697. Найдите синус, косинус и тангенс углов A и B треугольника ABC с прямым углом C, если: а) BC = 8, AB = 17; б) BC = 21, AC = 20; в) BC = 1, AC = 2; г) AC = 24, AB = 25.

a) В прямоугольном треугольнике ABC, где угол C - прямой, BC = 8 и AB = 17. Нужно найти синус, косинус и тангенс углов A и B. \(sin A = \frac{BC}{AB} = \frac{8}{17}\) \(cos A = \frac{AC}{AB}\). Чтобы найти AC, используем теорему Пифагора: \(AC^2 + BC^2 = AB^2\), отсюда \(AC = \sqrt{AB^2 - BC^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\). \(cos A = \frac{15}{17}\) \(tan A = \frac{BC}{AC} = \frac{8}{15}\) \(sin B = \frac{AC}{AB} = \frac{15}{17}\) \(cos B = \frac{BC}{AB} = \frac{8}{17}\) \(tan B = \frac{AC}{BC} = \frac{15}{8}\) б) BC = 21, AC = 20. Найдем AB по теореме Пифагора: \(AB = \sqrt{AC^2 + BC^2} = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29\). \(sin A = \frac{BC}{AB} = \frac{21}{29}\) \(cos A = \frac{AC}{AB} = \frac{20}{29}\) \(tan A = \frac{BC}{AC} = \frac{21}{20}\) \(sin B = \frac{AC}{AB} = \frac{20}{29}\) \(cos B = \frac{BC}{AB} = \frac{21}{29}\) \(tan B = \frac{AC}{BC} = \frac{20}{21}\) в) BC = 1, AC = 2. Найдем AB: \(AB = \sqrt{AC^2 + BC^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\). \(sin A = \frac{BC}{AB} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\) \(cos A = \frac{AC}{AB} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\) \(tan A = \frac{BC}{AC} = \frac{1}{2}\) \(sin B = \frac{AC}{AB} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\) \(cos B = \frac{BC}{AB} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\) \(tan B = \frac{AC}{BC} = \frac{2}{1} = 2\) г) AC = 24, AB = 25. Найдем BC: \(BC = \sqrt{AB^2 - AC^2} = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7\). \(sin A = \frac{BC}{AB} = \frac{7}{25}\) \(cos A = \frac{AC}{AB} = \frac{24}{25}\) \(tan A = \frac{BC}{AC} = \frac{7}{24}\) \(sin B = \frac{AC}{AB} = \frac{24}{25}\) \(cos B = \frac{BC}{AB} = \frac{7}{25}\) \(tan B = \frac{AC}{BC} = \frac{24}{7}\)