Question

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，详细过程

Answer 1

(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) - cos(A+B) = 2sin^2C
(cosAcosB+sinAsinB) - (cosAcosB+sinAsinB) = 2sin^2C
sinAsinB = sin^2C ......(2)

根据(1)、(2)：
(sinB+sinA)(sinB-sinA) = sin^2C
sin^2B = sin^2A+sin^2C，等效于b^2=a^2+c^2
直角三角形