Trigonometric Equations Question 142
Question: The length of the shadow of a pole inclined at 10o to the vertical towards the sun is 2.05 metres, when the elevation of the sun is 38o. The length of the pole is
[Roorkee 1976]
Options:
A) $ \frac{2.05\sin 38^{o}}{\sin ,42^{o}} $
B) $ \frac{2.05\sin 42^{o}}{\sin ,38^{o}} $
C) $ \frac{2.05\cos 38^{o}}{\cos 42^{o}} $
D) None of these
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Answer:
Correct Answer: A
Solution:
- $ \frac{\sin 38^{o}}{l}=\frac{\sin (SPO)}{2.05} $ = $ \frac{\sin (180^{o}-38^{o}-90^{o}-10^{o})}{2.05}\Rightarrow l=\frac{2.05\sin 38^{o}}{\sin 42^{o}} $ .
BETA
