∵A-C=90°,∴sinA=sina（90°+C）=cosC.∵A+B+C=180°,∴90°+C+B+C=180°,得,B=90°-2C.由三角形正弦定理,a/sinA=b/sinB=c/sinC 得,a/sinA=c/sinC=a/cosC,∴a=c·cosC/sinC；又得,b/sinB=c/sinC=b/sin（90°-2C）=b/cos2C.把a、b关于c和C角关系代入a+c=b×√2,得,c·cosC/sinC+c=（c·cos2C/sinC）×√2,c≠0,等式两边约去c,得到C角关系式,cosC/sinC+1=（cos2C）×√2,sinC≠0,等式两边同时乘以sinC得,cosC+sinC=√2·cos2C=√2·（cos²C-sin²C）=√2·（cosC+sinC）（cosC-sinC）,cosC+sinC≠0,等式两边约去cosC+sinC,得,1=√2·（cosC+sinC）,再两边平方,得,1=2（cos²C+sin²C+2cosCsinC）,1/2=1+2cosCsinC=1+sin2C,sin2C=1/2,∴2C=30°,即得,C=15°.解完.