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Chapter 10: Q31E (page 468)

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 109.5∘ .

Short Answer

Expert verified

Proved that Angle between Vertices of a tetrahedron is 109.50.

Step by step solution

01

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.

02

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

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

OD=1AD=12+12=2AO=AD2+OD2=(2)2+12=3

Using Trigonometry,

Cos∠AOD=13=33

∠AOD=cos−133=54.740

So, the Required Angle

∠AOB=2∠AOD=2×54.740=109.50

Hence it is proved that Angle between Vertices of a tetrahedron is 109.50.

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