In any triangle ABC, if `( sin A + sin B + sin C ) ( sin A + sin B - sin C ) = 3 sin A sin B`, then

Options

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

Answer

To solve the problem, we start with the given equation:

\[ (\sin A + \sin B + \sin C)(\sin A + \sin B - \sin C) = 3 \sin A \sin B \]

### Step 1: Rewrite Sine in Terms of Sides and Circumradius
Using the Law of Sines, we know that:

\[ \frac{A}{\sin A} = \frac{B}{\sin B} = \frac{C}{\sin C} = 2R \]

From this, we can express the sine values as:

\[ \sin A = \frac{A}{2R}, \quad \sin B = \frac{B}{2R}, \quad \sin C = \frac{C}{2R} \]

### Step 2: Substitute Sine Values into the Equation
Substituting these expressions into the original equation gives us:

\[ \left(\frac{A}{2R} + \frac{B}{2R} + \frac{C}{2R}\right)\left(\frac{A}{2R} + \frac{B}{2R} - \frac{C}{2R}\right) = 3 \left(\frac{A}{2R}\right)\left(\frac{B}{2R}\right) \]

This simplifies to:

\[ \frac{(A + B + C)(A + B - C)}{(2R)^2} = \frac{3AB}{(2R)^2} \]

### Step 3: Cancel Common Terms
Since \( (2R)^2 \) is common on both sides, we can cancel it out:

\[ (A + B + C)(A + B - C) = 3AB \]

### Step 4: Expand the Left Side
Now, we expand the left side:

\[ A^2 + B^2 + AB + A + B - AC - BC = 3AB \]

Rearranging gives:

\[ A^2 + B^2 - C^2 + 2AB = 3AB \]

### Step 5: Rearrange the Equation
This leads to:

\[ A^2 + B^2 - C^2 = AB \]

### Step 6: Use the Cosine Rule
Using the cosine rule, we know:

\[ C^2 = A^2 + B^2 - 2AB \cos C \]

Substituting this into our equation gives:

\[ A^2 + B^2 - (A^2 + B^2 - 2AB \cos C) = AB \]

This simplifies to:

\[ 2AB \cos C = AB \]

### Step 7: Solve for Cosine
Dividing both sides by \( AB \) (assuming \( AB \neq 0 \)) gives:

\[ 2 \cos C = 1 \implies \cos C = \frac{1}{2} \]

### Step 8: Determine the Angle
The angle \( C \) that satisfies \( \cos C = \frac{1}{2} \) is:

\[ C = 60^\circ \]

### Conclusion
Thus, the angle \( C \) in triangle \( ABC \) is \( 60^\circ \).

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