Question

Даны точки а(1; 1; 1) b(2; 13; 1) c(14; 1; 1) найти углы и стороны треугольника abc

Answer 1

AB = (1, 12, 0)
$ab = \sqrt{(2 - 1)^{2} + (13 - 1)^{2} + (1 - 1)^{2} } = \sqrt{1 + 144} = \sqrt{145}$
BC = (12,-12,0)
$bc = \sqrt{ {(14 - 2)}^{2} + {(1 - 13)}^{2} + {(1 - 1)}^{2} } = \sqrt{144 + 144} = 12 \sqrt{2}$
AC = (13,0,0)
$ac = \sqrt{(14 - 1)^{2} + \sqrt{(1 - 1)^{2} + {(1 - 1)}^{2} } } = \sqrt{169} = 13$
$\cos( \alpha ) = \frac{ab \times ac}{ \sqrt{145} \times 13} = \frac{13 }{ \sqrt{145} \times 13 } = \frac{1}{ \sqrt{145} }$
$\cos( \gamma ) = \frac{ac \times bc}{12 \sqrt{2} \times 13} = \frac{1}{ \sqrt{2} }$
$\cos( \beta ) = \frac{ab \times bc}{ \sqrt{145} \times 12 \sqrt{2} } = \frac{12 - 144}{ \sqrt{145} \times 12 \sqrt{2} } = - \frac{11}{ \sqrt{290} }$