Answer this it's urgent - Maths - Some Applications of Trigonometry

Question

Answer this it's urgent

Abhishek Kumar · Accepted Answer

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, ∠POM = α and ∠P'OM = β

Let PM = x

Then PN = PM + MN = PM + OA = x + h

In rt. ΔOPM, we have

$\tan α = \frac{PM}{OM} = \frac{x}{AN} \Rightarrow AN = x \cot α ... (1)$

In rt. ΔOMP', we have,

$\tan β = \frac{P'M}{OM} = \frac{x + 2 h}{AN} [∵ P'M= P'N + MN = (x + h) + h = x + 2 h]$

$\tan β = \frac{x + 2 h}{AN} \Rightarrow AN = (x + 2 h) \cot β ... (2)$

Equating (1) and (2):

$x \cot α = (x + 2 h) \cot β \Rightarrow x (\cot α - \cot β) = 2 h \cot β \Rightarrow x (\frac{1}{\tan α} - \frac{1}{\tan β}) = \frac{2 h}{\tan β} \Rightarrow \frac{x (\tan β - \tan α)}{\tan α . \tan β} = \frac{2 h}{\tan β} \Rightarrow x = \frac{2 h \tan α}{\tan β - \tan α}$

Hence, height of the cloud is given by PN = x + h

$= \frac{2 h \tan α}{\tan β - \tan α} + h = \frac{2 h \tan α + h (\tan β - \tan α)}{\tan β - \tan α} = \frac{h (\tan α + \tan β)}{\tan β - \tan α}$

Hence proved.