...
1
...
x = 16
...
18 mm 2
...
79 cm
3
...
25 cm (b) 4 cm
4
...
(a) 12
...
619 (c) 14
...
349
(e) 5
...
275
2
...
831 cm, ∠A = 59
...
96◦
(b) DE = 6
...
634 cm, GH = 10
...
810 cm, KM = 13
...
125 cm, NP = 8
...
346 cm, QS = 6
...
Constructions – see similar constructions in
worked problems 30 to 33 on pages 179–180
...
6
...
9
...

4
...

9
...

13
...
36
...
48 m
3
...
5 m 4
...
1 m
5
...
0 m
6
...
50 m 7
...
8 m
8
...
43 m, 10
...
60 m

9 cm
2
...
9
...
81 cm 5
...
21 m
6
...
18 cm
24
...
82 + 152 = 172
(a) 27
...
20
...
35 m, 10 cm
12
...
7 nautical miles
2
...
24 mm

Chapter 22
Exercise 87 (page 198)

Exercise 83 (page 185)
40
40
9
9
1
...
(a) 42
...
22◦ (b) 188
...
47◦
2
...
08◦ and 330
...
86◦ and 236
...
(a) 44
...
21◦ (b) 113
...
12◦

350 Basic Engineering Mathematics
4
...
α = 218◦41 and 321◦19
6
...
5
2
...
30
4
...
1, 120
6
...
3, 90
8
...
, 960◦ 10
...
4, 180◦ 12
...
40 Hz
14
...
1 ms
15
...
leading 17
...
p = 13
...
35◦, R = 78
...
7 cm2
2
...
127 m, Q = 30
...
17◦ ,
area = 6
...
X = 83
...
62◦, Z = 44
...
8 cm2
4
...
77◦, Y = 53
...
73◦ ,
area = 355 mm2

Exercise 92 (page 210)
Exercise 89 (page 203)
1
...
04 s or 40 ms (c) 25 Hz
(d) 0
...
62◦) leading 40 sin 50πt

1
...

5
...


193 km 2
...
6 m (b) 94
...
66◦, 44
...
4 m (b) 17
...
163
...
9 m, EB = 4
...
6
...
37 m
32
...
31◦

2
...
37 Hz (c) 0
...
54 rad (or 30
...
(a) 300 V (b) 100 Hz (c) 0
...
412 rad (or 23
...
(a) v = 120 sin100πt volts
(b) v = 120 sin (100πt + 0
...
i = 20 sin 80πt −
6
i = 20 sin(80πt − 0
...
3
...
488) m
7
...
75◦ lagging
(b) −2
...
363 A (d) 6
...
423 ms

Chapter 23
Exercise 90 (page 207)
1
...
1 mm, c = 28
...
A = 52◦2 , c = 7
...
152 cm,
area = 25
...
D = 19◦48 , E = 134◦12 , e = 36
...
E = 49◦ 0 , F = 26◦ 38 , f = 15
...
6 mm2
5
...
420 cm,
area = 6
...
811 cm, area = 0
...
K = 47◦ 8 , J = 97◦ 52 , j = 62
...
2 mm2 or K = 132◦52 , J = 12◦8 ,
j = 13
...
0 mm2

Exercise 93 (page 212)
1
...
42◦, 59
...
20◦ 2
...
23 m (b) 38
...
40
...
05◦
4
...
8 cm 5
...
2 m
6
...
3 mm, y = 142 mm 7
...
13
...

2
...

4
...

6
...

8
...
83, 59
...
83, 1
...
61, 20
...
61, 0
...
47, 116
...
47, 2
...
55, 145
...
55, 2
...
62, 203
...
62, 3
...
33, 236
...
33, 4
...
83, 329
...
83, 5
...
68, 307
...
68, 5
...
294, 4
...
(1
...
960)
(−5
...
500)
4
...
884, 2
...
353, −5
...
(−2
...
207)
(0
...
299)
8
...
252, −4
...
04, 12
...
04, 12
...
51, −32
...
51, −32
...
47
...

3
...

7
...


Answers to practice exercises
Exercise 103 (page 234)

Chapter 25
Exercise 96 (page 221)
1
...
t = 146◦

351

2
...
(i) rhombus (a) 14 cm2 (b) 16 cm (ii) parallelogram
(a) 180 mm2 (b) 80 mm (iii) rectangle (a) 3600 mm2
(b) 300 mm (iv) trapezium (a) 190 cm2 (b) 62
...
35
...
(a) 80 m (b) 170 m 4
...
2 cm2
5
...
1200 mm
7
...
560 m2
2
9
...
4 cm
10
...
43
...
32

1
...

