In triangle ABC,with usual notations,frac{b s | vedclass

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(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$b = k \sin B$ and $c = k \sin C$. Substituting the

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$a$

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(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$b = k \sin B$ and $c = k \sin C$. Substituting these into the expression: $\frac{b \sin B - c \sin C}{\sin (B - C)} = \frac{(k \sin B) \sin B - (k \sin C) \sin C}{\sin (B - C)}$ $= \frac{k(\sin^2 B - \sin^2 C)}{\sin (B - C)}$ Using the identity $\sin^2 B - \sin^2 C = \sin(B + C) \sin(B - C)$: $= \frac{k \sin(B + C) \sin(B - C)}{\sin (B - C)}$ $= k \sin(B + C)$ Since $A + B + C = 180^{\circ}$,then $B + C = 180^{\circ} - A$. $= k \sin(180^{\circ} - A) = k \sin A$ $= a$.

In $\triangle ABC$,with usual notations,$\frac{b \sin B - c \sin C}{\sin (B - C)} = $

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