Решение:
1. **Найдем сторону BC по теореме Пифагора:**
(AB^2 = AC^2 + BC^2)
(25^2 = 20^2 + BC^2)
(625 = 400 + BC^2)
(BC^2 = 225)
(BC = \sqrt{225} = 15) см
2. **Найдем sin A:**
(sin A = \frac{BC}{AB} = \frac{15}{25} = \frac{3}{5})
3. **Найдем cos A:**
(cos A = \frac{AC}{AB} = \frac{20}{25} = \frac{4}{5})
4. **Найдем tg A:**
(tg A = \frac{BC}{AC} = \frac{15}{20} = \frac{3}{4})
5. **Найдем sin B:**
(sin B = \frac{AC}{AB} = \frac{20}{25} = \frac{4}{5})
6. **Найдем cos B:**
(cos B = \frac{BC}{AB} = \frac{15}{25} = \frac{3}{5})
7. **Найдем tg B:**
(tg B = \frac{AC}{BC} = \frac{20}{15} = \frac{4}{3})
Ответ:
(sin A = \frac{3}{5}), (cos A = \frac{4}{5}), (tg A = \frac{3}{4})
(sin B = \frac{4}{5}), (cos B = \frac{3}{5}), (tg B = \frac{4}{3})