f 215 ] 
Sine of the Arch PK, the Area of the Sedor NSP 
fubtended by the given Arch NP, and comprehended 
in the Angle NSP made at the given Point S. 
From N and K let fail on the Diameter AP the 
Perpendiculars N M and KL, and join CN and CK. 
Then let t Hand for CP the Semidiameter of the 
Circle 5 f for C S the Diftance of the given Point S 
from the Centre 5 p for SP the Diftance of it from 
the Extremity of the Arch through which the Dia- 
meter A P paffes 5 and y for K L the Sine of the Arch 
KP in the given Circle. 
Thefe Subftitutions being prefuppofed, the Pro- 
blem is to be divided into two Cafes 5 one when SP 
is lefs, and the other when it is greater than the Semi- 
diameter CP. 
CASE L 
If SP be lefs than CP, then take an Area H equal 
to the Sum of the Redangles exprelfed by the feveral 
Terms of the following Series continued ad libitum: 
And the Area ~n*H will determine the Area of the 
2 
Sedor NSP ad libitum. 
For the Sedor P SN, being the Excefs of the Sedor 
NCP above the Triangle NCS, will be the Diffe- 
rence oft wo Redans;les: iCPxPN— iCSxNM; but 
c "2 2 
PN is the Multiple of the Arch PK, namely tzxPK 5 
and NM is the Sine of that multiple Arch : Where- 
fore if for CP be put t, for CS, f 9 according 
E e to 
