( 1040 ) 
In AABM is AB=2V; AM A34 DB BMS AI 
LZ BMA=15°55’; from this follows 2V—93°10’30"; V=—46°35/15". 
The extinction-angle with regard to the trace of (O01) in the 
secant-plane J/ (O10) is calculated as follows. If one takes 1 to the 
bisectrix QO. a plane Z, corresponding to the projection-plane in 
fig. 1, one finds for the planes (= WS) and P(—=V) the following 
coordinates of the poles: 
== 7 ANN =e ABaM. == A258! 
D= xa a= Nie, == 1G! 
=— /AaP and Arete igs 
f= 7 Aa a re a 
In order to find u and v one proceeds from the given equations: 
Me 8O ae Me EE Prise 
From these one calculates: 
oT 
cee = 63°37': OME = Baan GS Bi 20°20 
/AMP = 26°23' + 75°36' — 101°59'; 
so that now from AaMP ean be found: 
ea ge = 1016320" "atd p= 119530 5 Nen ree (ate 
Further one finds: 
ZMaP = 86°13'; w= /AaP = 20 — (/MaP + /BaM) = 
= Qa (BO 13! 709) — BOOAB Se (TE 
The planes M and E (1 «Q) cut each other according to the line 
OR; with regard to this line the extinction in Jf amounts to an 
angle y given by (cf. (2)): 
1 — sin? V cos? 7°2' — (1 — sin? V sin? 7°2’) sin? 1°6' 
cot 2y = — —— = NS SE 
sin 14°4 sin 1°6' sin? V 
lg cot 2y = 2,28980 (—). 
and as the pole of M(v—=172°58', o=1°6') lies in the Ied octant 
27a must be taken in the 2rd quadrant, so that 
2y4 == 1805 = hse. 
JT k 
Ya = 5 — SL LEDEN A 
The angle that the trace of P makes with the direction OR is 
found from (5) 
cos 1°6' tg 11°53'20" + sin 1°6' cos 86°13! 
sin 86°13! 
GOE 
lg cot h = 9.52678 (—) 
h =S LTE AAS OENE EEE AEN ee eee 
