Ta có : \(\widehat{B}+\widehat{C}=90^o\)
\(\Rightarrow\widehat{B}=90^o-\widehat{C}=90^o-40^o=50^o\)
\(sinB=\dfrac{AC}{BC}\Rightarrow sin60^o=\dfrac{6}{BC}\Rightarrow BC=4\sqrt{3}\left(cm\right)\)
\(tan60^o=\dfrac{AC}{AB}\Rightarrow\dfrac{6}{AB}=\sqrt{3}\Rightarrow AB=2\sqrt{3}\left(cm\right)\)