OK, I found an expression for the effective angular diameter of the first diagram (but not for the second one yet) using the

spherical law of cosines.

If the latitude of the sun is [itex]\theta[/itex], and the actual angular diameter of the sun ( straight line ) is [itex]d[/itex], the effective angular diameter ( curved line ) as shown in the first diagram ( when the direction the interferometer is pointing towards doesn't go through the center of the sun ) is,

[tex]d \phi = \cos (\theta) \arccos \left( \frac{\cos( d ) - \sin^2 ( \theta )}{\cos^2 ( \theta )} \right) [/tex]