由正弦定理可知(sinA+sinB+sinC)(sinA+sinB-sinC)=3sinAsinB ?(sinA+sinB) 2 -sin 2 C=3sinAsinB, ?sin 2 A+2sinAsinB+sin 2 B-sin 2 (A+B)=3sinAsinB, ?sin 2 A+sin 2 B-(sinAcosB+cosAsinB) 2 =sinAsinB, ?sin 2 A+sin 2 B-sin 2 A?cos 2 B-2sinAcosBcosAsinB-cos 2 A?sin 2 B=sinAsinB ?2sin 2 Asin 2 B-2sinAcosBsinBcosA=sinAsinB, ?cosAcosB-sinAsinB=-
∴cos(A+B)=-
∴A+B=
所以C=π-(A+B)=
故答案为:
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