ГДЗ по геометрии 9 класс Мерзляк Задание 11

\[1)\sin a = \frac{3}{5};\ \ 0{^\circ} \leq a \leq 90{^\circ}:\]

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

\[= \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}.\]

\[2)\sin a = \frac{1}{3};\ \ \ 90{^\circ} \leq a \leq 180{^\circ}:\]

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

\[= - \sqrt{1 - \frac{1}{9}} = - \sqrt{\frac{8}{9}} = - \frac{2\sqrt{2}}{3}.\]

\[3)\sin a = \frac{\sqrt{3}}{4}:\]

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

\[= \pm \sqrt{1 - \frac{3}{16}} = \pm \sqrt{\frac{13}{16}} = \pm \frac{\sqrt{13}}{4}.\]

\[4)\cos a = - 0,8:\]

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

\[= \pm \sqrt{1 - 0,64} = \pm \sqrt{0,36} = \pm 0,6.\]

\[5)\sin a = \frac{4}{5};\ \ \ 90{^\circ} < a \leq 180{^\circ}:\]

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

\[= - \sqrt{1 - \frac{16}{25}} = - \sqrt{\frac{9}{25}} = - \frac{3}{5}.\]

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

\[6)\cos a = \frac{12}{13};\ \ \ 0{^\circ} < a \leq 90{^\circ}:\]

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

\[= \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}.\]

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

\[Ответ:\ \ \frac{12}{5}.\]