112 Higher Engineering Mathematics

A section of the roof is shown in Fig. 11.30.

B

A 8.0 m C

338 408

Figure 11.30

Angle at ridge,B= 180 ◦− 33 ◦− 40 ◦= 107 ◦

From the sine rule:

8. 0

sin107◦

=

a

sin33◦

from which,

a=

8 .0sin33◦

sin107◦

= 4 .556m

Also from the sine rule:

8. 0

sin107◦

=

c

sin40◦

from which,

c=

8 .0sin40◦

sin107◦

= 5 .377m

Hence the roof slopes are 4.56m and 5.38m, correct

to the nearest centimetre.

Problem 32. Two voltage phasors are shown in

Fig. 11.31. IfV 1 =40V andV 2 =100V determine

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

angle the resultant makes withV 1.

45 

(^0) B
A
V 1 40 V
V 2 100 V
Figure 11.31
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 )cos135◦}
= 1600 + 10000 −{− 5657 }
= 1600 + 10000 + 5657 = 17257
The resultant
OA=
√
( 17257 )= 131 .4V
Applying the sine rule:
131. 4
sin135◦

100
sinAO B
from which, sinAO B=
100sin135◦
131. 4
= 0. 5381
Hence angle AO B=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 33. In Fig. 11.32,PRrepresents the
inclined jib of a crane and is 10.0 long.PQis 4.0m
long. Determine the inclination of the jib to the
vertical and the length of tieQR.
120 
4.0 m
10.0 m
R
Q
P
Figure 11.32
Applying the sine rule:
PR
sin120◦

PQ
sinR
from which,
sinR=
PQsin120◦
PR

( 4. 0 )sin120◦
10. 0
= 0. 3464
Hence∠R=sin−^10. 3464 = 20 ◦ 16 ′(or 159◦ 44 ′,which
is impossible in this case).
∠P= 180 ◦− 120 ◦− 20 ◦ 16 ′= 39 ◦ 44 ′, which is the
inclination of the jib to the vertical.
Applying the sine rule:
10. 0
sin120◦

QR
sin39◦ 44 ′