The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

The Angles of a Tetrahedron

Date: 07/08/98 at 16:04:20
From: Bryan J.
Subject: The tetrahedron

Hello Doctors,

If I'm not mistaken, the tetrahedron is a 3-D object that has 
equilateral triangles as its four faces. The angle from one vertex to 
the exact center of the tetrahedron to another vertex is around 
109.5 degrees. How is that? I have always wondered because the 
tetrahedral angle is encountered quite often in chemistry. Could you 
please show me how that angle measure is derived? Thanks a lot in 


Date: 07/22/98 at 10:46:53
From: Doctor Rick
Subject: Re: The tetrahedron

Hi, Bryan! Good question. Unlike a lot of numbers in chemistry (atomic
weights and such), that strange value of 109.5 degrees can be derived
geometrically. It isn't just an experimental measurement. I'll sketch 
out a derivation of the tetrahedral angle for you.

It isn't easy to visualize what goes on in the middle of a 
tetrahedron. You could start by making a model. Cut out this triangle 
and fold it up on the solid lines BC, CD, and DB. Tape the edges 
together so the points labeled A meet:

              / : \
             /  :  \
            /   D'  \
           /    :    \
         /\ .   :E  . /\
        /  \   .:.   /  \
       /    \ . A'. /    \
      /    .F\  :  /G.    \
     /  .C'   \ : /   B'.  \
   A            D            A

The center of the tetrahedron is the intersection of the lines A-A', 
B-B', C-C', and D-D' (where B', C', and D' are like A', the 
intersections of the medians of their triangles). Let's look at a 
cross-section through AED, since it contains the center (X in the 
figure below).

               / \
              /   \
             /     \
          D'/       \ A'
           /\       /\
          /:  \ X /   \
         / :    o      \
        /  :  /   \     \
       /   :/       \    \
      /   /:          \   \
     /  /  :             \ \
   A       H                 D

You can prove that the length of the sides AE and ED is sqrt(3)/2 
(that's the square root of 3), if the edge of the tetrahedron (AD) 
is 1. You can also prove that D' and A' cut AE and DE 2/3 of the way 
to E. From this you can show that if you drop a perpendicular from D' 
to AD, you get a right triangle D'HD with legs D'H = sqrt(2)/3 and 
HD = 2/3. Therefore the angle HDD' is arctan(sqrt(1/2)). 

Now, the tetrahedral angle is:

   AXD = 2*(angle HFD) = 180 - 2*arctan(sqrt(1/2))

Remembering that tan(a) = 1/cot(a) = 1/tan(90-a), this can also be 
written as:

   2*arctan(sqrt(2)) = 109.4712206...

So now you know where that mysterious number comes from. I left out a 
lot of steps. If I confused you, please write back. 

- Doctor Rick, The Math Forum
Check out our web site!   
Associated Topics:
High School Geometry
High School Physics/Chemistry
High School Polyhedra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.