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572 MAN OF SCIENCE. [Doc. 200 .
SWEDENBORG AS
145° 30’ + 15 °—6° 40’= 153° 50’, Sine 590 50’=993679
which is the true south angle Sine 650 29 = 995896
towards the west. Sine 51° 52’=989575.
6. For (o h) 16. For (a d)
1530 50’ — 90°=63° 50’ ; then by Sine compl. 65° 29’ = 961800
the sines; Tang. compl. 590 50ʻ=979435
Sine 22° 30’=958284 Tangent 33° 40’ =982365,
Sine 63° 50’=995304 17. For ( b c) has to be added
Sine 200 5’=953588. first 33° 40’ + 26° 37’ = 60 ° 17’ ;
7. For (
pc) and by logarithms:
59° 50’ + 6° 55’=66° 45’. Sine compl. 59° 50’=970115
8. 66° 45’ to be multiplied Sine compl. 60° 17’=969523
by 6 " -6’ 36 " . Sine compl. 75° 35’=939638.
9. 22’ 30 " +6’ 36’’=29’ 6 " . 18. For the angle ( b c a )
10. For ( p 0 ) will be the proportion : as
180 ° + 6 ° 55’ + 20 ° 5’ = 2070 75° 35’ the side already found
11. We require to use the is to 65° 29’, the complement
square : of the apparent north angle ;
as the square 1800-32400 so is 26 ° 37’, the apparent
is to the square 2070=32849, side of the calculus, to the
so is 29’ 6 " to 38’ 22" . angle required.
12. By decimal calculation Sine 26° 37’= 965129
38’ 22’ x 66° 45’=4° 7’. Sine 65° 29’= 995896
13. 22° 30’ + 4° 7’=26° 37’ ; 1961025.
26° 37’ is therefore the appar Sine 75° 35’= 998610
ent latus calculi. Sine 24 ° 53’= 962415.
14. 60° : 90° :: 4 ° 7’ : 6° 11’. 19. We must now find the
See the table of the calcu- side and angle at the south
lation at p. 326 (English edition, pole
Vol. II, p. 160) of the Principia. 590 50’—20° 5’=390 45’.
There are to be added 6° 11’ 20. By decimal calculation
to the true north angle 39° 45’ x 38’ 22" = 2 ° 28’.
108° 20’ + 6° 11’=114° 31’. 21. By reduction according
114 ° 31’ is thus the apparent to the rule at p. 320 (English
north angulus calculi towards edition, Vol. II, p. 160) of
the west ; the complement of the Principia :
which is 65° 29’. as 600 is to 900 so 20 28’
15. For (c d) is to 3° 7’.
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