ГДЗ по геометрии 9 класс Атанасян Задание 1019

\[\boxed{\mathbf{1019.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\textbf{а)}\ A(2;2);\]

\[\textbf{б)}\ A(0;3);\]

\[\textbf{в)}\ A( - \sqrt{3};1);\]

\[\textbf{г)}\ A\left( - 2\sqrt{2};2\sqrt{2} \right).\]

\[\mathbf{Найти:}\]

\[\angle\alpha - ?\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ OA^{2} = 4 + 4 = 8 \Longrightarrow\]

\[\Longrightarrow OA = 2\sqrt{2}.\]

\[\left\{ \begin{matrix} 2 = 2\sqrt{2} \bullet \cos\alpha \\ 2 = 2\sqrt{2} \bullet \sin\alpha \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \cos\alpha = \frac{\sqrt{2}}{2} \\ \sin\alpha = \frac{\sqrt{2}}{2} \\ \end{matrix} \right.\ \Longrightarrow \alpha = 45{^\circ}.\]

\[\textbf{б)}\ OA^{2} = 0 + 9 = 9 \Longrightarrow OA = 3.\]

\[\left\{ \begin{matrix} 0 = 3 \bullet \cos\alpha \\ 3 = 3 \bullet \sin\alpha \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \cos\alpha = 0 \\ \sin\alpha = 1 \\ \end{matrix} \right.\ \Longrightarrow \alpha = 90{^\circ}.\]

\[\textbf{в)}\ OA^{2} = 3 + 1 = 4 \Longrightarrow OA = 2.\]

\[\left\{ \begin{matrix} - \sqrt{3} = 2 \bullet \cos\alpha \\ 1 = 2 \bullet \sin\alpha\text{\ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \cos\alpha = - \frac{\sqrt{3}}{2} \\ \sin\alpha = \frac{1}{2}\text{\ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \alpha = 150{^\circ}.\]

\[\textbf{г)}\ OA^{2} = 8 + 8 = 16 \Longrightarrow OA = 4.\]

\[\left\{ \begin{matrix} - 2\sqrt{2} = 4 \bullet \cos\alpha \\ 2\sqrt{2} = 4 \bullet \sin\alpha\text{\ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \cos\alpha = - \frac{\sqrt{2}}{2} \\ \sin\alpha = \frac{\sqrt{2}}{2}\text{\ \ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \alpha = 135{^\circ}.\]

\[Ответ:а)\ 45{^\circ};\ б)\ 90{^\circ};\ в)\ 150{^\circ};\]

\[\textbf{г)}\ 135{^\circ}.\]

\[\mathbf{Параграф\ }2\mathbf{.\ Соотношения\ между\ сторонами\ и\ углами\ треугольника}\]