\[\boxed{\mathbf{699.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[S_{\text{ABCD}} = 12\ см^{2};\]
\[AB + CD = 10.\]
\[\mathbf{Найти:}\]
\[r - ?\]
\[\mathbf{Решение.}\]
\[1)\ В\ \text{ABCD\ }можно\ вписать\ \]
\[окружность:\]
\[AB + CD = BC + AD = 10\ см\ \]
\[(по\ свойству\ вписанной\ \]
\[окружности\ в\ \]
\[четырехугольник).\]
\[2)\ S_{\text{ABCD}} =\]
\[= \frac{1}{2} \bullet P_{\text{ABCD}} \bullet r\ \ (задача\ 697);\]
\[12 = \frac{1}{2}(10 + 10) \bullet r =\]
\[= \frac{1}{2} \bullet 20 \bullet r = 10r\]
\[r = 12\ :10 = 1,2\ см.\]
\[Ответ:r = 1,2\ см.\ \]
\[\boxed{\mathbf{699.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\cos\alpha = \frac{1}{2}:\]
\[\sin\alpha = \sqrt{1 - \cos^{2}\alpha} = \sqrt{1 - \frac{1}{4}} =\]
\[= \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2};\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{3}}{2} \bullet \frac{2}{1} = \sqrt{3}.\]
\[\textbf{б)}\cos\alpha = \frac{2}{3}:\]
\[\sin\alpha = \sqrt{1 - \cos^{2}\alpha} = \sqrt{1 - \frac{4}{9}} =\]
\[= \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3};\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{5}}{3} \bullet \frac{3}{2} = \frac{\sqrt{5}}{2}.\]
\[\textbf{в)}\sin\alpha = \frac{\sqrt{3}}{2}:\]
\[\cos\alpha = \sqrt{1 - \sin^{2}\alpha} = \sqrt{1 - \frac{3}{4}} =\]
\[= \sqrt{\frac{1}{4}} = \frac{1}{2};\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{3}}{2} \bullet \frac{2}{1} = \sqrt{3}.\]
\[\textbf{г)}\sin\alpha = \frac{1}{4}:\]
\[\cos\alpha = \sqrt{1 - \sin^{2}\alpha} = \sqrt{1 - \frac{1}{16}} =\]
\[= \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4};\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1}{4} \bullet \frac{4}{\sqrt{15}} = \frac{1}{\sqrt{15}} =\]
\[= \frac{\sqrt{15}}{15}.\]