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    Re: Basic question regarding Napier's triangles
    From: Bill Noyce
    Date: 2009 May 28, 19:02 -0400

    > 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...
    >
    
    I think you are applying the procedure correctly, and you have shown
    that there cannot be a spherical triangle with a right angle, adjacent
    side 70 degrees, and opposite side 40 degrees.
    
    Imagine placing your triangle with unknown side b along the equator,
    and side c extending 70 degrees northward.  No matter what length you
    choose for b, you cannot bring its endpoint within 40 degrees of c's
    endpoint.
    
    Certainly in a plane right triangle we expect the hypotenuse to be the
    longest side; I'm pretty sure the same is true of spherical triangles,
    at least until a side exceeds 90 degrees.
    
        -- Bill
    
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