116 
Mr. Whewell on ealculating 
Also r. cos. XY= cotan. 56® 45'. cotan. 47° 50' •^>'=43° 37' 
and YXE = 180 — 90° 40' = 84® 20' 
.% r. sin. YE=sin. XY. sin. 84^20' ^=sin. 43” 21'; .6864532 
The two planes of which we have to find the angle, are 
(2; o; i)(i; o; o).. 
Hence by the formula, Art. 8, 
a __ cos. $ 
— cos. 6= __ zf-dcos.e 
l/{4/*-4/<Jcos.g+.«} 
To find let tan. u = - ^7 ^ ^ — cotan. /3 ; and 
’ a sin. p d sm. 0 ' 
we shall have, — cos. 6 ~ sin. w . 6 = 90® 4- w. 
By the values above given, we shall find u = 60° 4S' and 
6 = 150® 43'. The value given by Mr. Phillips is 150° 12'. 
It may be observed, that (2 ; o ; 1) is the side adjacent to 
the primary plane (1 ; o ; o); and that we obtain sides adja- 
cent to other faces by taking corresponding co-existent planes 
from the formulse in Art. 32. 
Thus the primary faces (1 ; o ; o) have adjacent secondary 
faces (2 ; o ; 1) and (2 ; 1 ; o). 
The primary faces (o ; 1 ; o) have adjacent (1 ; 2 ; o) and (1 ; — 1 ; 
The primary faces (o ; o ; 1) have adjacent (1 ; o ; 2) and (1 ; 1 ; — 
The primary faces (1 ; 1 ; 1) have adjacent (2 ; 1 ; 2) and (2; 2; 1) 
Here instead of ( — 1 ; o: — 2) &c. we have written (1 ; o; 2) 
&c. which represents the same plane. 
§ 5- Inverse symmetrical Tetrahedron and rhombic Octahedron. 
36. Let A. X y z, fig. 16, be a tetrahedron ; and let its edges 
be bisected, and an octahedron formed as before. In this 
octahedron, let EFHK be the rhombic base; and the two 
