在△ABC中(b+a)/a=SinB/(SinB-SinA)三角形形状计算在△ABC中(b+a)/a=SinB/(SinB-Sin_数学解答

在△ABC中(b+a)/a=SinB/(SinB-SinA)三角形形状计算
在△ABC中(b+a)/a=SinB/(SinB-SinA)且Cos2C+CosC=1-Cos(A-B),试判断△ABC的形状,

人气：168 ℃　时间：2020-01-30 14:08:18

解答

(b+a)/a=sinB/(sinB-sinA)(sinB+sinA)/sinA = sinB/(sinB-sinA)(sinB+sinA)(sinB-sinA) = sinAsinB .(1)cos2C+cosC = 1-cos(A-B)1-2sin^2C + cos(180°-A-B) = 1-cos(A-B)1-2sin^2C - cos(A+B) = 1-cos(A-B)cos(A-B)...