ГДЗ по геометрии 8 класс Атанасян Задание 1015

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\[\textbf{а)}\ cos\ \alpha = 1:\]

\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]

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

\[= 0 \Longrightarrow \sin\alpha = 0;\]

\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{0}{1} = 0.\]

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

\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]

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

\[\sin\alpha = \pm \frac{1}{2}.\]

\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \pm \frac{1}{2}\ :\left( - \frac{\sqrt{3}}{2} \right) =\]

\[= \pm \frac{\sqrt{3}}{3}.\]

\[\textbf{в)}\ sin\ \alpha = \frac{\sqrt{2}}{2};\ 0{^\circ} < \alpha < 90{^\circ}:\]

\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]

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

\[= \frac{2}{4} \Longrightarrow \cos\alpha = \pm \frac{\sqrt{2}}{2};\]

\[0{^\circ} < \alpha < 90{^\circ};\ \]

\[\text{cos\ }\alpha > 0 \Longrightarrow \cos\alpha = \frac{\sqrt{2}}{2};\ \]

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

\[\textbf{г)}\ sin\ \alpha = \frac{3}{5};\ 90{^\circ} < \alpha < 180{^\circ}:\]

\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]

\[\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - \frac{9}{25} =\]

\[= \frac{16}{25} \Longrightarrow \cos\alpha = \pm \frac{4}{5};\]

\[90{^\circ} < \alpha < 180{^\circ};\ \]

\[\text{cos\ }\alpha < 0 \Longrightarrow \cos\alpha = - \frac{4}{5};\ \]

\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{3}{5}\ :\left( - \frac{4}{5} \right) = - \frac{3}{4}.\]