过点A作AD⊥BC于D
∵∠BAC=120,AB=AC
∴∠B=∠C=(180-∠BAC)/2=30
∵AD⊥BC
∴BD=CD=BC/2,AB=BD/(√3/2)=2BD/√3
∴AB=BC/√3
∴AB:BC=1:√3