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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...)
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[Sent from archive by: scheele-AT-telkomsa.net]


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