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° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Let AB be the tower and BC be the canal. C is the point on the other side of the canal directly opposite the tower.

Let AB = h, BC = x and CD = 20

From right △BAC,

$$\begin{gathered} \mathrm{co}t {60}^0=\frac{x}{h}. \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{h}. \\ \Rightarrow x=\frac{h}{\sqrt{3}}. \end{gathered}$$
From right △BAD,

$$\begin{gathered} cot {30}^0=\frac{BD}{AB}. \\ \Rightarrow \sqrt{3}=\frac{x+20}{h}. \\ \Rightarrow x=\sqrt{3}h-20. \end{gathered}$$
Now we have to equate the values of x.

$$\begin{gathered} h\sqrt{3}-20=\frac{h}{\sqrt{3}}. \\ \Rightarrow h=10\sqrt{3}. \end{gathered}$$
Replacing, the value of h in x we have, $x = \sqrt{3}\left(10\sqrt{3}\right)-20 = 10$

Therefore, the width of the canal is 10 m, and the height of the canal is $10\sqrt{3}$​ m.