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


1 Answers

Ramesh V
70 Points
13 years ago


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


Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult
to understand - post it here and we will get you the answer and detailed solution very quickly.We are all
IITians and here to help you in your IIT JEE preparation. All the best.

IIT Kgp - 05 batch

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy


Get your questions answered by the expert for free