Giải thích các bước giải:
Ta có:
$\begin{array}{l}
b = 6;c = 9;\widehat A = {30^0}\\
+ )a = \sqrt {{b^2} + {c^2} - 2bc\cos A} = \sqrt {{6^2} + {9^2} - 2.6.9.\cos {{30}^0}} \approx 4,84\\
+ )\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}}\\
\Rightarrow \sin B = \dfrac{{b\sin A}}{a} = \dfrac{{6\sin {{30}^0}}}{{4,84}} = 0,62\\
\Rightarrow \widehat B \approx {38^0}15'\\
\Rightarrow \widehat C = {180^0} - \widehat B - \widehat A = {111^0}45'\\
+ )S = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}.6.9.\sin {30^0} = \dfrac{{27}}{2}\\
+ )\dfrac{a}{{\sin A}} = 2R\\
\Rightarrow R = \dfrac{a}{{2\sin A}} = \dfrac{{4,84}}{{2\sin {{30}^0}}} = 4,84\\
+ )S = pr,p = \dfrac{{a + b + c}}{2} = \dfrac{{4,84 + 6 + 9}}{2} = 9,92\\
\Leftrightarrow r = \dfrac{S}{p} = \dfrac{{\dfrac{{27}}{2}}}{{9,92}} = 1,36\\
+ ){m_a} = \sqrt {\dfrac{{2\left( {{b^2} + {c^2}} \right) - {a^2}}}{4}} = \sqrt {\dfrac{{2\left( {{6^2} + {9^2}} \right) - 4,{{84}^2}}}{4}} \approx 7,26\\
{m_b} = \sqrt {\dfrac{{2\left( {{a^2} + {c^2}} \right) - {b^2}}}{4}} = \sqrt {\dfrac{{2\left( {4,{{84}^2} + {9^2}} \right) - {6^2}}}{4}} \approx 6,57\\
{m_c} = \sqrt {\dfrac{{2\left( {{a^2} + {b^2}} \right) - {c^2}}}{4}} = \sqrt {\dfrac{{2\left( {4,{{84}^2} + {6^2}} \right) - {9^2}}}{4}} \approx 3,08
\end{array}$
Vậy $a \approx 4,84;\widehat B = {38^0}15';\widehat C = {111^0}45';S = \dfrac{{27}}{2};R = 4,84;S = 9,92;r = 1,36;{m_a} \approx 7,26;{m_b} \approx 6,57;{m_c} \approx 3,08$
