So I am interested in finding out how solid angle trigonometry works. Specifically, in 3 dimensional space, if we have three vectors reaching out from the origin, when we link the tips of the vectors together, we should get a tetrahedron. The idea is that since a tetrahedron is in some sense the three dimensional analog of a triangle, is there some way which we can use a 3D analog of trigonometry in this situation? Perhaps we could define the ratios of the areas of the sides of the tetrahedra, or functions as the lengths as functions of the solid angle between the three vectors?

Upon googling, I haven't gotten much insight (all I got was things like the definition of solid angle). Perhaps this is due to there being a proper term for this concept which I am unaware of? If someone can enlighten me to the proper name of the concept, or better yet, if someone could provide me with links to important literature on the subject, it would be greatly appreciated. Thanks!

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    $\begingroup$ Why this downvote ? $\endgroup$ – Yves Daoust Jan 29 '18 at 8:42

There is a trigonometry of tetrahedra (separate from spherical trigonometry others mention), although it doesn't deal with solid angles. It's dimensionally-enhanced in a way that relates areas and dihedral angles. I call this field "hedronometry", and I've referenced some key results on Math.SE from time to time. For instance, I discuss the Law(s!) of Cosines here; for completeness, I'll include them here:

Given a tetrahedron $OABC$, with faces of area $W$, $X$, $Y$, $Z$ opposite respective vertices $O$, $A$, $B$, $C$, and with $\angle PQ$ denoting the dihedral angle along edge $\overline{PQ}$, we have

The First Law of Cosines

$$W^2 = X^2 + Y^2 + Z^2 - 2 Y Z \cos \angle OA - 2 Z X \cos \angle OB - 2 X Y \cos \angle OC$$

In particular, for a "right-corner" tetrahedron with hypotenuse-face $W$, we have a Pythagorean Theorem, aka de Gua's Theorem: $$W^2 = X^2 + Y^2 + Z^2$$

The Second Law of Cosines $$\begin{eqnarray*} W^2 + X^2 - 2 W X \cos \angle BC &= H^2 =& Y^2 + Z^2 - 2 Y Z \cos \angle OA \\ W^2 + Y^2 - 2 W Y \cos \angle CA &= J^2 =& Z^2 + X^2 - 2 Z X \cos \angle OB \\ W^2 + Z^2 - 2 W Z \cos \angle AB &= K^2 =& X^2 + Y^2 - 2 X Y \cos \angle OC \end{eqnarray*}$$

Here, $H$, $J$, $K$ are (what I call) "pseudo-faces" of the tetrahedron. They're related to projections of the tetrahedron in planes parallel to opposite edges. In any case, combining these Laws yields this tidy result:

The Sum-of-Squares Identity $$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2$$

Note that, whereas a triangle is determined by its three side-lengths, a tetrahedron —which admits six degrees of freedom— is not determined by its four face-areas. The three pseudo-faces fill the gap (and the Sum-of-Squares Identity prevents them from over-filling it!), so "hedronometry" is about how all the metric properties of a tetrahedron can be expressed in terms of face and pseudo-face areas.

You can read more about this stuff on my Hedronometry pages (which desperately need a style overhaul). The more-recent entries deal with the hedronometry of hyperbolic space, so you'll probably want to go back in time a bit for Euclidean stuff. I intend, someday, to compile that information into a unified document.

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    $\begingroup$ Great! Thank you very much for your help :) $\endgroup$ – YiFan Jan 29 '18 at 9:37
  • $\begingroup$ That's really interesting and intriguing! please keep up with your Hedronometry pages $\endgroup$ – G Cab Oct 24 '18 at 16:41

Interestingly, you added the spherical-trigonometry tag, which is precisely what you are after.

The generalization of planar trigonometry, which places the angles on the unit circle, places solid angles on the unit sphere. Much like a planar angle is measured by the arc length along the circle, a solid angle is measured by an area over the sphere.

The core of spherical trigonometry is devoted to the resolution of spherical triangles (delimited by three planes meeting at the origin and a portion of the sphere). The sides of such triangles are also (planar) angles.

Refer to the Wikipedia entry on spherical trigonometry https://en.wikipedia.org/wiki/Spherical_trigonometry.

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  • $\begingroup$ Interesting. Could you explain how that relates to the areas of the sides of the tetrahedra? As far as I know, spherical geometry deals only with entities on the surface of the sphere, and doesn't relate to the areas/lengths of the tetrahedra? Especially since it isn't specified that the vectors must be of equal length, which is what seems to be the case in spherical geometry. Thanks for the help! $\endgroup$ – YiFan Jan 29 '18 at 8:52
  • $\begingroup$ @user496634: it doesn't fit in the scope of my answer to explain how a tetrahedron is solved by spherical trigonometry. Bring home that spherical trigonometry deals with angles between intersecting planes/lines in 3D, just like ordinary trigonometriy deals with planar angles in 2D. $\endgroup$ – Yves Daoust Jan 29 '18 at 8:56
  • $\begingroup$ Can I at least confirm it can be done? $\endgroup$ – YiFan Jan 29 '18 at 9:05
  • $\begingroup$ @user496634: yes it can, if you provide enough data to make the solution unique. $\endgroup$ – Yves Daoust Jan 29 '18 at 9:06
  • $\begingroup$ could you give me a direction on how to search for it? Especially, how would one find the ratio between the areas of two sides of the tetrahedron? A link maybe? Thank you a lot for your patience! $\endgroup$ – YiFan Jan 29 '18 at 9:09

It is called spherical trigonometry.


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