# 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:

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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