Ta có: \(\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)\( = \dfrac{{8 + 6 + 2 - 2\sqrt {12} - 12}}{{4\sqrt 2 (\sqrt 6 - \sqrt 2 )}}\) \( = \dfrac{{4 - 4\sqrt 3 }}{{8\sqrt 3 - 8}}\) \( = \dfrac{{4(1 - \sqrt 3 )}}{{8(\sqrt 3 - 1)}} = - \dfrac{1}{2}\)
Do đó \(\widehat A = {120^0}\).
\(\cos B = \dfrac{{{c^2} + {a^2} - {b^2}}}{{2.ca}}\)\( = \dfrac{{6 + 2 - 2\sqrt {12} + 12 - 8}}{{2.(\sqrt 6 - \sqrt 2 ).2\sqrt 3 }}\) \( = \dfrac{{12 - 2\sqrt {12} }}{{4\sqrt {18} - 4\sqrt 6 }}\) \( = \dfrac{{4(3 - \sqrt 3 )}}{{4\sqrt 2 (3 - \sqrt 3 )}} = \dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}\).
Vậy \(\widehat B = {45^0}\).
Ta có: \({h_a} = \dfrac{{2S}}{a} = \dfrac{{ac\sin B}}{a} = c\sin B\)\( = \left( {\sqrt 6 - \sqrt 2 } \right)\dfrac{{\sqrt 2 }}{2} = \sqrt 3 - 1\)
\(\dfrac{b}{{\sin B}} = 2R\)\( \Rightarrow R = \dfrac{b}{{2\sin B}} = \dfrac{{2\sqrt 2 }}{{2.\dfrac{{\sqrt 2 }}{2}}} = 2\)
\(S = pr\)\( \Rightarrow r = \dfrac{S}{p} = \dfrac{{\dfrac{1}{2}ac\sin B}}{{\dfrac{1}{2}(a + b + c)}} = \dfrac{{ac\sin B}}{{a + b + c}}\) \( = \dfrac{{2\sqrt 3 \left( {\sqrt 6 - \sqrt 2 } \right)\dfrac{{\sqrt 2 }}{2}}}{{2\sqrt 3 + 2\sqrt 2 + \sqrt 6 - \sqrt 2 }}\)\( = \dfrac{{\sqrt 3 \left( {\sqrt 6 - \sqrt 2 } \right)}}{{\sqrt 6 + \sqrt 3 + 1}}\)
