# 11. 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 joing 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.

M manish

Suppose the $h$ is the height of the tower AB and BC = $x$ m
It is given that, width of CD is 20 m,
According to question,

In triangle $\Delta ADB$,
$\\\Rightarrow \tan 30^o = \frac{AB}{20+x}=\frac{h}{20+x}\\\\\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x}\\\\\Rightarrow 20+x = h\sqrt{3}\\\\\Rightarrow x = h\sqrt{3}-20$............(i)

In triangle ACB,
$\\\Rightarrow \tan 60^o = \frac{h}{x}=\sqrt{3}\\\\\Rightarrow x= \frac{h}{\sqrt{3}}$.............(ii)

On equating eq (i) and (ii) we get:

$h\sqrt{3}-20= \frac{h}{\sqrt{3}}$
from here we can calculate the value of  $h=10\sqrt{3}= 10 (1.732) = 17.32\: m$  and the width of the canal is 10 m.

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