Question

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

Answer 1

Let $AB$ be the tower and $C\mathrm{and}D$ be two points such that $AC=4$ m and $AD=9$ m.
Let:
$AB=h$ m, ∠$BCA=\theta$ and ∠$BDA={90}^{°}-\theta$

In the right ∆BCA, we have:
$$\begin{gathered} \tan \theta =\frac{AB}{AC} \\ \Rightarrow \tan \theta =\frac{h}{4}...\left(1\right) \end{gathered}$$
In the right ∆BDA, we have:
$$\begin{gathered} \tan \left(90°-\theta \right)=\frac{AB}{AD} \\ \Rightarrow \cot \theta =\frac{h}{9}\left[\tan \left(90°-\theta \right)=\cot \theta \right] \\ \Rightarrow \frac{1}{\tan \theta }=\frac{h}{9}...\left(2\right)\left[\cot \theta =\frac{1}{\tan \theta }\right] \end{gathered}$$
Multiplying equations (1) and (2), we get:
$$\begin{gathered} \tan \theta \times \frac{1}{\tan \theta }=\frac{h}{4}\times \frac{h}{9} \\ \Rightarrow 1=\frac{h^2}{36} \end{gathered}$$
⇒ 36 = h²
⇒ h = ±6

Height of a tower cannot be negative.
∴ Height of the tower = 6 m