\[\boxed{\mathbf{1040.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - ромб;\]
\[BD \cap CA = O;\]
\[BD = BC.\]
\[\mathbf{Найти:}\]
\[угол\ между\ векторами.\]
\[\mathbf{Решение.}\]
\[BC = BD = CD \Longrightarrow \ \]
\[\Longrightarrow \mathrm{\Delta}BCD - равносторонний;\]
\[BD = BA = AD \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}ABD - равносторонний.\]
\[\textbf{а)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{AD}}} \right) = 60{^\circ};\]
\[\textbf{б)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{DA}}} \right) = 120{^\circ};\]
\[\textbf{в)}\ \left( \widehat{\overrightarrow{\text{BA}};\overrightarrow{\text{AD}}} \right) = 120{^\circ};\]
\[\textbf{г)}\ \left( \widehat{\overrightarrow{\text{OC}};\overrightarrow{\text{OD}}} \right) = 90{^\circ};\]
\[\textbf{д)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{DA}}} \right) = 120{^\circ};\]
\[\textbf{е)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{CD}}} \right) = 180{^\circ}\]