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### 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

Bryan
```

```
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:

A
^
/:\
/ : \
/  :  \
/   D'  \
/    :    \
B/_____:_____\C
/\ .   :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).

E
^
/ \
/   \
/     \
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! http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Physics/Chemistry
High School Polyhedra

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