在△ABC中,设BC=a,AB=c,CA=b,若(b+a)/a=sinB/(sinB-sinA),cos2C+cosC=1-cos(A-B).试判断△ABC形状.
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/10 11:58:42
![在△ABC中,设BC=a,AB=c,CA=b,若(b+a)/a=sinB/(sinB-sinA),cos2C+cosC=1-cos(A-B).试判断△ABC形状.](/uploads/image/z/5274634-58-4.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%AE%BEBC%3Da%2CAB%3Dc%2CCA%3Db%2C%E8%8B%A5%EF%BC%88b%2Ba%EF%BC%89%2Fa%3DsinB%2F%EF%BC%88sinB-sinA%EF%BC%89%2Ccos2C%2BcosC%3D1-cos%28A-B%29.%E8%AF%95%E5%88%A4%E6%96%AD%E2%96%B3ABC%E5%BD%A2%E7%8A%B6.)
在△ABC中,设BC=a,AB=c,CA=b,若(b+a)/a=sinB/(sinB-sinA),cos2C+cosC=1-cos(A-B).试判断△ABC形状.
在△ABC中,设BC=a,AB=c,CA=b,若(b+a)/a=sinB/(sinB-sinA),cos2C+cosC=1-cos(A-B).试判断△ABC形状.
在△ABC中,设BC=a,AB=c,CA=b,若(b+a)/a=sinB/(sinB-sinA),cos2C+cosC=1-cos(A-B).试判断△ABC形状.
直角三角形
因为a/sinA=b/sinB=c/sinC
所以
(b+a)/a=sinB/(sinB-sinA)
1+b/a=sinB/(sinB-sinA)
1+sinB/sinA=sinB/(sinB-sinA)
(sinA+sinB)/sinA=sinB/(sinB-sinA)
sinAsinB=(sinB)^2-(sinA)^2.(1)
又
cos2C+cosC=1-cos(A-B)
cos2C=1+cos(A+B)-cos(A-B)
cos2C=1-2sinAsinB
因为(1)
所以cos2C=1-2(sinB)^2+2(sinA)^2
2(cosC)^2-1=1-2(sinB)^2+2(sinA)^2
(cosC)^2=1-(sinB)^2+(sinA)^2
1-(sinC)^2=1-(sinB)^2+(sinA)^2
(sinB)^2=(sinC)^2+(sinA)^2 .(2)
又由a/sinA=b/sinB=c/sinC得a^2/(sinA)^2=b^2/(sinB)^2=c^2/(sinC)^2
b^2/(sinB)^2=(a^2+c^2)/[(sinA)^2+(sinC)^2]
因为(2)所以b^2=a^2+c^2
所以是直角三角形
顶角是钝角的等腰三角形。