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Question

A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed, If the angle of depression of the car changes from 30 to 45 in 12 minutes, find the time taken by the car now to reach the tower.


Solution

S

Let AB be the vertical tower. Suppose D and C be the positions of the car when the angle of depression from the top of the tower is 30° and 45° respectively.

Consider the uniform speed of car be V m/min.

Time taken for the angle of depression to change from 30° to 45°  = 12 min 

∠EAD = ∠ADB = 30°  (Alternate angles)

∠EAC = ∠ACB = 45°  (Alternate angles)

Suppose AB = h m and BC = x m.

CD = Distance covered by car in 12 min=V m/min×12=12V min

In ΔABC,

tan 45o=ABBC1=hxh=x(1)

In ΔADB,

tan 30o=ABCD+BC13=h12V+x3h=12V+x3x=12V+x        [Using eqn (1)](31)x=12VxV=12(31) minxV=12(31)(31)(3+1) minxV=12(31)31 minxV=6(31) minxV=983 sec

Time taken by car to reach the tower from C=xV

∴ Time taken by car to reach the tower is 983 sec.

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