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Chapter 7

Powers, roots and laws of
indices
7
...
In this chapter, powers and
roots of numbers are explained, together with the laws
of indices
...


7
...
2
...
When written as 24 , 2 is
called the base and the 4 is called the index or power
...

Similarly, 35 is read as ‘three to the power of 5’
...
e
...

Thus,
42 is called ‘four squared’ rather than ‘4 to the power
of 2’ and
53 is called ‘five cubed’ rather than ‘5 to the power of 3’
When no index is shown, the power is 1
...

Problem 1
...
e
...
e
...
1016/B978-1-85617-697-2
...
e
...
e
...
Change the following to index form:
(a) 32 (b) 625
(a) (i) To express 32 in its lowest factors, 32 is initially
divided by the lowest prime number, i
...
2
...

(iii) 16 is also divisible by 2, i
...
16 = 2 × 8
...

(iv) 8 is also divisible by 2, i
...
8 = 2 × 4
...

(v) 4 is also divisible by 2, i
...
4 = 2 × 2
...

(vi) Thus, 32 = 25
...
e
...
The next prime number is 3 and 625
is not divisible by 3 either
...

(ii) 625 ÷ 5 = 125, i
...
625 = 5 × 125
...
e
...

Thus, 625 = 5 × 5 × 25
...
e
...
Thus,
625 = 5 × 5 × 5 × 5
...


48 Basic Engineering Mathematics
Problem 3
...


= 108

7
...
2 Square roots

1
...


27

When a number is multiplied by itself the product is
called a square
...

A square root is the reverse process; i
...
, the value of the
base which when multiplied by itself gives the number;
i
...
, the square
√ root of 9 is 3
...
Thus,
9 = 3
...

√
√
Because −3 × −3 = 9, 9 also equals √
−3
...
Simi√
larly, 16 = ±4 and 36 = ±6
...

√
1
9 2 ≡ 9 = ±3

3
...


24 × 32 × 2 ÷ 3

5
...


6
...


64 2

8
...


102 × 103
105

√
32 × 23 × 36
Problem 4
...
3

by cancelling

1

10
...


Laws of indices

There are six laws of indices
...
Evaluate
taking the
103
positive square root only

1

1

Hence,

22 × 23 = 25

or

22 × 23 = 22+3

This is the first law of indices, which demonstrates
that when multiplying two or more numbers
having the same base, the indices are added
...


(3) (35 )2 = 35×2 = 310 and (22 )3 = 22×3 = 26
This is the third law of indices, which demonstrates that when a number which is raised to
a power is raised to a further power, the indices
are multiplied
...

1
1
and −3 = 23
4
3
2
This is the fifth law of indices, which demonstrates
that a number raised to a negative power is the
reciprocal of that number raised to a positive
power
...

Here are some worked examples using the laws of
indices
...
Evaluate in index form 53 × 5 × 52
3

2

3

1

5 ×5×5 =5 ×5 ×5
=5

2

1

(Note that 5 means 5 )

3+1+2

Problem 9
...

2
2 5
But

Problem 10
...
001
= 10−3 =
8
10
1000

from law (2)

102 = 100, 103 = 1000, 104 = 10 000,
105 = 100 000, 106 = 1 000 000

=3

10−1 =

24
Problem 8
...
1, 10−2 = 2 =
= 0
...

= 24−4
=2

Hence,

from law (2)

Understanding powers of ten is important, especially
when dealing with prefixes in Chapter 8
...
Evaluate

But

3 × 32
34

3 × 32 31 × 32 31+2 33
=
= 4 = 4 = 33−4 = 3−1
34
34
3
3
from laws (1) and (2)

= 56

24

49

from law (2)

0

2 × 2 × 2 × 2 16
=
=1
2 × 2 × 2 × 2 16
20 = 1

Problem 11
...
For
example, 60 = 1, 1280 = 1, 137420 = 1 and so on
...
Simplify (a) (23 )4 (b) (32 )5 ,
expressing the answers in index form
From law (3):
(a)

(23 )4 = 23×4 = 212

(b)

(32 )5

= 32×5

Problem 13
...

