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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:

Heights and distances ncert solutions trigonometry class 10
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

 


 


Posted in : NCERT Solutions, Trigonometry


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