What is the angle subtended by an edge of a r | vedclass

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(C) Let the side length of the regular tetrahedron be $a$.Let $G$ be the centre (centroid) of the regular tetrahedron and $A, B, C$ be the vertices.Th

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$\cos ^{-1}\left(\frac{-1}{3}\right)$

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(C) Let the side length of the regular tetrahedron be $a$.Let $G$ be the centre (centroid) of the regular tetrahedron and $A, B, C$ be the vertices.The distance from the centre $G$ to any vertex of a regular tetrahedron with side $a$ is given by $R = \frac{\sqrt{6}}{4}a$.Consider the triangle formed by the centre $G$ and two vertices $A$ and $B$ of an edge of length $a$.In $	riangle GAB$,the sides are $GA = GB = R = \frac{\sqrt{6}}{4}a$ and $AB = a$.Using the law of cosines in $	riangle GAB$ for the angle $	heta = \angle AGB$:$\cos 	heta = \frac{GA^2 + GB^2 - AB^2}{2 \cdot GA \cdot GB}$$\cos 	heta = \frac{(\frac{\sqrt{6}}{4}a)^2 + (\frac{\sqrt{6}}{4}a)^2 - a^2}{2 \cdot (\frac{\sqrt{6}}{4}a) \cdot (\frac{\sqrt{6}}{4}a)}$$\cos 	heta = \frac{\frac{6}{16}a^2 + \frac{6}{16}a^2 - a^2}{2 \cdot \frac{6}{16}a^2}$$\cos 	heta = \frac{\frac{12}{16}a^2 - a^2}{\frac{12}{16}a^2} = \frac{\frac{3}{4}a^2 - a^2}{\frac{3}{4}a^2}$$\cos 	heta = \frac{-\frac{1}{4}a^2}{\frac{3}{4}a^2} = -\frac{1}{3}$Therefore,$	heta = \cos ^{-1}\left(\frac{-1}{3}ight)$.

What is the angle subtended by an edge of a regular tetrahedron at its centre?

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