70 PKOCEEDINGS OF THE AMERICAN ACADEMY. 



(Fa) (Fp) (Fy)iF8) ' ' 



21. Tetrahedron. The angles of a triangle will now be expressed 

 in terms of the sides. For the angle C A B oi the triangle A B C we 

 have 



AngleC AB = CA^AB = (^-4^-FC.4) 

 ^ {pAB){pCA) 



^ ((/> A) (FABC) 

 {pAB){pCA)' 



Replacing F by [(f) p] and applying (3) we have 



(0.4) {<j>p-ABC) 



CA,AB = 



(p A B) (p C A) 



(0 .4) 1(0 A) (pBC)+ (0 B) (pCA) + (0 (pAB)] 



ipAB){pCA) 

 Dividing numerator by (0 A)^ (0 B) (0 C), this becomes 



ABC A 



