Áp dụng định lý cos:
\(BC^2=AB^2+AC^2-2AB.AC.cos\widehat{A}\\ =>BC=\sqrt{AB^2+AC^2-2AB.AC.cos60}\\ =\sqrt{3^2+6^2-2.3.6.\dfrac{1}{2}}=3\sqrt{3}\left(cm\right)\)
Áp dụng định lý sin:
\(\dfrac{BC}{sin\widehat{A}}=2R=>R=\dfrac{BC}{sin60}.\dfrac{1}{2}=\dfrac{3\sqrt{3}}{\left(\dfrac{1}{2}\right)}.\dfrac{1}{2}=3\left(cm\right)\)