# Formula for the fourth side of a spherical quadrilateral

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

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