97 
the angles of crystals. 
be a rhomboid ; in which case, as has already been observed, 
two of the indices in the symbol are equal. 
Prop. Knowing the dihedral angles of the secondary rhom- 
boid, to find the symbol of its planes. 
Let (/), q^q) be the symbol of the planes, 0 the angle of 
{p-,q\ q) and{q;p; q). 
a — 2 pq + f—(p* + 2pq+ 3<7^) cos. « 
2 q * — 2 {z j) q q^) cos. a 
Here cos. 0 being known, we have a quadratic equation to 
determine q in terms of />, which as the proportion q \p only 
is wanted, is sufficient. 
The equation will be 
/)* (cos. ^ cos. a) q(l — cos. a 2 cos. a COS. &) -|- 
q^ 3 cos. a + 2 COS. 0 2 COS. a cos. 0) = 0 
There will be for each value of 5 two values of — , and there- 
p ’ 
fore two laws according to which the same secondary form 
may be produced. It is to be noticed however, that the direc- 
tion of the primitive faces, and consequently of the cleavage 
will be different in the two cases. 
iQ. Prop. It is required to find according to what law we 
shall have a rhomboid similar to the primary one. 
Here Q = u: therefore the first sum of the above equation 
vanishes, and the remaining part will be verified either by 
qz=zo, or by 
^(i — cos. a — 2 cos.* a) + ^(i — COS. a — 2cos.*a) = o, orq = — 
Therefore (i,o, o) and(i, — 2 , — 2 ) each give 0 = a. 
The first indicates the primary face, and the form is the pri- 
mary form. The other indicates a decrement by 2 in height 
on the inferior angle, which it appears gives a rhomboid iden- 
tical with the primary rhomboid. 
MDCCCXXV. O 
