96 
Mr. Whewell on calculating 
be three coexistent planes ; and if each of these planes be 
repeated, we shall have three pairs of parallel planes con- 
taining a rhomboid. 
If the three indices in the symbol (^, q,r] be all different, 
we shall have six planes, and repeating each of these, we 
shall have a dodecahedron consisting of two six-sided pyra- 
mids. To this case belongs the following example : 
Ew. 2. To find the angle of planes in carbonate of lime, 
resulting from the law (i, — 2,0). (Decrement on the 
lateral edges by two rows in breadth. Symbol Chaux 
Carhonatee Metastatique. Hauy.) 
Two adjacent* planes are (1 ; — 2 ; o) (1 ; o ; — 2), and 
preserving the same notation as before 
— ® = 5+4cos.» = — -2525 , 6 = 104“ 38'. 
By other laws we should find other dodecahedrons and 
their angles. But in many cases we have two laws, pro- 
ducing two sets of faces, and it may be required to find the 
angle between those of one set and of the other. 
Ex. 3 . To find the angles of planes (2, — 1,— 1) and 
( t , o, o). (Decrement by two rows in breadth on an inferior 
angle, combined with the primitive faces. Symbol P. Chaux 
Carhonatee Imitable. Hauy). 
Adjacent faces* are (2 ; — 1 ; — 1) and 1 ; o ; o) : and 
A 2 2 COS* £6 /I COS* ce^ a n g 
cos. ^ (6 ^ 6 cos. ») ^ —6— = • 7022 ; = 1 34° 37'- 
9. We proceed now to the inverse problem ; having given 
the angles of the secondary crystal to find the law of its 
planes. And we shall first suppose the secondary form to 
* It will be shown afterwards how we may determine of co-existent planes which 
are adjacent 
