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    Law of Cosines
    From: Jeff Gottfred
    Date: 1995 Dec 18, 17:11 -0700

    O.K., for those who haven't seen it, here is how to derive the classic
    formula for Hc from plain ol' high-school plane trig...
    
    First, we need to develop a formula for the law of cosines for spherical
    triangles.
    
    As a refresher, the law of cosines for plane triangles is this:
    
    For triangle abc,
    
    a^2 = b^2 + c^2 - 2bc * cos A
    
    With this, you can solve for the length of any side or any angle of a
    non-right triangle-- sure would be nice if we could find the same law
    for spherical triangles...
    
    Here's how:
    
    O.K. first draw a triangle on a sphere, and then connect the points of
    the triangle with the enter of the sphere. Cut out this wedge shaped
    piece: (lousy diagram follows)
    
    
    
                                         A
    X-----------------------------------------------
      \-._                               |\
        \ `-._                           | \
          \   `-._                       /  |
            \     `-._                  /   |
              \       `-._             /    |
                \         `-._       c/     |b
                  \           `-._   /      |
                    \             `-._    /
                      \           /   `-._/
                        \     __/_____/--C`-._
                         B\--/     a          `-._
                            \                     `-._
                              \
                                \
                                  \
                                    \
                                      \
                                        \
                                          \
                                            \
                                              \
                                                \
    
    X is the center of the sphere, and ABC is a spherical triangle on the
    surface of the sphere consisting of the arcs a, b, & c.
    
    (I would suggest that you grab a pencil and re-draw this diagram as we
    go along... in fact, you probably should do the same thing with the
    math..)
    
    Next, we construct a plane triangle AYZ which is perpendicular to the
    line XA.
    
    
    
    
                                         A
    X-----------------------------------------------
      \-._                               |\ \
        \ `-._                          || \ \
          \   `-._                      ||  ||
            \     `-._                  |   ||
              \       `-._             |    | \
                \         `-._      c /|    | |
                  \           `-._   / |    |b \
                    \             `-._ |  /    |
                      \           /   |-. /     \
                        \     __/_____|--C`-._   |
                         B\--/      a |       `-.|
                            \         |       ,' Z`-._
                              \       |     ,'
                                \     |   ,'
                                  \   | ,'
                                    \ |,'
                                     Y\
                                        \
                                          \
                                            \
                                              \
                                                \
    
    O.K., AY, AZ, and YZ are supposed to be straight lines forming a plane
    triangle AYZ that touches the sphere at point A, and sticks out through
    the surface of the sphere. The salient bit is that this thing forms two
    right triangles XAY and XAZ -- that is AY and AZ are perpendicular to
    XA.
    
    In this construction, the arc a is congruent with angle YXZ, and the
    angle A is congruent with the angle YAZ. We can therefore write two
    equations, one for a and one for A, using the law of cosines for plane
    triangles XYZ and AYZ:
    
    
    YZ^2 = XY^2 + XZ^2 - 2 * XY * XZ * cos a
    
    and
    
    YZ^2 = AY^2 + AZ^2 - 2 * AY * AZ * cos A
    
    
    As these are both equations for YZ^2, we can write:
    
    XY^2 + XZ^2 - 2 * XY * XZ * cos a = AY^2 + AZ^2 - 2 * AY * AZ * cos A
    
    Subtracting an AY and an AX from bpth sides:
    
    XY^2 - AY^2 + XZ^2 - AZ^2 - 2 * XY * XZ * cos a = - 2 * AY * AZ * cos A
    
    Now, here is the clever bit: because we are dealing with right
    triangles, we can write two new equations based on pythagoras:
    
    XY^2 = AX^2 + AY^2
    
       or AX^2 = XY^2 - AY^2
    
    and
    
    XZ^2 = AZ^2 + AX^2
    
       or AX^2 = XZ^2 - AZ^2
    
    Substituting these into the first equation we get:
    
    AX^2 + AX^2 - 2 * XY * XZ * cos a = - 2 * AY * AZ * cos A
    
    collecting AX^2's:
    
    2 * AX^2 - 2 * XY * XZ * cos a = - 2 * AY * AZ * cos A
    
    Subtract -2 AX^2 from both sides:
    
    - 2 * XY * XZ * cos a = -2 * AX^2 - 2 * AY * AZ * cos A
    
    Factoring by -2:
    
    - 2 (XY * XZ * cos a) = -2 (AX^2) - 2 (AY * AZ * cos A)
    
    Factoring cont...:
    
    - 2 (XY * XZ * cos a) = -2 (AX^2 + AY * AZ * cos A)
    
    Divide both sides by -2:
    
    XY * XZ * cos a = AX^2 + AY * AZ * cos A
    
    Subtract and AX^2 from both sides:
    
    XY * XZ * cos a - AX^2 = AY * AZ * cos A
    
    Now divide boths sides by AY * AZ:
    
    XY * XZ * cos a - AX^2
    ---------------------- = cos A
          AY * AZ
    
    Expanding:
    
    XY * XZ * cos a   AX * AX
    --------------- - ------- = cos A
       AY * AZ        AY * AZ
    
    Now, looking at our construction again, we can see that the following is
    true:
            AY           AZ
    tan c = --;  tan b = --
            AX           AX
    
    So, substituting, we can write:
    
    XY * XZ * cos a         1
    --------------- - ------------- = cos A
       AY * AZ        tan c * tan b
    
    Also, from plane trigonometry we know that
    
            sin X
    tan X = -----
            cos X
    
    
    So we can write:
    
    XY * XZ * cos a   cos c * cos b
    --------------- - ------------- = cos A
       AY * AZ        sin c * sin b
    
    
    Again, from plane trig, you can see that
    
            AY              AZ
    sin c = --; and sin b = --;
            XY              XZ
    
    so we can now write:
    
        cos a       cos c * cos b
    ------------- - ------------- = cos A
    sin c * sin b   sin c * sin b
    
    Grouping:
    
    cos a - cos c * cos b
    --------------------- = cos A
       sin c * sin b
    
    There you have it! the law of cosines for spherical triangles!
    If we just juggle it slightly then it looks like the textbook version:
    
            cos a - cos b * cos c
    cos A = ---------------------
                sin b * sin c
    
    
    O.K., so now let's use it to derive a formula for Hc:
    
    The nautical triangle (draw it yourself!!) consists of the three sides
    co-h (90 minus the height of the body above the horizon),
    co-L (90 minus the AP latitude),
    co-d (90 minus the declination of the body)
    t    (The meridian angle of the body: GHA ~ AP Longitude)
    
    Therefore, we can use the formula above to write:
    
    
            cos (co-h) - cos (co-L) * cos (co-d)
    cos t = ------------------------------------
                 sin (co-L) * sin (co-d)
    
    Now, because sin (90-x) = cos x; and cos (90-x) = sin x; we can write:
    
    
            sin Hc - sin L * sin d
    cos t = ---------------------
               cos L * cos d
    
    Multiplying both sides by cos L * cos d:
    
    cos t * cos L * cos d = sin Hc - sin L * sin d
    
    Now, adding a sin L * sin d from both sides:
    
    sin L * sin d + cos t * cos L * cos d = sin Hc
    
    or, in other words:
    
    Hc = sin^-1 [(sin L * sin d) + (cos L * cos d * cos t)]
    
    Voila! the classic formula for Hc from first principles!!!
    
    Cheers!
    
    Jeff
    
    
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