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

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

Grade:12

1 Answers

Ramesh V
70 Points
14 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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Ramesh 
IIT Kgp - 05 batch




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