A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of `30^@` . After some time, the angle of depression becomes `60^@` .The distance (in metres) travelled by the car during this time is

Options

A. `100sqrt3`
B. `(200sqrt3)/3`
C. `(100sqrt3)/3`
D. `200sqrt3`

Answer

B

Explanation

Let OT be the tower and A and B be the positions of the car.
In `triangle`'s AOT and BOT , we have
tan `30^@=(OT)/(OA)` and tan `60^@=(OT)/(OB)`

`rArr 1/sqrt3=100/(OA)` and `sqrt3=100/(OB) rArr OA = 100sqrt3` and OB =`100/sqrt3`
`therefore` Required distance `=AB=OA -OB=100sqrt3-100/sqrt3`
`=200/sqrt3=(200sqrt3)/3` metres.