Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{c^2} = {a^2} + {b^2} - 2ab.\cos C\\
\Leftrightarrow {c^2} = {5^2} + {7^2} - 2.5.7.\cos 120^\circ \\
\Leftrightarrow {c^2} = 109\\
\Rightarrow c = \sqrt {109} \\
{m_b}^2 = \dfrac{{{a^2} + {c^2}}}{2} - \dfrac{{{b^2}}}{4} = \dfrac{{{5^2} + 109}}{2} - \dfrac{{{7^2}}}{4} = \dfrac{{219}}{4} \Rightarrow {m_b} = \dfrac{{\sqrt {219} }}{2}\\
{S_{ABC}} = \dfrac{1}{2}.a.b.\sin C = \dfrac{1}{2}.5.7.\sin 120^\circ = \dfrac{{35\sqrt 3 }}{4}\\
{h_b} = \dfrac{{2{S_{ABC}}}}{b} = \dfrac{{5\sqrt 3 }}{2}\\
{S_{ABC}} = \dfrac{{abc}}{{4R}} \Rightarrow R = \dfrac{{abc}}{{4{S_{ABC}}}} = \dfrac{{5.7.\sqrt {109} }}{{35\sqrt 3 }} = \sqrt {\dfrac{{109}}{3}} \\
{S_{ABC}} = \dfrac{{a + b + c}}{2}.r \Rightarrow r = \dfrac{{2{S_{ABC}}}}{{a + b + c}} = \dfrac{{\dfrac{{35\sqrt 3 }}{2}}}{{12 + \sqrt {109} }} = \dfrac{{35\sqrt 3 }}{{24 + 2r109}}
\end{array}\)
