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    Basic question regarding Napier's triangles
    From: UNK
    Date: 2009 May 28, 12:16 -0700

    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 
    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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