ГДЗ по геометрии 8 класс Атанасян Задание 1012

\[\boxed{\mathbf{1012.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[1)\ M_{1}(0;1):\ \ \ \]

\[0^{2} + 1^{2} = 1 \Longrightarrow\]

\[\Longrightarrow M_{1} \in окружности\ (0;1).\]

\[M_{2}\left( \frac{1}{2};\frac{\sqrt{3}}{2} \right):\ \ \ \]

\[\left( \frac{1}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2} = 1\]

\[\frac{1}{4} + \frac{3}{4} = 1 \Longrightarrow\]

\[\Longrightarrow M_{2} \in окружности\ (0;1).\]

\[M_{3}\left( \frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2} \right):\ \ \ \]

\[\left( \frac{\sqrt{2}}{2} \right)^{2} + \left( \frac{\sqrt{2}}{2} \right)^{2} = 1\]

\[\frac{1}{2} + \frac{1}{2} = 1 \Longrightarrow\]

\[\Longrightarrow M_{3} \in окружности\ (0;1).\]

\[M_{4}\left( - \frac{\sqrt{3}}{2};\frac{1}{2} \right):\ \left( - \frac{\sqrt{3}}{2} \right)^{2} + \left( \frac{1}{2} \right)^{2} = 1\]

\[\frac{3}{4} + \frac{1}{4} = 1 \Longrightarrow\]

\[\Longrightarrow M_{4} \in окружности\ (0;1).\]

\[A(1;0):\ \ \ \]

\[1^{2} + 0^{2} = 1 \Longrightarrow\]

\[\Longrightarrow A \in окружности\ (0;1).\]

\[B( - 1;0):\ \ \ \]

\[( - 1)^{2} + 0^{2} = 1 \Longrightarrow\]

\[\Longrightarrow \text{\ B} \in окружности\ (0;1).\]

\[2)\ Рисунок\ по\ условию\ задачи:\]

\[sin\angle AOM_{1} = sin90{^\circ} = 1;\ \]

\[cos\angle AOM_{1} = 0;\ \]

\[tg\angle AOM_{1} - не\ существует.\]

\[sin\angle AOM_{2} = \frac{\sqrt{3}}{2};\ \]

\[cos\angle AOM_{2} = \frac{1}{2};\ \]

\[tg\angle AOM_{2} = \frac{sin\angle AOM_{2}}{cos\angle AOM_{2}} = \sqrt{3}.\]

\[sin\angle AOM_{3} = \frac{\sqrt{2}}{2};\ \]

\[cos\angle AOM_{3} = \frac{\sqrt{2}}{2};\ \]

\[tg\angle AOM_{3} = \frac{sin\angle AOM_{3}}{cos\angle AOM_{3}} = 1.\]

\[sin\angle AOM_{4} = \frac{1}{2};\ \]

\[cos\angle AOM_{4} = - \frac{\sqrt{3}}{3};\ \]

\[tg\angle AOM_{4} = \frac{sin\angle AOM_{4}}{cos\angle AOM_{4}} = - \sqrt{3}.\]

\[sin\angle AOB = 0;\ \]

\[cos\angle AOB = - 1;\ \]

\[tg\angle AOB = \frac{sin\angle AOB}{cos\angle AOB} = 0.\]