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.

$\cot\alpha=\cos\upsilon\:\tan\frac{\phi}{2}$

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

$\sin\frac{\lambda}{2}=\sin\frac{\phi}{2}\:\sin\upsilon$

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.