Figure 4. Dihedral Angle α

Referring to Fig. 4, since b and c both have endpoints on circle C, Δabc is an isosceles triangle with dihedral angle \phi between the two equal sides. If we construct a great circle arc from point SC to midpoint D of side a, we will bisect Δabc into two congruent right spherical triangles.

The bisected dihedral angle at SC=\frac{\phi}{2}, and side c = b = \upsilon. We can solve for the remaining dihedral angle α at P, or at SN, since these are equal. Since side c is perpendicular (Fig. 4) to the great circle arc made by plane B, this dihedral angle at P will be α.



Then, solving for side a (which was bisected) we get:



Also, because we are using a unit sphere,

{r}=\sin\upsilon=\sqrt{{1}-{s_z}^2}   (Unit sphere center SO can move up or down along the z-axis and the north pole will remain the north pole and circle C will still pass though the pole SN because \upsilon angle changes accordingly in order to keep this true.