Đáp án:
\(R = \dfrac{{10\sqrt 3 }}{3},b = 10,c = 10\)
Giải thích các bước giải:
Áp dụng định lí sin:
\(\dfrac{a}{{\sin A}} = 2R \Leftrightarrow R = \dfrac{a}{{2\sin A}} = \dfrac{{10}}{{2\sin {{60}^0}}} = \dfrac{{10\sqrt 3 }}{3}\)
Ta có:
\(\begin{array}{l}S = \dfrac{{abc}}{{4R}} = pr \Leftrightarrow \dfrac{{10bc}}{{4.\dfrac{{10\sqrt 3 }}{3}}} = \dfrac{{10 + b + c}}{2}.\dfrac{{5\sqrt 3 }}{3}\\ \Leftrightarrow 60bc = \dfrac{{40\sqrt 3 }}{3}.5\sqrt 3 \left( {10 + b + c} \right)\\ \Leftrightarrow 60bc = 200\left( {10 + b + c} \right)\\ \Leftrightarrow 3bc = 10\left( {10 + b + c} \right)\,\,\left( 1 \right)\end{array}\)
Áp dụng định lý cos:
\(\begin{array}{l}{a^2} = {b^2} + {c^2} - 2bc\cos A\\ \Leftrightarrow {10^2} = {b^2} + {c^2} - 2bc.\cos 60\\ \Leftrightarrow 100 = {b^2} + {c^2} - bc\,\,\left( 2 \right)\end{array}\)
Từ (1) và (2) suy ra
\(\begin{array}{l}\left\{ \begin{array}{l}3bc = 10\left( {10 + b + c} \right)\\100 = {b^2} + {c^2} - bc\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3bc = 10\left( {10 + b + c} \right)\\100 = {\left( {b + c} \right)^2} - 3bc\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3bc = 100 + 10\left( {b + c} \right)\\3bc = {\left( {b + c} \right)^2} - 100\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}100 + 10\left( {b + c} \right) = {\left( {b + c} \right)^2} - 100\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {b + c} \right)^2} - 10\left( {b + c} \right) - 200 = 0\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}b + c = 20\\b + c = - 10\left( {loai} \right)\end{array} \right.\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b + c = 20\\bc = 100\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 10\\c = 10\end{array} \right.\end{array}\)
Vậy \(R = \dfrac{{10\sqrt 3 }}{3},b = 10,c = 10\)
