Question

a player sitting on the top of the tower height 20m observes the angle of depressrion of ball as 60degrees find the distance between the foot of the tower and the ball

Answer 1

x is distance between foot of tower &ball

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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