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} $

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