The angle of elevation of a cloud from point h meters above the surface of a lake is b and angle of depression of its angle in lake is a.Prove that the height of the cloud above the lake is $= \frac{h (t a n α + t a n β)}{t a n β - t a n α}$

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Solution

Let AN be the surface of the lake and O be the point of observation such that OA = h metres.
Let P be the position of the cloud and P’ be its reflection in the lake
Then PN = P’N
Let OM $⊥$ PN
Also, $∠ P O M = α a n d ∠ P O M = β$
Let PM = x
Then PN = PM + MN = PM + OA = x + h
In rt. $Δ$ OPM, we have
$t a n α = \frac{P M}{O M} = \frac{x}{A N} \Rightarrow A N = x c o t α \dots (1)$
$I n r t . Δ O P M^{'}, w e h a v e,$
$t a n β = \frac{P^{'} M}{O M} = \frac{x + 2 h}{A N} [∵ P^{'} M = P^{'} N + M N = (x + h) + h = x + 2 h]$
$t a n β = \frac{x + 2 h}{A N}$
$\Rightarrow A N = (x + 2 h) c o t β \dots (2)$
Equating (1) and (2):
$x c o t α = (x + 2 h) c o t β$
$\Rightarrow x (c o t α - c o t β) = 2 h c o t β$
$\Rightarrow x (\frac{1}{t a n α} - \frac{1}{t a n β}) = \frac{2 h}{t a n β}$
$\Rightarrow \frac{x (t a n β - t a n α)}{t a n α . t a n β} = \frac{2 h}{t a n β}$
$\Rightarrow x = \frac{2 h t a n α}{t a n β - t a n α}$
Hence, height of the cloud is given by PN = x + h
$= \frac{2 h t a n α}{t a n β - t a n α}$
$= \frac{2 h t a n α + h (t a n β - t a n α)}{t a n β - t a n α}$
$= \frac{h (t a n α + t a n β)}{t a n β - t a n α}$
Hence proved.

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