of the Equinoxes. 79 
revolution ; or, which comes to the same, let the spherical angle 
FBL or ABG be equal to 360 x ^4 x mn x a ■ J - - . Let h =, 
f!r£. Then by spherical Trigonometry, sin. F : sin. BL : : sin. 
B : sin. FL ; and therefore, as FL and the angle FBL are ex- 
"xt t 
tremely small, the momentary precession FL = 360 x ~ x 
, sin. BL 
hmn * 
23. If FG be bisected in C, and the arc CE be perpendi- 
cular to FG, meeting LBA in E, and FBG in D ; then the arc 
CD is the measure of the angle at F. Also, as FL is ex- 
tremely small, CE may be taken for the measure of the angle 
at L. Hence as BDE is a right angle, radius : sin. BE or cos. 
BL: : sin. EBD : sin. ED. Consequently as ED is extremely 
small, 360 x x hmn x = ED, the momentary nuta- 
tion, or the momentary change in the inclination of the equator 
to the ecliptic. 
From the last proportion, and that concluding the preceding 
. , . n ,1 , r' T sin.BL cos. BL radius x sin. BL 
article, it follows that FL : ED : : — — ^ : — -r. — : : kt 
’ sin. F radius cos. BL 
: sin. F. But cos. BL : radius : : sin. BL : — tang. 
BL ; and therefore, the momentary precession FL is to the 
momentary nutation ED, as the tangent of the right ascension 
BL to the sine of the obliquity of the ecliptic. 
24. Let h = the sine of the obliquity of the ecliptic, c — its 
cosine ; z = the arc LS, x — its sine, and y = its cosine ; and 
let 2 .p = the circumference of the ecliptic. Then as LBS is a 
right angle, by the circular parts, cos. BS x cos. BL = radius 
x cos. LS; that is n x cos. BL=y, radius being 1. Again, 
radius : x : : b : bx = the sine of BS = m. Consequently, cos. 
BS x sin. BS x cos. BL = mn x cos. BL = bxy ; and therefore* 
