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Question

Mathematics Question on Boat and Stream

From the top of a light-house 60 m high with its base at the sea-level, the angle of depression of a boat is 1515\degree. The distance of the boat from the foot of the light-house is

A

(3131)×(60m)(\frac{\sqrt3-1}{\sqrt3-1})\times(60m)

B

(3+131)×(60m)(\frac{\sqrt3+1}{\sqrt3-1})\times(60m)

C

(3+131)×(30m)(\frac{\sqrt3+1}{\sqrt3-1})\times(30m)

D

(313+1)×(30m)(\frac{\sqrt3-1}{\sqrt3+1})\times(30m)

Answer

(3+131)×(60m)(\frac{\sqrt3+1}{\sqrt3-1})\times(60m)

Explanation

Solution

The correct option is (B): Top of a light house
Given AB = 60m60m ACB=15\angle ACB=15\degree To find BC:
Using identity tan(AB)=tanAtanB1tanAtanBtan(A-B)=\frac{tanA-tanB}{1-tanAtanB}

tan15=(tan60tan45)tan15\degree= (tan60\degree-tan45\degree)

(tan60tan45)1tan60tan45\frac{(tan60\degree-tan45\degree)}{1-tan60\degree tan45\degree}

311+3\frac{\sqrt3-1}{1+\sqrt3}……..(i)

InABCIn \bigtriangleup ABC
tan(15)=ABBCtan(15\degree)=\frac{AB}{BC}

313+1=60BC\frac{\sqrt3-1}{\sqrt3+1}=\frac{60}{BC}

BC=3+13160\frac{\sqrt3+1}{\sqrt3-1}60