Question

在△ABC中(b+a)/a=SinB/(SinB-SinA)三角形形状计算

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