The angle of elevation of the top of a building from the foot of a tower is 30°.

Solution:

Let AB be the building and PQ be the tower.

We have,

$\mathrm{PQ}=60 \mathrm{~m}, \angle \mathrm{APB}=30^{\circ}, \angle \mathrm{PAQ}=60^{\circ}$

In $\Delta \mathrm{APQ}$,

$\tan 60^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AP}}$

$\Rightarrow \sqrt{3}=\frac{60}{\mathrm{AP}}$

$\Rightarrow \mathrm{AP}=\frac{60}{\sqrt{3}}$

$\Rightarrow \mathrm{AP}=\frac{60 \sqrt{3}}{3}$

$\Rightarrow \mathrm{AP}=20 \sqrt{3} \mathrm{~m}$

Now, in $\triangle \mathrm{ABP}$,

$\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AP}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{AB}}{20 \sqrt{3}}$

$\Rightarrow \mathrm{AB}=\frac{20 \sqrt{3}}{\sqrt{3}}$

$\therefore \mathrm{AB}=20 \mathrm{~m}$

So, the height of the building is 20 m.