32 A NEW METHOD OF DETERMINING 
When the circumference is divided into thirty-two parts, each part is 11°. 25 
Let 
(OH hte, @O8j)eGOihe(s2) @.64=0.8)£6. 
Ch s 42%, Cosa) 69.28) C8) =C.9 46. 
(2.18)=¥,+¥, (2.10) =(2.18)4+ (10.26) (2.6)=(2.10)4 (6. 
. : @.H)=C.id)+@. 
G5 3l)—= v8 Ya (7.15) 2 (0-23) G5.3l) 2) = 04 hee 
(1-3) = 4.5 )+G: 
OsCij (2) Gsi0.8j)—¢. 
OSG Ci25)  O=a@.oj)—6. 
(q) = (2.10)—(6. 
Je) = (7.23) — (15.81) (2) = (8.11) —(7. 
Then 
8 (e+ 2.6) = (0.2) + (1.3) 
8 (¢o— 2.¢5) = (0.2) — (1.3) 
( 
( 
8(e+¢y) =(%)4 [ (2) —( f;) | cos 45° 
8 (¢,— 1) = [ (4) — (qs) | cos 22.°5 + [G4 — (,5;) | cos 67 5a 
8(a+ Ce) = (4) 
8(G—es) = [(a- (3) | cos 45° 
8(%+¢) =(%)—| Gr) — Gp | cos 45° 
8(¢—eC») = 4) —(5) | sin 22.° 5 — [Ga — (5s) | sin 67.°5 
16.¢; = (0.4) — (2.6) 
8(s,+sy4) = [@® + (75) | sin 22.°5 + [G+ + (;°5) | sin 67 .°5 
8@—su) =(G)— (8) | cos 45° + (;4;) 
8(s, + 82) =[(4) + () eos 45° 
8(%— sp) = (%) 
8 (s+ So) =[ (4) + Gs) | cos 22.°5—| (,3-) + (435) | cos 67 .° 5 
8 (s.—S») =[ (as) — (a) | cos 45° — (74). 
