Question

A point on the ground is 40 m away from the foot of a post, the angle of elevation of the top of the post is ${30}^0$. The angle of elevation of the top of a water reservoir (on the top of the post) is ${45}^0$. Find the depth of the reservoir.

• A. 30 m
• B. 45 m
• C. 17 m
• D. 25 m

Answer 1

The correct option is C

17 m

Let PC be the post of height `h' metres and CD be the water reservoir of height `h${}_1$' metres.

Let A be a point on the ground at a distance of 40 m away from the foot P of the post.

In $\Delta$ APD, we have,

$tan{45}^0$ = $\frac{PD}{AP}$

$\Rightarrow 1=\frac{h+h_1}{40}$

$\Rightarrow$ h + h${}_1$ = 40 m - (i)

In $\Delta$ ABC, we have,

$tan{30}^0$ = $\frac{PC}{AP}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{40}$

$\Rightarrow h=\frac{40}{\sqrt{3}}m=\frac{40\sqrt{3}}{3}m=23.1m$

Substituting the value of `h' in (i), we have -

23.1 + h${}_1$ = 40

$\Rightarrow$ h${}_1$ = (40 - 23.1) m = 16.9 m

Hence, the height of the post is h = 23.1 m and the depth of the reservoir is h${}_1$ = 16.9 m.