NavList:

David C, you wrote:
"I concur that the lat might be about 35° S but I am not sure that my logic makes sense."

Absolutely, that logic makes sense! Do you have any planispheres for different latitudes? Could you rule out 40° S or 30° S?? Well, of course, you do have planispheres for other latitudes available, but you may not think of it that way. Do you have a "planisphere" or "planetarium" sky simulation app installed on your computer? Stellarium is an excellent example, which happens to be free, but there are many other options. They fulfill the planisphere function and allow you to test out small differences in altitudes for specific latitudes. 

Here's an equation you may not have seen before: if Mintaka is at the western horizon, setting, then the position angle, θ, "theta", indicating the orientation angle up from horizontal to the line across the sky from Mintaka to Betelgeuse, yields latitude:
  Lat = θ - 36.8°.
The relationship is as accurate as need be. But how accurately can you observe that angle θ (theta)? We don't have instruments for this. That's why we don't do this as part of standard navigation! Yet we should consider visual estimation. Can you estimate that angle θ?

One way to assess this is by asking how accurately you can read the time from an analog clock without numbers on thre dial. The line from Mintaka to Betelgeuse is like the hour hand of a clock. Can you "call" its orientation? If you can measure or estimate that angle to the nearest degree, then you have your latitude by this procedure to comparable accuracy. If you can only measure that angle to the nearest five degrees, then that's the corresponding limit on the latitude from this process. By the way, that relationship was for the stars right at the horizon, which is a theoretical case, basically unobservable. If we bump Mintaka up to an altitude of 10°, then the relationship is closer to:
  Lat = θ - 35.4°.
So if the two stars are at equal altitudes, this implies that the latitude is -35.4°, very close to your estimate and a dead-on match to Stan's answer.

Frank Reed