(b)

Find the value of (a)

(32 )3
3 × 39

23 × 24
27 × 25

From the laws of indices:
(a)

(b)

23 × 24 2(3+4)
27
=
=
= 27−12
27 × 25 2(7+5) 212
1
1
= 2−5 = 5 =
2
32
(32 )3
32×3
36
= 1+9 = 10 = 36−10
9
3×3
3
3
= 3−4 =

Problem 15
...

1
...


35 × 33 × 3
in index form

3
...


33
35

5
...


23 × 2 × 26
27

7
...


104 ÷ 10

9
...


56 × 52 ÷ 57

11
...


(33)2

13
...


(9 × 32 )3
in
(3 × 27)2
index form

15
...


5−2
5−4

17
...


72 × 7−3
7 × 7−4

19
...


5−7 × 52
5−8 × 53

Here are some further worked examples using the laws
of indices
...


√
4 = ±2
√
4
3/4
(b) 16 = 163 = (2)3 = 8
(Note that it does not matter whether the 4th root
of 16 is found first or whether 16 cubed is found
first – the same answer will result
...
Grouping terms having the same base and
then applying the laws of indices to each of the groups
independently gives
33 × 57 33 57
=
×
= 3(3−4) × 5(7−3)
53 × 34 34 53
= 3−1 × 54 =

54
625
1
=
= 208
1
3
3
3

Powers, roots and laws of indices

Problem 17
...
Find the value of

32 × 55
2
3
32 × 55
= 4 34 × 5 3
4
4
3
3
3 ×5 +3 ×5
3 ×5
3 × 53
+
32 × 53 32 × 53

√
√
3
41
...
5 × 81/3
8×2
16
= 16
=
=
2
−2/5
1
1
2 × 32
4×
4
Alternatively,
Hence,

[(2)2 ]3/2 × (23 )1/3
41
...
Evaluate

32 × 55 + 33 × 53
34 × 54

Dividing each term by the HCF (highest common factor)
of the three terms, i
...
32 × 53 , gives
32 × 55 33 × 53
+
32 × 55 + 33 × 53
32 × 53 32 × 53
=
34 × 54
34 × 54
32 × 53
=

3(2−2) × 5(5−3) + 3(3−2) × 50
3(4−2) × 5(4−3)

=

30 × 52 + 31 × 50
32 × 51

=

1 × 25 + 3 × 1 28
=
9×5
45

32 × 55
34 × 54 + 33 × 53

To simplify the arithmetic, each term is divided by the
HCF of all the terms, i
...
32 × 53
...
5 × 81/3
Problem 18
...
Simplify

Since 7−3 =

1
1
1
,
= 32 and −2 = 52 , then
73 3−2
5

34 × 32 × 52
7−3 × 34
=
3−2 × 75 × 5−2
73 × 75
=

3(4+2) × 52 36 × 52
=
7(3+5)
78

162 × 9−2
4 × 33 − 2−3 × 82
expressing the answer in index form with positive
indices

Problem 22
...
e
...
Simplify
giving
 −3
2
5
the answer with positive indices
Raising a fraction to a power means that both the numerator and the denominator of the fraction are raised to that
 3
4
43
power, i
...

= 3
3
3
A fraction raised to a negative power has the same value
as the inverse of the fraction raised to a positive power
...

1
...


7−2 × 3−2
35 × 74 × 7−3

3
...


8−2 × 52 × 3−4
252 × 24 × 9−2

In Problems 5 to 15, evaluate the expressions given
...

6
...
25
32
1
 1/2
−
4
4
7
...

9
92 × 74
33 × 52
9
...

3 × 74 + 33 × 72
23 × 32 − 82 × 9
33 × 72 − 52 × 73
32 × 5 × 72
 3  −2
1
2
−
2
3
13
...


15
...
 2
2
9
12
...

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
...
£556 2
...
264 kg 4
...
14
...
(a) 0
...
(a) 440 K (b) 5
...
0
...
173
...
5
...
37
...
128
...
0
...
0
...
0
...
38
...
(a) 21
...
2% (c) 169%
13
5
9
11
...

