Show that $(b - c) sin A + (c - a) sin B + (a - b) sin C = 0$

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Solution

$(b - c) sin A + (c - a) sin B + (a - b) sin C$
$= \frac{a (b - c)}{2 R} + \frac{b (c - a)}{2 R} + \frac{c (a - b)}{2 R}$ [Using $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2 R$ ]
$= \frac{a b - a c + b c - a b + a c - b c}{2 R}$
$= 0$ .

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