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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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