CRYSTALLOGRAPHICAL EXAMINATIONS. 511 
From these angles, we can calculate the plane angles of P, m and T, by the formulae of 
spherical trigonometry : 
1 ^ \ / cos -g- (A + B + C) cos g (B + C — A) , 
sin 2 » _ V -sin B.sin C ’ 
sin a . sin C . ^ _ sin c . sin C 
^h = 
sin A ’ 
The plane angles a, 6, c, deduced from these formulae, are. 
For the face P = 92°59' and 87°0P 
. m = 93 14 and 86 46 
.T = 98 26 and 81 34 
The plane angles of a right section of the defect (z), are as follows : 
T on i (the supplement) = 43°54' 
m on i .. .. = 43 06 
T on m .. .. = 93 00 
The ratio of the three edges of the right section of the defect, are, 
a : b : c :: 1000 : 1014*69 ; 1462-5; 
for, b ~ 
sin A ’ 
and c = 
a.sin C , 
sin A ’ 
whence, a being = 1000, 
6 is = 1014*69, 
and c is = 1462*50. 
Fig. 29. 
The ratio of the edges of the right section of the defect, the plane angles, and the inclination 
of the primary planes being known, the ratio of the edges of the defect on the plane P is easily 
calculated. 
The triangles DCA and ECB are rightangled, on account 
of the parallelism of the primary edges CA', DD', EE', and 
the section ACB of the defect being a right section. 
DCA = DCA'—ACA'= 93oi4'-90o= 3^14' 
ECB = ECA'-BCA'= 98 26 —90 = 8 26 
In the triangle DCA, DC = 
R X CA 
cos DCA 
= 1015'57 
In the triangle ECB, EC = == 1010*93 
® cos ECB 
In the triangle CDE, two sides and the included angle are 
now known. To determine DE, we have the formula 
