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Question
If the angle of elevation of a cloud from a point h meters above a lake is a and the angle of depression of its reflection in the lake be b, prove that the distance of the cloud from the point of observation is `(2h sec alpha)/(tan beta - tan alpha)`
Solution
Let C′ be the image of cloud C. We have ∠CAB = α and ∠BAC' = β
Again let BC = x and AC be the distance of cloud from point of observation.
We have to prove that
`AC= (2h sec alpha)/(tan beta - tan alpha)`
The corresponding figure is as follows
We use trigonometric ratios.
in Δ ABC
`=> tan alpha = (BC)/(AB)`
`=> tan alpha = x/(AB)`
Again in Δ ABC'
`=> tan beta = (BC')/(AB)`
`=> tan beta = (x + 2h)/(AB)`
Now
`=> tan beta - tan alpha = (x + 2h)/(AB) - x/(AB)`
`=> tan bea - tan alpha = (2h)/(AB)`
`=> AB = (2h)/(tan beta - tan alpha)`
Again in Δ ABC
`=> cos alpha = (AB)/(AC)`
`=> AC = (AB)/(cos alpha)`
`=> (2h sec alpha)/(tan beta - tan alpha)`
Hence distance of cloud from points of observation is `(2h sec alpha)/(tan beta - tan alpha)`
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