4. Найдите синус, косинус, тангенс углов А и В прямоугольного треугольника АВС, если: а) АС = 3, AB = 5; 6) AC=10, ВС=8; в) ВС = 3√3, AB=6V2.

4.а) Дано: прямоугольный треугольник ABC, AC = 3, AB = 5. Найти sin A, cos A, tg A, sin B, cos B, tg B.

Решение:

По теореме Пифагора: $$BC^2 = AB^2 - AC^2 = 5^2 - 3^2 = 25 - 9 = 16$$

$$BC = \sqrt{16} = 4$$

$$sin A = \frac{BC}{AB} = \frac{4}{5} = 0.8$$

$$cos A = \frac{AC}{AB} = \frac{3}{5} = 0.6$$

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

$$sin B = \frac{AC}{AB} = \frac{3}{5} = 0.6$$

$$cos B = \frac{BC}{AB} = \frac{4}{5} = 0.8$$

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

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

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4.б) Дано: прямоугольный треугольник ABC, AC = 10, BC = 8. Найти sin A, cos A, tg A, sin B, cos B, tg B.

Решение:

По теореме Пифагора: $$AB^2 = AC^2 + BC^2 = 10^2 + 8^2 = 100 + 64 = 164$$

$$AB = \sqrt{164} = 2\sqrt{41} \approx 12.81$$

$$sin A = \frac{BC}{AB} = \frac{8}{2\sqrt{41}} = \frac{4}{\sqrt{41}} \approx 0.62$$

$$cos A = \frac{AC}{AB} = \frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}} \approx 0.78$$

$$tg A = \frac{BC}{AC} = \frac{8}{10} = 0.8$$

$$sin B = \frac{AC}{AB} = \frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}} \approx 0.78$$

$$cos B = \frac{BC}{AB} = \frac{8}{2\sqrt{41}} = \frac{4}{\sqrt{41}} \approx 0.62$$

$$tg B = \frac{AC}{BC} = \frac{10}{8} = 1.25$$

Ответ: sin A ≈ 0.62, cos A ≈ 0.78, tg A = 0.8, sin B ≈ 0.78, cos B ≈ 0.62, tg B = 1.25

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

Решение:

По теореме Пифагора: $$AC^2 = AB^2 - BC^2 = (6\sqrt{2})^2 - (3\sqrt{3})^2 = 36 \cdot 2 - 9 \cdot 3 = 72 - 27 = 45$$

$$AC = \sqrt{45} = 3\sqrt{5}$$

$$sin A = \frac{BC}{AB} = \frac{3\sqrt{3}}{6\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{6}}{4} \approx 0.61$$

$$cos A = \frac{AC}{AB} = \frac{3\sqrt{5}}{6\sqrt{2}} = \frac{\sqrt{5}}{2\sqrt{2}} = \frac{\sqrt{10}}{4} \approx 0.79$$

$$tg A = \frac{BC}{AC} = \frac{3\sqrt{3}}{3\sqrt{5}} = \frac{\sqrt{3}}{\sqrt{5}} = \frac{\sqrt{15}}{5} \approx 0.77$$

$$sin B = \frac{AC}{AB} = \frac{3\sqrt{5}}{6\sqrt{2}} = \frac{\sqrt{5}}{2\sqrt{2}} = \frac{\sqrt{10}}{4} \approx 0.79$$

$$cos B = \frac{BC}{AB} = \frac{3\sqrt{3}}{6\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{6}}{4} \approx 0.61$$

$$tg B = \frac{AC}{BC} = \frac{3\sqrt{5}}{3\sqrt{3}} = \frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{15}}{3} \approx 1.29$$

Ответ: sin A ≈ 0.61, cos A ≈ 0.79, tg A ≈ 0.77, sin B ≈ 0.79, cos B ≈ 0.61, tg B ≈ 1.29