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