5 
Annalen der Hydrographie und Maritimen Meteorologie, Dezember 1894. 
sin x = cos t/a Vcos # cos d 
sin 2/2 V cos (240 + x) cos (er? — x) 
= 5918 N log cos == 9,99814 
d= 10° 27 N log cos == 9,99274 
ve +d = 15° 45' — 9,99088 
5/2 = 9,99544 
log cos = 9,98501 
log sin = 9,98045 
Yo = Ost 59% 508 
x = 72° 56 
A @+0)= 7° 58 
s = 80° 49‘ 
u=— 65° 3 
log cos = 9,20302 
log cos = 9,62513 
s = 8,82815 
log sin = 9,41408 
ea = 15° 2 
z=: 30° 4 
h =— 59° 56‘ 
Ja 
sem X == cos @ cos d sem (12st — rt, 
o+d+x Y+d—x 
sem zZ == cos “— 1 008 —— a — 
log sem == 9,97006 
log cos = 9,99814 
log cos = 9,99274 
log sem =—" 9,96094- 
i2st — t == 10 0% 208 
= 5°18 N 
d= 10° 27 X 
x = 145° 53 
v+d — 15° 45° 
s =160° 98’ 
sa = 80° 49’ log cos = 9,20302 
9 = 65° 4 log cos = 9,62486 
z= 30° 4 log sem = 8,82788 
h — 59° 56' 
0 
3in x == sec 3 Vcos © cos d sem t 
co8 2/2 == cos az cos x 
t = 1st 59m 40° log sem = 8,82363 
= 5°18 N log cos = 9,99814 
d= 10° 27 N log cos = 9,99274 
—d= 5° 9 ss =— 8,81451 
sj2 =" 9,40726 
U @— = 2° 35 log sec = 10,00044 
log sin = 9,40770 
log cos == 9,99956 
log cos = 9,98531 
log cos = 9,98487 
9 — 15° 2 
z= 30° 4 
h = 59° 56' 
tg x = 2 cos @ cos d cosec (g — d) sem t 
€ Te ZZ 
ig 2/2 = Vie — tg E + x) 
t = 1%t 59m 40) log sem = 8,82363 
= 5° 18} log cos = 9,99814 
d = 10° 27 1 log cos = 9,99274 
p—d= 5° 9 log cosec = 11,04690 
log 2 = 0,30103 
log tg = 10,16244 
log tg = 8,65435 
log tg = 10,20505 
s = 8,85940 
log tg =" 9,42970 
x = 55° 28' 
he — 0 = 2° 30 
s— 58° 3 
6/2 vw 15° 3: 
z= 30° 6 
h = 59° 54