Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 407

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\[1)\ 40{^\circ} = \frac{\pi \bullet 40}{180} = \frac{4\pi}{18} = \frac{2\pi}{9}\]

\[2)\ 120{^\circ} = \frac{\pi \bullet 120}{180} = \frac{12\pi}{18} = \frac{2\pi}{3}\]

\[3)\ 150{^\circ} = \frac{\pi \bullet 150}{180} = \frac{15\pi}{18} = \frac{5\pi}{6}\]

\[4)\ 75{^\circ} = \frac{\pi \bullet 75}{180} = \frac{15\pi}{36} = \frac{5\pi}{12}\]

\[5)\ 32{^\circ} = \frac{\pi \bullet 32}{180} = \frac{16\pi}{90} = \frac{8\pi}{45}\]

\[6)\ 140{^\circ} = \frac{\pi \bullet 140}{180} = \frac{14\pi}{18} = \frac{7\pi}{9}\]