f(x)=根号3sin²x+sinxcosx-根号3/2=sqrt(3)(1/2-cos(2x)/2)+1/2*sin2x-sqrt(3)/2=
=1/2*(sin2x-sqrt(3)cos2x)=sqrt(1+3)/2*sin(2x-π/3)=sin(2x-π/3)
f(A)=f(B)=1/2
sin(2A-π/3)=sin(2B-π/3)=1/2
2A-π/3=2B-π/3=π/6
A=B=π/4
那么BC/AB=sin(A)=sqrt(2)/2