Xét ΔABD có
\(cosBAD=\dfrac{AB^2+AD^2-BD^2}{2\cdot AB\cdot AD}\)
=>\(8^2+6^2-BD^2=2\cdot8\cdot6\cdot cos60=48\)
=>\(BD^2=100-48=52\)
=>\(BD=2\sqrt{13}\left(cm\right)\)
Xét ΔBAC có \(cosABC=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(8^2+6^2-AC^2=2\cdot8\cdot6\cdot cos120=-48\)
=>\(AC^2=148\)
=>\(AC=2\sqrt{37}\left(cm\right)\)