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.

Answer 1

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)/sqrt 3`

In ΔABC,

`(AB)/(BC) = tan 30º`

`(AB)/(BD + DC) = 1/sqrt 3`

`AB sqrt3 = BD + DC`

`AB sqrt 3 = (AB)/sqrt 3 + DC`

`DC = AB sqrt 3 - (AB)/sqrt 3 = AB (sqrt 3 - 1/sqrt 3)`

= `(2 AB)/sqrt 3`

Time taken by the car to travel distance DC `("i.e"  (2AB)/(sqrt 3))` = 6 second

Time taken by the car to travel distance DB `("i.e" (AB)/sqrt 3) = 6/((2 AB)/sqrt 3) xx (AB)/sqrt 3`

= `6/2`

= 3 seconds

Answer 2

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)  (y sqrt (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.