[tex]\it AB-diametru \Rightarrow m(\widehat{BCA})=90^o \Rightarrow \Delta BCA\ -\ dreptunghic\ \^{i }n\ C[/tex]
[tex]\it m(\widehat{B})=\dfrac{m(\stackrel\frown{CA})}{2}=\dfrac{120^o}{2}=60^o \Rightarrow m(\widehat{A})=30^o\ (complementul\ lui\ 60^o)\\ \\ \\ AB=2R=2\cdot4=8\ cm\\ \\ Th.\ \angle30^o \Rightarrow BC=\dfrac{AB}{2}=\dfrac{8}{2}=4\ cm\\ \\ \\ BC\perp AC \Rightarrow d(B,\ \ AC)=BC=4\ cm[/tex]
