Question

A tower stands vertically on the ground. A man standing at the top of the tower observes his friend at an angle of depression of 30°, who is approaching the foot of the tower with a uniform speed. 30 seconds later, the angle of depression changes to 60°. Find the time taken by his friend to reach the foot of the tower from this point.

Answer 1

Let tower height 

= h, initial horizontal distance = x.

Using tan (opposite/adjacent)

`tan30° = h/x`

⇒ `x = h/tan30°`

= `hsqrt3`   ...[Use tan relation.]

After 30 s, the horizontal distance

= y where `tan60° = h/y`

⇒ `y = h/(tan60°)`

= `h/sqrt3`

Distance covered in 30s = x − y 

= `hsqrt3 − h/sqrt3`

= `(2h)/sqrt3`

Hence speed `v = (x − y)/30`

= `((2h)/sqrt3)/30`

= `h/(15sqrt3)`

Time to cover the remaining distance y at speed

`v: t = y/v = (h/sqrt3) ÷ (h/(15sqrt3)) = 15s`