question_answer

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?

A) \[\sqrt{2}\]                                

B) \[\sqrt{3}\]

C) \[\frac{1}{2}\]                          

D) \[\frac{1}{3}\]

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 \[\Delta CBD,\]\[\tan 30{}^\circ =\frac{h}{nh}\]

\[\Rightarrow \]\[\frac{1}{\sqrt{3}}=\frac{1}{n}\]\[\Rightarrow \]\[n=\sqrt{3}\]