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

问题描述:

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 s...