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    Re: FW: Mag. Variation
    From: Herbert Prinz
    Date: 2004 Feb 24, 15:54 -0500

    George Huxtable wrote:
    > hav (180-Z) = sec h sec L cos S cos (S - p)
    > I have quite a collection of navigational texts, but in none of them can I
    > find any referece to that same expression for azimuth. Although Bowditch is
    > usually rather good at explaining or deriving the expressions he uses, in
    > this case he doesn't seem to do so (or I can't find it).
    It's always the same boring old trick:
       2 cos((a+b)/2)*cos((a-b)/2) = cos(a) + cos(b)
       hav(a) = (1-cos(x))/2
    as the name suggests. So, from
       [1 - cos(180-z)] * cos(h) * cos(L)  =  2 * cos((h+L+p)/2) * cos((h+L-p)/2)
    we get
     [1+ cos(z)] * cos(h) * cos(L)  =  cos(h+L) + cos(p)  =  cos(h)*cos(L) -
    sin(h)sin(L) + cos(p)
    Eliminate bracket, move terms around and obtain the plain cosine formula for the
    PZX triangle:
    sin(h)*sin(L) + cos(h)*cos(L)* cos(z) = cos(p)
    Best regards
    Herbert Prinz

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