Question

已知三角形ABC中,（1-COSA)/(1-COSB)=a/b,试判断三角形形状.

Answer 1

（1-COSA)/(1-COSB)=sin(A/2)^2/sin(B/2)^2

a/b=sin(A/2)cos(A/2)/[sin(B/2)cos(B/2)]

tan(A/2)=tan(B/2)

de A=B