ГДЗ по геометрии 8 класс Атанасян ФГОС Задание 699

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\[\textbf{а)}\cos\alpha = \frac{1}{2}:\]

\[\sin\alpha = \sqrt{1 - \cos^{2}\alpha} = \sqrt{1 - \frac{1}{4}} =\]

\[= \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2};\]

\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{3}}{2} \bullet \frac{2}{1} = \sqrt{3}.\]

\[\textbf{б)}\cos\alpha = \frac{2}{3}:\]

\[\sin\alpha = \sqrt{1 - \cos^{2}\alpha} = \sqrt{1 - \frac{4}{9}} =\]

\[= \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3};\]

\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{5}}{3} \bullet \frac{3}{2} = \frac{\sqrt{5}}{2}.\]

\[\textbf{в)}\sin\alpha = \frac{\sqrt{3}}{2}:\]

\[\cos\alpha = \sqrt{1 - \sin^{2}\alpha} = \sqrt{1 - \frac{3}{4}} =\]

\[= \sqrt{\frac{1}{4}} = \frac{1}{2};\]

\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{3}}{2} \bullet \frac{2}{1} = \sqrt{3}.\]

\[\textbf{г)}\sin\alpha = \frac{1}{4}:\]

\[\cos\alpha = \sqrt{1 - \sin^{2}\alpha} = \sqrt{1 - \frac{1}{16}} =\]

\[= \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4};\]

\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1}{4} \bullet \frac{4}{\sqrt{15}} = \frac{1}{\sqrt{15}} =\]

\[= \frac{\sqrt{15}}{15}.\]