Question

The angle of elevation of a cloud from a point 60 m above a lake is ${30}^o$ and the angle of depression of Its image In lake is ${60}^o$. The height of the cloud is _____.

• A. 100 m
• B. 80 m
• C. 40 m
• D. 120 m

Answer 1

The correct option is D 120 m

$CD=OB=60$m

Say $AO=h$ units

$\therefore A^{{}^{\prime }}B=h+60=AB$ (image distance=object distance; it is a crude assumption)

$In△AOD,\tan 30°=\frac{h}{OD}=>h=\frac{OD}{\sqrt{3}}=>OD=\sqrt{3}h$

$In△A^{{}^{\prime }}OD,\tan 60°=\frac{h+60+60}{OD}=>\sqrt{3}=\frac{h+120}{\sqrt{3}h}=>2h=120°=>h=60°$

Therefore, height of the cloud above the lake is $60+60=120$ m.