Answer. Right ascension=198°.34'.32′′, and declination= 10°.4.31" South. 3. Required the right ascension and declination of Betelguese in the shoulder of Orion; latitude 16°.3'.3" South, longitude 25.25°.51'.46"; obliquity of the ecliptic 23°.28'? Answer. Right ascension 85°.59'.28", declination=7°.21′. 17" North. 4. Required the right ascension and declination of Antares in Scorpio; latitude 4°.32′.35" S., longitude 8.6°.51′.51′′; obliquity of the ecliptic 23°.28'? Answer. Right ascen.244°.11.3", declination=25°.57′.22′′ South. 5. Required the right ascension and declination of the moon, when her latitude is 4°.0'.34" N., longitude 78.14°.26'.21", and the obliquity of the ecliptic 23°.27′.48"? Answer. Right ascension=223°.11'.11", declin. -12°.21'.14" South. 6. Given the right ascension of Aldebaran=4b.25'.43", and its declination 16°.8'.36" N. in the year 1822, required its latitude and longitude; the obliquity of the ecliptic being 23°.28'? In the triangle nnc. nNc=156°.25′.45"AR+90°, AR being 4h.25.43". 1. To find the co-latitude nc. Draw the perpendicular CL, Then, in || In the oblique-angled triangle nnc the right-angled triangle CLN, rad. sine of 90° : COS CNL=RQ=23°.34'.15′′ ::tang NC=73°51′.24" : tang NL 72°.28.25". (B. 178.) · Then NL+Nn=nL=95°,56'.25" 9.01490 9.44417 8.98030 Hence the lat=5°.29.1" South. The latitude of this star, on the same day, in the year 1796, was 50.28′,27" S. making a difference of only a few seconds in 26 years. (See the note page 306.) 2. To find the co-longitude cnn. Sine nc 95°.28'.59"; sine nNc = 156°.25'.45":: sine NC = 73°.51′.24′′: sine cnN=22°.41′.53". Hence the longitude = 67°.18.7", or 2.70.18'.7". 7. Required the latitude and longitude of Sirius in the Great Dog, right ascension being 99.0.21", and declination=16°. 26.35′′ South; the obliquity of the ecliptic being 23°.28′? Answer. Latitude 39°.33.30" S. longitude 3.11°.13′.44′′. 8. Required the latitude and longitude of Procyon in the Little Dog; right ascension being 112°.6′.47′′, declination 5o. 45.3" N.; obliquity of the ecliptic 23°.28'? Answer. Latitude 15°.58.14" S., longitude 3s.22 .55'.42". 9. Required the latitude and longitude of Arcturus in Bootes: right ascension 211°.33′.17", declination 20°.15′.58" N.; the obliquity of the ecliptic being 23°.28'? Answer. Lat. 30°.52'.43′′ N., long. 6.21°.20.41′′. 10. Required the latitude and longitude of Fomalhaut, in the Southern Fish; right ascension being 341°.32′.34", declination 30°42′.51′′ S.; obliquity of the ecliptic 23°.28'? Answer. Lat. 21°.6.29" S., long. 11.0°.56'.42". 11. Required the latitude and longitude of the moon, on the 16th of June 1822, when her right ascension is 41°.24, declination 21°.21' N., and the obliquity of the ecliptic 23°27′53′′? Answer. Lat. 5.6′ N., longitude 45°.28'. PROBLEM XI. (Plate III. Fig. 8.) (W) The right ascensions and declinations of two stars, or the latitudes and longitudes of two stars given, to find their distance. Required the distance between Sirius in the Great Dog, right ascension=99°.0.21", and declination=16°.26'.35" S., and Procyon in the Little Dog, right ascension=112°.6′47′′, and declination 5°.45'.3" N.? * The longitude of this star in the year 1796 was found, by the same process, to be 66°.56′.24", making an increase of 21.43" in 26 years. By comparing together the places of the fixed stars deduced from observation, astronomers have found that their longitudes increase about 501" annually. This increase of longitude must necessarily cause an irregular motion in the same star with regard to the equinoctial. Hence the right ascensions and declinations of stars are continually varying, so that stars which formerly had north declination have now