在△ABC中,已知(a+b)/a=sinB/(sinB-sinA),且cos(A-B)+cosC=1-cos2C,试确定△ABC的形状.
问题描述:
在△ABC中,已知(a+b)/a=sinB/(sinB-sinA),且cos(A-B)+cosC=1-cos2C,试确定△ABC的形状.
答
ABC为RT三角形
答
∵cosC=cos(-C)=-cos(180°-C)=-cos(A+B)cos2C=1-2sin^2(C)∴cos(A-B)+cosC=1-cos2C可转换为cos(A-B)-cos(A+B)=2sin^2C∵cos(A+B)=cosA·cosB-sinA·sinBcos(A-B)=cosA·cosB+sinA·sinβB所以cos(A-B)-cos(...