  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+BC1√3=h12V+x⇒√3h=12V+x⇒√3x=12V+x        [Using eqn (1)]⇒(√3−1)x=12V⇒xV=12(√3−1) min⇒xV=12(√3−1)(√3−1)(√3+1) min⇒xV=12(√3−1)3−1 min⇒xV=6(√3−1) min⇒xV=983 sec Time taken by car to reach the tower from C=xV ∴ Time taken by car to reach the tower is 983 sec.  Suggest corrections   