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

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.

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 $30 \mathrm{~m}$.