Question

The angle of elevation of the top of a tower from two points distant $s$ and $t$ from its foot are complementary. Prove that the height of the tower is $\sqrt{st}$.

Answer 1

Step 1: Draw the diagram

Let $BC=s$; $PC=t$ be the distance of given points.

Let the height of the tower be $AC=h$

$\angle ABC=\theta$and $\angle APC=90°-\theta$

(∵ the angle of elevation of the top of the tower from two points P and B are complementary)

Step 2: Proof

$$\begin{gathered} In△ABC \\ \Rightarrow \tan \left(\theta \right)=\frac{AC}{BC} \\ \Rightarrow \tan \left(\theta \right)=\frac{h}{s}---------\left(i\right) \end{gathered}$$

Now,

$$\begin{gathered} In△ACP \\ \Rightarrow \tan \left(90°-\theta \right)=\frac{AC}{PC} \\ =\cot \left(\theta \right)=\frac{h}{t}--------\left(ii\right) \end{gathered}$$

By multiplying the equations (i) and (ii) we obtain

$$\begin{gathered} \Rightarrow \tan \left(\theta \right)\times \cot \left(\theta \right)=\frac{h}{s}\times \frac{h}{t} \\ \Rightarrow 1=\frac{h^2}{st}\because \tan \left(\theta \right)=\frac{1}{\cot \left(\theta \right)} \\ \Rightarrow st=h^2 \\ \Rightarrow h=\sqrt{st} \end{gathered}$$

Hence, the height of the tower is $\sqrt{st}$