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Question

In any triangle ABC if sinAsinC=sinABsinBC prove that a2,b2 and c2 are in A.P.

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Solution

In ABC,A+B+C=π
A=π(B+C) and C=π(A+B)
Now, sinAsinC=sin(AB)sin(BC)
sin(π(B+C))sin(π(A+B))=sin(AB)sin(BC)
sin(B+C)sin(A+B)=sin(AB)sin(BC)
sin(B+C)sin(B+C)=sin(AB)sin(A+B)
sin2Bsin2C=sin2Asin2B
By sine rule,
b2k2c2k2=a2k2b2k2
b2c2=a2b2
a2,b2,c2 are in A.P

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