Trigonometric Equations Question 325
Question: The shadow of a tower standing on a level ground is found to be 60 m longer when the sun’s altitude is $ 30^{o} $ than when it is $ 45^{o} $ . The height of the tower is
[EAMCET 2001]
Options:
A) 60 m
B) 30 m
C) $ 60\sqrt{3}m $
D) $ 30(\sqrt{3}+1)m $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \because ,AB=AM-BM\Rightarrow \frac{AB}{h}=\frac{AM}{h}-\frac{BM}{h} $ $ \frac{AB}{h}=\cot 30^{o}-\cot 45^{o}\Rightarrow h=\frac{60}{\sqrt{3}-1},=\frac{60(\sqrt{3}+1)}{3-1}, $
Þ $ h=30(\sqrt{3}+1),m $ .
BETA
