If a `DeltaABC, b=2, C=60^(@), c=sqrt(6)`, then a =

Options

A. `sqrt(3)-1`
B. `sqrt(3)`
C. `sqrt(3)+1`
D. none of these

Answer

To find the length of side \( a \) in triangle \( ABC \) where \( b = 2 \), \( C = 60^\circ \), and \( c = \sqrt{6} \), we can use the cosine rule. The cosine rule states:

\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]

### Step-by-Step Solution:

1. **Identify the known values**:
- \( b = 2 \)
- \( C = 60^\circ \)
- \( c = \sqrt{6} \)

2. **Substitute the known values into the cosine rule**:
\[ (\sqrt{6})^2 = a^2 + (2)^2 - 2 \cdot a \cdot 2 \cdot \cos(60^\circ) \]

3. **Calculate \( \cos(60^\circ) \)**:
\[ \cos(60^\circ) = \frac{1}{2} \]

4. **Substitute \( \cos(60^\circ) \) into the equation**:
\[ 6 = a^2 + 4 - 2 \cdot a \cdot 2 \cdot \frac{1}{2} \]

5. **Simplify the equation**:
\[ 6 = a^2 + 4 - 2a \]
Rearranging gives:
\[ a^2 - 2a + 4 - 6 = 0 \]
\[ a^2 - 2a - 2 = 0 \]

6. **Use the quadratic formula to solve for \( a \)**:
The quadratic formula is given by:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( A = 1 \), \( B = -2 \), and \( C = -2 \).

7. **Calculate the discriminant**:
\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot (-2) = 4 + 8 = 12 \]

8. **Substitute into the quadratic formula**:
\[ a = \frac{-(-2) \pm \sqrt{12}}{2 \cdot 1} = \frac{2 \pm 2\sqrt{3}}{2} \]
Simplifying gives:
\[ a = 1 \pm \sqrt{3} \]

9. **Determine the valid solution**:
Since lengths cannot be negative, we take:
\[ a = 1 + \sqrt{3} \]

### Final Answer:
\[ a = 1 + \sqrt{3} \]