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Let (sinA)/(sinB)=(sin(A-C))/(sin(C-B)), where A , B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a,b,c respectively, then
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sin(A-C)+2sinA+sin(A+C))/(sin(B-C)+2sinB+sin(B+C))` is equal to- - YouTube
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Let (sinA)/(sinB)=(sin(A-C))/(sin(C-B)), where A , B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a,b,c respectively, then
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