\[Схематический\ рисунок.\]
\[Дано:\]
\[\mathrm{\Delta}ABC;\]
\[\angle ABC = 60{^\circ};\]
\[\angle ADC = 45{^\circ};\]
\[R_{\mathrm{\Delta}ABC} = 4\ см.\]
\[Найти:\]
\[R_{\mathrm{\Delta}ACD}.\]
\[Решение.\]
\[1)\ В\ \mathrm{\Delta}ABC:\]
\[R = \frac{\text{AC}}{2\sin{\angle B}}\text{\ \ \ }\]
\[AC = 2R\sin{\angle B} =\]
\[= 2 \bullet 4 \bullet \sin{60{^\circ}} = 8 \bullet \frac{\sqrt{3}}{2} = 4\sqrt{3}.\]
\[2)\ В\ \mathrm{\Delta}\text{ACD}\]
\[R = \frac{\text{AC}}{2\sin{\angle D}} = \frac{4\sqrt{3}}{2\sin{45{^\circ}}} =\]
\[= 2\sqrt{3}\ :\frac{\sqrt{2}}{2} = 2\sqrt{3} \bullet \sqrt{2} = 2\sqrt{6}\ см.\]
\[Ответ:\ \ 2\sqrt{6}\ см.\]