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 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.
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