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To find the latitude and longitude of a star.

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in which the signs are used as in the preceding equation ; so that A and Dec. are always positive or negative at the same time. Instead of (458), its reciprocal may be used, which is

Ftang. A= cosec. R. A. tang. Dec. (459) Jl, then, B=E+A

(460) we have AP=FE - 90° F A= FB - 90° (461)

= 90° + A + E= 90° + B, in which the upper or lower signs are used, as in (457). Hence cos. PC: cos, AP= F sin. A: Fsin. B=sin. A : sin. B – sin. Dec. : sin. Lat.

(462) so that, since Dec. and A are both positive or both negative, B and Lat. must also be both positive or both negative. Again,

sin. PC: sio, PA = cos. A:+ cos. B (463) = + cotang: (R. A. + 6): + cotan. (Long. - 90°)

= + tang. R. A.: + tang. Long. in which the signs may be neglected, and Long. is to be found in the same quadrant with R. A., unless the foot P of the perpendicular falls within the triangle ; in which case the first value of AP (461) is used, so that B is obtuse. In this case, the longitude is in the adjacent quadrant on the same side of the solstitial colure with the right ascension. These results agree with the Rule in [B. p. 145.]

63. Corollary. The latitude and longitude of the zenith, that is, the zenith distance and longitude of the nonagesimal, might be found by the same method. But another rule

To find the latitude and longitude of a star.

can be used, which is of peculiar advantage, where these quantities are often to be calculated for the same place. We have by (369) and (370), calling B the zenith, and putting

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F=} (PZB-ZBP) or = 180°—} (PZB-ZBP) (461) G=} (PZB+ZBP) or=180°—}(PZB+ZPB) (465) tang. F=-cosec. 3 (PB+PZ) sin. 3 (PB-PZ) cotan.LT

= tang. (21" — F) (466) tang. G=-sec.3(PB+PZ)cos.3(PB_PZ)cot. ZT (467) 90° + F+G=PZB+90° or = 360°—PZB+90° (468)

= Long. or = 360° + Long. (469)

in which the first member of (466) is used when PB is greater than PZ, and the third when PB is less than PZ, that is, within the north polar circle; and the second members of (464, 465, 468) correspond to the position of the zenith at the east of the solstitial colure, but the third members to the west of the colure.

Again, by (354),

tang. } (90°—lat) = tang. j alt. nonagesimal

= cos. G. sec. F tang. 3 (PB + PZ), (470) and the preceding formulas correspond to the rule in [B. p. 402.]

64. Scholium. The rule with regard to the values of G appears to be a little different, but the difference is only apTo find the altitude of the nonagesimal.

parent; for it follows from (467), that G and 12 – IT are, at the same time, both acute or both obtuse, unless

(PB + PZ) > 90°, PB> 180° - PZ,

or

(471)

which corresponds to the south polar circle.

A=

65. The abridged method of calculating the altitude and longitude of the nonagesimal [B. p. 403], only consists in the previous computation of the values

log. [cos. (1 PB - PZ) sec. 1 (PB + PZ)] (472) C = log. tang. * (PB + PZ)

(473) B= log. tang. * (PB-PZ) - C

(474) = log. (tang. 1 (PB PZ) cotan. ] (PB+PZ)] =log. [cosec. }(PB + PZ) sin. 3 (PB - PZ)]— A,

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whence

Jog. [cosec. } (PB+PZ) sin. 2 (PB-PZ)]=B+A (475) and log. tang. G=A+log. (-cotan. T) (476) log. tang. F=A+B+log. (-cotan. 1 T) (477)

=log.tang. G + B

log. tang. alt, non=log. cos. G+log. sec. F+C. (478)

66. The rule in [B. p. 436] for finding right ascension and declination, when the longitude and latitude are given, may be obtained by a process precisely similar to that for the rule before it.

To find the latitude and longitude of a star.

67. EXAMPLES.

1. Calculate the latitude and longitude of the moon, when its right ascension is 4" 42° 56', and its declination 27° 21' 58" N., and the obliquity of the ecliptic 23° 27' 45'.

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2. Calculate the values of A, B, and C, for the obliquity 23° 27' 40", and the reduced latitude of 42° 12' 2" N. Solution.

Polar dist. = 47° 47' 58"
47° 47' 55"
23° 27' 40"

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3. Calculate the altitude and longitude of the nonagesimal, when the right ascension of the meridian is 19' 50m, the latitude 42° 12' 2". N., and the obliquity 23° 27' 40".

To find the latitude and longitude of a star.

Solution. T= 19 50m + 6 — 24 = 1* 50m į (1* 50m) cotan. 0.61137

A = 0.08015

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tang. 9.40671

long. = 315° 34' 5" 14° 18' 40"

alt. = 28° 37' 20".

4. Calculate the latitude and longitude of the moon, when its right ascension is 18" 27m 12, and its declination 27° 49' 38" S., and the obliquity of the ecliptic 23° 27' 45". Ans. The D's long. = 276° 1'44"

4° 30' 27" S.

Its lat. =

5. Calculate the values of A, B, and C, for Albany in reduced latitude 42° 27' 39", and for the obliquity 23° 27' 40".

Ans, A = 0.07965

1

B = 9.47565

C= 9.85327

6. Calculate the longitude and altitude of the nonagesimal, when the obliquity of the ecliptic is 23° 27' 40", the latitude 42° 12' 2" N., and the R. A. of the meridian 10% 10%. Ans. The long. = 138° 30' 25'

61° 18' 46".

alt. =

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