Question

The angle of elevation of a cloud from a point h meters above the surface of a lake is $\Theta$ and the angle of depression of its reflection in the lake is . prove that the height of the cloud above the lake surface is : $h\left(\frac{tan\varnothing +tan\theta }{tan\varnothing +tan\theta }\right).$

Answer 1

$Height=AC=x+h$

$\Rightarrow AB=x$

$△ABD$

$\Rightarrow \tan \theta =\frac{x}{OB}$

$OB=\frac{x}{\tan \theta }\to \left(1\right)$

$△DBO$

$\Rightarrow \tan \varphi =\frac{BD}{OB}=\frac{DC+CB}{OB}=\frac{x+h+h}{OB}$

$\tan \varphi =\frac{x+2h}{OB}$

$\therefore OB=\frac{x+2h}{\tan \varphi }\to \left(2\right)$

$\left(1\right)=\left(2\right)$

$\therefore \frac{x}{\tan \theta }=\frac{x+2h}{\tan \varphi }$

$\Rightarrow x\tan \varphi =x\tan \theta +2h\tan \theta$

$2h\tan \theta =x(\tan \varphi -\tan \theta )$

$\Rightarrow x=\frac{2h\tan \theta }{\tan \varphi -\tan \theta }$

$\Rightarrow$ Height of cloud $=x+h$

$=\frac{2h\tan \theta }{\tan \varphi -\tan \theta }+h$

$=\frac{2h\tan \theta +h\tan \varphi -h\tan \theta }{\tan \varphi -\tan \theta }$

$=\frac{h(\tan \theta +\tan \varphi )}{\tan \varphi -\tan \theta }$

Hence, proved.