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".