# NavList:

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

**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 --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To unsubscribe, email NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---