Question

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

• A. $h\left(\frac{tan\theta }{1+tan\theta }\right)$
• B. $h\left(\frac{1-tan\theta }{1+tan\theta }\right)$
• C. $h\left(\frac{1+tan\theta }{1-tan\theta }\right)$
• D. $h\left(\frac{1-tan\theta }{tan\theta }\right)$

Answer 1

The correct option is C

$h\left(\frac{1+tan\theta }{1-tan\theta }\right)$

Let H metre be the height of the could above water level. The distance of the reflection is same as that of the cloud from the lake surface.

Here, AB = h, AC = BQ = x

In $\Delta ACP,tan\theta =\frac{PC}{AC}=\frac{H-h}{x}$
$\Rightarrow x=\frac{H-h}{tan\theta }$ ... (1)

In $\Delta ACR,tan{45}^{\circ }=\frac{RC}{AC}=\frac{H+h}{x}$
$\Rightarrow x=\frac{H+h}{tan{45}^{\circ }}$

$\Rightarrow x=H+h$ ... (2)

From (1) and (2), we have
$(H+h)\left(tan\theta \right)=(H-h)\Rightarrow H(1-tan\theta )=h(1+tan\theta )$

$\Rightarrow H=h\left(\frac{1+tan\theta }{1-tan\theta }\right)$