若在△ABC中,(b+a)/a=sinB/(sinB-sinA),且cos2C+cosC=1-cos(A-B),则△ABC的形状为

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若在△ABC中,(b+a)/a=sinB/(sinB-sinA),且cos2C+cosC=1-cos(A-B),则△ABC的形状为

若在△ABC中,(b+a)/a=sinB/(sinB-sinA),且cos2C+cosC=1-cos(A-B),则△ABC的形状为
若在△ABC中,(b+a)/a=sinB/(sinB-sinA),且cos2C+cosC=1-cos(A-B),则△ABC的形状为

若在△ABC中,(b+a)/a=sinB/(sinB-sinA),且cos2C+cosC=1-cos(A-B),则△ABC的形状为
由正弦定理易得
(sinB+sinA)/sinA=(b+a)/a
(a+b)/a=sinB/(sinB-sinA)
因此sinBsinA=sin^2B-sin^2A-----(1)
cos(A-B)+cos((180-(A+B))=1-(1-2sin^2C)
化简得sinAsinB=sin^2C-----------(2)
联立等式(1)(2)得
sin^2B-sin^2A=sin^2C
sin^2B=sin^2A+sin^2C
即b^2=a^2+c^2
所以是直角三角形