In a `DeltaABC` if a = 2, `b=sqrt(6),c=sqrt(3)+1`, then cos A=

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. none of these

Answer

To find \( \cos A \) in triangle \( ABC \) where \( a = 2 \), \( b = \sqrt{6} \), and \( c = \sqrt{3} + 1 \), we will use the cosine rule. The cosine rule states:

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

### Step-by-Step Solution:

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

2. **Calculate \( b^2 \)**:
\[ b^2 = (\sqrt{6})^2 = 6 \]

3. **Calculate \( c^2 \)**:
\[ c^2 = (\sqrt{3} + 1)^2 = (\sqrt{3})^2 + 2 \cdot \sqrt{3} \cdot 1 + 1^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \]

4. **Calculate \( a^2 \)**:
\[ a^2 = 2^2 = 4 \]

5. **Substitute values into the cosine rule formula**:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{6 + (4 + 2\sqrt{3}) - 4}{2 \cdot \sqrt{6} \cdot (\sqrt{3} + 1)} \]

6. **Simplify the numerator**:
\[ \cos A = \frac{6 + 4 + 2\sqrt{3} - 4}{2 \cdot \sqrt{6} \cdot (\sqrt{3} + 1)} = \frac{6 + 2\sqrt{3}}{2 \cdot \sqrt{6} \cdot (\sqrt{3} + 1)} \]

7. **Factor out common terms in the numerator**:
\[ \cos A = \frac{2(\sqrt{3} + 3)}{2 \cdot \sqrt{6} \cdot (\sqrt{3} + 1)} \]

8. **Cancel the common factor of 2**:
\[ \cos A = \frac{\sqrt{3} + 3}{\sqrt{6} \cdot (\sqrt{3} + 1)} \]

9. **Further simplify**:
\[ \cos A = \frac{\sqrt{3}}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{2}} = \frac{1}{\sqrt{2}} \]

10. **Final result for \( \cos A \)**:
\[ \cos A = \frac{1}{\sqrt{2}} \]

### Conclusion:
Thus, \( \cos A = \frac{1}{\sqrt{2}} \), which corresponds to \( A = 45^\circ \).