The angle of elevation of the top of a tower from a point A on the ground is 30°.

Solution:

Let  be height of tower and the angle of elevation of the top of tower from a point  on the ground is  and on moving with distance m towards the foot of tower on the pointis.

Let  and 

Now we have to find height of tower and distance of tower from point A.

So we use trigonometrical ratios.

In,

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

$\Rightarrow \quad \tan 60^{\circ}=\frac{C D}{B C}$

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

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

Again in ,

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

$\Rightarrow \quad \tan 30^{\circ}=\frac{h}{x+20}$

$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{x+20}$

$\Rightarrow \quad x=\sqrt{3} h-20$

$\Rightarrow \frac{h}{\sqrt{3}}+20=\sqrt{3} h$

$\Rightarrow \frac{h}{\sqrt{3}}-\sqrt{3} h=-20$

$\Rightarrow \quad h-3 h=-20 \sqrt{3}$

$\Rightarrow \quad-2 h=-20 \sqrt{3}$

$\Rightarrow \quad h=10 \sqrt{3}$

$\Rightarrow \quad h=17.32$

$\Rightarrow \quad x=\frac{10 \sqrt{3}}{\sqrt{3}}$

$\Rightarrow \quad x=10$

So distance

$\Rightarrow \quad A C=x+20$

$\Rightarrow \quad A C=30$

Hence the required height is $17.32 \mathrm{~m}$ and distance is $30 \mathrm{~m}$.