Решебник по геометрии 9 класс Атанасян ФГОС Задание 1013

 

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

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

\[\text{si}n^{2}\alpha = 1 - cos^{2}\alpha = 1 - \frac{1}{4} = \frac{3}{4}\]

\[sin\ \alpha = \pm \frac{\sqrt{3}}{2}.\]

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

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

\[\text{si}n^{2}\alpha = 1 - cos^{2}\alpha = 1 - \frac{4}{9} = \frac{5}{9}\]

\[sin\ \alpha = \pm \frac{\sqrt{5}}{3}.\]

\[\textbf{в)}\ cos\ \alpha = - 1:\]

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

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

\[0 = sin\ \alpha = 0.\]

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\[\textbf{а)}\ \overrightarrow{a}\left\{ 2; - 5 \right\};\ \overrightarrow{b}\left\{ - 5;2 \right\}:\]

\[\overrightarrow{v} = 3\overrightarrow{a} - 3\overrightarrow{b} =\]

\[= 3\left\{ 2; - 5 \right\} - 3\left\{ - 5;2 \right\} =\]

\[= \left\{ 6; - 15 \right\} + \left\{ 15; - 6 \right\} =\]

\[= \left\{ 6 + 15; - 15 - 6 \right\} =\]

\[= \left\{ 21; - 21 \right\}.\]

\[\textbf{б)}\ \overrightarrow{a}\left\{ 4;1 \right\};\ \overrightarrow{b}\left\{ 1;2 \right\};\ \overrightarrow{c}\left\{ 2;7 \right\}:\]

\[\overrightarrow{v} = 2\overrightarrow{a} - 3\overrightarrow{b} + 4\overrightarrow{c} =\]

\[= 2\left\{ 4;1 \right\} - 3\left\{ 1;2 \right\} + 4\left\{ 2;7 \right\} =\]

\[= \left\{ 8;2 \right\} + \left\{ - 3; - 6 \right\} + \left\{ 8;28 \right\} =\]

\[= \left\{ 8 - 3 + 8;2 - 6 + 28 \right\} =\]

\[= \left\{ 13;24 \right\}.\]

\[\textbf{в)}\ \overrightarrow{a}\left\{ - 7; - 1 \right\};\ \overrightarrow{b}\left\{ - 1;7 \right\};\ \]

\[\overrightarrow{c}\left\{ 4; - 6 \right\}:\]

\[\overrightarrow{v} = 3\overrightarrow{a} - 2\overrightarrow{b} - \frac{1}{2}\overrightarrow{c} =\]

\[= 3\left\{ - 7; - 1 \right\} - 2\left\{ - 1;7 \right\} - \frac{1}{2}\left\{ 4; - 6 \right\} =\]

\[= \left\{ - 21; - 3 \right\} - \left\{ - 2;14 \right\} - \left\{ 2; - 3 \right\} =\]

\[= \left\{ - 19; - 17 \right\} - \left\{ 2; - 3 \right\} =\]

\[= \left\{ - 21; - 14 \right\}.\]

\[\textbf{г)}\ \overrightarrow{a}\left\{ 7; - 2 \right\};\ \overrightarrow{b}\left\{ 2;5 \right\};\ \overrightarrow{c}\left\{ - 3;3 \right\}:\]

\[\overrightarrow{v} = \overrightarrow{a} - \overrightarrow{b} - \overrightarrow{c} =\]

\[= \left\{ 7; - 2 \right\} - \left\{ 2;5 \right\} - \left\{ - 3;3 \right\} =\]

\[= \left\{ 5; - 7 \right\} - \left\{ - 3;3 \right\} = \left\{ 8; - 10 \right\}.\]