三角形ABC的内角ABC的对边分别为abc，且asin(A+B-C)=csin(B+C)求角C的值

由正弦定理得a=2RsinA,b=2RsinB,c=2RsinC, 故有 asin(B-C)+bsin(C-A)+csin(A-B) =2R(sinAsin(B-C)+sinBsin(C-A)+sinCsin(A-B)) =2R(sinA(sinBcosC-cosBsinC)+sinB(sinCcosA-cosCsinA)+sinC(sinAcosB-cosAsinB)) =2R(sinAsinBcosC-sinAcosBsinC+sinBsinCcosA-sinBcosCsinA+sinCsinAco