Question

The angle of elevation of a cloud from a point h meters above a take is
$\alpha$


and the angle of depression of its reflection in the take is
$\beta$


Prove that the height of the cloud is
$\frac{h \: (tan \beta + tan \alpha )}{tan \beta - tan \beta }$


​

Answer 1

SOLUTION:-

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

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

Let the height of the cloud be H meters

Then BC=BD=H m

$\bf Draw\:PQ\perp CD.\:Then$

$\bf \angle QPC=\alpha,\angle QPD=\beta,BQ=AP=h\:m$

$\bf CQ=(H-h)m \:and\:DQ=(H+h)m.$

$\bf From\:right\triangle CQP ,we\:have$

$\bf \frac{PQ}{CQ}=cot\alpha\implies\frac{PQ}{(H-h)}=cot\alpha$

$\bf PQ=(H-h)cot\alpha------[1]$

$\bf From \:right\triangle DQP,we\: have$

$\bf \frac{PQ}{DQ}=cot\beta$

$\bf \implies\frac{PQ}{(H+h)}=cot\beta$

$\bf \implies PQ=(H+h)cot\beta-----[2]$

From [1] and [2],we get...

$\bf (H-h)cot\alpha=(H+h)=cot\beta$

$\bf \implies \frac{(H+h)}{(H-h)}=\frac{tan\beta}{tan\alpha}$

$\bf \implies \frac{(H+h)+(H-h)}{(H+h)-(H-h)}=\frac{tan\beta+tan\alpha}{tan\beta-tan\alpha}-----|by\:componendo \:and\:dividendo|$

$\bf \implies \frac{2H}{2h}=\frac{tan\beta+tan\alpha}{tan\beta-tan\alpha}$

$\bf \implies \frac{H}{h}=\frac{tan\beta+tan\alpha}{tan\beta-tan\alpha}$

$\bf \implies H=\frac{h(tan\beta+tan\alpha)}{(tan\beta-tan\alpha)}$

$\bf Hence\:the\:height\:of\:the\:cloud\:is \frac{h(tan\beta+tan\alpha)}{tan\beta-tan\alpha}$

Answer 2

Answer:

see attachment picture,

Hope it helps