MR. R. T. GLAZEBROOK OR PLANE WAVES IN A BIAXAL CRYSTAL. 
325 
Similarly 
where 
Now 
cos 6= 
cos MO cos (PM + A) 
cos A 
( 5 ) 
tan A=tan MO cos x 
A'O = A'O / = 90°—9° 4' 58" [Section V. (6)]. 
= 80° 55' 2" 
From triangle A' M Q 
Whence 
Hence 
sin A'M — sin MQ sin Q 
A'M=1° 23' 52".(6) 
MO = 79° 31' 10". ' .... (7) 
MO / =82° 18' 54".(8) 
X—22° 33' 2" .( 9 ) 
A' = 29 0 39'5".(10) 
PM+X=f+QM+X 
= <f>- +40° 10' 16".(11) 
PM—X^f+QM-X' 
= </>' —12° 1'51".(12) 
Substituting these values of X, MO, PM-j-X, &c. (the values of <f> being taken from 
Tables XIV., XVII., and XX.), in equations (4), (5), we can get a series of values 
for 6 &. 
And as before [Section VI.], remembering that O' here is n—O' in Section VI., 
2v 1 2 = <x 2 +c 2 + (a 2 —c 2 ) cos (0—0') 
2v 2 2 =a 2 +c 2 -f-(a 2 —c 2 ) cos (0-{-$') 
Thus we obtain a series of values for v lt v 2 , and hence for g 1; /x 2 , given in 
Table XXIX. 
To obtain the distance of P from the minimum radius vector, we have to add to c/f 
the angle M Q=17° 37' 14" [Section VIII. (2)]. 
