Question

关于x的方程x^2-x*cosA*cosB-(cosC/2)^2=0有一个根为1,则△ABC一定是?

Answer 1

x^2-x*cosA*cosB-(cosC/2)^2=0

x1=1

1*x2=-(cosC/2)^2=-(cosC+1)/2

x2=-(cosC+1)/2

1+x2=cosA*cosB

1-(cosC+1)/2=[cos(A+B)+cos(A-B)]/2

1-cosC=-cosC+cos(A-B)

cos(A-B)=1

A-B=0

A=B

则△ABC一定是等腰△