LAWS OP DECREMENT. 329 



OCTAHEDRON, SQUARE BASE. 



We now require the plane angle o n t of the defect, 

 which being denoted by A 6) may be thus found, 

 cos. A 6 cot, i / 2 . cot. (180 J 5 ) 



~R~ 



From this angle and o t ft, which we have called 

 A z , the ratio may be found of t n : t o :: p : q, 

 and r may be determined as before. 



The planes of class h result from simple and mixed 

 decrements on the lateral angles. Let /^ represent 

 the inclination on P of one of the planes h adjacent 

 to it, and the law of decrement may be thus determined, 

 P _ sin. (/ t -}-7 6 180") 

 q " sin. (180 1 6 ) 



Let q r p s, fig. 342, represent a plane belonging 

 to class i or /:, whose general symbol would be 



(B P D q D'r B" s : B P Dr D' q B" s ) 

 and its inclination on P be called / 7 , and on P', / 8 . 

 A spherical triangle will give the plane angles r t q, 

 r t p, and hence the following ratios are known, 

 t r : t q :: p m : q n 

 t r : t p :: p m : r n 



And if we divided and ^ by sin '^% the par- 

 q n r n sin. A , 



ticular values of p, q, and r will be found. 



The fourth index s, may also be known from the 

 equation, 



L = 1 + I I 

 s q r p 



The decrements on the edges may be determined 

 by the general methods applied to analagous decre- 

 ments on the regular octahedron. 



