146 Dr. Robertson's demonstrations of the late 
PA, EF, and in this case SAR, in the first quadrant, is equal 
to B™qo°. Again in Fig. 3. let S be a star at the intersection 
of the arcs p A, e¥, and in this case SAQ, in the fourth qua- 
drant = 90° = B. In Fig. 6 . let S be a star at the intersection 
of the arcs PA, EF, and in this case SAR, in the second qua- 
drant, == 90° == B. Lastly, in Fig. 6 . let S be a star at the in- 
tersection of the arcs pA, rF, and in this case SAQ, in the 
third quadrant, = c)0 o =B. Hence it follows that if B = 90°, 
the star must be in PA />, and its right ascension as stated in 
the rule. 
'Demonstration of the third Rule. 
In Fip*. 2. let S be a star between the arcs EA, PA, whose 
longitude F is in the first quadrant, but its right ascension H 
in the fourth, and then according to the rule SAR, which is 
greater than PAR or 90° is equal to B. Again in Fig. 2. let 
S be a star between the arcs^A, <?A, whole longitude F is in 
the fourth quadrant, but its right ascension H in the first, and 
then according to the rule, SAG in the fourth, which is greater 
than ^>AQ or 90°, is equal to B. In Fig. 5. let S be a star be- 
tween the arcs PA, EA, whose longitude F is in the second 
quadrant, but its right ascension H in the third, and then ac- 
cording to the rule, SAR which is greater than PAR or 90° is 
equal to B. Lastly, in Fig. 5. let S be a star between the 
arcs pA, eA, whose longitude F is in the third quadrant, 
but its right ascension H in the secondhand then, accord- 
ing to the rule, SAQ, which is greater than pAQ, or 90°, is 
equal to B. 
It therefore follows, that if B be greater than go° the celes- 
tial object must be situated between EA and PA, or between 
|A and pA : for if it be not so situated, B will not be greater 
