If sides of a triangle are in the ratio `2:sqrt6:1+sqrt3` the what is the smallest angle of the triangle?

Options

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

Answer

To find the smallest angle of the triangle with sides in the ratio \(2 : \sqrt{6} : 1 + \sqrt{3}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the sides of the triangle**:
Let the sides of the triangle be:
- \( a = 2 \)
- \( b = \sqrt{6} \)
- \( c = 1 + \sqrt{3} \)

2. **Determine the smallest angle**:
The smallest angle in a triangle is opposite the shortest side. Here, we need to compare the lengths of the sides:
- \( a = 2 \)
- \( b = \sqrt{6} \approx 2.45 \)
- \( c = 1 + \sqrt{3} \approx 2.732 \)

Since \( a = 2 \) is the smallest side, the smallest angle is opposite side \( a \), which is angle \( A \).

3. **Use the cosine rule to find angle \( A \)**:
The cosine rule states:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Plugging in the values:
\[ \cos A = \frac{(\sqrt{6})^2 + (1 + \sqrt{3})^2 - (2)^2}{2 \cdot \sqrt{6} \cdot (1 + \sqrt{3})} \]

4. **Calculate \( b^2 \) and \( c^2 \)**:
- \( b^2 = 6 \)
- \( c^2 = (1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} \)

5. **Substituting back into the cosine formula**:
\[ \cos A = \frac{6 + (4 + 2\sqrt{3}) - 4}{2 \cdot \sqrt{6} \cdot (1 + \sqrt{3})} \]
Simplifying the numerator:
\[ \cos A = \frac{6 + 4 + 2\sqrt{3} - 4}{2 \cdot \sqrt{6} \cdot (1 + \sqrt{3})} = \frac{6 + 2\sqrt{3}}{2 \cdot \sqrt{6} \cdot (1 + \sqrt{3})} \]

6. **Simplifying further**:
The numerator can be simplified to:
\[ \cos A = \frac{3 + \sqrt{3}}{\sqrt{6}(1 + \sqrt{3})} \]

7. **Finding the angle**:
We can find \( A \) by calculating the inverse cosine:
\[ A = \cos^{-1}\left(\frac{3 + \sqrt{3}}{\sqrt{6}(1 + \sqrt{3})}\right) \]

8. **Final calculation**:
We find that \( A \) is approximately \( 45^\circ \).

### Conclusion:
The smallest angle of the triangle is \( 45^\circ \).