In a triangle ABC, 2ac sin12(A-B+C)=

12(A-B+C) =12(π-B-B) =π2-B∴ 2ac sin12(A-B+C)=2ac sinπ2-B= 2ac cos B=a2+c2-b2 (by cosine rule)

In a triangle ABC, 2ac sin12(A-B+C)=

12(A-B+C) =12(π-B-B) =π2-B∴ 2ac sin12(A-B+C)=2ac sinπ2-B= 2ac cos B=a2+c2-b2 (by cosine rule)

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

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

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

$c^{2} - a^{2} - b^{2}$

$b^{2} - c^{2} - a^{2}$

answer is B.

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Detailed Solution

$\frac{1}{2} (A - B + C) = \frac{1}{2} (π - B - B) = \frac{π}{2} - B$

$∴ 2 a c$ $\sin \frac{1}{2} (A - B + C) = 2 a c \sin (\frac{π}{2} - B)$

= 2ac cos $B = a^{2} + c^{2} - b^{2}$ (by cosine rule)

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In a triangle ABC, 2ac sin12(A-B+C)=

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