Navigation
posted by mike .
Just to see if we are talking the same language:
Navigation triangle PZX
LHA = 194.65 degrees (P)
Lat of Z 51.74 degrees
Lat of X 39.24 degrees
Solve for distance p
Thanks
Mike

I drew 194.65 deg X to the West of P which is of course 165.35 deg East (closer) so we are going to get two different answers depending on which way we go around the earth. The longer answer is the one you have specified with the 194.64 difference in longitude.
The angle XPZ is 90 degrees as I drew it so it is a right spherical triangle.
The side x, from P to Z is 51.74  39.24 =12.5 deg
The side z from X to P is 194.65 deg given
In this right triangle the side opposite the 90 deg satisfies:
cos p = cos x cos z
cos p = cos 194.65 cos 12.5
cos p = [.9675][.9763]
cos p = .9446
p = 160.83 deg.
of course 360  160.83 is also valid
so p can also be 199.17 deg which is the way you set it up [the cos of 199.17 is also .9446] 
By the way, my calculator has a special key with the  in parenthesis like () for doing things like times negative.

Whoops, that was much too simple. I took the LHA as a side, but it is not except at the equator!
I have to correct the distance z for the latitude of X (I am pretty rusty at this stuff)
the radius from a North south line through earth center perpendicular to that axis is = radius at equator *cos 39.24
Therefore I need to solve a triangle for z before I can do the one you want.
That triangle for z can be S at the equator due south of P , P and X 
The relationship between LHA and angle z is that they both have the same chord length between X and P
In other words 2 R sin z/2 = 2 r sin LHA/2
where R is the radius from earth center and r is the distance from the NS axis
We saw that r/R = cos Latitude = cos 39.24
so
sin (z/2) = cos 39.24 sin (194.65/2)
sin(z/2) = .7745 *.9918
sin(z/2)=.7682
z/2 = 50.19
z = 100.38 deg.

Now do the above with z corrected
cos p = cos x cos z
=cos 12.5 cos 100.38
= .9763 * .1802
=.1759
so p = 100.13 deg
check my numbers!
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