Which of the following is true in a triangle | 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,$a = k\sin A$,$b = k\sin B$,and $c = k\sin C$.Consid

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$(b - c)\cos \frac{A}{2} = a\sin \frac{B - C}{2}$

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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,$a = k\sin A$,$b = k\sin B$,and $c = k\sin C$.Consider the expression $\frac{b - c}{a} = \frac{\sin B - \sin C}{\sin A}$.Using the sum-to-product formula,$\sin B - \sin C = 2\sin \frac{B - C}{2} \cos \frac{B + C}{2}$.Since $A + B + C = \pi$,we have $\frac{B + C}{2} = \frac{\pi}{2} - \frac{A}{2}$,so $\cos \frac{B + C}{2} = \sin \frac{A}{2}$.Also,$\sin A = 2\sin \frac{A}{2} \cos \frac{A}{2}$.Substituting these into the expression:$\frac{b - c}{a} = \frac{2\sin \frac{B - C}{2} \sin \frac{A}{2}}{2\sin \frac{A}{2} \cos \frac{A}{2}} = \frac{\sin \frac{B - C}{2}}{\cos \frac{A}{2}}$.Rearranging gives $(b - c)\cos \frac{A}{2} = a\sin \frac{B - C}{2}$.

Which of the following is true in a triangle $ABC$?

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