在△ABC的内角A.B.C.对边为a.b.c.A-B=90°,a+c=根号2b,则C=?,

题目

在△ABC的内角A.B.C.对边为a.b.c.A-B=90°,a+c=根号2b,则C=?,

答案

A-C=90度,即A=C+90度所以sinA=cosC,cosA=-sinCa+c=根号2乘以b,则：(a+c)/b=根号2(sinA+sinC)/sinB=根号2(cosC+sinC)/sin(A+C)=根号2(cosC+sinC)/(sinAcosC+cosAsinC)=根号2(cosC+sinC)/(cos^2C-sin^2C)=根号2cosC-sinC=√2/2,两边平方得：1-sin2C=1/2sin2C=1/2因为A-C=90度,A+C