410 
-.T R 2 A R2 
Hence x = — , y=-j-, and z- 
a b c 
In triangle AGF, R 2 =AF~=FG 2 AAG~=z*-\-AG q . 
And in triangle AGH, AG 2 = AH 2 - J r HG 2 =x Q -\-y~. 
774 774 774 
Hence R*=x* + y* ^=^+-4-- 
CL~ 0 C 
And R* = - z. y 
i+v+i* 
106. In (fig. 33*, Plate IV .*), join OH and BF. Then because 
AF=R is perpendicular to the plane BCD , AF is perpendicular 
to OF and BF as well as JDF. 
1 
Therefore cos FAB 
R 
c 
also cos FAB=—- 
a 
and cos FAC—?-- 
b 
Where FAB, FAB, and FAG are the three angles which 
the normal makes with the three cubical axes which it cuts at 
the distances a, b, and c. 
107. Given the indices of any two faces of a crystal Oi the 
cubical system, find the angle between their two normals at 
the centre of cube, or the supplement of the angle of inclination 
of these two faces over the edge of their intersection. 
(In fig. 36*, Plate IV.*) 
Let AF—R be the normal to the plane whose indices are 
a, b, c. 
AF 1 =R 1 be the normal to the plane whose indices are 
<h> b v c v 
Let 03 = AH, y = HG, and z = FG be the rectanglar co-ordinates 
of the point F {see § 105) referred to the rectangular 
axes AX, AY, AZ. 
And a y 1 =S 1 G v z 1 =J’ 1 G ! 1 , similar co-ordinates for 
the point F v 
Pig. 36* is drawn in perspective. Fig. 37* is the plane 
