The angle of elevation of a cloud from a point h metres above a lake is $α$ and the angle of depression of its reflection in the lake is $β$ . Prove that the height of the cloud is $\frac{h (t a n β + t a n α)}{(t a n β - t a n α)} m e t r e s$ [4 MARKS]

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Solution

Concept: 2 Marks
Application: 2 Marks

Let AB be the surface of the lake and let P be a point vertically above A such that AP = h metres.

Let C be the position of the cloud and let D be its reflection in the lake.

Draw $P Q ⊥ C D$ . Then,

$∠ Q P C = α, ∠ Q P D = β$ ,

$B Q = A P = h m e t r e s$

Let $C Q = x m e t r e s$ . Then,

$B D = B C = (x + h) m e t r e s$

From right $Δ P Q C$ , we have

$\frac{P Q}{C Q} = c o t α \Rightarrow \frac{P Q}{x m} = c o t α$

$\Rightarrow P Q = x c o t α m e t r e s$ ........ (i)

From right $Δ P Q D$ , we have

$\frac{P Q}{Q D} = c o t β \Rightarrow \frac{P Q}{(x + 2 h) m} = c o t β$

$\Rightarrow P Q = (x + 2 h) c o t β m e t r e s$ ........ (ii)

From (i) and (ii), we get

$x c o t α = (x + 2 h) c o t β$

$\Rightarrow x (c o t α - c o t β) = 2 h c o t β \Rightarrow x (\frac{1}{t a n α} - \frac{1}{t a n β}) = \frac{2 h}{t a n β}$

$\Rightarrow x (\frac{t a n β - t a n α}{t a n α t a n β}) = \frac{2 h}{t a n β} \Rightarrow x = \frac{2 h t a n α}{(t a n β - t a n α)}$

$∴$ height of the cloud from the surface of the lake

$= (x + h) = {\frac{2 h t a n α}{(t a n β - t a n α)} + h} m$

$= \frac{h (t a n α + t a n β)}{(t a n β - t a n α)} m e t r e s$

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The angle of elevation of a cloud from a point h metres above a lake is $α$ and the angle of depression of its reflection in the lake is $β$ . Prove that the height of the cloud is $\frac{h (t a n β + t a n α)}{(t a n β - t a n α)} m e t r e s$ [4 MARKS]