the angles oj crystals, 115 
« 
from ED, parallel to xy^ and is in the negative direction, 
~ = ^ ox p — q = ^r. 
r r i 
Also since the section is parallel to Ajy we must have q=o. 
Hence (2 ; o ; 1) is the symbol required. And the co-ex- 
istent planes are 
(2 ; O, 1^(2; 1, 2)( 1 ; 0,-2} 1; 1, 1) 
each of the parentheses gives two planes, and hence we have 
8 arising from this law. 
35. To find the angles which these planes make with the 
planes of the octahedron. 
Example. Zircon unihinaire, Hauy. 
In the square octahedron, which has been considered as the 
primary form of zircon, the angle of two adjacent faces of a 
pyramid is 123° 15', and the angle of two opposite faces 
measured over the summit is 95° 40'. (Phillips). 
Hence the dihedral angle at A.r, which is («) the angle of 
the planes EFK, FDH, is 95° 40'. And (/ 3 ,) the angle at Ay 
is the angle of DEK, FEG, and is therefore the supplement 
of the angle of HFG, EFG, and it is therefore = 56^ 45'. In 
the same manner (7) the dihedral angle at hz is 56° 45'. 
In order to apply the formulae of Art. 8, we must find the 
values of dy e,J^. Let XYZ, fig. 15, be a spherical triangle 
made by describing a sphere with center A, meeting A x. 
Ay, A'z in X, Y, Z. Then if XD be drawn perpendicular 
to YZ, = sin. XD, similarly if YE be perpendicular on 
ZX, ez=: sin. YE, and/=^. 
Now by Napier’s rules, since XYD = 56° 45', and YXD 
= y 4<o') = 47 ° 50', r . cos. 56® 45' = cos. XD . sin. 47° 50 
/. d = sin. 42° 11 ; d=z .67301 25. 
