NAPIER'S RULES. 331 



7T 



Produce BG to F, and AC to A', making AA = B' = -', and 



join A'B'. 



Then .4'jB' G' is the secondary triangle ; and that it is a 

 right-angled triangle may be proved, independently of what 

 has been already said, by joining BA 1 . For since AA' is a 

 quadrant, and A a right angle, BA is as well as BB' a 

 quadrant, and therefore B is the pole of A'B', and therefore 

 B' a right angle. The hypothenuse of the secondary triangle 

 A G is the complement of A C, ft side of the first : the angle G is 

 common to the two triangles: GB' is the complement of the 

 hypothenuse BG : B'A = angle B'BA and is therefore the 

 complement of the angle ABG: lastly, the angle BAG has its 

 complement measured by the arc AB. 



Again producing B'A to (9, B' being a quadrant, and 

 similarly GA to (7, (7(7 being also a quadrant, we obtain a 

 third triangle OG'A\ and finally completing the figure (the 

 obvious details of demonstration are omitted) we get a reentering 

 pentagon, all the angles of which are right angles, each 

 being the right angle of one of the system of five right-angled 

 triangles. 



From this point of view it is plain that Napier's rules may 

 just as naturally be considered as the statement of similar pro- 

 perties of five associated triangles, as in the usual mode, namely, 

 as the statement of dissimilar properties in one triangle ; and 

 thus we obtain the rationale of their uniformity. Every relation 

 between a middle and opposites, is the relation between two 

 sides and the hypothenuse: every relation between a middle 

 and adjacents is the relation between two angles and the 

 hypothenuse. I only give these relations, of course, as instances, 

 for any other would do as well. 



A convenient mode of establishing these relations is afforded 

 by the Lemma. 



If we project OB in fig. 1, into OA, or if we project OB 

 into OG and then OG into OA, the coincidence of the two 

 results gives rise to the equation 



cos c = cos a cos b, 

 Again, OA tan c = AB, 



and ABmsA = AC= OAt&nb; 



