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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.

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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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