Question

The angle of elevation of the top of a tower from two points at distance the base and in the same straight line with it are complementary. Prove that the height of the tower is $\sqrt {ab} $ meters?

Answer 1

Hint: First, draw the diagram of the problem. Now, in $\Delta ABC$, get the value of $\tan \theta $. After that in $\Delta AMC$ again solve for $\tan \theta $. Now compare both the equation and do the calculation to get the height of the tower.

Complete step-by-step answer:
To prove:- The height of the tower is $\sqrt {ab} $ m.
Let the height of the tower be h m, the distance from the nearest point to the foot of the tower be b m and the distance from the other point to the foot of the tower be a m.
The figure of the problem is,

Now, in the $\Delta ABC$,
$\tan (90 - \theta ) = \dfrac{{AB}}{{BC}}$
Substitute the values of AB and BC in the above equation,
$\tan (90 - \theta ) = \dfrac{h}{b}$
As we know that, $\tan \left( {90 - \theta } \right) = \cot \theta $. Substitute $\cot \theta $ on the place of $\tan \left( {90 - \theta } \right)$.
$\cot \theta = \dfrac{h}{b}$
Now, we know that $\cot \theta = \dfrac{1}{{\tan \theta }}$. Replace $\cot \theta $ with $\dfrac{1}{{\tan \theta }}$.
$\dfrac{1}{{\tan \theta }} = \dfrac{b}{h}$
Reverse the equation to get the value of $\tan \theta $,
$\tan \theta = \dfrac{h}{b}$ …..(1)
In the$\Delta AMB,\angle B = 90^\circ $
$\tan \theta = \dfrac{{AB}}{{MB}}$
Substitute the values of AB and MB in the above equation,
$\tan \theta = \dfrac{h}{a}$ …..(2)
Compare both equation (1) and (2),
$\dfrac{h}{b} = \dfrac{a}{h}$
Cross multiply the equation we get,
${h^2} = ab$
Take the square root on both sides to get the final result,
$h = \sqrt {ab} $ m
Hence, it is proved that the height of the tower is $\sqrt {ab} $ m.

Note: If the sum of two angles is 90, then the two angles are called complementary angles. If the sum of two angles is 180, then the two angles are called supplementary angles.