MathematicsMedium19×since 2002Q5043Let sinAsinB=sin(A−C)sin(C−B){{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}sinBsinA=sin(C−B)sin(A−C), where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :Ab² −-− a² = a² + c²Bb², c², a² are in A.P.Cc², a², b² are in A.P.Da², b², c² are in A.P.Check answerSkip