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.
Let AB be the tower.
Initial position of the car is C, which changes to D after six seconds.
AB/DB = tan 60º
`DB = (AB)/sqrt3`
AB/BC = tan 30º
`(AB)/(BD + DC) = 1/sqrt3`
`ABsqrt3 = BD + DC`
`ABsqrt3 = (AB)/sqrt3 + DC`
`DC = ABsqrt3 - (AB)/sqrt3 = AB(sqrt3 - 1/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
Let PQ be the tower.
∠PBQ = 60° and ∠PAQ = 30°
Let PQ = h, AB = x and BQ = y
`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)`
= 3 sec
So, the time taken to reach the foot of the tower from the given point is 3 seconds.