$4)S_{\Delta ABC}=\dfrac{AB.AC.BC}{4R}\\ \Rightarrow R=\dfrac{AB.AC.BC}{4S_{\Delta ABC}}\\ =\dfrac{AB.AC.BC}{4\dfrac{1}{2}BC.AC\sin(\widehat{C})}\\ =\dfrac{AB}{2\sin(\widehat{C})}\\ =4\\ 6)\widehat{C}=90^o\\ a)\dfrac{BC}{\sin(\widehat{A})}=\dfrac{AC}{\sin(\widehat{B})}=\dfrac{AB}{\sin(\widehat{C})}\\ \Rightarrow BC=5\sqrt{3};AB=10\\ S_{\Delta ABC}=\dfrac{1}{2}BC.CA=\dfrac{25\sqrt{3}}{2}\\ b)S_{\Delta ABC}=\dfrac{AB.AC.BC}{4R}\\ \Rightarrow R=\dfrac{AB.AC.BC}{4S_{\Delta ABC}}=5\\ c)BK \equiv BC=5$
