\[\boxed{\mathbf{771.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[ABCD - параллелограмм;\ \]
\[AC \cap BD = O;\ \]
\[\overrightarrow{a} = \overrightarrow{\text{AB}};\ \]
\[\overrightarrow{b} = \overrightarrow{\text{AD}}.\]
\[Выразить:\ \]
\[\overrightarrow{\text{DC}} + \overrightarrow{\text{CB}};\ \]
\[\overrightarrow{\text{BO}} + \overrightarrow{\text{OC}};\ \]
\[\overrightarrow{\text{BO}} - \overrightarrow{\text{OC}};\ \]
\[\overrightarrow{\text{BA}} - \overrightarrow{\text{DA}}.\]
\[Решение.\]
\[1)\ \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} = \overrightarrow{\text{DB}} = \overrightarrow{\text{AB}} - \overrightarrow{\text{AD}} =\]
\[= \overrightarrow{a} - \overrightarrow{b}\ (по\ правилу\ треугольника).\]
\[2)\ \overrightarrow{\text{BO}} + \overrightarrow{\text{OC}} = \overrightarrow{\text{BC}} = \overrightarrow{\text{AD}} = \overrightarrow{b}.\]
\[3)\ \overrightarrow{\text{BO}} - \overrightarrow{\text{OC}} = \overrightarrow{\text{BO}} - \overrightarrow{\text{AO}} =\]
\[= \overrightarrow{\text{BO}} - \left( - \overrightarrow{\text{OA}} \right) = \overrightarrow{\text{BO}} + \overrightarrow{\text{OA}} =\]
\[= \overrightarrow{\text{BA}} = - \overrightarrow{\text{AB}} =\]
\[= - \overrightarrow{a}\ (по\ правилу\ треугольника).\]
\[4)\ \overrightarrow{\text{BA}} - \overrightarrow{\text{DA}} = \overrightarrow{\text{BA}} - \left( - \overrightarrow{\text{AD}} \right) =\]
\[= \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}} = - \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} =\]
\[= - \overrightarrow{a} + \overrightarrow{b} = \overrightarrow{b} - \overrightarrow{a}.\]
\[\boxed{\mathbf{771.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC} - вписанный\ \]
\[в\ окружность;\]
\[\text{AB} - диаметр;\]
\[\textbf{а)} \cup \text{BC} = 134{^\circ};\]
\[\textbf{б)} \cup \text{AC} = 70{^\circ}.\]
\[\mathbf{Найти:}\]
\[\angle CAB - ?\ \]
\[\angle ACB - ?\]
\[\angle CBA - ?\]
\[\mathbf{Решение.}\]
\[\angle ACB = 90{^\circ} - так\ как\ угол\ \]
\[опирается\ на\ диаметр.\]
\[\textbf{а)}\ \angle CAB = \frac{1}{2} \cup BC = \frac{134{^\circ}}{2} =\]
\[= 67{^\circ}\ (как\ вписанный\ угол);\]
\[\textbf{б)}\ \angle CBA = \frac{1}{2} \cup AC = \frac{70{^\circ}}{2} =\]
\[= 35{^\circ}\ (как\ вписанный\ угол);\]
\[Ответ:а)\ \angle ACB = 90{^\circ};\ \]
\[\angle CAB = 67{^\circ};\ \angle CBA = 23{^\circ};\]
\[\textbf{б)}\ \angle ACB = 90{^\circ};\ \angle CAB = 55{^\circ};\ \]
\[\angle CBA = 35{^\circ}.\ \]