If the vertices of a triangle are (1, 4, 2), (–2, 1, 2), (2, –3, 4) then the angles are

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$\cos^{- 1} \frac{1}{\sqrt{3}}, 90^{0}, \cos^{- 1} \sqrt{\frac{2}{3}}$

$90^{0}, 120^{0}, 30^{0}$

$\cos^{- 1} (\frac{1}{\sqrt[]{5}}), 90^{0}, \cos^{- 1} \frac{\sqrt{5}}{3}$

$30^{0}, 60^{0}, 90^{0}$

answer is A.

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Detailed Solution

$A (1, 4, 2) B (- 2, 1, 2) C (2, - 3, 4)$

dr's of $AB = (3, 3, 0) = (1, 1, 0)$

dr's of $BC = (- 4, 4, - 2) = (- 2, 2, - 1)$

dr's of $AC = (- 1, 7, - 2)$

$\begin{array}{rcl} cosA & = & \frac{AB \cdot AC}{| AB | | AC |} \\ = & \frac{- 1 + 7 - 0}{\sqrt{2} \sqrt{54}} \\ = & \frac{6}{6 \sqrt{3}} \\ cosA & = & \frac{1}{\sqrt{3}} \\ A & = & \cos^{- 1} \frac{1}{\sqrt{3}} \end{array}$

$\begin{array}{rcl} cosB & = & \frac{AB \cdot BC}{| AB | | BC |} \\ = & - 2 + 2 + 0 \\ = & 0 \\ B & = & \frac{π}{2} \end{array}$

$\begin{array}{rcl} cosC & = & \frac{AC \cdot BC}{| AC | | BC |} \\ = & \frac{2 + 14 + 2}{3 \sqrt{54}} \\ = & \frac{18}{9 \sqrt{6}} \\ = & \frac{\sqrt{2}}{3} \\ C & = & \cos^{- 1} \sqrt{\frac{2}{3}} \end{array}$