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Dear Peter,

Many thanks for your Kind and comprehensive reply.

It is interesting that you did visually check your initial result to be reasonable before publishing it .

I do agree that in most cases, problems similar to Dave's one generally yield solutions with Bodies LHA's inferior to 90°.

Still, for Observers close to the Poles, LHA's can be expected to easily exceed 90° even for reasonable altitudes.

As an example, with the following numbers rounded to integer values let's us start from :

- Tony's Position (i.e. from position N60° E030° , Hello Tony ! ) , and :

- Dave's Height ( 55° i.e. a sub-astral distance D = 2100 NM ),  and :

- an Azimuth Az = 030°,

we get a subastral point at N73°17.4' E124°06.6' hence with LHA = 94°06.6'

Solving this with the direct method, the Sine rule "1st quadrant LHA" indicates 85°53.4' , with a sub-astral point at E115°53.4' .

The distance between both points is 8°13'2 of Longitude at a Mid-Latitude close to N80°, representing about 90 NM.

The final check on Distance seems a simple way here to pick up the correct "2 nd Quadrant LHA" here, i.e. 94°06.6' .

Best Regards to all

Antoine