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 30^{º} and 45^{º} 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