340 
MR. R. T. GLAZEBROOK ON PLANE WAVES IN A BIAXAL CRYSTAL. 
we have seen already that 
For A R' P we have 
On substituting we get 
R'AC=1° 45' 28" 
sin AR'P= sin R'AP 
sm E P 
.‘.ART =59° 18' 37" 
AL=88° 38' 18" 
.\CL=1° 21' 42" 
For R' L the triangle R' A L gives 
sin R'L= 
sin R'AL sin AL 
whence 
sin AR'L 
R'L=2° 2' 37" 
For ALP the triangle ALP gives 
. sin AP . . 
sm AL 1 = ——— sm LAP 
sm LP 
LP=R'P — R'L = 3 5° 0' 19" 
Whence 
ALP = 59° 20' 11" . 
Thus we have determined the position of the plane R' P completely. 
(16) 
(17) 
Section III.— Determination of the Position of the Plane P Q with reference 
to the Principal Planes. 
Our next step will be to determine the position of P Q', Q' being the point in which 
Q O produced backwards cuts the sphere. 
Fig. 5. 
Let the direction angles of O Q' be a 3 /3 S y 3 
