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An observer at a point P on the top of a hill near the sea-shore notices that the angle of depression of a ship moving towards the hill in a straight line at a constant speed is 30°. After 45 minutes, this angle becomes 45°. If T (in minutes) is the total time taken by the ship to move to a point in the sea where the angle of depression from P of the ship is 60°, then T is equal to:

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$45 (1 + \frac{1}{\sqrt{3}})$

$45 (1 + \sqrt{3})$

$45 (1 + \frac{2}{\sqrt{3}})$

$45 (2 + \frac{1}{\sqrt{3}})$

answer is A.

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Detailed Solution

Let $OP = a,$ then

$\begin{aligned} \frac{O S_{1}}{O P} = \cot (30^{0}) = \sqrt{3} \\ \Rightarrow O S_{1} = \sqrt{3} a \end{aligned}$

Similarly ${OS}_{2} = a, {OS}_{3} = \frac{a}{\sqrt{3}}$

Let speed of the ship be v km/min, then

$\sqrt{3} a - a = 45 v \Rightarrow a = \frac{45 v}{\sqrt{3} - 1}$

Now , $S_{1} S_{3} = \sqrt{3} a - (\frac{a}{\sqrt{3}}) = \frac{2}{\sqrt{3}} a$

$\begin{array}{l} \Rightarrow T = \frac{S_{1} S_{3}}{v} = \frac{2}{\sqrt{3}} \frac{45}{\sqrt{3} - 1} = \frac{45 (\sqrt{3} + 1)}{\sqrt{3}} \\ = 45 (1 + \frac{1}{\sqrt{3}}) . \end{array}$

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