# Applications of Trigonometry Heights and Distances ncert solutions Chapter 9 Exercise 9.1 Question 11

Applications of Trigonometry Heights and Distances Chapter 9 Exercise 9.1 Question 11

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 $60^\circ$. 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 $30^\circ$. Find the height of the tower and the width of the canal.

Solution:

It is given that BC = 20 m

In $\triangle ADC$,

$\frac{AD}{DC}=tan 60^\circ$

$\Rightarrow AD = DC \times \sqrt{3}=DC\sqrt{3}$ m           (1)

In $\triangle ABD$

$\frac{AD}{BD}= tan 30^\circ$

$\Rightarrow \frac{AD}{BC+CD}=\frac{1}{\sqrt{3}}$

$\Rightarrow AD = (20+DC)\frac{1}{\sqrt{3}}$

Putting equation (1), we get

$DC\sqrt{3}=(20+DC)\frac{1}{\sqrt{3}}$
$\Rightarrow 3DC = 20+DC$

$\Rightarrow 2DC =20$

$\Rightarrow DC = \frac{20}{2}=10$ m

Using equation (1), we have

$AD =DC \times \sqrt{3}=10\sqrt{3}$ m

Therefore, Height of tower = AD = $10\sqrt{3}$ m

Width of Canal = DC = 10 m

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