The angles of elevation of the top of a tower from two points at a distance of $4 \mathrm{~m}$ and $9 \mathrm{~m}$ from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $6 \mathrm{~m}$.

Given:

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 complementary.

To do:

We have to prove that the height of the tower is $6\ m$.

Solution:

Let height of the tower be $AB$, $AC=4\ m$ and $AD=9\ m$.

Let $\angle BDA=\theta , \angle BCA=90^{o} -\theta$.

$\angle BDA+\angle BCA=90^{o}$   (Given)

In $\vartriangle BCA$

$tan 90^{o}-\theta =\frac{AB}{AC}=\frac{AB}{4}$

$\Rightarrow AB=4cot \theta$............(i)               [Since $tan 90^{o}-\theta=cot \theta$]

In $\vartriangle BDA$

$tan( 90^{o} -\theta ) =\frac{AB}{DA} =\frac{AB}{9}$

$\Rightarrow AB=9tan \theta$..............(ii)

On multiplying $( 1)$ and $( 2)$

$AB^{2} =4cot \theta \times 9tan \theta =36$     ($tan \theta \times cot \theta=1$)

$\Rightarrow AB=\sqrt{36}=6\ m$

Thus, the height of the tower is $6\ m$.

Hence proved.

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Updated on: 10-Oct-2022

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