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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
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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
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A ladder 3
...
0 m from the
wall
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8

21
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9,
opposite side
hypotenuse

sine θ =

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

BC
AC
C

12
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One
travels due west at 18
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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
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13(a)
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13(b), the horizontal and vertical
lines AC and BC are constructed
...
211 correct to 3
decimal places
...
e
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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
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sin 29◦ = 0
...
62◦ = 0
...
3907 cos 83
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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
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4541
43◦
= sin 67
...
◦ = 0
...
466666
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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
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65◦ = 0
...
Press sin

2
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Press◦ ”’

4
...
Press◦ ”’

6
...
Press =

Answer = 0
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Problem 9
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20◦ = 2
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Evaluate tan 2
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93 means 2
...

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

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

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