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    Re: Spherical triangle split by right triangles
    From: Henry Halboth
    Date: 2016 Jan 23, 02:50 -0500
    Don't know exactly what you are looking for, but hope the attached may be of some help.

    On Fri, Jan 22, 2016 at 2:02 PM, Mark Coady <NoReply_MarkCoady@fer3.com> wrote:

    Ok, I’m feeling stupid again.

    While I’ve pretty much used spherical law of cosines happily enough for my purposes, my introduction to HO208 and other methods of Nav triangle calculation had me attempt the split of  an oblique  spherical  triangle into two Right spherical triangles for an experimental calculation  on a circle route.  So I sketched out a simple great circle route with uneven legs.   Then debated how I would know where to split it. I know the included angle for the difference in Longitude,  I know the Latitude/co-latitude of my start (last fix) and of my destination. Now if it were isosceles...ok..midpoint of base/ split the vertex and away we go....  but oblique.....  I kept googling up the question various ways with everything coming back that you just draw a perpendicular to the line opposite my vertex.   Yup...ok..makes perfect sense.....thats exactly what last math site said...BUT HOW ??

    Its apparently so easy..nobody tells you ...so I feel dumb, but I just don’t see ...

     Geometric construction...ok...but.....What if I don’t want to guess drawing a circle course curved line perpendicular intersection on a flat chart....or don't have a plot and just  want math. Mathematically how do I proportion the values of the Difference  in longitude angle and course line?

    So I can then solve for the other unknowns with uncle napier?

    I feel like I am missing one more number I need to calculate using napier’s normal cos & tan the rules.

    I have the colatitude side from the N or S pole,  opposite the 90 degree angle for each of the two split triangles, but I don’t know the vertex split, don’t know intersection distance from the start or end point of the perpendicular?.

    is there a good explanation somewhere?   I looked up Napier's agian, but the relationships do not prescribe solutions with the 90 degree angle and only one side.

    What am I missing...... ? besides the obvious....

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