Introduction to trigonometry 115
Angle OA
B
= 180
◦
− 120
◦
− 16
◦ 47
= 43
◦ 13
.
Applying the sine rule:
30.0
sin 120 ◦ =
OB
sin 43 ◦ 13
from which,
OB
=
30.0 sin 43 ◦ 13
sin 120 ◦
= 23.72 cm
Since OB = 35.43 cm and OB = 23.72 cm then BB =
35.43 − 23.72 = 11.71 cm.
Hence B moves 11.71 cm when angle AOB changes
from 50 ◦ to 120 ◦ .
Problem 36. The area of a field is in the form of a
quadrilateral ABCD as shown in Fig. 11.39.
Determine its area.
62.3 m
21.4 m
39.8 m
568
1148
D
A
B
C
42 .5 m
Figure 11.39
A diagonal drawn from B to D divides the quadrilateral
into two triangles.
Area of quadrilateral ABCD
= area of triangle ABD + area of triangle BCD
=
1
2 (39.8)(21.4) sin 114
◦
+
1
2 (42.5)(62.3) sin 56
◦
= 389.04 + 1097.5 = 1487 m
2
Now try the following exercise
Exercise 52 Further problems on practical
situations involving trigonometry
1. PQ and QR are the phasors representing the
alternating currents in two branches of a circuit. Phasor PQ is 20.0 A and is horizontal.
Phasor QR (which is joined to the end of PQ
to form triangle PQR) is 14.0 A and is at an
angle of 35 ◦ to the horizontal. Determine the
resultant phasor PR and the angle it makes with
phasor PQ.
[32.48 A, 14 ◦ 19 ]
2. Three forces acting on a fixed point are represented by the sides of a triangle of dimensions
7.2 cm, 9.6 cm and 11.0 cm. Determine the
angles between the lines of action and the
three forces.
[80 ◦ 25 , 59 ◦ 23 , 40 ◦ 12 ]
3. Calculate, correct to 3 significant figures, the
co-ordinates x and y to locate the hole centre
at P shown in Fig. 11.40.
[x = 69.3 mm, y = 142 mm]
100 mm
x
P
y
116Њ
140Њ
Figure 11.40
4. An idler gear, 30 mm in diameter, has to be
fitted between a 70 mm diameter driving gear
and a 90 mm diameter driven gear as shown
in Fig. 11.41. Determine the value of angle θ
between the center lines.
[130 ◦ ]
70 mm dia
30 mm dia
90 mm dia
99.78 mm
Figure 11.41
5. A reciprocating engine mechanism is shown
in Fig. 11.42. The crank AB is 12.0 cm long
and the connecting rod BC is 32.0 cm long.
Angle OA
B
= 180
◦
− 120
◦
− 16
◦ 47
= 43
◦ 13
.
Applying the sine rule:
30.0
sin 120 ◦ =
OB
sin 43 ◦ 13
from which,
OB
=
30.0 sin 43 ◦ 13
sin 120 ◦
= 23.72 cm
Since OB = 35.43 cm and OB = 23.72 cm then BB =
35.43 − 23.72 = 11.71 cm.
Hence B moves 11.71 cm when angle AOB changes
from 50 ◦ to 120 ◦ .
Problem 36. The area of a field is in the form of a
quadrilateral ABCD as shown in Fig. 11.39.
Determine its area.
62.3 m
21.4 m
39.8 m
568
1148
D
A
B
C
42 .5 m
Figure 11.39
A diagonal drawn from B to D divides the quadrilateral
into two triangles.
Area of quadrilateral ABCD
= area of triangle ABD + area of triangle BCD
=
1
2 (39.8)(21.4) sin 114
◦
+
1
2 (42.5)(62.3) sin 56
◦
= 389.04 + 1097.5 = 1487 m
2
Now try the following exercise
Exercise 52 Further problems on practical
situations involving trigonometry
1. PQ and QR are the phasors representing the
alternating currents in two branches of a circuit. Phasor PQ is 20.0 A and is horizontal.
Phasor QR (which is joined to the end of PQ
to form triangle PQR) is 14.0 A and is at an
angle of 35 ◦ to the horizontal. Determine the
resultant phasor PR and the angle it makes with
phasor PQ.
[32.48 A, 14 ◦ 19 ]
2. Three forces acting on a fixed point are represented by the sides of a triangle of dimensions
7.2 cm, 9.6 cm and 11.0 cm. Determine the
angles between the lines of action and the
three forces.
[80 ◦ 25 , 59 ◦ 23 , 40 ◦ 12 ]
3. Calculate, correct to 3 significant figures, the
co-ordinates x and y to locate the hole centre
at P shown in Fig. 11.40.
[x = 69.3 mm, y = 142 mm]
100 mm
x
P
y
116Њ
140Њ
Figure 11.40
4. An idler gear, 30 mm in diameter, has to be
fitted between a 70 mm diameter driving gear
and a 90 mm diameter driven gear as shown
in Fig. 11.41. Determine the value of angle θ
between the center lines.
[130 ◦ ]
70 mm dia
30 mm dia
90 mm dia
99.78 mm
Figure 11.41
5. A reciprocating engine mechanism is shown
in Fig. 11.42. The crank AB is 12.0 cm long
and the connecting rod BC is 32.0 cm long.