Question

已知三角形abc中(b+a)(sinB-sinA)=asinB

已知三角形abc中(b+a)(sinB-sinA)=asinB,又cos2C+cosc=1-cos(A-B).1.试判断三角形的形状2。求cosC的值

Answer 1

1)正弦定理:a/sinA=b/sinB=2R∵(b+a)(sinB-sinA)=asinB∴(b+a)(b/2R-a/2R)=ab/2R, 即(b+a)(b-a)=ab∴b�0�5-a�0�5=ab ①∵cos2C+cosC=1-cos(A-B)∴1-2sin�0�5C-cos(A+B)=1-cos(A-B)∴-2sin�0�5C-cosAcosB+sinAsinB=-cosAcosB-sinAsinB∴sinAsinB=sin�0�5C∴(a/2R)(b/2R)=(c/2R)�0�5,即ab=c�0�5②由①②得 b�0�5-a�0�5=c�0�5, b�0�5=a�0�5+c�0�5, △ABC是直角三角形,且B是直角2)cosC=a/b由①得 a�0�5+ab-b�0�5=0两边同除b�0�5得 (a/b)�0�5+(a/b)-1=0, 即cos�0�5C+cosC-1=0解得cosC=(-1±√5)/2而B是直角,则C是锐角,cosC>0, ∴cosC=(-1+√5)/2