36 Application of Napier’s rules. 
which gives(as in Fig. 2) sine (co.C)=tang. PCxtang.(co. AC), whence 
tang. PC=cos. C x tang. AC. Thenin the triangles ABP, APC, we 
have AB, AC and PC to find BP, and if to these we join the perpen. 
dicular AP, we shall find, in the triangle APC, that the complement of 
AC isthe middle part and PC an opposite part. The triangle ABP 
is to be marked ina similar manner. Then the rule, Sine middle 
part « cosines of the opposite parts, gives Sine (co. AC) : 
PE-:2ssind{co: — cos. BP. — found BP we have a 
= BPS PC? -7* 
EXAMPLE 2. ae ee tig 
Giyen BC, AC, and the angle C, to find AB. Fig. — 
As in the last example w find. PC, thence BP — BC ¥ PC. 
‘Then in i ‘APC, ABP we have AC, PC and PB to find 
AB, which requires that the triangles should be marked. as in the last 
example, and the rule 
Sine middle part « cosines of Opposite parts, gives sine (co..AC) : 
cos. PC :: sine (co. AB) : cos. BP, whence we obtain cos. PC : 
cos. AC :: cos. BP; cos, AB. 
paten 2 
ae 
EXAMPLE. 3 
‘Given AC, ay aay find the angle B. Fig. 4. : 
Find PC, PB, as in the preceding example, then the rule, sine mid- 
dle part & tangents of ‘adjacent parts gives sine PC : Raxco C= 
sine BP : tang. (co. B). 
eee omen 
EXAMPLE 4. 
Gingn, the angles dnsindChaddbisaidesAC 40 GietiATs Fig. 5. 
In the right angled triangle ACP we have AC and the angle C, © 
whence we may find the angle PAC by the rule, Sine middle part” 
== 5 tangents adjacent parts, which gives, (as in fig. 1.) Sine (co, AC } 
