A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man’s eye is 30 (Ignore man’s height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is . $45^{0}$ Then the time taken (in seconds) by the boat from B to reach the base of the tower is

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$10 \sqrt{3}$

$10 (\sqrt{3} + 1)$

$10 (\sqrt{3} - 1)$

answer is A.

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Detailed Solution

$\begin{array}{l} v = \frac{d}{t} \Rightarrow t = \frac{d}{v} \Rightarrow 20 = \frac{x}{v} \\ It gives x = 20 v \\ From triangle A B P, it gives Tan 30 ° = \frac{h}{x + y} \\ \frac{1}{\sqrt{3}} = \frac{h}{x + y} \Rightarrow x + y = \sqrt{3} h \end{array}$

$\begin{array}{l} From triangle Δ P Q B, the value of Tan 45^{0} = \frac{h}{y}, it implies y = h \\ x + y = \sqrt{3} y \\ x = (\sqrt{3} - 1) y \\ 20 V = (\sqrt{3} - 1) y \\ \frac{y}{v} = \frac{20}{\sqrt{3} - 1} = 10 (\sqrt{3} + 1) \sec \end{array}$

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