A TV tower stands vertically on a bank of canal. From a point on other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°.

Solution:

Let PQ=h m be the height of the TV tower and BQ=x m be the width of the canal.

We have,

$\mathrm{AB}=20 \mathrm{~m}, \angle \mathrm{PAQ}=30^{\circ}, \angle \mathrm{PBQ}=60^{\circ}, \mathrm{BQ}=x$ and $\mathrm{PQ}=h$

In $\Delta$ PBQ,

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

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

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

Again, in $\triangle \mathrm{APQ}$

$\tan 30^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AQ}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{\mathrm{AB}+\mathrm{BQ}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{x \sqrt{3}}{20+x} \quad[\operatorname{Using}(\mathrm{i})]$

$\Rightarrow 3 x=20+x$

$\Rightarrow 3 x-x=20$

$\Rightarrow 2 x=20$

$\Rightarrow x=\frac{20}{2}$

$\Rightarrow x=10 \mathrm{~m}$

Substituting $x=10$ in (i), we get

$h=10 \sqrt{3} \mathrm{~m}$

So, the height of the TV tower is $10 \sqrt{3} \mathrm{~m}$ and the width of the canal is $10 \mathrm{~m}$.