ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1268

\[\boxed{\mathbf{1268}\mathbf{.}}\]

\[0 < a < \frac{\pi}{2}.\]

\[1)\cos a = 0,8:\]

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

\[= \sqrt{1 - {0,8}^{2}} = \sqrt{1 - 0,64} =\]

\[= \sqrt{0,36} = 0,6;\]

\[tg\ a = \frac{\sin a}{\cos a} = \frac{0,6}{0,8} = \frac{6}{8} = \frac{3}{4};\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = \frac{4}{3}.\]

\[2)\sin a = \frac{5}{13}:\]

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

\[= \sqrt{1 - \left( \frac{5}{13} \right)^{2}} = \sqrt{\frac{169}{169} - \frac{25}{169}} =\]

\[= \sqrt{\frac{144}{169}} = \frac{12}{13};\]

\[tg\ a = \frac{\sin a}{\cos a} = \frac{5}{13} \bullet \frac{13}{12} = \frac{5}{12};\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = \frac{12}{5}.\]

\[3)\ tg\ a = 2,4:\]

\[\cos a = \sqrt{\frac{1}{1 + tg^{2}\text{\ a}}} = \sqrt{\frac{1}{1 + {2,4}^{2}}} =\]

\[= \sqrt{\frac{1}{1 + 5,76}} = \sqrt{\frac{100}{676}} = \frac{10}{26} = \frac{5}{13};\]

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

\[= \sqrt{1 - \left( \frac{5}{13} \right)^{2}} = \sqrt{\frac{169}{169} - \frac{25}{169}} =\]

\[= \sqrt{\frac{144}{169}} = \frac{12}{13};\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = \frac{1}{2,4} = \frac{10}{24} = \frac{5}{12}.\]

\[4)\ ctg\ a = \frac{7}{24}:\]

\[\sin a = \sqrt{\frac{1}{1 + ctg^{2}\text{\ a}}} =\]

\[= \sqrt{\frac{1}{1 + \left( \frac{7}{24} \right)^{2}}} = \sqrt{\frac{1}{\frac{576}{576} + \frac{49}{576}}} =\]

\[= \sqrt{\frac{576}{625}} = \frac{24}{25};\]

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

\[= \sqrt{1 - \left( \frac{24}{25} \right)^{2}} = \sqrt{\frac{625}{625} - \frac{576}{625}} =\]

\[= \sqrt{\frac{49}{625}} = \frac{7}{25};\]

\[tg\ a = \frac{1}{\text{ctg\ a}} = \frac{24}{7}.\]