If in a triangle`ABC` ,
(sinA+sinB+sinC)(sinA+sinB-sinC)=3sinAsinB` then

Options

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

Answer

To solve the problem step by step, we will start from the given equation and use the sine rule to express the sines in terms of the sides of the triangle.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ (sinA + sinB + sinC)(sinA + sinB - sinC) = 3sinAsinB \]

2. **Use the sine rule:**
According to the sine rule, we have:
\[ \frac{A}{sinA} = \frac{B}{sinB} = \frac{C}{sinC} = 2R \]
From this, we can express \(sinA\), \(sinB\), and \(sinC\) in terms of the sides \(A\), \(B\), and \(C\):
\[ sinA = \frac{A}{2R}, \quad sinB = \frac{B}{2R}, \quad sinC = \frac{C}{2R} \]

3. **Substitute these values into the equation:**
\[ \left(\frac{A}{2R} + \frac{B}{2R} + \frac{C}{2R}\right)\left(\frac{A}{2R} + \frac{B}{2R} - \frac{C}{2R}\right) = 3\left(\frac{A}{2R}\right)\left(\frac{B}{2R}\right) \]

4. **Simplify the left-hand side:**
Factor out \(\frac{1}{(2R)^2}\):
\[ \frac{1}{(2R)^2}\left(A + B + C\right)\left(A + B - C\right) = \frac{3AB}{4R^2} \]

5. **Multiply both sides by \(4R^2\):**
\[ 4R^2 \cdot \frac{1}{(2R)^2}\left(A + B + C\right)\left(A + B - C\right) = 3AB \]
This simplifies to:
\[ (A + B + C)(A + B - C) = 3AB \]

6. **Expand the left-hand side:**
\[ A^2 + B^2 + C^2 + 2AB - C^2 = 3AB \]
This simplifies to:
\[ A^2 + B^2 - C^2 + 2AB = 3AB \]

7. **Rearranging gives:**
\[ A^2 + B^2 - C^2 = AB \]

8. **Use the cosine rule:**
Recall that from the cosine rule:
\[ cosC = \frac{A^2 + B^2 - C^2}{2AB} \]
Substitute \(A^2 + B^2 - C^2\):
\[ cosC = \frac{AB}{2AB} = \frac{1}{2} \]

9. **Find angle \(C\):**
The angle \(C\) for which \(cosC = \frac{1}{2}\) is:
\[ C = cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \text{ or } 60^\circ \]

### Conclusion:
Thus, we find that \(C = 60^\circ\).