Question

The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the Sun is 60°. Find the angle of elevation of the Sum at the time of the longer shadow.​

Answer 1

Answer:

$30^{\circ}$.

Step-by-step explanation:

Let $\theta$ denote the unknown angle of elevation. Let $h$ denote the height of the tower.

Refer to the diagram attached. In this diagram, ${\sf A}$ denotes the top of the tower while ${\sf B}$ denote the base of the tower; ${\sf BC}$ and ${\sf BD}$ denote the shadows of the tower when the angle of elevation of the sun is $60^{\circ}$ and $\theta$, respectively. The length of segment ${\sf AB}$ is $h$; $\angle {\sf ACB} = 60^{\circ}$, $\angle {\sf ADB} = \theta$, and ${\sf BD} = 3\, {\sf BC}$..

Note that in right triangle $\triangle {\sf ABC}$, segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ACB}$. At the same time, segment ${\sf BC}$ (shadow of the tower when the angle of elevation of the sun is $60^{\circ}$) is adjacent to $\angle {\sf ACB}$.

Thus, the ratio between the length of these two segments could be described with the tangent of $m\angle {\sf ACB} = 60^{\circ}$:

$\begin{aligned}\tan(\angle {\sf ACB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AB}}{{\sf BC}}\end{aligned}$.

$\begin{aligned}\frac{{\sf AB}}{{\sf BC}} = \tan(60^{\circ}) = \sqrt{3}\end{aligned}$.

The length of segment ${\sf AB}$ is $h$. Rearrange this equation to find the length of segment ${\sf BC}$:

$\begin{aligned} {\sf BC} &= \frac{{\sf AB}}{\tan(\angle ACB)} \\ &= \frac{h}{\tan(60^{\circ})}\\ &= \frac{h}{\sqrt{3}} \\ &\end{aligned}$.

Therefore:

$\begin{aligned}{\sf BD} &= 3\, {\sf BC} \\ &= \frac{3\, h}{\sqrt{3}} \\ &= (\sqrt{3})\, h\end{aligned}$.

Similarly, in right triangle ${\sf ABD}$, segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ADB}$. Segment ${\sf BD}$ (shadow of the tower, with $\theta$ as the angle of elevation of the sun) is adjacent to $\angle {\sf ADB}$.

$\begin{aligned}\tan(\angle {\sf ADB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AD}}{{\sf BD}}\end{aligned}$.

$\begin{aligned}\frac{{\sf AB}}{{\sf BD}} = \tan(\theta) \end{aligned}$.

Since ${\sf AB} = h$ while ${\sf BD} = (\sqrt{3})\, h$:

$\begin{aligned}\tan(\theta) &= \frac{{\sf AB}}{{\sf BD}} \\ &= \frac{h}{(\sqrt{3})\, h} \\ &= \frac{1}{\sqrt{3}}\end{aligned}$.

Therefore:

$\begin{aligned}\theta &= \arctan\left(\frac{1}{\sqrt{3}}\right) \\ &= 30^{\circ}\end{aligned}$.

In other words, the angle of elevation of the sun at the time of the longer shadow would be $30^{\circ}$.