Question

If the angle of elevation of a cloud from a point ‘h’ metres above a lake is ${\theta }_1$ and the angle of depression of its reflection in the lake is ${\theta }_2$. Find the height at which the cloud is located from the ground.

• A. $\frac{h(tan{\theta }_1+tan{\theta }_2)}{tan{\theta }_2-tan{\theta }_1}$
• B. $\frac{h(tan{\theta }_1-tan{\theta }_2)}{tan{\theta }_2+tan{\theta }_1}$
• C. $\frac{h(tan{\theta }_1-tan{\theta }_2)}{tan{\theta }_2-tan{\theta }_1}$
• D. $\frac{h(tan{\theta }_1+tan{\theta }_2)}{tan{\theta }_1-tan{\theta }_2}$

Answer 1

The correct option is A $\frac{h(tan{\theta }_1+tan{\theta }_2)}{tan{\theta }_2-tan{\theta }_1}$

Let P be the cloud and Q be its reflection.
Let A be the point of observation such that AB = h
Let the height of the cloud be x. (PS = x)
PR = x – h and QR = x + h


Let AR = y
In the right ∆ARP,
$tan{\theta }_1=\frac{PR}{AR}=\frac{x-h}{y}\dots .\left(1\right)$
In the ∆ AQR, $tan{\theta }_2=\frac{QR}{AR}=\frac{x+h}{y}\dots .\left(2\right)$

Add (1) and (2),
$tan{\theta }_1+tan{\theta }_2=\frac{x-h}{y}+\frac{x+h}{y}=\frac{2x}{y}$

Subtracting (1) from (2),
$tan{\theta }_2-tan{\theta }_1=\frac{x+h}{y}-\frac{x-h}{y}=\frac{2h}{y}$
$\frac{tan\theta 1+tan\theta 2}{tan\theta 2-tan\theta 1}=\frac{2x}{y}÷\frac{2h}{y}=\frac{2x}{y}\times \frac{y}{2h}=\frac{x}{h}$
$\therefore x=\frac{h(tan{\theta }_1+tan{\theta }_2)}{tan{\theta }_2-tan{\theta }_1}$