114 Higher Engineering Mathematics
448
488
30.0 m
D
C
B
A
Figure 11.36
For triangleABC, using Pythagoras’ theorem:
BC^2 =AB^2 +AC^2
(
DC
tan44◦
) 2
=( 30. 0 )^2 +
(
DC
tan48◦
) 2
DC^2
(
1
tan^244 ◦
−
1
tan^248 ◦
)
= 30. 02
DC^2 ( 1. 072323 − 0. 810727 )= 30. 02
DC^2 =
30. 02
0. 261596
= 3440. 4
Hence, height of aerial,
DC=
√
3440.4=58.65m
Problem 35. A crank mechanism of a petrol
engine is shown in Fig. 11.37. ArmOAis 10.0cm
long and rotates clockwise about O. The connecting
rodABis 30.0cm long and endBis constrained to
move horizontally.
30.0 cm
A
B
O
10.0 cm
508
Figure 11.37
(a) For the position shown in Fig. 11.37 determine
the angle between the connecting rodABand
the horizontal and the length ofOB.
(b) How far doesBmove when angleAOB
changes from 50◦to 120◦?
(a) Applying the sine rule:
AB
sin50◦
=
AO
sinB
from which,
sinB=
AOsin50◦
AB
=
10 .0sin50◦
30. 0
= 0. 2553
Hence B=sin−^10. 2553 = 14 ◦ 47 ′ (or 165◦ 13 ′,
which is impossible in this case).
Hence the connecting rodABmakes an angle
of 14◦ 47 ′with the horizontal.
AngleOAB= 180 ◦− 50 ◦− 14 ◦ 47 ′= 115 ◦ 13 ′.
Applying the sine rule:
30. 0
sin50◦
=
OB
sin115◦ 13 ′
from which,
OB=
30 .0sin115◦ 13 ′
sin50◦
=35.43cm
(b) Figure11.38showstheinitial andfinalpositionsof
thecrank mechanism.In triangleOA′B′, applying
the sine rule:
30. 0
sin120◦
=
10. 0
sinA′B′O
from which,
sinA′B′O=
10 .0sin120◦
30. 0
= 0. 2887
A A
BB O
50
30.0 cm
10.0 cm
120
Figure 11.38
Hence A′B′O=sin−^10. 2887 = 16 ◦ 47 ′ (or 163◦ 13 ′
which is impossible in this case).