The Angles of a TetrahedronDate: 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 advance. 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/