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

6. Найдем косинусы углов треугольника с вершинами: \(A\{0;2\}, B\{3;7\}, C\{-1;5\}\)

Сначала найдем векторы, соответствующие сторонам треугольника:

$$\overrightarrow{AB} = B - A = \{3-0; 7-2\} = \{3; 5\}$$ $$\overrightarrow{BC} = C - B = \{-1-3; 5-7\} = \{-4; -2\}$$ $$\overrightarrow{CA} = A - C = \{0-(-1); 2-5\} = \{1; -3\}$$

Теперь найдем косинусы углов:

Угол A:

$$cos(A) = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| \cdot |\overrightarrow{AC}|}$$

\(\overrightarrow{AC} = -\overrightarrow{CA} = \{-1; 3\}\)

$$\overrightarrow{AB} \cdot \overrightarrow{AC} = 3 \cdot (-1) + 5 \cdot 3 = -3 + 15 = 12$$ $$|\overrightarrow{AB}| = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$$ $$|\overrightarrow{AC}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$$ $$cos(A) = \frac{12}{\sqrt{34} \cdot \sqrt{10}} = \frac{12}{\sqrt{340}} = \frac{12}{2\sqrt{85}} = \frac{6}{\sqrt{85}}$$

Угол B:

$$cos(B) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| \cdot |\overrightarrow{BC}|}$$

\(\overrightarrow{BA} = -\overrightarrow{AB} = \{-3; -5\}\)

$$\overrightarrow{BA} \cdot \overrightarrow{BC} = (-3) \cdot (-4) + (-5) \cdot (-2) = 12 + 10 = 22$$ $$|\overrightarrow{BA}| = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$$ $$|\overrightarrow{BC}| = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$ $$cos(B) = \frac{22}{\sqrt{34} \cdot \sqrt{20}} = \frac{22}{\sqrt{680}} = \frac{22}{2\sqrt{170}} = \frac{11}{\sqrt{170}}$$

Угол C:

$$cos(C) = \frac{\overrightarrow{CB} \cdot \overrightarrow{CA}}{|\overrightarrow{CB}| \cdot |\overrightarrow{CA}|}$$

\(\overrightarrow{CB} = -\overrightarrow{BC} = \{4; 2\}\)

$$\overrightarrow{CB} \cdot \overrightarrow{CA} = 4 \cdot 1 + 2 \cdot (-3) = 4 - 6 = -2$$ $$|\overrightarrow{CB}| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$ $$|\overrightarrow{CA}| = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$$ $$cos(C) = \frac{-2}{\sqrt{20} \cdot \sqrt{10}} = \frac{-2}{\sqrt{200}} = \frac{-2}{10\sqrt{2}} = -\frac{1}{5\sqrt{2}} = -\frac{\sqrt{2}}{10}$$

Ответ:

• $$cos(A) = \frac{6}{\sqrt{85}}$$
• $$cos(B) = \frac{11}{\sqrt{170}}$$
• $$cos(C) = -\frac{\sqrt{2}}{10}$$