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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
Suppose AB be the tower of height h meters. Let C be the initial position of the car and let after 12 minutes the car be at D. It is given that the angles of depression at C and D are 30º and 45º respectively.
Let the speed of the car be v meter per minute. Then,
CD = distance travelled by the car in 12 minutes
CD = 12v meters
Suppose the car takes t minutes to reach the tower AB from D. Then DA = vt meters
In ∆DAB, we have
`tan 45^@ = (AB)/(AD)`
`=> 1= h/(vt)`.....(i)
In ∆CAB, we have
`tan 30^@ = (AB)/(AB)`
`=> 1/sqrt3 = h/(vt + 12v)`
`=> sqrt3t = vt + 12v` .....(ii)
Substituting the value of h from equation (i) in equation (ii), we get
`=> sqrt3t = t + 12`
`=> t = 12/(sqrt3 - 1) = 12/((sqrt3 - 1)) xx ((sqrt3 + 1))/((sqrt3 + 1)) = 6(sqrt3 + 1)`
t = 16.39 minutes
t = 16 minutes 23 seconds
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