#### Question

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as 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.

#### Solution 1

Let PQ be the tower.

We have,

∠PBQ = 60° and ∠PAQ = 30°

Let PQ = h, AB = x and BQ = y

In ΔAPQ,

`tan 30° = (PQ)/(AQ)`

`⇒ 1/ sqrt(3) = h/(x+y) `

`⇒ x+y = h sqrt(3)` ..................(1)

Also, in ΔBPQ,

`tan 60° = ( PQ)/(BQ)`

`⇒ sqrt(3) = h/y`

`⇒ h = y sqrt(3) ` ...............(2)

Substituting `h = y sqrt(3)` in (i), we get

`x +y = sqrt(3) (ysqrt(3))`

⇒ x + y = 3y

⇒ 3y - y = x

⇒ 2y = x

`⇒ y = x/2`

`"As, speed of the car from "A to B = (AB) /6 = x/6 units/ sec`

So, the time taken to reach the foot of the tower i.e. Q from B `(BQ)/(speed)`

`=y/((x/6))`

`=((x/2))/((x/6))`

`=6/2`

= 3 sec

So, the time taken to reach the foot of the tower from the given point is 3 seconds.

#### Solution 2

Let AB be the tower.

Initial position of the car is C, which changes to D after six seconds.

In ΔADB,

AB/DB = tan 60º

`(AB)/(DB) =sqrt3`

`DB = (AB)/sqrt3`

In ΔABC,

AB/BC = tan 30º

`(AB)/(BD + DC) = 1/sqrt3`

`ABsqrt3 = BD + DC`

`ABsqrt3 = (AB)/sqrt3 + DC`

`DC = ABsqrt3 - (AB)/sqrt3 = AB(sqrt3 - 1/sqrt3)`

`= (2AB)/sqrt3`

Time taken by the car to travel distance DC `("i.e" "2AB"/sqrt3)` = 6 second

Time taken by the car to travel distance DB `("i.e" (AB)/sqrt3) = 6/((2AB)/sqrt3)xx(AB)/sqrt3`

= 6/2 = 3 seconds