MR. R. T. G-LAZEBROOK OR PLARE WAVES IR A BIAXAL CRYSTAL. 
339 
We have now sufficient data to calculate the angle R' P. 
CP = yi 
AP = a 1 
BP=A 
CRL=7r— y 2 
AR' = 7r — a. : = 89° 40' 50" 
B'R'=/A. 
From triangle B' A IP 
whence 
From triangle CAP 
cos B'R' = cos BAR' sin AR' 
cos B'AR'= 
sill AR 
_ cos /3 3 
sin a„ 
B'AR' = 88° 14' 32" 
R'AC= 1° 45' 28" 
cos CP — cos CAP sin AP 
~, -r, cos CP 
cos CAP= - -- 
sin AP 
CAP= 31° 17' 20" 
R'AP = 33° 2' 48" 
From triangle R' A P 
cos RT = cos R'A cos AP + sin R'A sin AP cos R'AP 
= cos (tt — a 3 ) cos a x + sin (rr — a 3 ) sin a x cos R'AP 
whence 
R'P = 37° 2' 53" 
But R'P can he measured directly, the mean of twelve observations, none of which 
differed from the mean by 10", gave 
R'P = 37° 2' 56" 
The close agreement between these results confirms the accuracy of the values of 
°h, A, &c - 
Let P R' cut C A in L, we proceed to determine C L, R' L, and the angle ALP. 
From triangle AR'L 
cot AL sin APt'= cot ART sin R'AC-j- cos APf cos R'A.C 
cot AL sin a 2 = cot AR P sin R'AC+ cos a 2 cos R'AC 
2x2 
