In a triangle ABC,2ac sin frac{1}{2}(A-B+C) i | vedclass

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(C) In a $	riangle ABC$,we know that $A+B+C = 180^{\circ}$.We need to evaluate $2ac \sin \frac{1}{2}(A-B+C)$.Since $A+C = 180^{\circ}-B$,we substitut

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$a^2+c^2-b^2$

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(C) In a $	riangle ABC$,we know that $A+B+C = 180^{\circ}$.We need to evaluate $2ac \sin \frac{1}{2}(A-B+C)$.Since $A+C = 180^{\circ}-B$,we substitute this into the expression:$2ac \sin \frac{1}{2}(180^{\circ}-B-B) = 2ac \sin \frac{1}{2}(180^{\circ}-2B)$$= 2ac \sin (90^{\circ}-B)$$= 2ac \cos B$Using the cosine rule,$\cos B = \frac{a^2+c^2-b^2}{2ac}$.Substituting this value,we get:$= 2ac \left( \frac{a^2+c^2-b^2}{2ac} ight) = a^2+c^2-b^2$.

In a $\triangle ABC$,$2ac \sin \frac{1}{2}(A-B+C)$ is equal to

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