Higher Engineering Mathematics

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

]

12.11 Practical situations involving

trigonometry

There are a number ofpractical situationswhere
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◦

=

a

sin 33◦

from which,

a=

8 .0 sin 33◦

sin 107◦

= 4 .556 m

Also from the sine rule:

8. 0

sin 107◦

=

c

sin 40◦

from which,

c=

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.

Problem 26. Two voltage phasors are shown in

Fig. 12.26. IfV 1 =40 V andV 2 =100 V deter-

mine the value of their resultant (i.e. lengthOA)

and the angle the resultant makes withV 1.

Figure 12.26

AngleOBA= 180 ◦− 45 ◦= 135 ◦

Applying the cosine rule:

OA^2 =V 12 +V 22 − 2 V 1 V 2 cosOBA

= 402 + 1002 −{2(40)(100) cos 135◦}

= 1600 + 10000 −{− 5657 }

= 1600 + 10000 + 5657 = 17257

The resultant

OA=

√

(17257)= 131 .4V

Applying the sine rule:

131. 4

sin 135◦

=

100

sinAOB

from which, sinAOB=

100 sin 135◦

131. 4

= 0. 5381

Hence angle AOB=sin−^10. 5381 = 32 ◦ 33 ′ (or

147 ◦ 27 ′, which is impossible in this case).

Hence the resultant voltage is 131.4 volts at 32◦ 33 ′

toV 1.

Problem 27. In Fig. 12.27,PRrepresents the

inclined jib of a crane and is 10.0 long.PQis

4.0 m long. Determine the inclination of the jib

to the vertical and the length of tieQR