ГДЗ по геометрии 8 класс Атанасян Задание 1069

\[\boxed{\mathbf{1069.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC - равнобедренный;\]

\[AC = BC;\]

\[AA_{1},\ BB_{1} - медианы;\]

\[\angle BCA = 90{^\circ}.\]

\[\mathbf{Найти:}\]

\[\angle AOB - ?\]

\[\angle BOA_{1} - ?\]

\[\mathbf{Решение.}\]

\[1)\ Пусть\ BC = CA = 2a;\]

\[2)\ В\ \mathrm{\Delta}BCB_{1}:\]

\[BB_{1} = \sqrt{BC^{2} + \left( CB_{1} \right)^{2}} =\]

\[= \sqrt{4a^{2} + a^{2}} = a\sqrt{5};\]

\[AA_{1} = a\sqrt{5}.\]

\[3)\ \overrightarrow{BB_{1}} = \overrightarrow{CB_{1}} - \overrightarrow{\text{CB}};\]

\[\overrightarrow{AA_{1}} = \overrightarrow{CA_{1}} - \overrightarrow{\text{CA}}.\]

\[4)\ \angle BCA \Longrightarrow \ \overrightarrow{CB_{1}} \bullet \overrightarrow{CA_{1}} = 0:\]

\[\overrightarrow{BB_{1}} \bullet \overrightarrow{AA_{1}} =\]

\[5)\cos{\angle AOB} = \frac{\left| \overrightarrow{BB_{1}} \bullet \overrightarrow{AA_{1}} \right|}{\overrightarrow{BB_{1}} \bullet \overrightarrow{AA_{1}}} =\]

\[= \frac{4a^{2}}{a\sqrt{5} \bullet a\sqrt{5}} = \frac{4a^{2}}{5a^{2}} = \frac{4}{5};\]

\[\angle AOB = 36{^\circ}51^{'}.\]

\[6)\ \angle BOA_{1} = 180{^\circ} - 36{^\circ}51^{'} =\]

\[= 143{^\circ}9^{'}\ (как\ смежные).\]

\[Ответ:\ \angle AOB = 36{^\circ}51^{'};\ \]

\[\angle BOA_{1} = 143{^\circ}9^{'}.\]