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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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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. –  nbubis Jul 19 '12 at 11:51
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1 Answer

up vote 1 down vote accepted

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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