In a triangle ABC if `2a=sqrt(3)b+c`, then possible relation is

Options

A. `c^(2)=a^(2)+b^(2)-ab`
B. `a^(2)=b^(2)+c^(2)`
C. `b^(2)=a^(2)+c^(2)-ac sqrt(3)`
D. a = b = c

Explanation

To solve the problem, we need to establish a relationship between the sides \(a\), \(b\), and \(c\) of triangle \(ABC\) given the equation \(2a = \sqrt{3}b + c\).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ 2a = \sqrt{3}b + c \]

2. **Rearrange the equation**:
\[ 2a - c = \sqrt{3}b \]
Now, we can express \(b\) in terms of \(a\) and \(c\):
\[ b = \frac{2a - c}{\sqrt{3}} \]

3. **Use the Sine Rule**:
According to the sine rule in triangle \(ABC\):
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
where \(k\) is a constant.

4. **Express \(a\), \(b\), and \(c\) in terms of \(k\)**:
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

5. **Substitute these expressions into the rearranged equation**:
\[ 2(k \sin A) - (k \sin C) = \sqrt{3}(k \sin B) \]
Dividing the entire equation by \(k\) (assuming \(k \neq 0\)):
\[ 2\sin A - \sin C = \sqrt{3}\sin B \]

6. **Rearranging gives**:
\[ 2\sin A = \sqrt{3}\sin B + \sin C \]

7. **Using trigonometric identities**:
We can relate the angles using known values. If we assume angles \(B\) and \(C\) correspond to specific values, we can use the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

8. **Assuming \(B = 60^\circ\) and \(C = 30^\circ\)**:
This gives us:
\[ \sin B = \frac{\sqrt{3}}{2}, \quad \sin C = \frac{1}{2} \]
Therefore, substituting these values:
\[ 2\sin A = \sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \]
Simplifying:
\[ 2\sin A = \frac{3}{2} + \frac{1}{2} = 2 \implies \sin A = 1 \implies A = 90^\circ \]

9. **Using the cosine rule**:
Since \(A = 90^\circ\), we apply the cosine rule:
\[ a^2 = b^2 + c^2 \]

### Conclusion:
The relationship we derived is:
\[ a^2 = b^2 + c^2 \]
This corresponds to the Pythagorean theorem, confirming that triangle \(ABC\) is a right triangle.