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.

Options

A. `60^0`
B. `30^0`
C. `45^0`
D. `15^0`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the problem
We have a tower (let's denote its height as \( h \)) and two scenarios for the shadow of the tower: one when the angle of elevation of the sun is \( 60^\circ \) and another when the angle of elevation is unknown (let's denote it as \( \theta \)). The shadow at the angle of elevation \( 60^\circ \) has a length \( x \), and the shadow at angle \( \theta \) is three times longer, which is \( 3x \).

### Step 2: Set up the first triangle
Using the angle of elevation \( 60^\circ \):
- In triangle \( ABC \) (where \( A \) is the top of the tower, \( B \) is the base of the tower, and \( C \) is the tip of the shadow), we can use the tangent function:
\[ \tan(60^\circ) = \frac{h}{x} \]
From trigonometric values, we know that \( \tan(60^\circ) = \sqrt{3} \). Therefore, we can write:
\[ \sqrt{3} = \frac{h}{x} \]
Rearranging gives us:
\[ h = x \sqrt{3} \quad \text{(Equation 1)} \]

### Step 3: Set up the second triangle
Now, using the angle of elevation \( \theta \) when the shadow is \( 3x \):
- In triangle \( ABD \) (where \( D \) is the tip of the longer shadow), we can again use the tangent function:
\[ \tan(\theta) = \frac{h}{3x} \]
Substituting \( h \) from Equation 1 into this equation gives:
\[ \tan(\theta) = \frac{x \sqrt{3}}{3x} \]
This simplifies to:
\[ \tan(\theta) = \frac{\sqrt{3}}{3} \]

### Step 4: Solve for \( \theta \)
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \text{and} \quad \tan(60^\circ) = \sqrt{3} \]
Thus, we can rewrite:
\[ \tan(\theta) = \frac{1}{\sqrt{3}} \quad \Rightarrow \quad \theta = 30^\circ \]

### Conclusion
The angle of elevation of the sun at the time of the longer shadow is \( \theta = 30^\circ \).

---