4. Найдите синус, косинус, тангенс углов А и В прямоугольного треугольника АВС, если: а) AC = 3, AB = 5; 6) AC=10, BC=8; в) ВС = 3√3, AB=6√2.

Ответ: a) sin(A) = 3/5, cos(A) = 4/5, tan(A) = 3/4; sin(B) = 4/5, cos(B) = 3/5, tan(B) = 4/3; б) sin(A) = 8/\(2\sqrt{41}\), cos(A) = 10/\(2\sqrt{41}\), tan(A) = 4/5; sin(B) = 5/\(\sqrt{41}\), cos(B) = 4/\(\sqrt{41}\), tan(B) = 5/4; в) sin(A) = 1/2, cos(A) = \(\sqrt{3}\)/2, tan(A) = 1/\(\sqrt{3}\); sin(B) = \(\sqrt{3}\)/2, cos(B) = 1/2, tan(B) = \(\sqrt{3}\)

a) Дано: \(AC = 3\), \(AB = 5\). Треугольник ABC прямоугольный, поэтому BC - катет, AB - гипотенуза. Сначала найдем BC по теореме Пифагора:

\[BC = \sqrt{AB^2 - AC^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\]

• \(\sin(A) = \frac{AC}{AB} = \frac{3}{5}\)
• \(\cos(A) = \frac{BC}{AB} = \frac{4}{5}\)
• \(\tan(A) = \frac{AC}{BC} = \frac{3}{4}\)
• \(\sin(B) = \frac{BC}{AB} = \frac{4}{5}\)
• \(\cos(B) = \frac{AC}{AB} = \frac{3}{5}\)
• \(\tan(B) = \frac{BC}{AC} = \frac{4}{3}\)

б) Дано: \(AC = 10\), \(BC = 8\). Сначала найдем AB по теореме Пифагора:

\[AB = \sqrt{AC^2 + BC^2} = \sqrt{10^2 + 8^2} = \sqrt{100 + 64} = \sqrt{164} = 2\sqrt{41}\]

• \(\sin(A) = \frac{BC}{AB} = \frac{8}{2\sqrt{41}} = \frac{4}{\sqrt{41}}\)
• \(\cos(A) = \frac{AC}{AB} = \frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}\)
• \(\tan(A) = \frac{BC}{AC} = \frac{8}{10} = \frac{4}{5}\)
• \(\sin(B) = \frac{AC}{AB} = \frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}\)
• \(\cos(B) = \frac{BC}{AB} = \frac{8}{2\sqrt{41}} = \frac{4}{\sqrt{41}}\)
• \(\tan(B) = \frac{AC}{BC} = \frac{10}{8} = \frac{5}{4}\)

в) Дано: \(BC = 3\sqrt{3}\), \(AB = 6\sqrt{2}\). Сначала найдем AC по теореме Пифагора:

\[AC = \sqrt{AB^2 - BC^2} = \sqrt{(6\sqrt{2})^2 - (3\sqrt{3})^2} = \sqrt{72 - 27} = \sqrt{45} = 3\sqrt{5}\]

Проверим, что треугольник прямоугольный

• \(\sin(A) = \frac{BC}{AB} = \frac{3\sqrt{3}}{6\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}}\)
• \(\cos(A) = \frac{AC}{AB} = \frac{3\sqrt{5}}{6\sqrt{2}} = \frac{\sqrt{5}}{2\sqrt{2}}\)

Стороны указаны неверно, т.к. \(AB^2
eq AC^2 + BC^2\) Треугольник не прямоугольный.

Предположим, что AB - гипотенуза, а BC и AC - катеты, тогда:

\[AC=\sqrt{(6\sqrt{2})^2 - (3\sqrt{3})^2} = \sqrt{72 - 27} = \sqrt{45} = 3 \sqrt{5}\]

Вычислим углы в предположении, что угол C прямой:

\[\sin B = \frac{3 \sqrt{5}}{6 \sqrt{2}} = \frac{\sqrt{5}}{2 \sqrt{2}}\]

\[\cos B = \frac{3 \sqrt{3}}{6 \sqrt{2}} = \frac{\sqrt{3}}{2 \sqrt{2}}\]

Данных недостаточно, чтобы решить задачу.

Предположим, что \(AC = 3\), \(BC = 3\sqrt{3}\), \(AB = 6\), где AB - гипотенуза:

• \(\sin(A) = \frac{BC}{AB} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}\)
• \(\cos(A) = \frac{AC}{AB} = \frac{3}{6} = \frac{1}{2}\)
• \(\tan(A) = \frac{BC}{AC} = \frac{3\sqrt{3}}{3} = \sqrt{3}\)
• \(\sin(B) = \frac{AC}{AB} = \frac{3}{6} = \frac{1}{2}\)
• \(\cos(B) = \frac{BC}{AB} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}\)
• \(\tan(B) = \frac{AC}{BC} = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}\)

Ответ: a) sin(A) = 3/5, cos(A) = 4/5, tan(A) = 3/4; sin(B) = 4/5, cos(B) = 3/5, tan(B) = 4/3; б) sin(A) = 8/\(2\sqrt{41}\), cos(A) = 10/\(2\sqrt{41}\), tan(A) = 4/5; sin(B) = 5/\(\sqrt{41}\), cos(B) = 4/\(\sqrt{41}\), tan(B) = 5/4; в) sin(A) = 1/2, cos(A) = \(\sqrt{3}\)/2, tan(A) = 1/\(\sqrt{3}\); sin(B) = \(\sqrt{3}\)/2, cos(B) = 1/2, tan(B) = \(\sqrt{3}\)