The shadow of a pole of height `(sqrt3+1)` metres standing on the ground is found is found to be 2 metres longer when the elevation is `30^@` than when elevation was `alpha`.Then , `alpha`=

Options

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

Explanation

To solve the problem step by step, we will use the properties of right triangles and the tangent function.

### Step 1: Understanding the problem
We have a pole of height \( h = \sqrt{3} + 1 \) meters. The shadow of the pole is longer when the angle of elevation is \( 30^\circ \) than when it is \( \alpha \) degrees. The shadow at \( 30^\circ \) is 2 meters longer than the shadow at angle \( \alpha \).

### Step 2: Setting up the equations
Let the length of the shadow when the angle of elevation is \( \alpha \) be \( x \) meters. Therefore, the length of the shadow when the angle of elevation is \( 30^\circ \) will be \( x + 2 \) meters.

### Step 3: Using the tangent function
For the angle of elevation \( 30^\circ \):
\[ \tan(30^\circ) = \frac{\text{height of the pole}}{\text{length of the shadow at } 30^\circ} \]
Substituting the values:
\[ \tan(30^\circ) = \frac{\sqrt{3} + 1}{x + 2} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \). Therefore, we can write:
\[ \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{x + 2} \]

### Step 4: Cross-multiplying
Cross-multiplying gives us:
\[ 1 \cdot (x + 2) = \sqrt{3} \cdot (\sqrt{3} + 1) \]
This simplifies to:
\[ x + 2 = 3 + \sqrt{3} \]

### Step 5: Solving for \( x \)
Now, we can solve for \( x \):
\[ x = 3 + \sqrt{3} - 2 \]
\[ x = 1 + \sqrt{3} \]

### Step 6: Finding \( \tan(\alpha) \)
Now, we will use the triangle formed by the pole and the shadow when the angle of elevation is \( \alpha \):
\[ \tan(\alpha) = \frac{\text{height of the pole}}{\text{length of the shadow at } \alpha} \]
Substituting the values:
\[ \tan(\alpha) = \frac{\sqrt{3} + 1}{x} \]
Substituting \( x = 1 + \sqrt{3} \):
\[ \tan(\alpha) = \frac{\sqrt{3} + 1}{1 + \sqrt{3}} \]

### Step 7: Simplifying \( \tan(\alpha) \)
Notice that:
\[ \tan(\alpha) = 1 \]
This implies that:
\[ \alpha = 45^\circ \]

### Final Answer
Thus, the angle \( \alpha \) is:
\[ \alpha = 45^\circ \]