Решение задачи 591
В прямоугольном треугольнике ABC (угол C = 90°):
- \( \sin A = \frac{BC}{AB} \)
- \( \cos A = \frac{AC}{AB} \)
- \( \operatorname{tg} A = \frac{BC}{AC} \)
- \( \sin B = \frac{AC}{AB} \)
- \( \cos B = \frac{BC}{AB} \)
- \( \operatorname{tg} B = \frac{AC}{BC} \)
а) BC = 8, AB = 17
Найдем AC по теореме Пифагора: \( AC^2 = AB^2 - BC^2 = 17^2 - 8^2 = 289 - 64 = 225 \), \( AC = \sqrt{225} = 15 \).
- \( \sin A = \frac{8}{17} \)
- \( \cos A = \frac{15}{17} \)
- \( \operatorname{tg} A = \frac{8}{15} \)
- \( \sin B = \frac{15}{17} \)
- \( \cos B = \frac{8}{17} \)
- \( \operatorname{tg} B = \frac{15}{8} \)
б) BC = 21, AC = 20
Найдем AB по теореме Пифагора: \( AB^2 = AC^2 + BC^2 = 20^2 + 21^2 = 400 + 441 = 841 \), \( AB = \sqrt{841} = 29 \).
- \( \sin A = \frac{21}{29} \)
- \( \cos A = \frac{20}{29} \)
- \( \operatorname{tg} A = \frac{21}{20} \)
- \( \sin B = \frac{20}{29} \)
- \( \cos B = \frac{21}{29} \)
- \( \operatorname{tg} B = \frac{20}{21} \)
в) BC = 1, AC = 2
Найдем AB по теореме Пифагора: \( AB^2 = AC^2 + BC^2 = 2^2 + 1^2 = 4 + 1 = 5 \), \( AB = \sqrt{5} \).
- \( \sin A = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \)
- \( \cos A = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \)
- \( \operatorname{tg} A = \frac{1}{2} \)
- \( \sin B = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \)
- \( \cos B = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \)
- \( \operatorname{tg} B = \frac{2}{1} = 2 \)
г) AC = 24, AB = 25
Найдем BC по теореме Пифагора: \( BC^2 = AB^2 - AC^2 = 25^2 - 24^2 = 625 - 576 = 49 \), \( BC = \sqrt{49} = 7 \).
- \( \sin A = \frac{7}{25} \)
- \( \cos A = \frac{24}{25} \)
- \( \operatorname{tg} A = \frac{7}{24} \)
- \( \sin B = \frac{24}{25} \)
- \( \cos B = \frac{7}{25} \)
- \( \operatorname{tg} B = \frac{24}{7} \)