a) Ta có \(a = 6, b = 5, c = 4\)
\(\eqalign{
& \cos A = {{{b^2} + {c^2} - {a^2}} \over {2bc}} = {{{5^2} + {4^2} - {6^2}} \over {2.5.4}} = {1 \over 8}\cr& \Rightarrow \widehat A \approx {83^0} \cr
& \cos B = {{{a^2} + {c^2} - {b^2}} \over {2ac}} = {{{6^2} + {4^2} - {5^2}} \over {2.6.4}} = {9 \over {16}}\cr& \Rightarrow \widehat B \approx {56^0} \cr
& \Rightarrow \,\,\widehat C \approx {41^0} \cr} \)
b) Ta có
\(\eqalign{
& m_a^2 = {1 \over 4}\left( {2{b^2} + 2{c^2} - {a^2}} \right) \cr&\;\;\;\;\;\;= {1 \over 4}\left( {50 + 32 - 36} \right) = {{46} \over 4}\,\, \Rightarrow \,\,{m_a} = {{\sqrt {46} } \over 2} \cr
& m_b^2 = {1 \over 4}\left( {2{a^2} + 2{c^2} - {b^2}} \right) = {{79} \over 4}\,\, \Rightarrow \,\,{m_b} = {{\sqrt {79} } \over 2} \cr
& \Rightarrow \,\,{m_c} = {{\sqrt {106} } \over 2} \cr} \)