trigonometry

128 GEOMETRY AND TRIGONOMETRY
4. x = 21 mm, y = 34 mm, z = 42 mm
[ Z = 29 ◦ 46 ′ , Y = 53 ◦ 30 ′ ]
,
Z = 96 ◦ 44 ′ , area = 355 mm 2
Problem 26. Two voltage phasors are shown in
Fig. 12.26. If V 1 = 40V and V 2 = 100V determine
the value of their resultant (i.e. length OA)
and the angle the resultant makes with V 1 .
12.11 Practical situations involving
trigonometry
There are a number of practical situations where
the use of trigonometry is needed to find unknown
sides and angles of triangles. This is demonstrated
in Problems 25 to 30.
Problem 25. A room 8.0 m wide has a span
roof which slopes at 33 ◦ on one side and 40 ◦
on the other. Find the length of the roof slopes,
correct to the nearest centimetre.
A section of the roof is shown in Fig. 12.25.
Figure 12.25
Angle at ridge, B = 180 ◦ − 33 ◦ − 40 ◦ = 107 ◦
From the sine rule:
8.0
sin 107 ◦ =
from which,
a =
a
sin 33 ◦
8.0 sin 33◦
sin 107 ◦ = 4.556 m
Also from the sine rule:
8.0
sin 107 ◦ =
from which,
c =
c
sin 40 ◦
8.0 sin 40◦
sin 107 ◦ = 5.377 m
Hence the roof slopes are 4.56 m and 5.38 m,
correct to the nearest centimetre.
Figure 12.26
Angle OBA = 180 ◦ − 45 ◦ = 135 ◦
Applying the cosine rule:
OA 2 = V 2 1 + V 2 2 − 2V 1V 2 cos OBA
The resultant
= 40 2 + 100 2 −{2(40)(100) cos 135 ◦ }
= 1600 + 10000 −{−5657}
= 1600 + 10000 + 5657 = 17257
OA = √ (17257) = 131.4V
Applying the sine rule:
from which,
131.4
sin 135 ◦ = 100
sin AOB
100 sin 135◦
sin AOB =
131.4
= 0.5381
Hence angle AOB = sin −1 0.5381 = 32 ◦ 33 ′ (or
147 ◦ 27 ′ , which is impossible in this case).
Hence the resultant voltage is 131.4 volts at 32 ◦ 33 ′
to V 1 .
Problem 27. In Fig. 12.27, PR represents the
inclined jib of a crane and is 10.0 long. PQ is
4.0 m long. Determine the inclination of the jib
to the vertical and the length of tie QR