# Applications of Trigonometry Heights and Distances ncert solutions Chapter 9 Exercise 9.1 Question 10

Applications of Trigonometry Heights and Distances Chapter 9 Exercise 9.1 Question 10

10.   Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^\circ$ and $30^\circ$, respectively. Find the height of the poles and the distances of the point from the poles.

Solution:

It is given that poles (AB and ED) are of equal height.

It is also given that BD = 80 m

Let CD = x m

And, let BC = (80 - x) m

In $\triangle ECD$, we have

$\frac{ED}{CD}= tan 30^\circ$

$\Rightarrow ED = \frac{x}{\sqrt{3}}$ m         (1)

In $\triangle ABC$, we have

$\frac{AB}{BC}=tan 60^\circ$

$\Rightarrow AB=(80-x)\sqrt{3}$        (2)

We have ED = AB because both the poles are of equal height.

Therefore, $\frac{x}{\sqrt{3}} =(80-x)\sqrt{3}$

$\Rightarrow x = (80-x)3$

$\Rightarrow x=240-3x$

$\Rightarrow 4x=240$

$\Rightarrow x =\frac{240}{4}=60$ m

BC=80-x = 80-60=20 m

Putting value of x in equation (1), we get

$ED =\frac{60}{\sqrt{3}}=\frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{60\sqrt{3}}{3}=20\sqrt{3}$ m

Therefore, Distances of  the point C from the poles are 60 m and 20 m.

And, the height of poles is $20\sqrt{3}$ m

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