Найдите : a) cos A, tg A, если sin A=3. 5. б) sin A, tg A, если cos A=√13 7. в) cos A, tg А, если sin A=5 13 г) cos A, tg А, если sin A=0,6 д) cos A, tg A, если sin A=√7 4 e)sin A, tg A, если cos A=7 25

Ответ: a) cos A = \(\frac{4}{5}\), tg A = \(\frac{3}{4}\); б) sin A = \(\frac{6}{\sqrt{13}}\), tg A = \(\frac{6}{\sqrt{13}}\); в) cos A = \(\frac{12}{13}\), tg A = \(\frac{5}{12}\); г) cos A = 0.8, tg A = 0.75; д) cos A = \(\frac{3}{4}\), tg A = \(\frac{\sqrt{7}}{3}\); e) sin A = \(\frac{24}{25}\), tg A = \(\frac{24}{7}\)

1. а) Дано: sin A = \(\frac{3}{5}\). Необходимо найти cos A и tg A. \(cos A = \sqrt{1 - sin^2 A} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}\) \(tg A = \frac{sin A}{cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}\)
2. б) Дано: cos A = \(\frac{\sqrt{13}}{7}\). Необходимо найти sin A и tg A. \(sin A = \sqrt{1 - cos^2 A} = \sqrt{1 - (\frac{\sqrt{13}}{7})^2} = \sqrt{1 - \frac{13}{49}} = \sqrt{\frac{36}{49}} = \frac{6}{7}\) \(tg A = \frac{sin A}{cos A} = \frac{\frac{6}{7}}{\frac{\sqrt{13}}{7}} = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}\)
3. в) Дано: sin A = \(\frac{5}{13}\). Необходимо найти cos A и tg A. \(cos A = \sqrt{1 - sin^2 A} = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}\) \(tg A = \frac{sin A}{cos A} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}\)
4. г) Дано: sin A = 0.6. Необходимо найти cos A и tg A. \(cos A = \sqrt{1 - sin^2 A} = \sqrt{1 - 0.6^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8\) \(tg A = \frac{sin A}{cos A} = \frac{0.6}{0.8} = 0.75\)
5. д) Дано: sin A = \(\frac{\sqrt{7}}{4}\). Необходимо найти cos A и tg A. \(cos A = \sqrt{1 - sin^2 A} = \sqrt{1 - (\frac{\sqrt{7}}{4})^2} = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}\) \(tg A = \frac{sin A}{cos A} = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} = \frac{\sqrt{7}}{3}\)
6. e) Дано: cos A = \(\frac{7}{25}\). Необходимо найти sin A и tg A. \(sin A = \sqrt{1 - cos^2 A} = \sqrt{1 - (\frac{7}{25})^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}\) \(tg A = \frac{sin A}{cos A} = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7}\)

Ответ: a) cos A = \(\frac{4}{5}\), tg A = \(\frac{3}{4}\); б) sin A = \(\frac{6}{\sqrt{13}}\), tg A = \(\frac{6}{\sqrt{13}}\); в) cos A = \(\frac{12}{13}\), tg A = \(\frac{5}{12}\); г) cos A = 0.8, tg A = 0.75; д) cos A = \(\frac{3}{4}\), tg A = \(\frac{\sqrt{7}}{3}\); e) sin A = \(\frac{24}{25}\), tg A = \(\frac{24}{7}\)