If one supposes 
ard 0 + f/°— e?—y? —h’) 
(c—h)?— b? 
i il wat Cae 
— p 
(c—h)?—b? maa 
(9) 
Alae— fg) + 2b(hHe) 
(« —h)" — b? = 
2b(e—h) | 
(c—h)j?—B? 
then becomes 
cos*20 + pcos 20 + ¢= — (r+ scos2o)sm2o. . . (10) 
and 
OW ay tne RO EN 
cos* 2o + Sr cos* 20 + . aah — cos? 29 + 
1+s? 3 143? > 
Dj pa—rs ab. = a2 
See Ve 05) cos 29 + erge ke ae (11) 
1+ 3s’ Ls? 
This equation can now be solved; the value found for o, intro- 
duced into (6) furnishes 6. 
If however in the equation (6) 
acoso +esnga=—gqvsoa+s/sno=0, 
then sino becomes indefinite. In that case is 
a q 
ON rs == ET, 
af == él) 
consequently 
(cot ve, coth, cos y ~— cot a, cot h,) (cot we, — cot u, cos y) = 
= cot? , cot h, sin? y = cot? «, cot h, (1—cos*y) 
cot «, cot h, (cot w, cos y — cot «,) + cot re, cot h, (cot «, cos y — cot u,) ="! 
cote, coth, cot, cosy — cota, 12 
EEN Te) 
o is then found by introducing the value of y in (1). 
What is said here finds an application to the determination of the 
plane that cuts the octahedron planes (111), A11) and (411) in such 
a way that the traces of the planes (111): (441) and (411) : A11) 
enclose right angles. If one supposes (111) to be the equator plane, 
then becomes 
út, = 180° — 70°31'43", {V, = (A11; «, = 70°31'43", §V, = (111)}: 
P= 6) oh, Sh Wy oS HO”. 
