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    Re: Math behind horizontal sextant angles
    From: Mike Wescott
    Date: 1997 Mar 04, 15:46 EST

    > I've known since I was a beginner that if you measure the horizontal
    > angle between two objects, it gives a circle of position.  I even
    > plotted a few points on a sheet of paper once to convince myself.  But
    > every time I sit down to try to generate a mathematical proof of this,
    > I come up smack against my relatively poor memory of trig identities.
    > Can anybody point me to a reference work where I could look up a
    > proof or derivation?
    Can't come up with a pointer but I can sketch out the proof.  We can,
    WLOG*, put the two objects on the cartesian plane at A=(0,0) and
    B=(1,0). Now consider a point X=(x,y) such that the angle AXB is T.
    Drop a perpendicular from X to the x-axis and decompose Tinto
    T1 and T2:
              /    T1 | T2    \
           /          |           \
        /             |               \
      .               .                 .
    (0,0)           (x,0)             (1,0)
    (Sorry for the poor diagram).
    We get:
            T = T1 + T2
            tan(T1) = x/y
            tan(T2) = (1-x)/y
            tan(T) = tan(T1 + T2) = (tan(T1) + tan(T2))/(1 - tan(T1)tan(T2))
    substituting and simplifying we get
            tan(T) = y/(y^2 - x + x^2)
    let tan(T) be the constant k and we can rearrange into
            y^2 - y/k + x^2 - x = 1
    which can, in turn, be arranged:
            (x - 1/2)^2 + (y - 1/(2k))^2 = 5/4 + 1/(4k^2)
    i.e. (x,y) lies on a circle. A little more work shows that all such
    points on the circle and above the x-axis have the same angle. A
    symmetry argument shows that there is another circle centered below
    the x-axis with the same property.
    *WLOG - Without Loss of Generality - a mathematical term meaning that
    I've figured out that I can make this simplification but I'm not going
    to explain it to you. :-)
    	-Mike Wescott

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