Tetrahedron

Geometry Level 4

The angle subtended by an edge of a regular tetrahedron at its center is

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3 π 8 \frac{3\pi}{8} cos 1 ( 1 3 ) \cos^{-1}\left(\frac{-1}{3}\right) π 4 \frac{\pi}{4} cos 1 ( 1 3 ) \cos^{-1}\left(\frac{-1}{\sqrt 3}\right)

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1 solution

WLOG a regular tetrahedron can be looked upon as formed by connecting six face diagonals of an unit cube, each length= 2 . \sqrt2 .
The cube and the tetrahedron share the same center. Thus the distances of the center and all cube vertices= 3 2 \dfrac{\sqrt3} 2 .
This is also true for tetrahedron, and is the half solid diagonal.
Thus angle 'A' by an tetrahedron edge at the center is the vertex angle of the isosceles Δ 3 2 , 3 2 , 2 . \Delta~~\dfrac{\sqrt3} 2,\dfrac{\sqrt3} 2, \sqrt2.
By Cos Rule, A = C o s 1 ( 3 2 ) 2 + ( 3 2 ) 2 ( 2 ) 2 2 ( 3 2 ) ( 3 2 ) = C o s 1 ( 1 3 ) . A=Cos^{-1}\dfrac {\Big (\dfrac{\sqrt3} 2 \Big )^2+ \Big (\dfrac{\sqrt3} 2 \Big )^2- \Big (\sqrt2 \Big )^2}{2* \Big (\dfrac{\sqrt3} 2 \Big )* \Big (\dfrac{\sqrt3} 2 \Big ) }= Cos^{-1} \Big (\dfrac {-1} 3 \Big ).

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