a) +) [tex]cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{AC^2+AB^2-BC^2}{2.AC.AB}=\frac{(4\sqrt{2})^2+2^2-6^2}{2.4\sqrt{2}.2}=0[/tex]
+) tương tự [tex]cosB=\frac{1}{3}[/tex]
Mà [tex]cos^2B+sin^2B=1[/tex] [tex]\Rightarrow sinB=\frac{2\sqrt{2}}{3}[/tex]
+) [tex]S=\frac{1}{2}.ac.sinB=\frac{1}{2}.BC.AB.\frac{2\sqrt{2}}{3}=4\sqrt{2} \Rightarrow r=\frac{S}{p}=\frac{4\sqrt{2}}{\frac{1}{2}.(6+2+4\sqrt{2})}=-2+2\sqrt{2}[/tex]