a) Дано: AC = 3, AB = 5. Так как не указано, какой угол прямой, предположим, что угол C прямой (угол C = 90°). Тогда AB - гипотенуза. Найдем BC по теореме Пифагора:
$$BC = \sqrt{AB^2 - AC^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$$
Теперь найдем синус, косинус и тангенс углов A и B:
$$\sin A = \frac{BC}{AB} = \frac{4}{5}$$
$$\cos A = \frac{AC}{AB} = \frac{3}{5}$$
$$\tan A = \frac{BC}{AC} = \frac{4}{3}$$
$$\sin B = \frac{AC}{AB} = \frac{3}{5}$$
$$\cos B = \frac{BC}{AB} = \frac{4}{5}$$
$$\tan B = \frac{AC}{BC} = \frac{3}{4}$$
б) Дано: AC = 10, BC = 8. Предположим, что угол C прямой.
$$AB = \sqrt{AC^2 + BC^2} = \sqrt{10^2 + 8^2} = \sqrt{100 + 64} = \sqrt{164} = 2\sqrt{41}$$
$$\sin A = \frac{BC}{AB} = \frac{8}{2\sqrt{41}} = \frac{4}{\sqrt{41}} = \frac{4\sqrt{41}}{41}$$
$$\cos A = \frac{AC}{AB} = \frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}} = \frac{5\sqrt{41}}{41}$$
$$\tan A = \frac{BC}{AC} = \frac{8}{10} = \frac{4}{5}$$
$$\sin B = \frac{AC}{AB} = \frac{10}{2\sqrt{41}} = \frac{5\sqrt{41}}{41}$$
$$\cos B = \frac{BC}{AB} = \frac{8}{2\sqrt{41}} = \frac{4\sqrt{41}}{41}$$
$$\tan B = \frac{AC}{BC} = \frac{10}{8} = \frac{5}{4}$$
в) Дано: BC = 3√3, AB = 6√2. Предположим, что угол C прямой.
$$AC = \sqrt{AB^2 - BC^2} = \sqrt{(6\sqrt{2})^2 - (3\sqrt{3})^2} = \sqrt{72 - 27} = \sqrt{45} = 3\sqrt{5}$$
$$\sin A = \frac{BC}{AB} = \frac{3\sqrt{3}}{6\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{6}}{4}$$
$$\cos A = \frac{AC}{AB} = \frac{3\sqrt{5}}{6\sqrt{2}} = \frac{\sqrt{5}}{2\sqrt{2}} = \frac{\sqrt{10}}{4}$$
$$\tan A = \frac{BC}{AC} = \frac{3\sqrt{3}}{3\sqrt{5}} = \frac{\sqrt{3}}{\sqrt{5}} = \frac{\sqrt{15}}{5}$$
$$\sin B = \frac{AC}{AB} = \frac{3\sqrt{5}}{6\sqrt{2}} = \frac{\sqrt{10}}{4}$$
$$\cos B = \frac{BC}{AB} = \frac{3\sqrt{3}}{6\sqrt{2}} = \frac{\sqrt{6}}{4}$$
$$\tan B = \frac{AC}{BC} = \frac{3\sqrt{5}}{3\sqrt{3}} = \frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{15}}{3}$$