⑥ В $$\triangle ABC$$: A(1;2) B(-3;4) C(5;-2). Найдите cos $$\angle$$A, cos $$\angle$$B, cos $$\angle$$C.

Даны координаты вершин треугольника ABC: A(1;2), B(-3;4), C(5;-2).

Найдем длины сторон треугольника:

$$AB = \sqrt{(-3-1)^2 + (4-2)^2} = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$ $$BC = \sqrt{(5-(-3))^2 + (-2-4)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ $$AC = \sqrt{(5-1)^2 + (-2-2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$$

Теперь найдем косинусы углов, используя теорему косинусов:

$$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle A)$$ $$\cos(\angle A) = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} = \frac{20 + 32 - 100}{2 \cdot 2\sqrt{5} \cdot 4\sqrt{2}} = \frac{-48}{16\sqrt{10}} = \frac{-3}{\sqrt{10}} = -\frac{3\sqrt{10}}{10}$$ $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle B)$$ $$\cos(\angle B) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{20 + 100 - 32}{2 \cdot 2\sqrt{5} \cdot 10} = \frac{88}{40\sqrt{5}} = \frac{11}{5\sqrt{5}} = \frac{11\sqrt{5}}{25}$$ $$AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\angle C)$$ $$\cos(\angle C) = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{32 + 100 - 20}{2 \cdot 4\sqrt{2} \cdot 10} = \frac{112}{80\sqrt{2}} = \frac{14}{10\sqrt{2}} = \frac{7}{5\sqrt{2}} = \frac{7\sqrt{2}}{10}$$

Ответ: $$\cos(\angle A) = -\frac{3\sqrt{10}}{10}$$, $$\cos(\angle B) = \frac{11\sqrt{5}}{25}$$, $$\cos(\angle C) = \frac{7\sqrt{2}}{10}$$