(1)
(sinA-sinB)/sin(A+B)=(√2sinA-sinC)/(sinA+sinB)
∴(sinA-sinB)(sinA+sinB)=sin(A+B)(√2sinA-sinC)
sin²A-sin²B=sin(A+B)[√2sinA-sin(A+B)]
sin²A-sin²B=√2sinAsin(A+B)-sin²(A+B)
sin²A-sin²B=√2sinAsinC-sin²C
根据正弦定理,易得
a²-b²=√2ac-c²
根据余弦定理,得
cosB=(a²+c²-b²)/(2ac)=√2/2
∴B=45°
(2)若cosA=3/5,则sinA=4/5
sinC
=sin(A+B）
=sinAcosB+sinBcosA
=7√2/10
此即所求