№3. Найдите: а) cos A, tg A, если sin A = 3/5.

Для решения используем основное тригонометрическое тождество $\sin^2 A + \cos^2 A = 1$ (откуда $\sin A = \sqrt{1 - \cos^2 A}$ или $\cos A = \sqrt{1 - \sin^2 A}$) и формулу $\text{tg } A = \frac{\sin A}{\cos A}$. а) $\cos A = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$; $\text{tg } A = \frac{3/5}{4/5} = \frac{3}{4}$. **Ответ: $\cos A = 0,8; \text{tg } A = 0,75$** б) $\sin A = \sqrt{1 - (\frac{\sqrt{13}}{7})^2} = \sqrt{1 - \frac{13}{49}} = \sqrt{\frac{36}{49}} = \frac{6}{7}$; $\text{tg } A = \frac{6/7}{\sqrt{13}/7} = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}$. **Ответ: $\sin A = \frac{6}{7}; \text{tg } A = \frac{6\sqrt{13}}{13}$** в) $\cos A = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$; $\text{tg } A = \frac{5/13}{12/13} = \frac{5}{12}$. **Ответ: $\cos A = \frac{12}{13}; \text{tg } A = \frac{5}{12}$** г) $\cos A = \sqrt{1 - 0,6^2} = \sqrt{1 - 0,36} = 0,8$; $\text{tg } A = \frac{0,6}{0,8} = 0,75$. **Ответ: $\cos A = 0,8; \text{tg } A = 0,75$** д) $\cos A = \sqrt{1 - (\frac{\sqrt{7}}{4})^2} = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$; $\text{tg } A = \frac{\sqrt{7}/4}{3/4} = \frac{\sqrt{7}}{3}$. **Ответ: $\cos A = 0,75; \text{tg } A = \frac{\sqrt{7}}{3}$** е) $\sin A = \sqrt{1 - (\frac{7}{25})^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}$; $\text{tg } A = \frac{24/25}{7/25} = \frac{24}{7}$. **Ответ: $\sin A = 0,96; \text{tg } A = 3\frac{3}{7}$** ж) $\sin A = \sqrt{1 - (\frac{\sqrt{10}}{10})^2} = \sqrt{1 - \frac{10}{100}} = \sqrt{\frac{90}{100}} = \frac{3\sqrt{10}}{10}$; $\text{tg } A = \frac{3\sqrt{10}/10}{\sqrt{10}/10} = 3$. **Ответ: $\sin A = \frac{3\sqrt{10}}{10}; \text{tg } A = 3$** з) $\sin A = \sqrt{1 - (\frac{2\sqrt{6}}{5})^2} = \sqrt{1 - \frac{24}{25}} = \frac{1}{5}$; $\text{tg } A = \frac{1/5}{2\sqrt{6}/5} = \frac{1}{2\sqrt{6}} = \frac{\sqrt{6}}{12}$. **Ответ: $\sin A = 0,2; \text{tg } A = \frac{\sqrt{6}}{12}$** и) $\sin A = \sqrt{1 - (\frac{\sqrt{19}}{10})^2} = \sqrt{1 - \frac{19}{100}} = \sqrt{\frac{81}{100}} = \frac{9}{10}$; $\text{tg } A = \frac{9/10}{\sqrt{19}/10} = \frac{9}{\sqrt{19}} = \frac{9\sqrt{19}}{19}$. **Ответ: $\sin A = 0,9; \text{tg } A = \frac{9\sqrt{19}}{19}$** к) $\sin A = \sqrt{1 - (\frac{3\sqrt{11}}{10})^2} = \sqrt{1 - \frac{99}{100}} = \frac{1}{10}$; $\text{tg } A = \frac{1/10}{3\sqrt{11}/10} = \frac{1}{3\sqrt{11}} = \frac{\sqrt{11}}{33}$. **Ответ: $\sin A = 0,1; \text{tg } A = \frac{\sqrt{11}}{33}$**