\[\boxed{\mathbf{1055.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равнобедренный;\]
\[AA_{1},\ BB_{1} - медианы;\]
\[AA_{1}\bot BB_{1}.\]
\[\mathbf{Найти:}\]
\[\angle C - ?\]
\[\mathbf{Решение.}\]
\[1)\ Введем\ обозначения:\]
\[\overrightarrow{CA_{1}} = \overrightarrow{a};\ \overrightarrow{CB_{1}} = \overrightarrow{b};\]
\[CA_{1} = CB_{1} = a.\]
\[Тогда:\ \]
\[\overrightarrow{AA_{1}} = \overrightarrow{CA_{1}} - \overrightarrow{\text{CA}} = \overrightarrow{a} - 2\overrightarrow{b};\]
\[\overrightarrow{BB_{1}} = \overrightarrow{CB_{1}} - \overrightarrow{\text{CB}} = \overrightarrow{b} - 2\overrightarrow{a};\]
\[\overrightarrow{AA_{1}} \bullet \overrightarrow{BB_{1}} = \left( \overrightarrow{a} - 2\overrightarrow{b} \right)\left( \overrightarrow{b} - 2\overrightarrow{a} \right) =\]
\[= 5\overrightarrow{a} \bullet \overrightarrow{b} - 2\overrightarrow{a} \bullet \overrightarrow{a} - 2\overrightarrow{b} \bullet \overrightarrow{b}.\]
\[2)\ \overrightarrow{AA_{1}}\bot\overrightarrow{BB_{1}} \Longrightarrow \ \overrightarrow{AA_{1}} \bullet \overrightarrow{BB_{1}} = 0.\]
\[3)\ \overrightarrow{a} \bullet \overrightarrow{b} = a^{2}\cos{\angle C};\ \]
\[\overrightarrow{a} \bullet \overrightarrow{a} = a^{2};\ \]
\[\overrightarrow{b} \bullet \overrightarrow{b} = a^{2};\]
\[5a^{2}\cos{\angle C} - 4a^{2} = 0\]
\[\cos{\angle C} = \frac{4}{5} \Longrightarrow \angle C \approx 36{^\circ}52^{'}.\]
\[\mathbf{Ответ:\ }36{^\circ}52'.\]