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Question
If the angle of elevation of a cloud from a point h m above a lake is α, and the angle of depression of its reflection in the lake be β, prove that distance of the cloud from the point of observation is `("2h"secα)/(tanα - tanβ)`.
Solution
Let AB be the surface of the lake and let P be an point of observation such that AP = h meters . Let c be the position of the cloud and C' be its reflection in the lake . Then ∠CPM = α and ∠MPC = β. Let CM = x.
Then , CB = CM + MB = CM + PA = x + h
In ΔCPM,
tan α = `"CM"/"PM"`
⇒ `tan α = "x"/"AB"` [∵ PM = AB]
⇒ AB = x cotα ...(1)
In ΔPMC',
`tanβ = (C'M)/(PM)`
⇒ `tanβ = ("x" + "2h")/"AB"`
⇒ AB = (x + 2h)cotβ ....(2)
From (1) and (2),
x cotα = (x + 2h) cot β
⇒ `"x"(1/tanα - 1/tanβ) = "2h"/tanβ`
⇒ `"x"((tanβ - tanα) /(tanα tanβ)) = "2h"/tanβ`
⇒ x = `("2h"tanα)/(tanβ- tanα)`
Again, in Δ CPM,
`sin α = "CM"/"PC"` = `"x"/("PC")`
⇒ PC = `"x"/sin α`
⇒ `"PC" = ("2h"tanα)/(tanβ - tanα) xx 1/sin α`
⇒ `"PC" = ("2h"secα)/(tanβ - tanα)`
Hence , the distance of the cloud from the point of observation is `("2h"secα)/(tanβ - tanα)`.
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