Application of Napier’s rules. (35 
signify, as before, that the sine of the middle part is equal to the rectan- 
gle of the tangents of the adjacent parts, or to the rectangle of the co- 
sines of the opposite parts; but when applied to an Oblique- Angled 
Triangle, they signify, that the sines of the middle parts’ are res- 
pectively proportional to the tangents of the adjacent parts, or to the 
cosines of the opposite parts of the same triangle ; observing, that the 
perpendicular being common to both triangles APB, APC, and hav- 
ing the same name in each of them cannot be made use of in the anal- 
ogies, and must not therefore be counted as a middle part. This cir- 
Pye can produce no embarrassment, because all the | cases 
in wan ee Be teen may be solved without calcu 
ie These "1 a2 for anne oe, cases of F Oblique Spherié etic 
are easily deduced from those given by Lord Napier. For if es oe 
M for the middle part, A for the adjacent part, and B for the « 
part of the triangle APC (Fig. 3, 4, or 5), m, a, 6, for the correspond- 
ing parts of the triangle APB, and P forthe perpendicular AP. Then, 
if P is an adjacent part, we ‘shall have satis ad ameag wats Pex 
= and tang. P= ae Sine m- bhai 
Sine M i tang. Az sine m: tang. a; mh Pie an oppo pat, we 
Seis M fc ees M_ 
shall have, cos: Poe F and cos- r= = 9 % He = 
gts SF ot wy oust i‘ ar ae 
sine m 
Soap? Consequently sine M : eee sine m : 0s. Which are 
the two propositions to be demonstrated. Jit}: 4 = 
To illustrate these rules the following examples a are oe: 
Examli2 higtet) scinee, niin 
> Given AB, AC, and the angle C, to find BC, Fie. gt 
ea ‘right-angled triangle ACP we have AC and Cv ar 
may find PC by the rule, sine of middle part = ta 
