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

`30^(\circ)`.

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)) = √(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)} \\ &= (h)/(\tan(60^(\circ)))\\ &= (h)/(√(3)) \\ &\end{aligned}`.

Therefore:

`\begin{aligned}{\sf BD} &= 3\, {\sf BC} \\ &= (3\, h)/(√(3)) \\ &= (√(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} = (√(3))\, h`:

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

Therefore:

`\begin{aligned}\theta &= \arctan\left((1)/(√(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)`.