在△ABC中,若a-b=c(cosB-cosA),判断△ABC的形状a-b=(a^2+c^2-b^2)/2a-(b^2+c^2-a^2)/2b到2(a-b)(a^2+b^2-c^2)/2ab=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/08 14:34:04
![在△ABC中,若a-b=c(cosB-cosA),判断△ABC的形状a-b=(a^2+c^2-b^2)/2a-(b^2+c^2-a^2)/2b到2(a-b)(a^2+b^2-c^2)/2ab=0](/uploads/image/z/6847972-52-2.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%8B%A5a-b%3Dc%28cosB-cosA%29%2C%E5%88%A4%E6%96%AD%E2%96%B3ABC%E7%9A%84%E5%BD%A2%E7%8A%B6a-b%3D%28a%5E2%2Bc%5E2-b%5E2%29%2F2a-%28b%5E2%2Bc%5E2-a%5E2%29%2F2b%E5%88%B02%28a-b%29%28a%5E2%2Bb%5E2-c%5E2%29%2F2ab%3D0)
xŒJ@_% ȆvBcH;{b_A%UEIZX$Eۇ)&l?hE/^sZ*&qκ/<9~Jv^I32L>RûMSLnN.WpSA *IHtriH7<8tXB wT |I^=Ɖ]Y0f^_Isr0)V1G@RqاYVJBA?EXw@5ΓA+yDw