Найдите синус, косинус и тангенс углов А и В треугольника АВС с прямым углом С, если: а) ВС=8, AB=17; 6) BC=21, АС = 20; в) ВС = 1, АС = 2; г) АС = 24, АВ = 25.

Ответ: а) sin A = 20/29, cos A = 21/29, tg A = 20/21, sin B = 21/29, cos B = 20/29, tg B = 21/20

\[AC = \sqrt{AB^2 - BC^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\]\[sin A = \frac{BC}{AB} = \frac{8}{17}\]\[cos A = \frac{AC}{AB} = \frac{15}{17}\]\[tg A = \frac{BC}{AC} = \frac{8}{15}\]\[sin B = \frac{AC}{AB} = \frac{15}{17}\]\[cos B = \frac{BC}{AB} = \frac{8}{17}\]\[tg B = \frac{AC}{BC} = \frac{15}{8}\]\[AB = \sqrt{AC^2 + BC^2} = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29\]\[sin A = \frac{BC}{AB} = \frac{21}{29}\]\[cos A = \frac{AC}{AB} = \frac{20}{29}\]\[tg A = \frac{BC}{AC} = \frac{21}{20}\]\[sin B = \frac{AC}{AB} = \frac{20}{29}\]\[cos B = \frac{BC}{AB} = \frac{21}{29}\]\[tg B = \frac{AC}{BC} = \frac{20}{21}\]\[AB = \sqrt{AC^2 + BC^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\]\[sin A = \frac{BC}{AB} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\]\[cos A = \frac{AC}{AB} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\]\[tg A = \frac{BC}{AC} = \frac{1}{2}\]\[sin B = \frac{AC}{AB} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\]\[cos B = \frac{BC}{AB} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\]\[tg B = \frac{AC}{BC} = \frac{2}{1} = 2\]\[BC = \sqrt{AB^2 - AC^2} = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7\]\[sin A = \frac{BC}{AB} = \frac{7}{25}\]\[cos A = \frac{AC}{AB} = \frac{24}{25}\]\[tg A = \frac{BC}{AC} = \frac{7}{24}\]\[sin B = \frac{AC}{AB} = \frac{24}{25}\]\[cos B = \frac{BC}{AB} = \frac{7}{25}\]\[tg B = \frac{AC}{BC} = \frac{24}{7}\]

Ответ: а) sin A = 20/29, cos A = 21/29, tg A = 20/21, sin B = 21/29, cos B = 20/29, tg B = 21/20