Question

The shadow of a tower is three times as long as the shadow of the tower when the sun's rays met the ground at an angle of $60°$. Find the angle of elevation of the sun at the time of the long shadow.

Answer 1

Step 1: Drawing the diagram of the situation

Let $A$ be the position of the sun.

Let the height of the tower be $hmeters$ and the angle at the time of longer shadow be $\theta .$
Now, let $BC$ and $BD$ be the lengths of the shadow of the tower when the angles are $60°$ and $\theta ,$ respectively.

Step 2: Finding the angle of elevation of the sun at the time of the long shadow.

In $∆ABC$
$$\begin{gathered} \tan \left(C\right)=\frac{\text{side opposite to∠C}}{\text{side adjacent to ∠C}} \\ \Rightarrow \tan \left({60}^{\circ }\right)=\frac{AB}{BC} \\ \Rightarrow \sqrt{3}=\frac{h}{BC}(\because \tan ({60}^{\circ })=\sqrt{3}) \\ \Rightarrow h=\sqrt{3}BC...\left(i\right) \end{gathered}$$

And, in $∆ABD$

$$\begin{gathered} \tan \left(D\right)=\frac{\text{side opposite to ∠D}}{\text{side adjacent to ∠D}} \\ \Rightarrow \tan \left(\theta \right)=\frac{AB}{BD} \\ \Rightarrow \tan \left(\theta \right)=\frac{h}{BD} \\ \Rightarrow h=BD\tan \left(\theta \right)...\left(ii\right) \end{gathered}$$
From the equation $\left(i\right)$ and $\left(ii\right),$ we have

$$\begin{gathered} \sqrt{3}BC=BD\tan \left(\theta \right) \\ \Rightarrow \sqrt{3}BC=3BC\tan \left(\theta \right)(\because BD=3BC) \\ \Rightarrow \tan \left(\theta \right)=\frac{1}{\sqrt{3}} \\ \Rightarrow \theta ={30}^{\circ }\left(\because \tan (30°)=\frac{1}{\sqrt{3}}\right) \end{gathered}$$

Hence, the angle of elevation of the sun at the time of long shadow is ${30}^{\circ }.$