Application of Napier’s rules. 37 
= tang. (co. C) x tang. (co. PAC), or cotang. PAC = cos. AC x 
tang. C, whence we have BAP = BAC $ PAC. Then in the trian- 
gles ABP, APC we have the angles PAC, BAP, and the side AC, to 
find AB; and if to these we join the perpendicular AP we shall find, in 
the triangle APC, that the complement of the angle PAC is the middle 
part and the complement of AC is an adjacent part ; and the trian- 
gle APB is to be marked ina similar manner. Whence -the rule, 
Sine middle part « tangents adjacent parts, gives sine (co. PAC): 
tang. (co, AC) :: sine (co. PAB) : tang. (co. AB). | 
In the same manner we may proceed with any other oa 
Seen See to cxaiude tha endicular from th a 
epee 
‘ing two ca cog question is ee two sides 
and the ica angles of an oblique angled triangle, or where three 
sides are given to find an angle. The first of these may be solved by : 
te Soe rule that the sine of a side is proportional to the sine of its 
opposite angle. Thus if AB, AC, and snipe given to find C, we 
i gees sine B:: sine AB : sine C. ~ _ The case 
where three sides are given includes that where nee sfigles are piver 
by faking the supplementary triangle. The usual rule for finding one 
gles fed Fare ot three ges AB, AC, BC are given, is by 
Log. cos. }B=}x or sine Fee sie EAC tog nce, AB gon BC} 
Both these excepted cases occur so frequently im practice that it be- 
comes quite easy to remember the rules for solving them, and if to 
these we join the rules of Napier, altered in the imantier- we have here 
suggested, Bee ate cece eee ecessar 
