Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 415

\[\boxed{\mathbf{415.}}\]

\[\mathbf{Первый\ столбец:}\]

\[a = 30{^\circ} = \frac{\pi \bullet 30}{180} = \frac{\pi}{6}\]

\[l = aR = \frac{\pi}{6} \bullet 2 = \frac{\pi}{3}\]

\[S = \frac{R^{2}}{2} \bullet a = \frac{2^{2}}{2} \bullet \frac{\pi}{6} = 2 \bullet \frac{\pi}{6} = \frac{\pi}{3}\]

\[Второй\ столбец:\]

\[a = \frac{\pi}{5} = \left( \frac{180}{\pi} \bullet \frac{\pi}{5} \right)^{{^\circ}} = 36{^\circ}\]

\[R = \frac{l}{a} = 2\ :\frac{\pi}{5} = 2 \bullet \frac{5}{\pi} = \frac{10}{\pi}\]

\[S = \frac{R^{2}}{2} \bullet a = \left( \frac{10}{\pi} \right)^{2}\ :2 \bullet \frac{\pi}{5} =\]

\[= \frac{100}{2\pi^{2}} \bullet \frac{\pi}{5} = \frac{10}{\pi}\]

\[Третий\ столбец:\]

\[a = \frac{l}{R} = \frac{5}{10} = \frac{1}{2}\]

\[a = \left( \frac{180}{\pi} \bullet \frac{1}{2} \right)^{{^\circ}} = \left( \frac{90}{\pi} \right)^{{^\circ}}\]

\[S = \frac{R^{2}}{2} \bullet a = \frac{10^{2}}{2} \bullet \frac{1}{2} = \frac{100}{4} = 25\]

\[Четвертый\ столбец:\]

\[a = \frac{2S}{R^{2}} = \frac{2 \bullet 50}{5^{2}} = \frac{100}{25} = 4\]

\[a = \left( \frac{180}{\pi} \bullet 4 \right)^{{^\circ}} = \left( \frac{720}{\pi} \right)^{{^\circ}}\]

\[l = aR = 4 \bullet 5 = 20\]

\[Пятый\ столбец:\]

\[a = \left( \frac{180}{\pi} \bullet 2 \right)^{{^\circ}} = \left( \frac{360}{\pi} \right)^{{^\circ}}\]

\[R = \sqrt{\frac{2S}{a}} = \sqrt{\frac{2 \bullet 25}{2}} = \sqrt{25} = 5\]

\[l = aR = 2 \bullet 5 = 10\]

\[Шестой\ столбец:\]

\[R = \frac{l}{a}\]

\[S = \frac{R^{2}}{2} \bullet a = \frac{l^{2}}{2a^{2}} \bullet a = \frac{l^{2}}{2a}\text{\ \ }\]

\[a = \frac{2S}{l^{2}} = \frac{2 \bullet 50}{10^{2}} = 1\]

\[a = \left( \frac{180}{\pi} \right)^{{^\circ}}\]

\[R = 10\]