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Question
From a point on the ground, 20 m away from the foot of a vertical tower, the angle elevation of the top of the tower is 60°, What is the height of the tower?
Solution
Let AB be the height of tower is h meters.
Given that: angle of elevation is 60° and `BC=20` meters.
Here we have to find height of tower.
So we use trigonometric ratios.
In a triangle `ABC`,
`⇒ tan C=(AB)/(BC)`
`⇒ tan 60°=h/20 [∵ tan 60°=sqrt3]`
`⇒ sqrt 3=h/20`
`⇒ h= 20sqrt3`
Hence height of tower is `20sqrt3`
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