\[\boxed{\mathbf{1071.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[\angle A < 90{^\circ};\]
\[S_{\text{ABC}} = 3\sqrt{3};\]
\[AB = 4\sqrt{3};\]
\[AC = 3.\]
\[\mathbf{Найти:}\]
\[R - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABC}} = \frac{1}{2}AB \bullet AC \bullet \sin{\angle A};\]
\[3\sqrt{3} = \frac{1}{2} \bullet 4\sqrt{3} \bullet 3\sin{\angle A}\]
\[3\sqrt{3} = 6\sqrt{3}\sin{\angle A}\]
\[\sin{\angle A} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2}\]
\[\angle A = 30{^\circ}.\]
\[2)\ По\ теореме\ косинусов:\]
\[BC^{2} =\]
\[= AC^{2} + AB^{2} - 2AC \bullet AB \bullet \cos{30{^\circ}}\]
\[BC^{2} = 9 + 48 - 24\sqrt{3} \bullet \frac{\sqrt{3}}{2} =\]
\[= 57 - 12 \bullet 3 = 57 - 36 = 21\]
\[BC = \sqrt{21}.\]
\[3)\ По\ теореме\ синусов:\]
\[\frac{\text{BC}}{\sin{\angle A}} = 2R\]
\[\frac{\sqrt{21}}{\sin{30{^\circ}}} = 2R\]
\[\sqrt{21}\ :\frac{1}{2} = 2R\]
\[R = \frac{2\sqrt{21}}{2} = \sqrt{21}.\]
\[Ответ:\ R = \sqrt{21}.\]