13
...

20
16
16
1
15
...
25, D = 25%, E = 0
...
60, H = 60%, I = 0
...
(a) 2 mA (b) 25 V 2
...
685
...
83 lb10 oz
5
...
1 litres (b) 16
...
29
...
584
...
$1012

Exercise 28 (page 46)
Exercise 22 (page 36)
1
...

5
...

10
...


21
...
9
...
4 t (b) 8
...
67%
14 minutes 57 seconds
37
...
39
...
7%
15
...
60 m

(c) 20
...

(c) 5
...

8
...

12
...

16
...
5%

2
...
8 g
£611
38
...
3
...
20 days
3
...
18 (b) 6
...
3375 4
...
(a) 300 × 103 (b) 0
...

5
...

13
...

15
...


2
...
18%
3
...
£175 000
£260
6
...
£9116
...
£50
...
60 10
...
70 11
...
7
...
6 kg, B 0
...
5 kg
54%, 31%, 15%, 0
...
5 mm, 11
...
600 kW

Chapter 6

1
...
±5

2
...
±8

3
...
100

5
...
64

4
...
1

Exercise 30 (page 50)
1
5
...
8
3
...

9
1
or 0
...
5 11
...
100 8
...

100
12
...
36
14
...
1 16
...
5 or
18
...

or 0
...
1
3
243
2
1
...
39

Exercise 24 (page 41)
1
...
3
...
47 : 3
1
4
...
5 hours or 5 hours 15 minutes
4
6
...
12 cm
8
...


1
3 × 52

5
...
1 : 15 2
...
25% 4
...
6 kg
5
...
3 kg 6
...
−
18
9
...


1
3
7 × 37

6
...
−1
14
...
±
2

3
...
−3
15
...


1
210 × 52

8
...

45
9

2
3

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 =

Answers to practice exercises
I
PR
E
11
...
C = (F − 32)
9
9
...
L =

XL
2π f

12
...
x = a(y − 3)
14
...
64 mm

1
2π CX C

*

1
√

Z2 −

14
...
1 × 10−6

aμ
ρCZ 4 n

2

Chapter 13
Exercise 47 (page 87)

Exercise 49 (page 92)

S −a
a
1
...
x =

yd
d
(y + λ) or d +
λ
λ

3
...
D =

AB 2
5E y

5
...
R2 =

R R1
R1 − R

E −e
E − e − Ir
or R =
−r
I
I

y

ay
8
...
x = 
2
4ac
(y 2 − b2 )

7
...
R =
πθ
√
Z 2 − R2
, 0
...
L =
2π f
10
...
u =




xy
1
...
r =
(1 − x − y)
c
5
...
b =
2( p2 + q 2 )
9
...
L =

8S 2
3d

Q
, 55
mc
+ d, 2
...
L =
μ−m


x−y
6
...
R = 4

uf
, 30
u− f


2dgh
, 0
...
v =
0
...
v =

x = 4, y = 2
x = 2, y = 1
...
5, n = 0
...

4
...

8
...

12
...

16
...
a = N 2 y − x

Exercise 48 (page 89)

1
...

5
...

9
...

13
...


1
...

5
...


p = −1, q = −2
a = 2, b = 3
x = 3, y = 4
x = 10, y = 15

2
...

6
...


x = 4, y = 6
s = 4, t = −1
u = 12, v = 2
a = 0
...
40

Exercise 51 (page 96)
1
1
1
...
p = , q =
4
5
5
...
x = 5, y = 1
4

1
1
2
...
x = 10, y = 5
1
6
...
1

Exercise 52 (page 99)
1
...

5
...

8
...
2, b = 4
u = 12, a = 4, v = 26
m = −0
...
00426, R0 = 22
...
I1 = 6
...
62
4
...
a = 12, b = 0
...
F1 = 1
...
5

Exercise 53 (page 100)
1
...
x = 5, y = −1, z = −2

2
...
x = 4, y = 0, z = 3

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)

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