[tex]AB=AC=6cm\\ \\ Din\ \ teorema\ lui \ \ Pitagora\ generalizat\breve a,\ \ rezult\breve a:[/tex]
[tex]\it BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot cosA=6^2+6&^2-2\cdot6\cdot6\cdot cos120^o=\\ \\ =36+36-2\cdot36\cdot\Big(-\dfrac{1}{2}\Big)=36+36+36=36\cdot3 \Rightarrow BC=\sqrt{36\cdot3}=6\sqrt3\ cm[/tex]