\[\boxed{\mathbf{1055.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равнобедренный;\]
\[AA_{1},\ BB_{1} - медианы;\]
\[AA_{1}\bot BB_{1}.\]
\[\mathbf{Найти:}\]
\[\angle C - ?\]
\[\mathbf{Решение.}\]
\[1)\ Введем\ обозначения:\]
\[\overrightarrow{CA_{1}} = \overrightarrow{a};\ \overrightarrow{CB_{1}} = \overrightarrow{b};\]
\[CA_{1} = CB_{1} = a.\]
\[Тогда:\ \]
\[\overrightarrow{AA_{1}} = \overrightarrow{CA_{1}} - \overrightarrow{\text{CA}} = \overrightarrow{a} - 2\overrightarrow{b};\]
\[\overrightarrow{BB_{1}} = \overrightarrow{CB_{1}} - \overrightarrow{\text{CB}} = \overrightarrow{b} - 2\overrightarrow{a};\]
\[\overrightarrow{AA_{1}} \bullet \overrightarrow{BB_{1}} = \left( \overrightarrow{a} - 2\overrightarrow{b} \right)\left( \overrightarrow{b} - 2\overrightarrow{a} \right) =\]
\[= 5\overrightarrow{a} \bullet \overrightarrow{b} - 2\overrightarrow{a} \bullet \overrightarrow{a} - 2\overrightarrow{b} \bullet \overrightarrow{b}.\]
\[2)\ \overrightarrow{AA_{1}}\bot\overrightarrow{BB_{1}} \Longrightarrow \ \overrightarrow{AA_{1}} \bullet \overrightarrow{BB_{1}} = 0.\]
\[3)\ \overrightarrow{a} \bullet \overrightarrow{b} = a^{2}\cos{\angle C};\ \]
\[\overrightarrow{a} \bullet \overrightarrow{a} = a^{2};\ \]
\[\overrightarrow{b} \bullet \overrightarrow{b} = a^{2};\]
\[5a^{2}\cos{\angle C} - 4a^{2} = 0\]
\[\cos{\angle C} = \frac{4}{5} \Longrightarrow \angle C \approx 36{^\circ}52^{'}.\]
\[\mathbf{Ответ:\ }36{^\circ}52'.\]
\[\boxed{\mathbf{1055.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[Окружность\ (O;R).\]
\[\textbf{а)}\ M( - 3;5);\]
\[N(7; - 3).\]
\[\textbf{б)}\ M(2; - 1);\ \]
\[N(4;3).\]
\[\mathbf{Написать:}\]
\[уравнение\ окружности.\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ MN - диаметр \Longrightarrow\]
\[\Longrightarrow R = \frac{\text{MN}}{2};MO = ON.\]
\[\left\{ \begin{matrix} x_{0} = \frac{x_{M} + x_{N}}{2} \\ y_{0} = \frac{y_{M} + y_{N}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{0} = \frac{- 3 + 7}{2} \\ y_{0} = \frac{5 - 3}{2}\text{\ \ \ } \\ \end{matrix}\text{\ \ } \right.\ \]
\[\left\{ \begin{matrix} x_{0} = 2 \\ y_{0} = 1 \\ \end{matrix} \right.\ \Longrightarrow O(2;1).\]
\[2)\ R = MO =\]
\[= \sqrt{(2 + 3)^{2} + (1 - 5)^{2}} = \sqrt{41}.\]
\[3)\ (x - 2)^{2} + (y - 1)^{2} = 41.\]
\[\textbf{а)}\ 1)\ MN - диаметр \Longrightarrow\]
\[\Longrightarrow R = \frac{\text{MN}}{2};MO = ON.\]
\[\left\{ \begin{matrix} x_{0} = \frac{x_{M} + x_{N}}{2} \\ y_{0} = \frac{y_{M} + y_{N}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{0} = \frac{2 + 4}{2}\text{\ \ \ } \\ y_{0} = \frac{- 1 + 3}{2} \\ \end{matrix}\text{\ \ \ } \right.\ \]
\[\left\{ \begin{matrix} x_{0} = 3 \\ y_{0} = 1 \\ \end{matrix} \right.\ \Longrightarrow O(3;1).\]
\[2)\ R = MO =\]
\[= \sqrt{(3 - 2)^{2} + (1 + 1)^{2}} = \sqrt{5}.\]
\[3)\ (x - 3)^{2} + (y - 1)^{2} = 5.\]
\[Ответ:а)\ (x - 2)^{2} + (y - 1)^{2} =\]
\[= 41;\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (x - 3)^{2} + (y - 1)^{2} =\]
\[= 5\ .\]