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Question
The angle of elevation of a stationary cloud from a point 2500 m above a lake is 15° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level? (Use tan 15° = 0.268)
Solution
Let AB be the surface of lake and P be the point of observation such that AP = 2500 m. Let C be the position of cloud and C' be the reflection in the lake. Then CB = C'B
Let PQ be the perpendicular from P on CB.
Let PQ = x m, CQ = h, QB = 2500 m. then CB = h + 2500 consequently C'B = h + 2500 m. and ∠CPQ - 15°, .∠QPC' = 45°
Here we have to find the height of cloud.
We use trigonometric ratios.
In ΔPCQ
`=> tan 15° = (CQ)/(PQ)`
`=> 2 - sqrt3 = h/x`
`=> x = h/(2 - sqrt3)`
Again in Δ PQC
`=> tan 45^@ = (QB + BC')/(PQ)`
`=> 1 = (2500 + h + 2500)//x`
`=> x = 5000 } h`
`=> h /()2 - sqrt3 = 5000 + h`
`=> h = 2500 (sqrt3 - 1)`
`=> CB = 2500 + 2500 (sqrt3 - 1)`
`=> CB = 2500sqrt3`
Hence the height of cloud is `2500sqrt3` m
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