the angles of crystals 
119 
The calculations .would .be nearly the same as in the case 
of the square octahedron, article 35. We should have to 
calculate f e,f from the angles of the octahedron. Thus in 
sulphur, according to Mr. Phillips (p. 361) we have inci- 
dence of 
GEF on GEK = 106° 30 ; angle at A jc = 73® 30 = y 
GFHonGFE= 85 * 5 ; angle at Ay = 94® 55 = 0 
GHF on DHF = 143° 25 ; angle at A 2; = 36° 3^ == a, 
And if we construct a triangle, of which the three angles are 
a, 0 , y, and draw arcs from these angles perpendicular on 
the opposite sides, the sines of these arcs will be respectively 
d, e,f. And by first finding the sides of the triangle by 
spherical trigonometry, these may be calculated. 
§ 6 . The regular' triangular Prism. Fig. 19- 
39 - This is a right prism, having for its base an equilateral 
triangle. It includes the regular hexagonal prism by re- 
peating the lateral faces. 
Prop. To find the co-existent planes. 
By the law of symmetry, for every plane on one angle A, 
we must have co-existent planes onx,y. Let/>^rbeany 
plane whose symbol is (/>; q\ r), .and the lines Kp = h, 
Aq = k, Ar=l, when p = -^, g = r=-i. Then we 
shall have a plane POR where xF = hy x Q=: x R=: 1. 
Draw a: M, y O parallel to PQ. 
Similarly if a: N be parallel to RP, AN = Aa:.~ = -^=:-^. 
'' X r n r 
Hence the equation of the plane x MN is 
xO — xy. ^ = Aif Ax = X3/ = 1. 
k 
I - 
P 
