6. Найдите косинус угла \(ACB\) треугольника \(ABC\) с координатами точек \(A(2; 8)\), \(B(-1; 5)\), \(C(3; 1)\).

\(\cos \angle ACB = \frac{\mathbf{CA} \cdot \mathbf{CB}}{\|\mathbf{CA}\| \|\mathbf{CB}\|}\). \(\mathbf{CA} = (2 - 3; 8 - 1) = (-1; 7)\), \(\mathbf{CB} = (-1 - 3; 5 - 1) = (-4; 4)\). \(\mathbf{CA} \cdot \mathbf{CB} = -1 \cdot -4 + 7 \cdot 4 = 4 + 28 = 32\). \(\|\mathbf{CA}\| = \sqrt{1^2 + 7^2} = \sqrt{50}\), \(\|\mathbf{CB}\| = \sqrt{4^2 + 4^2} = \sqrt{32}\). \(\cos \angle ACB = \frac{32}{\sqrt{50} \cdot \sqrt{32}} = \frac{32}{\sqrt{1600}} = \frac{32}{40} = 0.8\). Ответ: \(б\).