Question

The angle of elevation of a cloud from a point 100 meter above the surface of a lake is ${30}^0$ and the angle of depression of its image in the lake is ${60}^0$ then height of the cloud above the lake is

• A. 100 m
• B. 50 m
• C. 200 m
• D. 150 m

Answer 1

Answer: (C)

200 m

$tan{30}^0=\frac{h-100}{d}=\frac{1}{\sqrt{3}}$......(1)
$tan{60}^0=\frac{h+100}{d}=\sqrt{3}$......................(2)
Dividing equation (1) by (2) , we get
$\frac{h-100}{h+100}=\frac{1}{3}$
$\Rightarrow 3h-300=h+100$
$\Rightarrow 2h=400$
$h=200$ metre
$\therefore$ Height above the lake=200 m