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The angle of elevation of a stationary c...

The angle of elevation of a stationary cloud from a point 2500 feet above a lake is `30^@` and the angle of depression of its reflection in the lake is `45^@`.Find the height of cloud above the lake water surface .

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In the figure ,observer is at point B and cloud is at point D.

E is the reflection of colud in the lake .
`therefore` FD =FE =H
In triangle BCD ,
`tan 30 ^@ (H- 2500)/(BC)`
`rArr H- 2500 =BC tan 30 ^@`
In triangle BCE,
`tan 45 ^@=(H+2500)/(BC)`
`rArr H+ 2500 = BC tan 45^@`
`therefore (H-2500)/(H+ 2500)=(BC tan 30^@)/(BC tan 45^@)=1/sqrt(3)`
`rArr sqrt(3)(H-2500 )= H+ 2500`
`rArr (sqrt(3)-1)H=2500 (sqrt(3)+1)`
`rArr H=2500 xx (sqrt(3)+1)/(sqrt(3)-1)` feet
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