question_answer

A man from the top of a 100 meters high tower sees a car moving towards the tower at an angle of depression of 30 °. After some time, the angle of depression becomes\[{{60}^{o}}\]. The distance (in meters) travelled by the car during the time is [IIT Screening 2001]

A) \[100\sqrt{3}\]

B) \[\frac{200\sqrt{3}}{3}\]

C) \[\frac{100\sqrt{3}}{3}\]

D) \[200\sqrt{3}\]

Correct Answer: B

Solution :

\[x=QR-QS=100\cot {{30}^{o}}-100\cot {{60}^{o}}\] \[=100\sqrt{3}-\frac{100}{\sqrt{3}}=\frac{200}{\sqrt{3}}=\frac{200\sqrt{3}}{3}\].