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The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is \( \Large 60 ^{\circ} \), and the angle of elevation of the top of the second tower from the foot of the first tower is \( \Large 30 ^{\circ} \). The distance between the two towers is n times the height of the shorter tower. What is n equal to?


A) \( \Large \sqrt{2} \)

B) \( \Large \sqrt{3} \)

C) \( \Large \frac{1}{2} \)

D) \( \Large \frac{1}{3} \)

Correct Answer:
B) \( \Large \sqrt{3} \)

Description for Correct answer:
Let h be the height of shorter tower. Then, the distance between the two towers is given by nh m.



In \( \Large \triangle BCD, \tan 30 ^{\circ} = \frac{h}{nh} \)

\( \Large \frac{1}{\sqrt{3}}=\frac{1}{n} => n=\sqrt{3} \)

Part of solved Height and Distance questions and answers : >> Elementary Mathematics >> Height and Distance




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