Question

A player sitting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60 degrees. Find the distance between the foot of the tower and the hall.

Answer 1

see the answer in the photo uploaded above
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Answer 2

Height of the tower is BC = 20m and D is the ball

DC is the distance from foot of the ball to the tower.

∠ABD =∠BDC = 60°

Let DC =x m

then   tan60°= BC/DC= 20/x    [tan theta= perpendicular/base ]

⇒x =20/√3m (tan60° =√3)

So the distance between the foot of the tower and the ball is 20/√3m= 20* 1.732  = 11.54 m ANS

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