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In a triangle ABC, 2ac sin 1/2 (A – B + C) =
(A) a2+b2-c2 (B) c2+a2-b2
(C) b2-c2-a2 (D) c2-a2-b2

pallavi pradeep bhardwaj , 15 Years ago
Grade 12
anser 1 Answers
Ramesh V

Last Activity: 15 Years ago

A+B+C=Pi

so A+C = Pi - B

so here it will be :  2ac.sin(Pi/2 -B) = 2ac.cosB

by cosine rule, we have: c2+a2-b2 = 2ac cos B

option B

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Regards,
Ramesh 
IIT Kgp - 05 batch




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