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Given two sides $a,b$ of a spherical triangle and the angle $C$ between them, the spherical law of cosines gives an elegant formula for the missing edge length $c$:

$$\cos c = \cos a \cos b+ \sin a \sin b \cos C.$$

I have a spherical quadrilateral and know the lengths of three consecutive edges $a,b,c$, and the angles between them $\theta_{ab}$ and $\theta_{bc}$. The last edge length (and its two adjacent angles) are unknown.

Is there an elegant formula for $\cos d$ in terms of the known lengths and angles, and preferably needing only the cosine (and not the sine) of $\theta_{ab}$ and $\theta_{bc}$?

I know it is possible to solve for $d$, for instance by arbitrarily diagonalizing the quad and then "cutting the ear" by solving for the new angles. But when I do this I get a horrible mess of trig, and am hoping that a simplified formula is known.

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  • $\begingroup$ It's a nice question, and unfortunately I don't have an answer. You could try the TrigSimplify command in Mathematica though and see if that helps "find" the missing formula. $\endgroup$ – nbubis Jul 19 '12 at 11:51
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After some manipulations, the nicest formula I've found so far is

$$\begin{align*} \cos d = \qquad &\cos a \cos b \cos c\\ +\ &\sin b\,\left(\sin a \cos c \cos \theta_{ab} + \cos a \sin c \cos \theta_{bc}\right)\\ +\ &\sin a \sin c\, \left(\sin \theta_{ab}\sin \theta_{bc} - \cos b \cos \theta_{ab} \cos \theta_{bc} \right).\end{align*}$$

I haven't checked my work too closesly, but the above does have the right symmetries, and reduces to the right formulas as any side length goes to zero.

I'm still hoping there are simplifications of the above, or a way of removing the sines of $\theta_{ab}$ and $\theta_{bc}$.

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  • $\begingroup$ You can't express this using only cosines because that would imply the same distance $d$ when one of the angles is negated, or replaced with $2\pi-\theta$ (resulting in a non-convex polygon); but clearly the distance is not the same in general. Consider a triangle, where $d=0$, deforming into a 'Z' shape, where $d\neq0$, while keeping all three lengths and one angle constant. $\endgroup$ – mr_e_man Aug 7 '20 at 1:26
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I was able to get the same result and then verify it using the tool GeoGebra. I haven't found any simpler expression either. I didn't understand the lingo "cutting the ear" at first but now see what you meant. For anyone else having trouble, here are the steps I used:

  1. Use the spherical law of cosines for sides to get $\cos l$, where $l$ is the diagonal opposite the angle $\theta_{ab}$.
  2. Use the spherical law of cosines for sides to get $\cos \theta_{bl}$, where $\theta_{bl}$ is the angle between the central edge, $b$, and the diagonal, $l$.
  3. Use the angle-difference relation to get $\cos \theta_{lc}$, where $\theta_{lc}$ is the angle between the diagonal, $l$, and the third edge, $c$.
  4. Use the spherical law of cosines for sides to get $\cos d$, where $d$ is the unknown edge. (Use spherical law of sines to eliminate $\sin l$ and $\sin\theta_{bl}$.)
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  • $\begingroup$ Hi WMich, Welcome to Maths SE! $\endgroup$ – Andrea May 24 '16 at 15:07

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