4. Найдите синус, косинус, тангенс углов А и В прямоугольного треугольника АВС, если: а) AC = 4, AB = 5; б) АС=15, ВС=8; в) ВС = 6√3, AB=9√2.

4.а) Дано: прямоугольный треугольник ABC, AC = 4, AB = 5. Найти sin A, cos A, tg A, sin B, cos B, tg B.

Решение:

По теореме Пифагора: $$BC^2 = AB^2 - AC^2 = 5^2 - 4^2 = 25 - 16 = 9$$

$$BC = \sqrt{9} = 3$$

$$sin A = \frac{BC}{AB} = \frac{3}{5} = 0.6$$

$$cos A = \frac{AC}{AB} = \frac{4}{5} = 0.8$$

$$tg A = \frac{BC}{AC} = \frac{3}{4} = 0.75$$

$$sin B = \frac{AC}{AB} = \frac{4}{5} = 0.8$$

$$cos B = \frac{BC}{AB} = \frac{3}{5} = 0.6$$

$$tg B = \frac{AC}{BC} = \frac{4}{3} = 1.(3)$$

Ответ: sin A = 0.6, cos A = 0.8, tg A = 0.75, sin B = 0.8, cos B = 0.6, tg B = 1.(3)

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4.б) Дано: прямоугольный треугольник ABC, AC = 15, BC = 8. Найти sin A, cos A, tg A, sin B, cos B, tg B.

Решение:

По теореме Пифагора: $$AB^2 = AC^2 + BC^2 = 15^2 + 8^2 = 225 + 64 = 289$$

$$AB = \sqrt{289} = 17$$

$$sin A = \frac{BC}{AB} = \frac{8}{17} \approx 0.47$$

$$cos A = \frac{AC}{AB} = \frac{15}{17} \approx 0.88$$

$$tg A = \frac{BC}{AC} = \frac{8}{15} \approx 0.53$$

$$sin B = \frac{AC}{AB} = \frac{15}{17} \approx 0.88$$

$$cos B = \frac{BC}{AB} = \frac{8}{17} \approx 0.47$$

$$tg B = \frac{AC}{BC} = \frac{15}{8} = 1.875$$

Ответ: sin A ≈ 0.47, cos A ≈ 0.88, tg A ≈ 0.53, sin B ≈ 0.88, cos B ≈ 0.47, tg B = 1.875

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4.в) Дано: прямоугольный треугольник ABC, BC = 6√3, AB = 9√2. Найти sin A, cos A, tg A, sin B, cos B, tg B.

Решение:

По теореме Пифагора: $$AC^2 = AB^2 - BC^2 = (9\sqrt{2})^2 - (6\sqrt{3})^2 = 81 \cdot 2 - 36 \cdot 3 = 162 - 108 = 54$$

$$AC = \sqrt{54} = 3\sqrt{6}$$

$$sin A = \frac{BC}{AB} = \frac{6\sqrt{3}}{9\sqrt{2}} = \frac{2\sqrt{3}}{3\sqrt{2}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3} \approx 0.82$$

$$cos A = \frac{AC}{AB} = \frac{3\sqrt{6}}{9\sqrt{2}} = \frac{\sqrt{6}}{3\sqrt{2}} = \frac{\sqrt{3}}{3} \approx 0.58$$

$$tg A = \frac{BC}{AC} = \frac{6\sqrt{3}}{3\sqrt{6}} = \frac{2\sqrt{3}}{\sqrt{6}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.41$$

$$sin B = \frac{AC}{AB} = \frac{3\sqrt{6}}{9\sqrt{2}} = \frac{\sqrt{6}}{3\sqrt{2}} = \frac{\sqrt{3}}{3} \approx 0.58$$

$$cos B = \frac{BC}{AB} = \frac{6\sqrt{3}}{9\sqrt{2}} = \frac{2\sqrt{3}}{3\sqrt{2}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3} \approx 0.82$$

$$tg B = \frac{AC}{BC} = \frac{3\sqrt{6}}{6\sqrt{3}} = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \approx 0.71$$

Ответ: sin A ≈ 0.82, cos A ≈ 0.58, tg A ≈ 1.41, sin B ≈ 0.58, cos B ≈ 0.82, tg B ≈ 0.71