Đáp án:
a) $R = \dfrac{{10}}{{\sqrt 3 }}$; b) $b = c = 10$
Giải thích các bước giải:
$\begin{array}{l}
a)\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\\
\Rightarrow 2R = \dfrac{{10}}{{\sin {{60}^o}}} = \dfrac{{10}}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{{20}}{{\sqrt 3 }} \Rightarrow R = \dfrac{{10}}{{\sqrt 3 }}\\
b) + )\dfrac{{abc}}{{4R}} = \dfrac{{a + b + c}}{2}.r\\
\Leftrightarrow \dfrac{{10bc}}{{4.\dfrac{{10}}{{\sqrt 3 }}}} = \dfrac{{10 + b + c}}{2}.\dfrac{{5\sqrt 3 }}{3} \Leftrightarrow b + c + 10 = \dfrac{3}{{10}}bc\left( 1 \right)\\
+ ){a^2} = {b^2} + {c^2} - 2bc\cos A\\
\Leftrightarrow 100 = {b^2} + {c^2} - 2bc\cos {60^0}\\
\Leftrightarrow {b^2} + {c^2} - bc = 100\left( 2 \right)\\
\left( 1 \right),\left( 2 \right) \Rightarrow \left\{ \begin{array}{l}
b + c + 10 = \dfrac{3}{{10}}bc\\
{b^2} + {c^2} - bc = 100
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b + c = \dfrac{3}{{10}}bc - 10\\
{\left( {b + c} \right)^3} - 3bc = 100
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b + c = \dfrac{3}{{10}}bc - 10\\
{\left( {\dfrac{3}{{10}}bc - 10} \right)^2} - 3bc = 100
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b + c = \dfrac{3}{{10}}bc - 10\\
\dfrac{9}{{100}}{\left( {bc} \right)^2} - 9bc = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b + c = \dfrac{3}{{10}}bc - 10\\
bc = 100
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b + c = 20\\
bc = 100
\end{array} \right. \Leftrightarrow b = c = 10
\end{array}$
