174. Den Ort eines Gestirns ans beobachteten Alignements zu finden. 601 
* 
Zweites Alignement. 
a = 2 s 22° 53' 22" 
a' = 3 s 14° 34' 25" 
a—a = 21° 41' 3" 
10° 50' 31|" 
ß = 66° 3' 50" 
ß' = io 0 4' 35" 
ß ,J rß= .76° 8'25" 
ß' — ß — — 55° 59' 15" 
log tang a -- 2 a = 9,748 40 log tang = 9,282 22 
log sin (b' + b) = 9,937 89 log sin /8) = 9,987 17 
compì, log sin (5' — b) = 0,198 75 compì, log sin {ft—fi) = 0,081 49 n 
log tang (co -f = 9,885 04 log tang ((o' + = 9,350 88 M 
CO 4~ 
37° 30' 14" 
29° 15' 38|" 
co = 8° 14' 35-J-" 
a = 2 s 10° 58' 42" 
“' + 4-^=1670 21' 23" 
u 
10° 50'34" 
co’= 156° 30'54" 
a = 2 s 22° 53'22" 
N= 2 s 2° 44' 6^" iV" = 9 s 16°22'30£" 
N' — N 
E = - = 111049' 12" 
u 
long tang b = 9,264 31 
log sin co = 9,156 47 
log tang i] — 0,107 84 
long tang ß = 0,352 72 
log sin co' = 9,600 45 
log tang # = 0,752 27 
g = 52° 2' 30" 
log tang E = 0,397 53 n 
log sin (# g) = 9,871 02 
compì, log sin (# — t}) = 0,329 43 
log tang x = 0,597 98 w 
# = 79° 58' 6" 
g 52° 2' 30" 
#4- 7? = 132° 0' 36" 
# —?/ = 27° 55'36" 
ic = 284° 9'48" log tang77 = 0,107 84 
E =_ 111° 49' 12" log sin {E+x) = 9,769 04 
E-\-x= 35° 59' 0" log tangB = 9,876 88 
N= 2 s 2° 44' 6|" .5 = 36° 59' 7" 
= 3 s 8° 43' 6£" 
De Lambre findet A = 3 s 8° 43' 8" und B = 36° 59' 8".