已知△ABC的周长为√2+1,且SinA+Sin B=√2*SinC 若BC*AC=1/3,求△ABC的面积

运用正弦定理a/sinA=b/ainB=c/sinC(a+b)/(sinA+sinB)=(a+b)/(√2*SinC )=c/sinCa+b=√2ca+b+c=√2+1c=1a+b=√2BC*AC=1/3所以BC＝√2/4,AC＝3√2/4AC^2=BC^2+c^2所以S△ABC＝1/2*bc*c=√2/8注意AB=c,AC=b,BC=a