QUADRATURE of the CONIG SECTIONS, &c. 34% 
Exampre III. The length of an arch of 90°, calculated 
from the fourth feries, (Art. 28.). ’ . 
cofa=o cof} a= 0.980 785 280.... 
cof$a= 0.707106 781 1865 coft;a=0.905 1847...... 
colt a= 0.923 879532 51.-. cof -a=o0.998 80.5... 
I 13—cofa+12cofta _ 
“316° 3-4 cola—4colia — -163 053 go2 0108 
Cee = .O0I 215 2777778 
Amount of pofitive terms, .164 269 179 788 6 
a 13 —cof+4a—t12cofta 
3.16° 3+cofsa+4colta 
iene 13—cofita—tz2colita 
~ 3.164 3+ colta+4colz,a 
I 13—cofsa—i200f,4 
3-165 3+ colzga+ 4colysa 
Pika 13 — cof, a— 12 cof 74 
3.16 3-4 cols-a+4colz,a 
Each of the remaining terms, being tn} 
-000 013 261 796 5 
= .000 000 198 794 2 
= -000 000 003 074 4 
= .000 000 000 047 8 
ly z'5th of the term before it, their fum ¢ .002 000 000 000 8 
will be nearly #5 of the laft term, or 
itt Amount of negative terms, /000013 463 7137 
Difference between the “ar Ua agi seabed th Slam 
and negative terms, or gi — *164 255 716 074.9 
I 
= = +405 284 734 569 3 
Aj = .636 619 772 3676 
Arch of 90°, or 4 =.1.570 796 326 795. 
Uu2 EXAMPLE 
