题目内容：

已知三角形ABC周长为根2+1,且sinA+sinB=根2sinC,若面积为6分之1sinC,求C

优质解答

设三边长分别为a、b、c,已知a+b+c=√2+1sinA+sinB=√2sinCS=(SinC)/6正弦定理：sinA/a=sinB/b=sinC/c 变形得sinA/a=sinB/b=sinC/c=(sinA+sinB+sinC)/(a+b+c)=(√2sinC+sinC)/(√2+1)=sinC(sinC)/c=sinC c=1毕!...