在△ABC中,设BC=a,AB=c,CA=b,若(b+a)/a=sinB/(sinB-sinA),cos2C+cosC=
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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
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
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