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Applications of Trigonometry Heights and Distances ncert solutions Chapter 9 Exercise 9.1 Question 16

Applications of Trigonometry Heights and Distances Chapter 9 Exercise 9.1 Question 16


 

16.    The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complimentary. Prove that the height of the tower is 6 m.

 

Solution:

Ncert solutions Applications of trigonometry class 10

 

It is given that \angle ADB and \angle ACB are complementary which means their sum is equal to 90^\circ.

Let \angle ADB = x^\circ and \angle ACB = (90-x)^\circ

It is given that BD = 4 m and BC = 9m which means DC = 9-4 = 5 m

 

In \triangle ABC

\frac{AB}{BC}= tan (90-x)

\Rightarrow \frac{AB}{9}=cot x                    {tan(90-x) = cot x}

\Rightarrow AB = 9cot x              (1)

 

In \triangle ABD

\frac{AB}{BD}=tan x

\Rightarrow \frac{AB}{4}=tanx

\Rightarrow AB=4tan x                           (2)

 

From (1) and (2), we can say that 9cot x=4tan x

\Rightarrow \frac{9}{tan x}=4tan x
\Rightarrow tan^2 x =\frac{9}{4}

\Rightarrow tan x =\frac{3}{2}             (3)

 

Putting equation (3) in equation (2), we get

AB = 4 \times \frac{3}{2}=6 m

 

Therefore, Height of the Tower is equal to 6 m


 


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