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. Prove that the height of the tower is 6 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 $TF$, $CF=4\ m$ and $DF=9\ m$.

Let $\angle TCF=\theta , \angle TDF=90^{o} -\theta$.

$\angle TCF+\angle TDF=90^{o}$   (Given)

In a right angled triangle $\vartriangle TCF$

$tan\theta =\frac{TF}{CF}=\frac{TF}{4}$

$\Rightarrow TF=4tan\theta \ \ \ \ \ \ \ .................( 1)$

In $\vartriangle TDF$

$tan( 90^{o} -\theta ) =\frac{TF}{DF} =\frac{TF}{9}$

$\Rightarrow TF=9tan( 90^{o} -\theta )=9cot\theta \ \ \ \ ....................( 2)$

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

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

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

Thus, the height of the tower is 6 m.

Hence proved.

Updated on: 10-Oct-2022

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