In a triangle $A B C$ , if $(sin A + sin B + sin C) (sin A + sin B - sin C) = 3 sin A sin B$ is equal to

\frac{π}{2}

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\frac{π}{3}

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\frac{π}{4}

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\frac{π}{6}

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Solution

The correct option is B $\frac{π}{3}$
$\Rightarrow$ $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2 R$

$\Rightarrow$ $(sin A + sin B + sin C) (sin A + sin B - sin C) = 3 sin A sin B$

$\Rightarrow$ $(\frac{a}{2 R} + \frac{b}{2 R} + \frac{c}{2 R}) (\frac{a}{2 R} + \frac{b}{2 R} - \frac{c}{2 R}) = 3 \times \frac{a}{2 R} \times \frac{b}{2 R}$

$\Rightarrow$ $(\frac{a + b + c}{2 R}) (\frac{a + b - c}{2 R}) = \frac{3 a b}{4 R^{2}}$

$\Rightarrow$ $(a + b + c) (a + b - c) = 3 a b$
$\Rightarrow$ $(a + b)^{2} - c^{2} = 3 a b$
$\Rightarrow$ $a^{2} + 2 a b + b^{2} - c^{2} = 3 a b$
$\Rightarrow$ $a^{2} + b^{2} - c^{2} = a b$

$\Rightarrow$ $cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

$\Rightarrow$ $cos C = \frac{a b}{2 a b}$

$\Rightarrow$ $cos C = \frac{1}{2}$

$\Rightarrow$ $C = {cos}^{- 1} (\frac{1}{2})$

$\Rightarrow$ $C = \frac{π}{3}$

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