If the shadow of a pole of height `(sqrt(3)+ 1)` m standing on the ground is found to be 2 m longer, when the elevation is `30^(@)` than when elevation was `alpha`, then `alpha` is equal to

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `75^(@)`

Answer

To solve the problem, we need to find the angle of elevation \( \alpha \) given the height of the pole and the relationship between the lengths of the shadows at two different angles of elevation.

### Step-by-step Solution:

1. **Understanding the Problem:**
- We have a pole of height \( h = \sqrt{3} + 1 \) meters.
- When the angle of elevation is \( 30^\circ \), the shadow is 2 meters longer than when the angle of elevation is \( \alpha \).

2. **Setting Up the Diagram:**
- Let \( D \) be the top of the pole, \( A \) be the point on the ground where the shadow ends when the angle is \( 30^\circ \), and \( B \) be the point where the shadow ends when the angle is \( \alpha \).
- The length of the shadow at \( 30^\circ \) is \( AB = x + 2 \) meters, and the length of the shadow at \( \alpha \) is \( AB = x \) meters.

3. **Using Trigonometric Ratios:**
- For the angle \( 30^\circ \):
\[ \tan(30^\circ) = \frac{\text{height of the pole}}{\text{length of the shadow at } 30^\circ} \]
\[ \tan(30^\circ) = \frac{\sqrt{3} + 1}{x + 2} \]
- We know \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{x + 2} \]

4. **Cross-Multiplying:**
- Cross-multiplying gives:
\[ x + 2 = \sqrt{3}(\sqrt{3} + 1) \]
- Simplifying the right side:
\[ x + 2 = 3 + \sqrt{3} \]
- Therefore:
\[ x = 1 + \sqrt{3} \]

5. **Finding \( \alpha \):**
- Now, we will use triangle \( DBC \) to find \( \alpha \):
\[ \tan(\alpha) = \frac{\text{height of the pole}}{\text{length of the shadow at } \alpha} \]
\[ \tan(\alpha) = \frac{\sqrt{3} + 1}{1 + \sqrt{3}} \]
- Simplifying:
\[ \tan(\alpha) = 1 \]
- This implies:
\[ \alpha = 45^\circ \]

### Conclusion:
The value of \( \alpha \) is \( 45^\circ \).