the angle of elevation of a cloud from a point h m above a lake is alpha the angle - Maths - Some Applications of Trigonometry

Question

the angle of elevation of a cloud from a point h m above a lake
is alpha &the angle of depression of it's reflection in the
lake is bita prove that the height of the cloud is h(tan bita+ tan
alpha )/(tan bita-tan alpha)metres

Harleen Birdi · Accepted Answer

Dear Student!! Here is the answer to your question

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 âŠ¥ PN Also, âˆ POM = Î± and âˆ P'OM = Î² Let PM = x Then PN = PM + MN = PM + OA = x + h In rt Î”OPM, we have $\tan α = \frac{PM}{OM} = \frac{x}{AN} \Rightarrow AN = x \cot α (1)$ In rt Î”OMP', 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 Cheers !!

the angle of elevation of a cloud from a point h m above a lake is alpha &the angle of depression of it's reflection in the lake is bita. prove that the height of the cloud is h(tan bita+ tan alpha )/(tan bita-tan alpha)metres.