In any `DeltaABC`, if `(sinA+sinB+sinC)xx(sinA+sinB-sinC)=3sinAsinB`, then

Options

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

Answer

To solve the problem, we start with the given equation in triangle \( ABC \):

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

### Step 1: Expand the left-hand side

Using the distributive property (also known as the FOIL method for binomials), we can expand the left-hand side:

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

### Step 2: Simplify the expression

Now, we simplify \((\sin A + \sin B)^2\):

\[ (\sin A + \sin B)^2 = \sin^2 A + 2 \sin A \sin B + \sin^2 B \]

Thus, we can rewrite the left-hand side as:

\[ \sin^2 A + 2 \sin A \sin B + \sin^2 B - \sin^2 C \]

### Step 3: Set the equation

Now, we set the expanded left-hand side equal to the right-hand side:

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

### Step 4: Rearranging the equation

Rearranging gives us:

\[ \sin^2 A + \sin^2 B - \sin^2 C + 2 \sin A \sin B - 3 \sin A \sin B = 0 \]

This simplifies to:

\[ \sin^2 A + \sin^2 B - \sin^2 C - \sin A \sin B = 0 \]

### Step 5: Using the sine rule

Using the sine rule, we know that:

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

Substituting these into the equation gives:

\[ \left(\frac{a^2}{(2R)^2} + \frac{b^2}{(2R)^2} - \frac{c^2}{(2R)^2} - \frac{ab}{(2R)^2}\right) = 0 \]

### Step 6: Simplifying further

Multiplying through by \((2R)^2\) to eliminate the denominator:

\[ a^2 + b^2 - c^2 - ab = 0 \]

### Step 7: Rearranging for \(c^2\)

Rearranging gives:

\[ c^2 = a^2 + b^2 - ab \]

### Step 8: Applying the cosine rule

By the cosine rule, we have:

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

Setting the two expressions for \(c^2\) equal gives:

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

### Step 9: Solving for \(\cos C\)

This simplifies to:

\[ -ab = -2ab \cos C \]

Dividing both sides by \(-ab\) (assuming \(a, b \neq 0\)) gives:

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

### Step 10: Finding angle \(C\)

The angle \(C\) for which \(\cos C = \frac{1}{2}\) is:

\[ C = 60^\circ \]

Thus, we conclude that:

\[ A = 60^\circ, B = 60^\circ, C = 60^\circ \]

### Final Answer

The angles of triangle \(ABC\) are \(A = 60^\circ\), \(B = 60^\circ\), \(C = 60^\circ\).

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