Question

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer 1

here is your answer what you want

Answer 2

SOLUTION:

Let AB be the tower of height h metres . Let C be the initial position of the car and let after 6 seconds the car be at D . It is given that the angles of depression at C and D are 30° and 60° respectively.

Let the speed of the car be v metre per second . Then,

CD = Distance travelled by car in 6 seconds

CD = speed * time

CD = 12 v metres

Suppose the car takes t seconds to reach the tower AB from D . Then,

DA = speed * time

DA = vt metres

In ∆DAB, we have

Tan 60° = AB/AD

√3 = h / vt

h = √3 VT .............(1.)

In ∆CAB, we have

Tan 30° = AB / AC

1/√3 = h / vt + 6v

√3 h = vt + 6v .........(2.)

Substitute the value of h from eq(1.) in eq(2.)

√3 ( √3 vt ) = vt + 6v

3vt = v ( t + 6 )

3t = v ( t + 6 ) / v

3t = t + 6

3t - t = 6

2t = 6

t = 6/2

t = 3 seconds

Thus, the car will reach the tower from D point in 3 seconds