Question

The angle of elevation of a cloud from a point h metres above a lake is $\theta$. The angle of depression of its reflection in the lake is ${45}^o$. The height of the cloud is __________.

• A. $htan({45}^o+\theta )$
• B. $2hcot({45}^o-\theta )$
• C. $hcot({45}^o-\theta )$
• D. $hcot({45}^o+\theta )$

Answer 1

The correct option is A $htan({45}^o+\theta )$

$\angle ADO=\theta$

Say $AO=p$ units

$\therefore$ In $△AOD$,

$\tan \theta =\frac{p}{OD}$

$OD=\frac{p}{\tan \theta }$

$A^{\prime }B=AB=p+h$

Now, In $△A^{\prime }OD$

$\tan {45}^{\circ }=\frac{A^{\prime }O}{OD}=\frac{A^{\prime }B+OB}{OD}=\frac{p+h+h}{OD}$

$\Rightarrow 1=\frac{p+2h}{p\times \cot \theta }$

$\Rightarrow p\times (\cot \theta -1)=2h$

$\Rightarrow -p\times (\tan \theta -1)=2h\times \tan \theta$

$\Rightarrow p\times (1-\tan \theta )=2h\times \tan \theta$

$\Rightarrow p=\frac{2h\times \tan \theta }{(1-\tan \theta )}$

Therefore, height of the cloud above the lake $=\frac{2h\times \tan \theta }{(1-\tan \theta )}+h$
$=\left(\frac{1+\tan \theta }{1-\tan \theta }\right)\times h$
$=h\times \tan (\theta +{45}^{\circ })$