MR. R. T. GLAZEBROOK ON PLANE WAVES IN A BIAXAL CRYSTAL. 303 
MQ /= 57° 51' 30" = 9 say; 
M/Q^SO 0 9'=9' say ; 
and the triangles M A Q, M/A Q, give, 
cos AQ = 
cos 6 + cos 0’ 
2 cos 
where /x=AM=M/A. 
Whence substituting for 9 and 9' 
AQ, = 2° 43' 20" 
( 2 ) 
In the triangle M A Q / the three sides are known; we can thence find the angle 
M A Q / and obtain 
MAQ 7 = 76° 24" 45'.(3) 
In the triangle Q A Q / the sides A Q, A Q, and the angle Q A Q / are known 
AQ = 45° 30' 
AQ y = 2° 43' 20" 
QAQ,= 103° 35' 15" 
Whence from the formula 
cos QQ,= cos AQ cos AQ,+ sin AQ sin AQ / cos QAQ, 
we get 
QQ / = 46° 12' (or rather less). 
The mean of several experiments had already given for the value of QQ / 
46° 10' 
These results, differing as they do by less than 2/ afford a verification of the 
accuracy of the measurements. 
