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

a) Дано: \( BC = a = 8 \), \( AB = c = 17 \). Тогда \( AC = b = \sqrt{AB^2 - BC^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \). \( \sin A = \frac{a}{c} = \frac{8}{17} \), \( \cos A = \frac{b}{c} = \frac{15}{17} \), \( \tan A = \frac{a}{b} = \frac{8}{15} \). \( \sin B = \frac{b}{c} = \frac{15}{17} \), \( \cos B = \frac{a}{c} = \frac{8}{17} \), \( \tan B = \frac{b}{a} = \frac{15}{8} \). б) Дано: \( BC = a = 20, AC = b = 21 \). Тогда \( AB = c = \sqrt{BC^2 + AC^2} = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29 \). \( \sin A = \frac{a}{c} = \frac{20}{29} \), \( \cos A = \frac{b}{c} = \frac{21}{29} \), \( \tan A = \frac{a}{b} = \frac{20}{21} \). \( \sin B = \frac{b}{c} = \frac{21}{29} \), \( \cos B = \frac{a}{c} = \frac{20}{29} \), \( \tan B = \frac{b}{a} = \frac{21}{20} \). в) Дано: \( BC = a = 1, AC = b = 2 \). Тогда \( AB = c = \sqrt{BC^2 + AC^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \). \( \sin A = \frac{a}{c} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \), \( \cos A = \frac{b}{c} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \), \( \tan A = \frac{a}{b} = \frac{1}{2} \). \( \sin B = \frac{b}{c} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \), \( \cos B = \frac{a}{c} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \), \( \tan B = \frac{b}{a} = \frac{2}{1} = 2 \). г) Дано: \( AC = b = 24, AB = c = 25 \). Тогда \( BC = a = \sqrt{AB^2 - AC^2} = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7 \). \( \sin A = \frac{a}{c} = \frac{7}{25} \), \( \cos A = \frac{b}{c} = \frac{24}{25} \), \( \tan A = \frac{a}{b} = \frac{7}{24} \). \( \sin B = \frac{b}{c} = \frac{24}{25} \), \( \cos B = \frac{a}{c} = \frac{7}{25} \), \( \tan B = \frac{b}{a} = \frac{24}{7} \).