Car Approaching a Tower

CAT 2003 Slot 1 · QA · Medium · Trigonometry

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A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 min for the angle of elevation to change from 45° to 60°. After how much more time will this car reach the base of the tower?

  1. A.

    5(3+1)5 (\sqrt{3} + 1)

  2. B.

    6(3+2)6 (\sqrt{3} + \sqrt{2})

  3. C.

    7(3−1)7 (\sqrt{3} - 1)

  4. D.

    8(3−2)8 (\sqrt{3} - 2)

Answer

A

Explanation

Let tower height h=1h = 1. Initial distance C=cot⁡45∘=1C = \cot 45^\circ = 1. Distance after 10 min D=cot⁡60∘=1/3D = \cot 60^\circ = 1/\sqrt{3}. Distance covered in 10 min = 1−1/31 - 1/\sqrt{3}. Remaining distance = 1/31/\sqrt{3}. Time required = 10×1/31−1/3=103−1=5(3+1)10 \times \frac{1/\sqrt{3}}{1 - 1/\sqrt{3}} = \frac{10}{\sqrt{3} - 1} = 5(\sqrt{3} + 1) minutes.

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