Решение:
Для прямоугольного треугольника ABC с прямым углом C, синус, косинус и тангенс углов A и B определяются следующим образом:
- \( \sin A = \frac{BC}{AB} \), \( \cos A = \frac{AC}{AB} \), \( \operatorname{tg} A = \frac{BC}{AC} \)
- \( \sin B = \frac{AC}{AB} \), \( \cos B = \frac{BC}{AB} \), \( \operatorname{tg} B = \frac{AC}{BC} \)
Необходимо найти гипотенузу AB, если она не дана, по теореме Пифагора: \( AB = \sqrt{AC^2 + BC^2} \).
а) BC = 8, AB = 17
- \( AC = \sqrt{AB^2 - BC^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \)
- \( \sin A = \frac{8}{17} \), \( \cos A = \frac{15}{17} \), \( \operatorname{tg} A = \frac{8}{15} \)
- \( \sin B = \frac{15}{17} \), \( \cos B = \frac{8}{17} \), \( \operatorname{tg} B = \frac{15}{8} \)
б) BC = 21, AC = 20
- \( AB = \sqrt{AC^2 + BC^2} = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29 \)
- \( \sin A = \frac{21}{29} \), \( \cos A = \frac{20}{29} \), \( \operatorname{tg} A = \frac{21}{20} \)
- \( \sin B = \frac{20}{29} \), \( \cos B = \frac{21}{29} \), \( \operatorname{tg} B = \frac{20}{21} \)
в) BC = 1, AC = 2
- \( AB = \sqrt{AC^2 + BC^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \)
- \( \sin A = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \), \( \cos A = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \), \( \operatorname{tg} A = \frac{1}{2} \)
- \( \sin B = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \), \( \cos B = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \), \( \operatorname{tg} B = \frac{2}{1} = 2 \)
г) AC = 24, AB = 25
- \( BC = \sqrt{AB^2 - AC^2} = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7 \)
- \( \sin A = \frac{7}{25} \), \( \cos A = \frac{24}{25} \), \( \operatorname{tg} A = \frac{7}{24} \)
- \( \sin B = \frac{24}{25} \), \( \cos B = \frac{7}{25} \), \( \operatorname{tg} B = \frac{24}{7} \)
Ответ: а) \( \sin A = \frac{8}{17}, \cos A = \frac{15}{17}, \operatorname{tg} A = \frac{8}{15}; \sin B = \frac{15}{17}, \cos B = \frac{8}{17}, \operatorname{tg} B = \frac{15}{8} \). б) \( \sin A = \frac{21}{29}, \cos A = \frac{20}{29}, \operatorname{tg} A = \frac{21}{20}; \sin B = \frac{20}{29}, \cos B = \frac{21}{29}, \operatorname{tg} B = \frac{20}{21} \). в) \( \sin A = \frac{\sqrt{5}}{5}, \cos A = \frac{2\sqrt{5}}{5}, \operatorname{tg} A = \frac{1}{2}; \sin B = \frac{2\sqrt{5}}{5}, \cos B = \frac{\sqrt{5}}{5}, \operatorname{tg} B = 2 \). г) \( \sin A = \frac{7}{25}, \cos A = \frac{24}{25}, \operatorname{tg} A = \frac{7}{24}; \sin B = \frac{24}{25}, \cos B = \frac{7}{25}, \operatorname{tg} B = \frac{24}{7} \).