6 В ДАВС: А(1:2) B(3;4) C(5; - 2). Найдите cos cos LA, cos cos ∠B, Cos Cos ∠C.

Дано: $$\triangle ABC$$, $$A(1; 2)$$, $$B(-3; 4)$$, $$C(5; -2)$$. Найти: $$\cos(\angle A)$$, $$\cos(\angle B)$$, $$\cos(\angle C)$$. Решение: Найдем длины сторон треугольника: $$AB = \sqrt{(-3-1)^2 + (4-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$ $$BC = \sqrt{(5-(-3))^2 + (-2-4)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ $$AC = \sqrt{(5-1)^2 + (-2-2)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$$ По теореме косинусов: $$\cos(\angle A) = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} = \frac{20 + 32 - 100}{2 \cdot 2\sqrt{5} \cdot 4\sqrt{2}} = \frac{-48}{16\sqrt{10}} = \frac{-3}{\sqrt{10}} = -\frac{3\sqrt{10}}{10}$$ $$\cos(\angle B) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{20 + 100 - 32}{2 \cdot 2\sqrt{5} \cdot 10} = \frac{88}{40\sqrt{5}} = \frac{11}{5\sqrt{5}} = \frac{11\sqrt{5}}{25}$$ $$\cos(\angle C) = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{32 + 100 - 20}{2 \cdot 4\sqrt{2} \cdot 10} = \frac{112}{80\sqrt{2}} = \frac{14}{10\sqrt{2}} = \frac{7}{5\sqrt{2}} = \frac{7\sqrt{2}}{10}$$ Ответ: $$\cos(\angle A) = -\frac{3\sqrt{10}}{10}$$, $$\cos(\angle B) = \frac{11\sqrt{5}}{25}$$, $$\cos(\angle C) = \frac{7\sqrt{2}}{10}$$