6. Найдите косинусы углов треугольника с вершинами: A(1;2); B(-3;4); C(5;-2)

Найдем стороны треугольника: \[AB = \sqrt{(-3-1)^2 + (4-2)^2} = \sqrt{16 + 4} = \sqrt{20}\] \[BC = \sqrt{(5-(-3))^2 + (-2-4)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\] \[AC = \sqrt{(5-1)^2 + (-2-2)^2} = \sqrt{16 + 16} = \sqrt{32}\] Теперь найдем косинусы углов: Для угла A: \[\cos A = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} = \frac{20 + 32 - 100}{2 \cdot \sqrt{20} \cdot \sqrt{32}} = \frac{-48}{2 \sqrt{640}} = \frac{-24}{\sqrt{640}} = \frac{-24}{8\sqrt{10}} = \frac{-3}{\sqrt{10}}\] Для угла B: \[\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{20 + 100 - 32}{2 \cdot \sqrt{20} \cdot 10} = \frac{88}{20\sqrt{20}} = \frac{22}{5\sqrt{20}} = \frac{22}{10\sqrt{5}} = \frac{11}{5\sqrt{5}}\] Для угла C: \[\cos C = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{32 + 100 - 20}{2 \cdot \sqrt{32} \cdot 10} = \frac{112}{20\sqrt{32}} = \frac{28}{5\sqrt{32}} = \frac{28}{20\sqrt{2}} = \frac{7}{5\sqrt{2}}\] Ответ: \(\cos A = \frac{-3}{\sqrt{10}}\), \(\cos B = \frac{11}{5\sqrt{5}}\), \(\cos C = \frac{7}{5\sqrt{2}}\).