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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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