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Applications of Trigonometry Heights and Distances Chapter 9 Exercise 9.1 Question 7

Applications of Trigonometry Heights and Distances Chapter 9 Exercise 9.1 Question 7

 


 

7.      From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45^\circ and 60^\circ respectively. Find the height of the tower.

 

Solution:

ncert solutions heights and distances class 10

It is given that tower (AD) is fixed at the top of 20 m high building (DC).

 

In \triangle BDC, we have

\frac{DC}{BC} =tan 45^\circ

\Rightarrow DC = BC

\Rightarrow BC = 20 m

 

In \triangle ABC, we have

\frac{AC}{BC}=tan 60^\circ

\Rightarrow AC= BCtan 60^\circ = 20\sqrt{3} m

 

Length of tower (AD) = AC - DC = 20\sqrt{3} - 20 = 20(\sqrt{3}-1) m


 


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