Introduction to trigonometry 115
AngleOA′B′= 180 ◦− 120 ◦− 16 ◦ 47 ′= 43 ◦ 13 ′.
Applying the sine rule:
30. 0
sin120◦
=
OB′
sin43◦ 13 ′
from which,
OB′=
30 .0sin43◦ 13 ′
sin120◦
= 23 .72cm
SinceOB= 35 .43cm andOB′= 23 .72cm thenBB′=
35. 43 − 23. 72 = 11 .71cm.
HenceBmoves 11.71cm when angleAOBchanges
from 50◦to 120◦.
Problem 36. The area of a field is in the form of a
quadrilateralABCDas 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 fromBtoDdivides the quadrilateral
into two triangles.
Area of quadrilateral ABCD
=area of triangleABD+area of triangleBCD
=^12 ( 39. 8 )( 21. 4 )sin114◦+^12 ( 42. 5 )( 62. 3 )sin 56◦
= 389. 04 + 1097. 5 =1487m^2
Now try the following exercise
Exercise 52 Further problemson practical
situations involving trigonometry
- PQandQRare the phasors representing the
alternating currents in two branches of a cir-
cuit. PhasorPQis 20.0A and is horizontal.
PhasorQR(which is joined to the end ofPQ
to form trianglePQR) is 14.0A and is at an
angle of 35◦to the horizontal. Determine the
resultant phasorPRand theangleit makes with
phasorPQ. [32.48A, 14◦ 19 ′]
- Three forces acting on a fixed point are repre-
sented by the sides of a triangle of dimensions
7.2cm, 9.6cm and 11.0cm. Determine the
angles between the lines of action and the
three forces. [80◦ 25 ′, 59 ◦ 23 ′, 40 ◦ 12 ′] - Calculate, correct to 3 significant figures, the
co-ordinatesxandyto locate the hole centre
atPshown in Fig. 11.40.
[x=69.3mm,y=142mm]
x 100 mm
P
y
116 ^140
Figure 11.40
- An idler gear, 30mm in diameter, has to be
fitted between a 70mm diameter driving gear
and a 90mm diameter driven gear as shown
in Fig. 11.41. Determine the value of angleθ
between the center lines. [130◦]
70 mm dia
30 mm dia
90 mm dia
99.78 mm
Figure 11.41
- A reciprocating engine mechanism is shown
in Fig. 11.42. The crankABis 12.0cm long
and the connecting rodBCis 32.0cm long.