Question

The angle of elevation of the top of a tower from a certain point is $30°$. If the observer moves $20m$ towards the tower, the angle of elevation of the top increases by $15°$. Find the height of the tower.

Answer 1

Finding the height of the tower:

Let $PR=hmeter$, be the height of the tower.

The observer is standing at a point $Q$ such that, the distance between the observer and the tower is $QR=(20+x)m$, where

$$\begin{gathered} QR=QS+SR=\left(20+x\right)m \\ \angle PQR=30° \\ \angle PSR=\theta \end{gathered}$$

$$\begin{gathered} In△PQR \\ \Rightarrow \tan \left(30°\right)=\frac{PR}{QR} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x} \\ \Rightarrow 20+x=\sqrt{3}h \\ \Rightarrow x=\sqrt{3}h-20--------\left(i\right) \end{gathered}$$

$$\begin{gathered} In△PSR \\ \Rightarrow \tan \left(\theta \right)=\frac{h}{x} \end{gathered}$$

The angle of elevation increases by $15°$ when the observer moves $20m$ toward the tower.

Hence

$\theta =30°+15°=45°$

$$\begin{gathered} \Rightarrow \tan \left(45°\right)=\frac{h}{x} \\ \Rightarrow 1=\frac{h}{x} \\ \Rightarrow h=x \end{gathered}$$

Substituting $x=hin\left(i\right)$, we get

$$\begin{gathered} \Rightarrow h=\sqrt{3}h-20 \\ \Rightarrow \left(\sqrt{3}-1\right)h=20 \\ \Rightarrow h=\frac{20}{\sqrt{3}-1}m \end{gathered}$$

On rationalizing the denominator we get

$$\begin{gathered} \Rightarrow h=\frac{20}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ \Rightarrow h=\frac{20(\sqrt{3}+1)}{2} \\ \Rightarrow h=10(\sqrt{3}+1)m \end{gathered}$$

Hence, the required height of the tower is $10(\sqrt{3}+1)m$.