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°. 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°. Find the height of the tower and the width of the canal.

In △ABD,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow BD = AB\sqrt{3} \text{ m}.$

In △ABC,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = \dfrac{AB}{\sqrt{3}} \text{ m}.$

From figure,

CD = BD - BC

$\Rightarrow 20 = AB\sqrt{3} - \dfrac{AB}{\sqrt{3}} \\[1em] \Rightarrow 20 = \dfrac{3AB - AB}{\sqrt{3}} \\[1em] \Rightarrow 20 = \dfrac{2AB}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{20\sqrt{3}}{2} \\[1em] \Rightarrow AB = 10\sqrt{3} \text{ m}.$

Width of canal (BC) = $\dfrac{AB}{\sqrt{3}} = \dfrac{10\sqrt{3}}{\sqrt{3}}$ = 10 m.

Hence, height of tower = $10\sqrt{3}$ m and width of canal = 10 m.