Find the angular elevation of the Sun when the shadow of a 15 metre long pole is `(15)/(sqrt(3))` metre .

Options

A. `45^(@)`
B. `60^(@)`
C. `30^(@)`
D. `90^(@)`

Answer

To find the angular elevation of the Sun when the shadow of a 15-meter long pole is \( \frac{15}{\sqrt{3}} \) meters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- The height of the pole (perpendicular) = 15 meters.
- The length of the shadow (base) = \( \frac{15}{\sqrt{3}} \) meters.

2. **Set Up the Tangent Ratio**:
- The tangent of the angle of elevation \( \theta \) can be expressed as:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{Height of the pole}}{\text{Length of the shadow}} \]
- Substituting the values:
\[ \tan(\theta) = \frac{15}{\frac{15}{\sqrt{3}}} \]

3. **Simplify the Expression**:
- To simplify \( \tan(\theta) \):
\[ \tan(\theta) = 15 \times \frac{\sqrt{3}}{15} = \sqrt{3} \]

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

5. **Final Answer**:
- The angular elevation of the Sun is \( 60^\circ \).