Question

A player is sitting on the top of the tower of Height 20m observes the angle of depression of a ball lying on the ground is 60° find the distance between the foot of the tower and the ball

Answer 1

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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Answer 2

h = 20 m

base = x m

angle = 60

tan 60 = p/b

root 3 = 20/ x

x = 20 /root3  

x = 20 x root3 / root3 x root3

x = 20 root3 / 3

x = 11.54 m ans