Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 409

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\[\mathbf{а)\ }a_{d} = \frac{180{^\circ}}{3} = 60{^\circ}\]

\[a_{r} = \frac{\pi \bullet 60}{180} = \frac{\pi}{3}\]

\[\textbf{б)}\ a_{d1} = 90{^\circ}\ \ и\ \ \]

\[a_{d2} = \frac{180{^\circ} - 90{^\circ}}{2} = \frac{90{^\circ}}{2} = 45{^\circ}\]

\[a_{r1} = \frac{\pi \bullet 90}{180} = \frac{\pi}{2}\text{\ \ }и\ \]

\[a_{r2} = \frac{\pi \bullet 45}{180} = \frac{\pi}{4}\]

\[\textbf{в)}\ a_{d} = \frac{1}{4} \bullet (4 - 2) \bullet 180{^\circ} =\]

\[= 2 \bullet 45 = 90{^\circ}\]

\[a_{r} = \frac{\pi \bullet 90}{180} = \frac{\pi}{2}\]

\[\textbf{г)}\ a_{d} = \frac{1}{6} \bullet (6 - 2) \bullet 180 =\]

\[= 4 \bullet 30 = 120{^\circ}\]

\[a_{r} = \frac{\pi \bullet 120}{180} = \frac{2\pi}{3}\]