In a triangle ABC, 2ac sin12(A-B+C)= 

In a triangle ABC, 2ac sin12(A-B+C)= 

  1. A

    a2+b2-c2 

  2. B

    c2+a2-b2

  3. C

    b2-c2-a2

  4. D

    c2-a2-b2

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    Solution:

    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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