# Dihedral angles between tetrahedron faces from triangles' angles at the tip

Say I have a general tetrahedron, or in fact only the tip of a tetrahedron, being three triangles in 3-space sharing a single vertex and sharing edges pairwise.

Given that I only know the three angles inside those triangles at the single common vertex, is there a simple way to compute the three dihedral angles between each of the neighbouring triangles?

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I don't think so - to find the dihedral angles, you need edge lengths (or at least their ratios), and you need more than just the top angle –  SSumner Feb 26 '13 at 16:15
@SSummer Really? In the end the only thing I'm uncertain of is the bottom plane of the supposed tetrahedron and this plane's location shouldn't change the dihedral angles between the other three faces. The tip's angles should completely define the tip triangles' locations relative to each other, shouldn't it? If I'm wrong on this, you could make the counter-proof an answer. –  Christian Rau Feb 26 '13 at 16:19
Actually, I think you're right, and I found this calculator: had2know.com/academics/… But I have no idea how it works –  SSumner Feb 26 '13 at 16:22
@SSumner Thanks for this link. "But I have no idea how it works" - Well, it has the formulas written on the site (though I'm also not able to directly derive this from spherical trigonometry, but at least I got a formula). Maybe somebody comes up with a nice answer showing the actual derivation of those formulas. –  Christian Rau Feb 26 '13 at 16:26
As somebody who studied spherical trigonometry in high school (shows you how old I am!), I can say that all the basic facts you learned in plane geometry (Law of Sines, Law of Cosines, etc.) drop out of the basic facts for the trigonometry of a right spherical triangle. There are ten of them, sometimes called Napier’s Rules, and they’re harder to discover than to prove. For instance, Pythagorean Theorem comes out as $\cos c=\cos a\cos b$. If the Rules really go back to Napier, he was a really smart gentleman. –  Lubin Feb 26 '13 at 16:55

This is to address one of the comments in how to derive the formula for the dihedral angles.

Given a tip of a tetrahedron, label the 3 edges attached to it as $1, 2, 3$ in such a way when you view the tip within the tetrahedron, the edges $1, 2, 3$ are arranged in counterclockwise manner.

Let $\vec{v}_1$, $\vec{v}_2$, $\vec{v}_3$ be the 3 unit vectors pointing away from the tip along the direction of edge $1, 2, 3$ respectively. Let us look at one of edge, say edge $2$. It is in contact with two faces. One bounded by vectors $\vec{v}_1$ and $\vec{v}_2$. Another bounded by vectors $\vec{v}_2$ and $\vec{v}_3$. The inward normal vectors of these two faces are given by:

$$\vec{n}_{12} = \frac{\vec{v}_1 \times \vec{v}_2}{|\vec{v}_1 \times \vec{v}_2|} \,\,\,\text{ and }\,\,\, \vec{n}_{23} = \frac{\vec{v}_2 \times \vec{v}_3}{|\vec{v}_2 \times \vec{v}_3|}$$ In terms of the normal vectors, the dihedral angle at edge $2$, $\phi_2$, satisfies:

\begin{align} \cos(\phi_2) &= -\vec{n}_{12}\cdot\vec{n}_{23}\\ &= -\frac{( \vec{v}_1 \times \vec{v}_2 )\cdot(\vec{v}_2 \times \vec{v}_3)}{ |\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\\ &= -\frac{(\vec{v}_1\cdot\vec{v}_2)(\vec{v}_2\cdot\vec{v}_3) - (\vec{v}_1\cdot\vec{v}_3)|\vec{v}_2|^2}{|\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\\ &= \frac{\vec{v}_1\cdot\vec{v}_3 - (\vec{v}_1\cdot\vec{v}_2)(\vec{v}_2\cdot\vec{v}_3)}{|\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\tag{*} \end{align} Let $\theta_{ij}$ be the vertex angle between edge $i$ and $j$. We can simplify R.H.S of (*) to get: $$\cos(\phi_2) = \frac{\cos(\theta_{13}) - \cos(\theta_{12})\cos(\theta_{23})}{\sin(\theta_{12})\sin(\theta_{23})}$$

The formula of other dihedral angles can be derived in same manner.

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+1 Man, of course, the normal vectors with some vector arithmetics, nothing 'bout spherical trigonometry, me stupid! –  Christian Rau Feb 26 '13 at 23:32

With the tetrahedron having three triangular faces $A$, $B$, and $C$ with vertex angles $a$, $b$, and $c$, respectively, and dihedral angles between each face $\angle AB$, $\angle BC$, and $\angle AC$, the dihedral angles are given by:

$\angle AB = cos^{-1}(\frac{cos(c) - cos(b) cos(a)}{sin(b) sin(a)})$

$\angle BC = cos^{-1}(\frac{cos(a) - cos(c) cos(b)}{sin(c) sin(a)})$

$\angle AC = cos^{-1}(\frac{cos(b) - cos(c) cos(a)}{sin(c) sin(a)})$

Source

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Since you were the first one to come up with a solution, it hurts not giving you the credit of acceptance. But I have to make a decision and achille's answer is more elaborate. But +1, of course. –  Christian Rau Feb 27 '13 at 17:50
Of course! The more thorough answer should be at the top! I didn't come up with it anyway, just found it on another site. –  SSumner Feb 27 '13 at 23:26