$A$ $TV$ tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower,the angle of elevation of the top of the tower is $60^{\circ}$. From another point $20 \, m$ away from this point on the line joining this point to the foot of the tower,the angle of elevation of the top of the tower is $30^{\circ}$ (see figure). Find the height of the tower and the width of the canal.

(N/A) Let the height of the $TV$ tower be $AB = h$ and the width of the canal be $BC = x$.
In $\triangle ABC$,we have:
$\frac{AB}{BC} = \tan 60^{\circ}$
$\frac{h}{x} = \sqrt{3} \implies h = x\sqrt{3} \quad \dots(1)$
In $\triangle ABD$,we have:
$\frac{AB}{BD} = \tan 30^{\circ}$
$\frac{h}{x + 20} = \frac{1}{\sqrt{3}} \implies h\sqrt{3} = x + 20 \quad \dots(2)$
Substituting the value of $h$ from equation $(1)$ into equation $(2)$:
$(x\sqrt{3})\sqrt{3} = x + 20$
$3x = x + 20$
$2x = 20 \implies x = 10 \, m$
Now,substituting $x = 10$ into equation $(1)$:
$h = 10\sqrt{3} \, m$
Therefore,the height of the tower is $10\sqrt{3} \, m$ and the width of the canal is $10 \, m$.