MR. R. T. GLAZEBROOK ON PLANE WAVES IN A BIAXAL CRYSTAL. 297 
To determine the angles 
since AB= -j 
whence 
B A P, B A P, the triangles BAP and B' A P, give 
cos PAB= 
cos /3 
sin a 
similarly from B' A P y 
BAP =88° 46' 40" 
B'AP, =88° 49' 10" 
Eig. 3. 
From triangle PAP ( 
cos PP, = cos AP y cos AP+ sin AP, sin AP cos PAP 
Whence 
PP =137° 9' 55" 
✓ 
The mean of numerous experiments gave 
PP =137° 9' 30" 
The smallness of the difference affords a strong presumption in favour of the 
measurements. 
Let P P / cut A C' in L. 
Then 
• a sin PAP. . . 
sm AP P= . - sin AP 
' sm PP, 
Whence 
APP=0° 3' 23".(7) 
Again 
cot LP, sin AP,= cos AP, cos AP,L-f- sin AP,L cot LAP, 
LP, = 67° 27' 25".(8) 
sin ALP= 
sin AP, sin LAP, 
sin LP, 
ALP=1° 12' 4" 
2 Q 
( 9 ) 
MDCCCLXXIX. 
