Trigonometric Equations Question 47
Question: The shadow of a tower is found to be 60 metre shorter when the sun?s altitude changes from $ 30^{o} $ to $ 60^{o} $ . The height of the tower from the ground is approximately equal to
[Kerala (Engg.) 2005]
Options:
A) 62m
B) 301m
C) 101m
D) 75m
Show Answer
Answer:
Correct Answer: E
Solution:
- $ \tan 30^{o}=\frac{h}{x+60} $ , $ \frac{1}{\sqrt{3}}=\frac{h}{x+60} $ $ x+60=\sqrt{3}h $ , $ x=\sqrt{3}h-60 $ $ \tan 60^{o}=\frac{h}{x} $ , $ x=\frac{h}{\sqrt{3}} $
Þ $ \sqrt{3}h-60=\frac{h}{\sqrt{3}} $ Þ $ 3h-60\sqrt{3}=h $
Þ $ h=\frac{60\sqrt{3}}{2}=30\sqrt{3} $ $ =51.96\approx 52m $ .
BETA
