342 
MR. R. T. GLAZEBROOK ON PLAYE WAVES IN A BIAXAL CRYSTAL. 
Let the figure, as before, represent the sphere as seen by an eye on O C produced. 
Fig. 6. 
Let Q' P produced cut C A in L' we proceed to determine C Lb P L', and P L' A. 
From the triangle L' A P 
cot AL' sin AP= cot APL' sin L'AP-j- cos AP' cos L'AP 
From triangle P Q' A. 
sin APL' = 
sin AQ' 
sin PQ' 
sin PAQ' 
Also 
7r-APL' = 64° 41' 28" 
L'AP = CAP=31° 17' 20" 
[Section II. from previous work.] 
Substituting these values we get 
AL'=88° 44' 20" 
CL'=1° 15' 40" 
The triangle A L' P gives 
a t /T-. sin AP . * t~,t / 
sin AL P= ——— sin APL 
nin A I ' 
whence 
ALT- 59° 13' 2" . . 
Also 
. T sin AL' . tmt, 
sill L P= -— sm L AP 
sm APL 
whence 
L'P —35° 3' 14" . . 
(7) 
( 8 ) 
(9) 
To corroborate these results draw L K perpendicular to P L', then L K L' is 
approximately a right-angled triangle, and angle LL'K=60° nearly. 
Hence 
L'K=^LL' approximately 
Now L'K=PL'—PL = 3' approximately 
LL' = CL — CL' = 6' approximately 
.•.L'K is equal to ^LL' 
