# A person observed the angle of elevation of the top of a tower as 30°.

Question:

A person observed the angle of elevation of the top of a tower as 30°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.

Solution:

Let  be the tower of height. And person makes an angle of elevation of top of tower is 30°, he walks m towards the foot of tower then makes an angle of elevation 60°

Let $B C=x, C D=50$, and $\angle A C B=60^{\circ}, \angle A D B=30^{\circ}$

Now we have to find height of tower.

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle ABC,

$\Rightarrow \quad \tan C=\frac{A B}{B C}$

$\Rightarrow \quad \tan 60^{\circ}=\frac{h}{x}$

$\Rightarrow \quad \sqrt{3}=\frac{h}{x}$

$\Rightarrow \quad x=\frac{h}{\sqrt{3}}$

Again in a triangle,

$\Rightarrow \quad \tan D=\frac{A B}{B C+C D}$

$\Rightarrow \quad \tan 30^{\circ}=\frac{h}{x+50}$

$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{x+50}$

$\Rightarrow \quad \sqrt{3} h=x+50$

$\Rightarrow \sqrt{3} h=\frac{h}{\sqrt{3}}+50$

$\Rightarrow \sqrt{3} h=\frac{h}{\sqrt{3}}+50$

$\Rightarrow 3 h=h+50 \sqrt{3}$

$\Rightarrow 2 h=50 \sqrt{3}$

$\Rightarrow h=25 \sqrt{3}$

$\Rightarrow h=25 \times 1.73$

$\Rightarrow h=43.25$

Hence the height of tower is $43.25 \mathrm{~m}$.