( 724 ) 
The equation (12) changes into 0 = 0; in (6) not only the nume- 
rator becomes =O because «=e=0: but also the denominator 
because 
g cos vy + f sing = 
= cot a, sin y cos Q + (cot a, — cot a, cos y) sin 9 — 
Li 
cot a, bsin y cos @ — (1 +4 cos y) sin 95 = 
s 
i 
=) fe sin? 60° + (: EE + sin sor = 
1 a eo 
== tobe, (- (ove) al nt =) == 
From (1) one finds on the contrary 
| 
cot h, cos @ = 0 = cos 6 cot a, + sin 6 sin Oo 
cot a, cot 70°31'43" 
(GO = TT ESE 
sin Q sin 30 
. . ‘ zo im epi ie / JT 
from which o= 35°15'53”, the angle between (001) : (111) — = 
The secant plane is consequently (OOL). 
A single example may suffice to demonstrate the applicability of 
the formulas deducted above; I choose for it a problem in which 
the result obtained can easily be controlled. 
The three rhombododecahedron planes (101), (1410) and (011) are 
cut by a plane in such a way, that their traces include angles of 
60°. What is the orientation of the secant-plane ? 
