427 
ft where hC 2 , JcC 3 , and gG 1 meet, will be the pole of l mu. 
Through h, in the plane NAG 3 , draw KE perpendicular to AC 3 . 
Let angle AA0 3 =A 9 . Then since angle AHN= 90°, angle 
anh=\ 2 . - 6 
In triangle NAG 3 tan ANG 3 =^A or tan A 9 = i 
AN " n 
In triangle AKE tan liAE= tan X 2 = 
KE 1 
KE 
AE 
AE_AC*-20 3 __l-EG. 
Hence — , =- and hE= = 
AE n n n 
But by similar triangles KEG 3 , CL4CL 
llE GyA 1 rpv, ? 7 77j -rn n 
m^ACri TWfore ™=EC t ; 
and EG s 
1 -EC. 
n 
3 and nEC 3 =l-EC 3 . 
Whence EG 3 = 
71 -f- 1 
But by similar triangles C.AGo, hECo, 
A P P P i 
= ^ but AC 3 =1 and EC., — — — 
EG 3 hC 3 3 3 74+1 
Hence ~-^ 3 = n A 1, and CLfe = — 3 . 
7*0 3 3 7i + 1 
Hence 7 i, the pole of 1 oo 71, is found by taking the point h 
in CL CL so that GJi- — - 1 ^ 3 - 
71+1 
Again since tan \ 2 = - and angle hAC 3 =\ 2 , if the angular 
elements be given, C 1 C 3 is the chord of 90° and h is the point 
where the angle X 2 protracted from A meets G-fi^ considering 
C 3 as zero. 
The chord of 90° marked as a protractor is obtainable from 
any mathematical instrument maker, or may be readily marked 
on the chord of 90° by using any form of protractor. 
p p 
Similarly it may be shown that aC 3 = —. 2 3 , and that q is 
m+1 J 
the point where the angle X 3 is marked on C 2 C 3 as the chord of 
90°, C 3 being zero ; and tan X 3 =— . Also ~kC 2 — — h being* 
771 m +l 5 
71 
the point where the angle X x is marked on the chord of 90°, 
C 2 being zero, and tan X,=— . 
71 
Join C^g, C 2 h, and C 3 l c. f, the point where these three 
lines meet, is the pole of the face of the six-faced octahedron 
