\[\boxed{\mathbf{1129.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Дано:\ \]
\[\angle\alpha - внешний;\ \]
\[\angle\alpha_{n} - внутренний.\]
\[Найти:\]
\[n - количество\ сторон\ \]
\[многоугольника.\]
\[Решение.\]
\[1)\ \alpha + \alpha_{n} = 180{^\circ}\ \]
\[(как\ смежные) \Longrightarrow \alpha_{n} =\]
\[= 180{^\circ} - \alpha.\]
\[2)\ \alpha_{n} = \frac{n - 2}{n} \bullet 180{^\circ}.\]
\[3)\ \frac{n - 2}{n} \bullet 180{^\circ} = 180{^\circ} - \alpha\]
\[(n - 2) \bullet 180{^\circ} = (180{^\circ} - \alpha)n\]
\[180{^\circ}n - 360{^\circ} = 180{^\circ}n - \alpha n\]
\[\alpha n = 360\]
\[n = \frac{360{^\circ}}{\alpha}.\]
\[\textbf{а)}\ \alpha = 18{^\circ}:\ \ \]
\[n = \frac{360{^\circ}}{18{^\circ}} = 20;\]
\[\textbf{б)}\ \alpha = 40{^\circ}:\]
\[n = \frac{360{^\circ}}{40{^\circ}} = 9;\]
\[\textbf{в)}\ \alpha = 72{^\circ}:\]
\[\ n = \frac{360{^\circ}}{72{^\circ}} = 5;\]
\[\textbf{г)}\ \alpha = 60{^\circ}:\ \]
\[n = \frac{360{^\circ}}{60{^\circ}} = 6.\]
\[\boxed{\mathbf{1129.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - ромб;\]
\[BD \cap CA = O;\]
\[BD = BC.\]
\[\mathbf{Найти:}\]
\[угол\ между\ векторами.\]
\[\mathbf{Решение.}\]
\[BC = BD = CD \Longrightarrow \ \]
\[\Longrightarrow \mathrm{\Delta}BCD - равносторонний;\]
\[BD = BA = AD \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}ABD - равносторонний.\]
\[\textbf{а)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{AD}}} \right) = 60{^\circ};\]
\[\textbf{б)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{DA}}} \right) = 120{^\circ};\]
\[\textbf{в)}\ \left( \widehat{\overrightarrow{\text{BA}};\overrightarrow{\text{AD}}} \right) = 120{^\circ};\]
\[\textbf{г)}\ \left( \widehat{\overrightarrow{\text{OC}};\overrightarrow{\text{OD}}} \right) = 90{^\circ};\]
\[\textbf{д)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{DA}}} \right) = 120{^\circ};\]
\[\textbf{е)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{CD}}} \right) = 180{^\circ}\]