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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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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
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...

(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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