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МатематикаРешебник по геометрии 7 класс Атанасян ФГОС Задание 1015
Задание 1015
\[\boxed{\mathbf{1015.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\textbf{а)}\ cos\ \alpha = 1:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\sin^{2}\alpha = 1 - \cos^{2}\alpha = 1 - 1 =\]
\[= 0 \Longrightarrow \sin\alpha = 0;\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{0}{1} = 0.\]
\[\textbf{б)}\ cos\ \alpha = - \frac{\sqrt{3}}{2}:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\sin^{2}\alpha = 1 - \cos^{2}\alpha = 1 - \frac{3}{4} = \frac{1}{4}\]
\[\sin\alpha = \pm \frac{1}{2}.\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \pm \frac{1}{2}\ :\left( - \frac{\sqrt{3}}{2} \right) =\]
\[= \pm \frac{\sqrt{3}}{3}.\]
\[\textbf{в)}\ sin\ \alpha = \frac{\sqrt{2}}{2};\ 0{^\circ} < \alpha < 90{^\circ}:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - \frac{2}{4} =\]
\[= \frac{2}{4} \Longrightarrow \cos\alpha = \pm \frac{\sqrt{2}}{2};\]
\[0{^\circ} < \alpha < 90{^\circ};\ \]
\[\text{cos\ }\alpha > 0 \Longrightarrow \cos\alpha = \frac{\sqrt{2}}{2};\ \]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{2}}{2}\ :\frac{\sqrt{2}}{2} = 1.\]
\[\textbf{г)}\ sin\ \alpha = \frac{3}{5};\ 90{^\circ} < \alpha < 180{^\circ}:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - \frac{9}{25} =\]
\[= \frac{16}{25} \Longrightarrow \cos\alpha = \pm \frac{4}{5};\]
\[90{^\circ} < \alpha < 180{^\circ};\ \]
\[\text{cos\ }\alpha < 0 \Longrightarrow \cos\alpha = - \frac{4}{5};\ \]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{3}{5}\ :\left( - \frac{4}{5} \right) = - \frac{3}{4}.\]
\[\boxed{\mathbf{1015.еуроки - ответы\ на\ пятёрку}}\]
\[\overrightarrow{a}\ \left\{ 3;7 \right\};\ \overrightarrow{b}\left\{ - 2;1 \right\};\ \overrightarrow{c}\ \left\{ 6;14 \right\};\ \]
\[\overrightarrow{d}\ \left\{ 2;\ - 1 \right\};\ \overrightarrow{e}\ \left\{ 2;4 \right\}\]
\[По\ лемме\ о\ коллинеарных\ \]
\[векторах:\]
\[1)\ \overrightarrow{a}\ и\ \overrightarrow{c}\ :\ \overrightarrow{c} = 2 \bullet \overrightarrow{a}.\]
\[2)\ \overrightarrow{b}\ и\ \overrightarrow{d}\ :\ \overrightarrow{b} = - 1 \bullet \overrightarrow{d}.\ \]
