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)`.

Answer 1

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)`.