Ta có \(\overrightarrow{AB}.\overrightarrow{CB}=AB.BC.cos\widehat{B}=AB.BC.\dfrac{AB}{BC}=AB^2=4\)
\(\Rightarrow AB=2\)
\(\overrightarrow{AC}.\overrightarrow{BC}=AC.BC.cos\widehat{C}=AC.BC.\dfrac{AC}{BC}=AC^2=9\)
\(\Rightarrow AC=3\)
\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{2^2+3^2}=\sqrt{13}\)