\[\boxed{\mathbf{1126.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[R = 10\ см;\]
\[\cup AB = 60{^\circ}.\]
\[\mathbf{Найти:}\]
\[S_{ост} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{ост} = S_{кр} - S_{\text{AB}}.\]
\[2)\ S_{кр} = \pi R^{2} = 3,14 \bullet 10^{2} =\]
\[= 3,14 \bullet 100 = 314\ см^{2}.\]
\[3)\ S_{\text{AB}} = \frac{\pi R^{2}}{360{^\circ}} \bullet \alpha =\]
\[= \frac{3,14 \bullet 100}{360{^\circ}} \bullet 60{^\circ} = 52,33\ см^{2}.\]
\[4)\ S_{ост} = 314 - 52,33 =\]
\[= 261,67\ см^{2}.\]
\[Ответ:S_{ост} = 261,67\ см^{2}.\]
\[\boxed{\mathbf{1126.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\angle\text{CAB} = 12{^\circ}30';\]
\[\angle ABC = 72{^\circ}42';\]
\[AB = 70\ м;\]
\[DC\bot AB.\]
\[\mathbf{Найти:}\]
\[CD - ?\]
\[\mathbf{Решение.}\]
\[1)\ В\ треугольнике\ ADC:\]
\[tg\ 12{^\circ}30^{'} = \frac{\text{DC}}{\text{AD}}\]
\[CD = AD \bullet tg\ 12{^\circ}30^{'}.\]
\[В\ треугольнике\ BDC:\]
\[tg\ 72{^\circ}42^{'} = \frac{\text{CD}}{\text{DB}}\]
\[CD = DB \bullet tg\ 72{^\circ}42^{'}.\]
\[2)\ Пусть\ AD = x;\ DB = 70 - x:\]
\[x \bullet tg\ 12{^\circ}30^{'} =\]
\[= (70 - x) \bullet tg\ 72{^\circ}42^{'}\]
\[x \bullet 0,2217 = (70 - x) \bullet 3,21\]
\[x \bullet 0,2217 = 224,7 - 3,21x\]
\[3,4317x = 224,7\]
\[x = 65,48\ м.\]
\[3)\ CD = 65,48 \bullet 0,2217 =\]
\[= 14,52\ м.\]
\[Ответ:14,52\ м.\]