\[\boxed{\mathbf{768.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC;\ \]
\[AM = MB;\]
\[AN = NC;\ \]
\[\overrightarrow{a} = \overrightarrow{\text{AM}};\ \]
\[\overrightarrow{b} = \overrightarrow{\text{AN}}.\]
\[Выразить:\]
\[\overrightarrow{\text{BM}};\ \overrightarrow{\text{NC}};\ \overrightarrow{\text{MN}};\ \overrightarrow{\text{BN}}.\]
\[Решение.\]
\[1)\ \overrightarrow{\text{BM}} = \overrightarrow{\text{MA}} = - \overrightarrow{a}.\]
\[2)\ \overrightarrow{\text{NC}} = \overrightarrow{\text{AN}} = \overrightarrow{b}.\]
\[3)\ \overrightarrow{\text{MN}} = \overrightarrow{\text{AN}} - \overrightarrow{\text{AM}} =\]
\[= \overrightarrow{b} - \overrightarrow{a}\ (по\ правилу\ треугольника).\]
\[4)\ \overrightarrow{\text{BN}} = \overrightarrow{\text{BM}} + \overrightarrow{\text{MN}} =\]
\[= - \overrightarrow{a} + \left( \overrightarrow{b} - \overrightarrow{a} \right) = - \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{a} =\]
\[= - 2\overrightarrow{a} + \overrightarrow{b}\]
\[(по\ правилу\ треугольника).\]
\[\boxed{\mathbf{768.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[окружность\ (O,\ R);\ \]
\[\angle AOB = \angle ACB + 30{^\circ}.\]
\[\mathbf{Найти:}\]
\[\angle AOB - ?\]
\[\ \angle ACB - ?\]
\[\mathbf{Решение.}\]
\[1)\ Пусть\ \angle ACB = x;\ \]
\[\angle AOB = x + 30{^\circ}.\]
\[2)\ \angle AOB = \cup AB;\ \ \ \]
\[\angle ACB = \frac{1}{2} \cup AB\ \]
\[(по\ теореме\ о\ вписанном\ угле).\]
\[Отсюда:\ \]
\[\angle ACB = \frac{1}{2}\angle AOB.\]
\[3)\ x = \frac{1}{2}(x + 30{^\circ}) =\]
\[= \frac{1}{2}x + 15{^\circ} = > \frac{1}{2}x = 15{^\circ}.\]
\[Отсюда:\]
\[x = 30{^\circ} \Longrightarrow \angle ACB = 30{^\circ} \Longrightarrow\]
\[\Longrightarrow \angle AOB = 60{^\circ}.\]
\[Ответ:\angle ACB = 30{^\circ};\ \]
\[\angle AOB = 60{^\circ}.\]