From the top of a building 15 m high the angle of elevation of the top of a tower is found to be 30°.

Solution:

In the figure let OD = h and AD be the tower. The angle of elevation from the top of building to the top of tower is to be found 30°. Height of building ism and an angle of elevation from the bottom of same building is found to be 60°.

Let DC = x and , , 

Here we have to find height of tower and distance between the tower and building.

The corresponding diagram is as follows

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

$\Rightarrow \quad \tan 30^{\circ}=\frac{O D}{D C}$

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

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

Again in a triangle $O A B$,

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

$\Rightarrow \quad \tan 60^{\circ}=\frac{h+15}{x}$

$\Rightarrow \quad \sqrt{3}=\frac{h+15}{x}$

$\Rightarrow \quad \sqrt{3}=\frac{h+15}{\sqrt{3 h}}$

$\Rightarrow \quad 3 h=h+15$

$\Rightarrow \quad 2 h=15$

$\Rightarrow \quad h=7.5$

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

$\Rightarrow \quad x=7.5 \times 1.732$

$\Rightarrow \quad x=12.9$

So height of the tower is as follows:

$\Rightarrow \quad O A=h+15$

$\Rightarrow \quad O A=7.5+15$

$\Rightarrow \quad O A=22.5$

Hence the required height is $22.5$ meter and distance is $12.9$ meter.