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Question
The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of depression from the top of the tower to the foot of the hill is 30°. If the tower is 50 m high, find the height of the hill ?
Solution
Let AB be the hill and CD be the tower.
Angle of elevation of the hill at the foot of the tower is 60°, i.e., ∠ADB = 60° and the angle of depression of the foot of hill from the top of the tower is 30°, i.e., ∠CBD = 30°.
Now in right angled ΔCBD:
`tan30^@=(CD)/(BD)`
`rArrBD=(CD)/tan30^@`
`rArrBD=50/((1/sqrt3))`
`rArrBD =50sqrt3`
In right ΔABD:
`tan60^@=(AD)/(BD)`
`rArr AB=BDxxtan60^@`
`=50sqrt3xxsqrt3m`
`=50xx3m`
`=150m`
Hence, the height of the hill is 150 m.
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