What is the angle of elevation of the Sun, when the shadow of a pole of height xm is `x/sqrt3`m?

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `75^@`

Answer

To solve the problem of finding the angle of elevation of the Sun when the shadow of a pole of height \( x \) meters is \( \frac{x}{\sqrt{3}} \) meters, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- We have a pole of height \( x \) meters.
- The length of the shadow of the pole is \( \frac{x}{\sqrt{3}} \) meters.
- We need to find the angle of elevation \( \theta \) of the Sun.

2. **Draw a Right Triangle**:
- Let point A be the top of the pole, point B be the base of the pole, and point C be the tip of the shadow.
- In triangle ABC:
- AB (the height of the pole) = \( x \) meters (perpendicular).
- BC (the length of the shadow) = \( \frac{x}{\sqrt{3}} \) meters (base).

3. **Use the Tangent Function**:
- The tangent of the angle of elevation \( \theta \) is given by the formula:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \]
- Substituting the values:
\[ \tan(\theta) = \frac{x}{\frac{x}{\sqrt{3}}} \]

4. **Simplify the Expression**:
- Simplifying the right-hand side:
\[ \tan(\theta) = \frac{x \cdot \sqrt{3}}{x} = \sqrt{3} \]

5. **Find the Angle**:
- We know that \( \tan(60^\circ) = \sqrt{3} \).
- Therefore, we can conclude that:
\[ \theta = 60^\circ \]

### Final Answer:
The angle of elevation of the Sun is \( 60^\circ \).

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