# 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 point is .

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 $30 \mathrm{~m}$.