\[\boxed{\mathbf{1069.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равнобедренный;\]
\[AC = BC;\]
\[AA_{1},\ BB_{1} - медианы;\]
\[\angle BCA = 90{^\circ}.\]
\[\mathbf{Найти:}\]
\[\angle AOB - ?\]
\[\angle BOA_{1} - ?\]
\[\mathbf{Решение.}\]
\[1)\ Пусть\ BC = CA = 2a;\]
\[2)\ В\ \mathrm{\Delta}BCB_{1}:\]
\[BB_{1} = \sqrt{BC^{2} + \left( CB_{1} \right)^{2}} =\]
\[= \sqrt{4a^{2} + a^{2}} = a\sqrt{5};\]
\[AA_{1} = a\sqrt{5}.\]
\[3)\ \overrightarrow{BB_{1}} = \overrightarrow{CB_{1}} - \overrightarrow{\text{CB}};\]
\[\overrightarrow{AA_{1}} = \overrightarrow{CA_{1}} - \overrightarrow{\text{CA}}.\]
\[4)\ \angle BCA \Longrightarrow \ \overrightarrow{CB_{1}} \bullet \overrightarrow{CA_{1}} = 0:\]
\[\overrightarrow{BB_{1}} \bullet \overrightarrow{AA_{1}} =\]
\[5)\cos{\angle AOB} = \frac{\left| \overrightarrow{BB_{1}} \bullet \overrightarrow{AA_{1}} \right|}{\overrightarrow{BB_{1}} \bullet \overrightarrow{AA_{1}}} =\]
\[= \frac{4a^{2}}{a\sqrt{5} \bullet a\sqrt{5}} = \frac{4a^{2}}{5a^{2}} = \frac{4}{5};\]
\[\angle AOB = 36{^\circ}51^{'}.\]
\[6)\ \angle BOA_{1} = 180{^\circ} - 36{^\circ}51^{'} =\]
\[= 143{^\circ}9^{'}\ (как\ смежные).\]
\[Ответ:\ \angle AOB = 36{^\circ}51^{'};\ \]
\[\angle BOA_{1} = 143{^\circ}9^{'}.\]
\[\boxed{\mathbf{1069.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Решение\ задачи\ дано\ }\]
\[\mathbf{в\ учебнике.}\]