⑥. Найдите косинусы углов треугольника с вершинами: A(0;2); B(3;7); C(-1;5)

Дано:

• A(0;2)
• B(3;7)
• C(-1;5)

Найдем длины сторон треугольника:

\[AB = \sqrt{(3-0)^2 + (7-2)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}\]

\[BC = \sqrt{(-1-3)^2 + (5-7)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}\]

\[AC = \sqrt{(-1-0)^2 + (5-2)^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}\]

Теперь найдем косинусы углов, используя теорему косинусов:

\[\cos(\angle A) = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} = \frac{34 + 10 - 20}{2 \cdot \sqrt{34} \cdot \sqrt{10}} = \frac{24}{2 \cdot \sqrt{340}} = \frac{12}{\sqrt{340}}\]

\[\cos(\angle B) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{34 + 20 - 10}{2 \cdot \sqrt{34} \cdot \sqrt{20}} = \frac{44}{2 \cdot \sqrt{680}} = \frac{22}{\sqrt{680}}\]

\[\cos(\angle C) = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{10 + 20 - 34}{2 \cdot \sqrt{10} \cdot \sqrt{20}} = \frac{-4}{2 \cdot \sqrt{200}} = \frac{-2}{\sqrt{200}}\]

Упростим:

• \[\cos(\angle A) = \frac{12}{\sqrt{340}} = \frac{12}{2\sqrt{85}} = \frac{6}{\sqrt{85}} \approx 0.65\]
• \[\cos(\angle B) = \frac{22}{\sqrt{680}} = \frac{22}{2\sqrt{170}} = \frac{11}{\sqrt{170}} \approx 0.84\]
• \[\cos(\angle C) = \frac{-2}{\sqrt{200}} = \frac{-2}{10\sqrt{2}} = -\frac{1}{5\sqrt{2}} = -\frac{\sqrt{2}}{10} \approx -0.14\]

Ответ: \[\cos(\angle A) = \frac{6}{\sqrt{85}}\]; \[\cos(\angle B) = \frac{11}{\sqrt{170}}\]; \[\cos(\angle C) = -\frac{\sqrt{2}}{10}\]