In an ambiguoa ambiguous case of solving a triangleshen `a = sqrt5,b = 2, /_A=pi/6` and the two possible values of third side are `c_1 and c_2`, then

Options

A. `|c_(1)-c_(2)|=2sqrt6`
B. `|c_(1)-c_(2)|=4sqrt6`
C. `|c_(1)-c_(2)|=4`
D. `|c_(1) -c_(2)|=6`

Answer

To solve the given problem, we will use the Law of Cosines to find the two possible values of the third side \( c \) in triangle \( ABC \) where \( a = \sqrt{5} \), \( b = 2 \), and \( \angle A = \frac{\pi}{6} \) (which is equivalent to 30 degrees).

### Step-by-Step Solution:

1. **Identify the known values**:
- \( a = \sqrt{5} \)
- \( b = 2 \)
- \( \angle A = \frac{\pi}{6} \)

2. **Apply the Law of Cosines**:
The Law of Cosines states:
\[ c^2 = a^2 + b^2 - 2ab \cos A \]
We need to find \( c \) using the values we have.

3. **Calculate \( \cos A \)**:
Since \( A = \frac{\pi}{6} \):
\[ \cos A = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

4. **Substitute the values into the formula**:
\[ c^2 = (\sqrt{5})^2 + (2)^2 - 2 \cdot \sqrt{5} \cdot 2 \cdot \frac{\sqrt{3}}{2} \]
Simplifying this:
\[ c^2 = 5 + 4 - 2\sqrt{5} \cdot \sqrt{3} \]
\[ c^2 = 9 - 2\sqrt{15} \]

5. **Set up the quadratic equation**:
To find the possible values of \( c \), we rearrange the equation:
\[ c^2 - 2\sqrt{15}c + 9 = 0 \]

6. **Use the quadratic formula**:
The quadratic formula is given by:
\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2\sqrt{15} \), and \( c = 9 \):
\[ c = \frac{2\sqrt{15} \pm \sqrt{(-2\sqrt{15})^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \]

7. **Calculate the discriminant**:
\[ (-2\sqrt{15})^2 = 4 \cdot 15 = 60 \]
\[ 4 \cdot 1 \cdot 9 = 36 \]
Therefore,
\[ \text{Discriminant} = 60 - 36 = 24 \]

8. **Substitute back into the formula**:
\[ c = \frac{2\sqrt{15} \pm \sqrt{24}}{2} \]
Since \( \sqrt{24} = 2\sqrt{6} \):
\[ c = \frac{2\sqrt{15} \pm 2\sqrt{6}}{2} \]
\[ c = \sqrt{15} \pm \sqrt{6} \]

9. **Final values of \( c \)**:
Thus, the two possible values of the third side \( c \) are:
\[ c_1 = \sqrt{15} + \sqrt{6} \]
\[ c_2 = \sqrt{15} - \sqrt{6} \]