Question

The vertices of a tetrahedron are four vertices of a cube symmetrically chosen so that no two are adjacent. Show that the angle between the vertices of a tetrahedron is $\mathbf{109}{\mathbf{.5}}^{\mathbf{\circ }}$ .

Short answer

Proved that Angle between Vertices of a tetrahedron is ${109.5}^0$.

Step-by-step solution

Definition of tetrahedron

A tetrahedron also referred to as a triangle pyramid in geometry, is a polyhedron with four triangular faces, six straight edges, and four vertex corners.

Step-2: Basic use of trigonometry and Pythagoras Theorem.

In triangle AOD, assuming the Cube has two-unit-length-sides.

$\begin{array}{c}\mathrm{OD}=1\\ \mathrm{AD}=\sqrt{1^2+1^2}\\ =\sqrt{2}\\ \mathrm{AO}=\sqrt{{\mathrm{AD}}^2+{\mathrm{OD}}^2}\\ =\sqrt{{\left(\sqrt{2}\right)}^2+1^2}\\ =\sqrt{3}\end{array}$

Using Trigonometry,

$\begin{array}{c}\mathrm{Cos}\angle \mathrm{AOD}=\frac{1}{\sqrt{3}}\\ =\frac{\sqrt{3}}{3}\end{array}$

$\begin{array}{c}\angle \mathrm{AOD}={\cos }^{-1}\frac{\sqrt{3}}{3}\\ =54{.74}^0\end{array}$

So, the Required Angle

$\begin{array}{c}\angle \mathrm{AOB}=2\angle \mathrm{AOD}\\ =2\times 54{.74}^0\\ =109{.5}^0\end{array}$

Hence it is proved that Angle between Vertices of a tetrahedron is ${109.5}^0$.