7
...

11
...

16
...

20
...
2376 mm2
3
...
1709 mm
6
...
(a) 106
...
9 cm2
2
21
...
17
...
07 cm2
(a) 59
...
8 mm
12
...
2 cm
8
...
48 cm 14
...
5◦ 15
...
698 rad (b) 804
...
10
...
24%
19
...
8 mm
7
...
(a) 2 (b) (3, −4)
2
...
Circle, centre (0, 1), radius 5
4
...

2
...

5
...


482 m2
(a) 50
...
9 mm2 (c) 3183 mm2
2513 mm2
4
...
19 mm (b) 63
...
01 cm2 (b) 129
...
5773 mm2
2
1
...
1932 mm2 2
...
(a) 0
...

4
...

8
...

12
...

15
...

19
...
2 m3
2
...
8 cm3
3
2
(a) 3840 mm (b) 1792 mm
972 litres
6
...
500 litres
3
9
...
3 cm3 (b) 61
...
44 m
(a) 2400 cm3 (b) 2460 cm2 11
...
04 m
1
...
8796 cm3
4
...
9 cm2
2
...
28060 cm3 , 1
...
22 m by 8
...
62
...
4
...
80 ha

2
...
3
...
45
...
259
...
2
...
47
...
38
...
12730 km 7
...
13 mm

Exercise 106 (page 246)
1
...
1 cm3 , 159
...
7
...
81 cm2
3
...
1 cm3 , 113
...
5
...
3 cm
6
...
(a) 268 083 mm3 or 268
...
06 cm2
8
...
53 cm
9
...
09 × 1012 km3 10
...
(a) 0
...
481 (c) 4
...
(a) 210◦ (b) 80◦ (c) 105◦
4
...

2
...

6
...
(a)

5890 mm2 or 58
...
55 cm3 (b) 84
...
13
...
393
...
32 cm3
(i) (a) 670 cm3 (b) 523 cm2 (ii) (a) 180 cm3
(b) 154 cm2 (iii) (a) 56
...
8 cm2
(iv) (a) 10
...
0 cm2 (v) (a) 96
...

9
...

13
...
5 cm3 (b) 142 cm2
(vii) (a) 805 cm3 (b) 539 cm2
(a) 17
...
0 cm
8
...
3 m , 25
...
6560 litres
12
...
7 cm3
657
...
77 m (c) £140
...
69 cm
5
...
72 N at −14
...
15 m/s at 29
...
28 N at 16
...
6
...
56◦
◦
15
...
33 to the 10 N force
21
...
22◦ S

Exercise 115 (page 276)

Exercise 108 (page 255)
1
...

5
...


6
...

8
...

11
...
403 cm3 , 337 cm2
4
...
55910 cm3 , 6051 cm2

1
...
0 N at 78
...
64 N at 4
...
(a) 31
...
81◦ (b) 19
...
63◦

Exercise 116 (page 277)
1
...
5 km/h at 71
...
4 minutes 55 seconds, 60◦
3
...
79 km/h, E 9
...
8 : 125

2
...
2 g

Chapter 28

Exercise 117 (page 277)

Exercise 110 (page 259)
1
...
5 square units 2
...
7 square units 3
...
33 m
4
...
70 ha
5
...
42
...
147 m3

3
...
42 m3

Exercise 112 (page 263)
1
...
5 A
3
...
093 As, 3
...
49
...
5 kPa

1
...

5
...

9
...
5j − 4k
3
...
4j − 6
...

4
...

8
...


4i + j − 6k
5i − 10k
−5i + 10k
20
...
(a) 2
...
(a) 31
...
4
...
5◦ )
2
...
9 sin(ωt + 0
...
5 sin(ωt − 1
...
13 sin(ωt + 0
...
A scalar quantity has magnitude only; a vector
quantity has both magnitude and direction
...
scalar
3
...
vector 5
...
scalar
7
...
scalar
9
...

2
...

4
...


17
...
00◦ to the 12 N force
13 m/s at 22
...
40 N at 37
...
43 N at 129
...
31 m at 21
...
4
...
5◦ )
2
...
9 sin(ωt + 0
...
5 sin(ωt − 1
...
13 sin(ωt + 0
...
4
...
5◦ )
2
...
9 sin(ωt + 0
...
5 sin(ωt − 1
...
13 sin(ωt + 0
...
11
...
324)
5
...
73 sin(ωt − 0
...
−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

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