Решебник по геометрии 9 класс Мерзляк Задание 12

\[1)\sin a = \frac{5}{13}:\]

\[\cos a = \pm \sqrt{1 - \sin^{2}a} =\]

\[= \pm \sqrt{1 - \frac{25}{169}} = \pm \sqrt{\frac{144}{169}} = \pm \frac{12}{13}.\]

\[2)\cos a = \frac{1}{6}:\]

\[\sin a = \pm \sqrt{1 - \cos^{2}a} =\]

\[= \pm \sqrt{1 - \frac{1}{36}} = \pm \sqrt{\frac{35}{36}} = \pm \frac{\sqrt{35}}{6}.\]

\[3)\sin a = \frac{5}{13};\ \ \ 0{^\circ} \leq a < 90{^\circ}:\]

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

\[= \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}.\]

\[tg\ a = \frac{\sin a}{\cos a} = \frac{5}{13}\ :\frac{12}{13} = \frac{5}{12}.\]

\[4)\cos a = - \frac{8}{17}:\]

\[\sin a = \pm \sqrt{1 - \cos^{2}a} =\]

\[= \pm \sqrt{1 - \frac{64}{289}} = \pm \sqrt{\frac{225}{289}} = \pm \frac{15}{17}.\]

\[ctg\ a = \frac{\cos a}{\sin a} = - \frac{8}{17}\ :\left( \pm \frac{15}{17} \right) =\]

\[= \pm \frac{8}{15}.\]