\[\boxed{\mathbf{783.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[M \in BC;\ \ \]
\[\ BM\ :MC = 3\ :1.\]
\[\overrightarrow{a} = \overrightarrow{\text{AD}};\ \ \ \]
\[\overrightarrow{b} = \overrightarrow{\text{MD.}}\]
\[Выразить:\]
\[\overrightarrow{\text{AM}}\ \ и\ \ \overrightarrow{\text{MD}}.\]
\[\mathbf{Решение.}\]
\[1)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{AM}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BM}} = \overrightarrow{b} + \frac{3}{4}\overrightarrow{\text{BC}} =\]
\[= \overrightarrow{b} + \frac{3}{4}\overrightarrow{\text{AD}} = \overrightarrow{b} + \frac{3}{4}\overrightarrow{\text{a.}}\]
\[2)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{MD}} = \overrightarrow{\text{MC}} + \overrightarrow{\text{CD}} = \frac{1}{4}\overrightarrow{\text{BC}} + \frac{\overrightarrow{\text{BA}}}{4} =\]
\[= \frac{1}{4}\overrightarrow{\text{Bc}} + \left( - \overrightarrow{\text{AB}} \right) = \frac{1}{4}\overrightarrow{\text{AD}} - \overrightarrow{\text{AB}} =\]
\[= \frac{1}{4}\overrightarrow{a} - \overrightarrow{b}.\]
\[\boxed{\mathbf{783.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Памятка:как\ определить\ }\]
\[\mathbf{градусную\ меру\ угла\ }\]
\[\mathbf{по\ клеткам.}\]
\[\textbf{а)}\ \angle ABC = \frac{1}{2}\angle AOC = \frac{1}{2} \cdot 90{^\circ} =\]
\[= 45{^\circ}.\]
\[\textbf{б)}\ \angle ABC = \frac{1}{2}\angle AOC = \frac{1}{2} \cdot 90{^\circ} =\]
\[= 45{^\circ}.\]
\[\textbf{в)}\ \angle AOC = 90{^\circ} + 30{^\circ} = 120{^\circ};\]
\[\angle ABC = \frac{1}{2}\angle AOC = 60{^\circ}.\]
\[\textbf{г)}\ \alpha = 360{^\circ} - 90{^\circ} = 270{^\circ};\]
\[\angle ABC = \frac{1}{2}\alpha = \frac{1}{2} \cdot 270{^\circ} = 135{^\circ}.\]
\[\textbf{д)}\ \angle ABO = 45{^\circ};\]
\[\angle ABC = 180{^\circ} - (90{^\circ} + 45{^\circ}) =\]
\[= 45{^\circ}.\]
\[\textbf{е)}\ \angle ABC = 45{^\circ}.\]
\[\textbf{ж)}\ \angle ABC = 45{^\circ}.\]