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Height and Distance
Height and Distance
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Question

A man is observing, from the top of a tower, a boat speeding towards the lower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30∘^\circ (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∘^\circ. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :

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Solution

hx+y=tan⁡30∘{h \over {x + y}} = \tan 30^\circ x+y=3hx + y = \sqrt 3 h ...... (1) Also, hy=tan⁡45∘{h \over y} = \tan 45^\circ h=yh = y ..... (2) put in (1) x+y=3yx + y = \sqrt 3 y x=(3−1)yx = \left( {\sqrt 3 - 1} \right)y x20=′v′{x \over {20}} = 'v' speed ∴\therefore time taken to reach Foot from B =yV = {y \over V} =x(3−1).x×20 = {x \over {\left( {\sqrt 3 - 1} \right).x}} \times 20 =10(3+1) = 10\left( {\sqrt 3 + 1} \right)

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