Advertisement Remove all ads

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. - Mathematics

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.

Advertisement Remove all ads


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

  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

View all notifications

      Forgot password?
View in app×