The angle of elevation of the top of a vertical tower from a point on the ground is 60°.

Solution:

Let PQ be the tower.

We have,

$\mathrm{AB}=10 \mathrm{~m}, \angle \mathrm{MAP}=30^{\circ}$ and $\angle \mathrm{PBQ}=60^{\circ}$

Also, $\mathrm{MQ}=\mathrm{AB}=10 \mathrm{~m}$

Let $\mathrm{BQ}=x$ and $\mathrm{PQ}=h$

So, $\mathrm{AM}=\mathrm{BQ}=x$ and $\mathrm{PM}=\mathrm{PQ}-\mathrm{MQ}=h-10$

In $\Delta \mathrm{BPQ}$,

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

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

$\Rightarrow x=\frac{h}{\sqrt{3}} \quad \ldots \ldots$ (i)

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

$\tan 30^{\circ}=\frac{\mathrm{PM}}{\mathrm{AM}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-10}{x}$

$\Rightarrow h \sqrt{3}-10 \sqrt{3}=x$

$\Rightarrow h \sqrt{3}-10 \sqrt{3}=\frac{h}{\sqrt{3}} \quad[$ Using (i) $]$

$\Rightarrow 3 h-30=h$

$\Rightarrow 3 h-h=30$

$\Rightarrow 2 h=30$

$\Rightarrow h=\frac{30}{2}$

$\therefore h=15 \mathrm{~m}$

So, the height of the tower is 15 m.