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    Re: Basic question regarding Napier's triangles
    From: Peter Hakel
    Date: 2009 May 28, 16:04 -0700
    Hi Christian,

    From the spherical law of sines we have

    sin a / sin A = sin c / sin C,   or,
    sin C = sin c * sin A / sin a

    Your values are A = 90 (i.e. sin A = 1) and c > a, which leads to sin C > 1.
    This triangle does not exist.


    Peter


    From: "NavList@fer3.com" <NavList@fer3.com>
    To: NavList@fer3.com
    Sent: Thursday, May 28, 2009 12:16:23 PM
    Subject: [NavList 8432] Basic question regarding Napier's triangles


    Hello my name is Christian Scheele and I live in Cape Town. I am by no means a professional navigator or mathematician and hope that I am not gatecrashing any party by making myself heard in here without much ado.

    I would much appreciate it if somebody could help me understand the application of Napier's triangles to specific scenarios. I have obviously run afoul of some very basic premise governing Napier's rules.

    Consider a spherical triangle superimposed on the following parts of an ideally spherical rather than spheroid globe, where capitals denote angles and small letters denote sides: A = 90 degrees, c = 70 degrees , a = 40 degrees. 

    Applying Napier's rules, (90~A) is the "middle part" which is opposite b and c.

                          Sin (90~A) = Cos b. Cos C
    removing complement:  Cos a = Cos b. Cos c
    transposing formula:  Cos b = Cos a/ Cos c

    substituting values:  Cos b = Cos 40/ Cos 70
                          Cos b = 0.766.../ 0.342...
                          Cos b = 2.24 ( 2 d.p.)
                         
    But b is not defined for values > 1. It would appear that certain right-angles spherical triangles, namely such yielding Cos values > 1, cannot be solved as illustrated in the above problem and application.

    I have obviously made a blunder somewhere...

    Either way, could somebody please help me approach this problem? I would be most grateful.



    (This problem can be complicated should one assume that the triangle lies in the plane of a spheroid, such as the earth, rather than a sphere. For example, one could assume that b lies on 20 degrees N and a lies on 0 degrees longitude. This implies that c can be broken up into a component of 20 N and 50 S. In that case corrections must be made to any results arrived at by having assumed perfectly spherical properties initially. Perhaps Inman's Tables could come in handy here...)
    -----------------------------------------------
    [Sent from archive by: scheele-AT-telkomsa.net]



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