在△ABC中,已知a+b/a=sinB/sinB-sinA,且cos(A-B)+cosC=1-cos2C.(1)试确定△ABC的形状;(2)求a+c/b的范围.
问题描述:
在△ABC中,已知
=a+b a
,且cos(A-B)+cosC=1-cos2C.sinB sinB-sinA
(1)试确定△ABC的形状;
(2)求
的范围.a+c b
答
(1)由a+ba=sinBsinB-sinA,可得cos2C+cosC=1-cos(A-B)得cosC+cos(A-B)=1-cos2C,cos(A-B)-cos(A+B)=2sin2C,即sinAsinB=sin2C,根据正弦定理,ab=c2,①,又由正弦定理及(b+a)(sinB-sinA)=asinB可知b2...