Quantitative Aptitude Ques 1788
Question: The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is $ 60{}^\circ $ and the angle of elevation of the top of the second tower from the foot of the first tower is $ 30{}^\circ . $ The distance between the two towers is n times the height of the shorter tower. What is n equal to?
Options:
A) $ \sqrt{2} $
B) $ \sqrt{3} $
C) $ \frac{1}{2} $
D) $ \frac{1}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Let h be the height of shorter tower, Then, the distance between the two towers is given by nh m. In $ \Delta CBD, $ $ \tan 30{}^\circ =\frac{h}{nh} $
$ \Rightarrow $ $ \frac{1}{\sqrt{3}}=\frac{1}{n} $
$ \Rightarrow $ $ n=\sqrt{3} $
BETA
