
- 
| 
| 
| 
a 

Vol. IV, No.8.] Geometrical Theory of a Plune Non-Oyelic Arc. 401 
[N.S.] 
For, if R’ be the radius of the circle PPQ, then, by Theo- 
rem XT, 
= R(1+ 2a tan 8). 
Therefore chord PQ=2R’ sin a=2R’a=2R(a+ 2a? tan 8). 
But the arc PQ differs from chord PQ by ‘an infinitesimal of 
the third order. 
herefore s=l=2R tee tan 8). 
Theorem XIIf—I€ O,, O,, O; be any three points on the 
non-cyclic ee are PO,0,0,Q, the angle 0,0,0, is 
equal to (1—2a,; tan 8) (a, -a,), where a), a,, ag are the angles 
which PO,, PO,, PO, make with PT. 
Let angle 0,030, =2. 


0,0, dO 
Then sin 2=oR. and sin (41— 01) 5p where Ryo, and 
Rg mean the radii of the circles 0,0,0, and PO,O, respec- 
tively. 
eee sine fy, _ R{1+2(a,+a,) tan n 8}. 
sin(az—a)) Rios ~ RU + 2(a, +a, + a3) tan 6 } 
=]1—2a, tan 6, 


Therefore x=(a,—a,) (1—2a, tan 0). 
Cor. A.—Angle PO, ,0,=a,(1~2a, tan 4). 
Theorem XIV,—In any non-cyclic infinitesimal arc PO,0,Q:, 
chord 0,0,=(PO,—PO,)+ Rajag (a,—a,), neglecting canis auaanae 
of fifth order, where a,, a, are the angles which PO,, PO, 
make with the tangent at P, and Ris the radius of curvature 
at P. 
We have, by Heigtapinsicy, 
0,P0, . 0,0,P . 0,P0,+0,0,P 
3 2 sin S sin 2 i‘ 

0,0,+ PO, ~PO,=8R,; sin 

But Rig= R{1 + 2(a, +2) tan 3} 
O,PO,_ @,—,  A—4a) 
2". 
sin = 9m 



O,0.P 
sin -_ =5(1- 2a, tan §) 

sin O\P oat O,0,P = 25% 5 $1 (1-20, tan 8) 
= 3 (1-2a, tun 8). 
Therefore, ee 
0,0,+ PO,—PO,=R(a,—a,) aa, (1+0 tand) 
