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.
人气:368 ℃ 时间:2020-04-10 08:36:51
解答
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(...
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