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    Re: spherical right triangles, napiers circle
    From: George Huxtable
    Date: 2004 Jan 28, 19:32 +0000

    Patrick Stanistreet asked-
    
    >I have been going through a book
    >Navigational applications of plane and
    >spherical trigonometry by Carol
    >Congleton, just  trying to learn the math
    >background involved.  On a chapter
    >concerning spherical right triangles
    >are two rules which I would like to
    >find a source for their derivation
    >or a proof.
    >
    >Rule 1:  the sine of a part is equal to
    >the product of the tangents of the
    >adjacent parts.
    >
    >Rule 2:  the sine of a part is equal to
    >the product of the cosines of the
    >opposite parts.
    >
    >Also if anyone knows how Napiers circle
    >was derived I would like understand the
    >process.
    
    Response from George-
    
    It's not a book I am aware of.
    
    Patrick does well to ask where those rules are derived from. I haven't seen
    them before, and wonder if they are true. Is Patrick really quoting from
    the book word-for-word? Has he left something out? I'm not very conversant
    with "Napier's rules".
    
    Take a sphere and chop it in three planes at right angles. This leaves 8
    segments, bounded, if you like by the equator, North pole, long = 0 deg.,
    and long = 90 deg.. Just consider the surface of one segment. It's a
    spherical triangle. It's a right-angled spherical triangle. In fact, it's
    as right-angled a spherical triangle as it's possible to get, because all
    three sides, a, b, c, and all three angles, A, B, C are right angles. So it
    should conform to those "rules", if those rules are true. Are they? I doubt
    it.
    
    Consider "rule 1". Take an angle A (at the North pole, say).
    
    sin A = 1 because it's a right-angle
    
    Presumably, the adjacent parts are sides b and c. It doesn't matter whether
    we have chosen the right "parts" because all the parts are 90 deg. So Tan a
    = Tan b = infinity.  The product of the tangents is (infinity) squared,
    which does not equal 1. Therefore "rule 1" appears to be untrue.
    
    Consider "rule 2". Take an angle A
    
    sin A = 1 because it's a right-angle.
    
    What are the "opposite parts"? In a spherical triangle, there are six
    parts, so there's only one "opposite part". But in the definition of the
    "parts" in Napier's rules for a right-angled triangle, by convention the
    right-angle is omitted, so that there are then only five parts, in which
    case it IS possible to have two "opposite parts". But in this triangle, ALL
    the parts are 90 deg. So the cosines of the "opposite parts", whichever
    they are, will be zero. The product of those cosines will be (zero)
    squared, which does not equal 1. Therefore "rule 2" also appears to be
    untrue.
    
    Have I got it wrong, or misunderstood something? If so, someone will put me
    right. Has Patrick got it wrong?
    
    Otherwise, perhaps we might start a Nav-l collection of "books to avoid",
    and put this one right at the top.
    
    George.
    
    ================================================================
    contact George Huxtable by email at george@huxtable.u-net.com, by phone at
    01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
    Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ================================================================
    
    
    

       
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