30 A NEW METHOD OF DETERMINING 
.4)=(0.8) +(4.12) @©.2)=(0.4)+4 2.6) 
=(,9) 26.8) €.3jS0.54 6,9) 
(2.6) = (2.10) + (6.14) 
G.0) = Goll) -4 (15): 
eS 
iS © 
Then 
A(c + 2. ¢) = (0.2) 
Ale) —2.¢) = (1.3) 
4(e,-+¢) = (0.8)—(4.12) 
4(ce,—¢) = {[(1.9)—(.18)]—[(8.11)— (7.15) | cos 45° 
A(s, +s) = §[(1.9)—(.13)|+[(8.11)—(7.15)]} cos 45° 
4(s,—s,) =(2.10)— (6.14) 
8.c,= (0.4) — (2.6) 
8.s,=(1.5)— (8.7) 
Aleit) | = (4) G4) — Gs) cos 452 
4(¢q,—c,) = L@ _ (5) | cos 22 .°5 + [GD — (4°5 )| cos 67 .°5 
Ale +65) = (t)—| (Ps) — Gy) | eos 45° 
4(e,—e,) =| (4)—(z5) | sin 22.°5—[ (,8,) — (5) | sin 67.°5 
4(, +s) = L@ + (5) | sin 22.°5 + [a rae a sin 67 .°5 
A(s,— 5) =|’) + Gip)| cos 45°=- G aA 
A(s, +s) =| (4) + Gs) ] cos 22.°5—| (8) + Gy) | cos 67° 5 
A(s,—ss) =| (5) + Gi) | cos 45°— (G4). 
When the circumference is divided into twenty-four parts, each part is 15°. 
Let 
(0.12)=¥%+ Y¥, (0.6)=(0.12)4+ (6.18)  (2)=(0.12)—(6.18) 
C= 4 ale) 1) E\=O1)=—@.) 
(2.14)= ¥,+ Yi, Ca) 2) OR Cae) 
(il By= 42, GansG. neoan 23) a y= Geil) — O23) 
