Question

A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car change from `30^(@)" and "45^(@)` in 12 minutes, find the time taken by the car now toreach the tower.

Answer 1

Let AB = h be the height of the tower.
Since, time taken from D to C = 12 min.
Now, in right `DeltaABC`,
`tan45^(@)=h/yrArr1=h/yrArrh=y …(1)`
In right `DeltaABD`,
`tan30^(@)=h/(x+y)rArr1/sqrt3=h/(x+y)`
`rArr x+y=hsqrt3` ,brgt `rArr x+y=ysqrt3" [from(1)]"`
`rArr y(sqrt3-1)=x ...(2)`
Now time taken by car in moving x m = 12 min
`:.` TIme taken by car in moving `y(sqrt3-1)m=12 "min"`
`:.` Time taken by car in moving y m`=12/(sqrt3-1)"min"`
`=(12(sqrt3+1))/((sqrt3-1)(sqrt3+1))"min"=(12(1.732+1))/2`
`=6xx2.732=16.39" min "`
Hence, required time = 16.39 minutes.