Question

The angle of elevation of a cloud from a point $h$ metres above the lake water level is $\theta$ and the angle of depresion of its image in the lake is $\varphi$. The height of the cloud is

• A. $\frac{h\sin (\varphi +\theta )}{\sin (\varphi -\theta )}$
• B. $\frac{h\sin (\varphi -\theta )}{\sin (\varphi +\theta )}$
• C. $\frac{h\sin (\theta +\varphi )}{\sin (\theta -\varphi )}$
• D. $\frac{h\sin (\theta -\varphi )}{\sin (\theta +\varphi )}$

Answer 1

Answer: (A)

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

$\tan \theta =\frac{h_1-h}{d}$

$\tan \varphi =\frac{h_1+h}{d}$

$\Rightarrow \frac{\tan \theta }{\tan \varphi }=\frac{h_1-h}{h_1+h}$

$\Rightarrow \frac{\tan \theta +\tan \varphi }{\tan \theta -\tan \varphi }=\frac{h_1-h+h_1+h}{h_1-h-h_1-h}$

Using C&D property, we get

$\Rightarrow \frac{h_1}{h}=\frac{\tan \theta +\tan \varphi }{\tan \theta -\tan \varphi }$

$\Rightarrow h_1=h\left(\frac{\cos \theta \sin \varphi +\sin \theta \cos \varphi }{\sin \varphi \cos \theta -\cos \varphi \sin \theta }\right)$

$\Rightarrow h_1=h\frac{\sin (\varphi +\theta )}{\sin (\varphi -\theta )}$