Question

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

Answer 1

Let $AB$ be the tower.

$C$ and $D$be the two points with a distance $4m$ and $9m$

In $△ABC$

$$\begin{gathered} \Rightarrow \tan x=\frac{AB}{BC} \\ \Rightarrow \tan x=\frac{AB}{4} \\ \Rightarrow AB=4\tan x------\left(i\right) \end{gathered}$$

Also,

$$\begin{gathered} \Rightarrow \tan \left(90-x\right)=\frac{AB}{BD} \\ \Rightarrow \cot x=\frac{AB}{9} \\ \Rightarrow AB=9cotx------\left(ii\right) \end{gathered}$$

Multiplying equation $\left(i\right)$ and $\left(ii\right)$

$$\begin{gathered} {AB}^2=4\tan x\times 9\cot x \\ {AB}^2=36\times \tan x\times \cot x\because \tan x=\frac{1}{\cot x} \\ {AB}^2=36\times \tan x\times \frac{1}{\tan x} \\ {AB}^2=36 \\ AB=6m \end{gathered}$$

Therefore, the height of the tower is $6m$

Hence proved