Question

The vertices of a triangle are $(2,3,5),(-1,3,2),(3,5,-2)$, then the angles are

• A. ${30}^{\circ },{30}^{\circ },{120}^{\circ }$
• B. ${\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right),{90}^{\circ },{\cos }^{-1}\left(\frac{\sqrt{5}}{\sqrt{3}}\right)$
• C. ${30}^{\circ },{60}^{\circ },{90}^{\circ }$
• D. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right),{90}^{\circ },{\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right),{90}^{\circ },{\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)$

Given vertices: $A(2,3,5);B(-1,3,2);C(3,5,-2)$

$AB=\sqrt{9+0+9}=3\sqrt{2}$

$BC=\sqrt{16+4+16}=6$

$AC=\sqrt{1+4+49}=3\sqrt{6}$

$\cos A=\frac{{CA}^2+{BA}^2-{BC}^2}{2CA\times BA}=\frac{18+54-36}{2\times 3\sqrt{2}\times 3\sqrt{6}}=\frac{36}{18\times 2\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\cos B=\frac{{AB}^2+{BC}^2-{AC}^2}{2AB\times BC}=\frac{18+36-54}{2.3\sqrt{2}\times 6}=0$

$\cos C=\frac{54+36-18}{2\times 3\sqrt{6}\times 6}=\frac{72}{36\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}$

$\therefore A={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right);B=90°;C={\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)$