south declination, et contra. Their latitudes are also subject to a small variation, See the Tables, page 153, &c. Naut. Alm. 1822. In the triangle xST let the star at T represent Procyon, because its right ascension is the greatest, and the star at x represent Sirius. The XST 13°.6'.26" diff. of their right ascensions=RQ. sx=73°.33.25′′ the complement of Sirius's declination RX. ST 95°.45.3" Procyon's declination +90° sQ+QT. Hence, here are two sides and the contained angle given, to find the third side xT, by any of the methods in the preceding problems. Answer. XT=25°.42′.10". EXAMPLE II. The latitude and longitude of Sirius being 39°.33.30′′ Si, and 3o.11°.13′.44′′; and the latitude and longitude of Procyon =15°.58′.14′′ South, and 3o.22°.55′.42". Required their distance? In the triangle xmт as before, we have xm=50°.26′.30" the complement of Sirius's lat. vx. TMm=74°.1′.46′′ the complement of Procyon's lat. T. The xmT=110.41.58" the difference of their longitudes V3. Hence, as before, here are two sides and the contained angle given, to find the third side, XT=25°.42′ Answer. EXAMPLE III. Required the distance between Capella in the Goat, right ascension=75°.21′.19", declination=45°.46′.15" North, and Procyon in the Little Dog, right ascension=112°.6′.47′′, and declination=5°.45'.3" North? In the triangle TNC, let the star at T represent Procyon, and c Capella. NC 440.13.45" the co-declination of Capella. TN 84°.14.57" the co-declination of Procyon. TNC 36°.45′.28" the diff. of their right ascensions=RQ. Here we have two sides and the contained angle given, to find the third side, as in the preceding example. Answer, CT 51°.6'.56". EXAMPLE IV. Required the distance between Capella, whose latitude is 220.51.57" North, longitude 2o.18°.57.57", and Procyon, latitude 150.58'.14" South, longitude 3o.22°.55.42′′? In the triangle тnc. Let the star at T represent Procyon, that at c, Capella. nc пт =67°. 8'. 3"comp. Capella's lat. oc. ▲ Tnc= 33°.57′.45"=diff. of their longitudes=0. Here we have two sides and the contained angle given, to find the third side cT-51°.6'.39" Answer. PROBLEM XII. (Plate III. Fig. 8.) (X) The places of two stars being given, and their distances from a third star, to find the place of this third star. EXAMPLE 1. Suppose the distance of a comet or new star c, was measured by a sextant to be 65°.47.42′′ from Sirius the dog star, lat. 39°.33.30" South, longitude 3.110.13.44"; and 51°.6′.56", from Procyon the Little Dog, lat. 15°.58.14" South, longitude 38.22°.55′.42"; it is required to find the latitude and longitude of this comet or star. Let x represent the place of Sirius, or one of the known stars, T that of Procyon the other known star, having the greatest longitude, and c the place of the comet or unknown star. First. In the triangle xnT, we have given xn=129°.33.30" Sirius's distance from the pole of the ecliptic nearest the comet or star; Tn=105°.58′.14′′ Procyon's distance from the same pole. xNT 110.41'.58" the diff. of longitude between the given stars. To find XT the distance between the two known stars= 25°.42′.10", and the angle nxT 26°.42′.39": To find the former of these quantities, there are two sides and the included angle given, and to find the latter the three sides and one angle are given. Secondly. In the triangle XTC, we have given XT=25°.42′.10") To find the CXT at the star x. By TC-51°. 6.56" any of the methods (W.231. or P. 248,) and xc 65°.47.42" you will find cxT=49°.14′. This angle being greater than nxt shews the longitude of the star at c to be less than at x. Hence CXT ~ nxt = 22°.31'.21" =nxc. Thirdly. In the triangle xnc, we have given xn=129°.33'.30" xc= 65°.47'.42" dnxc= 22°.31.21" To find the side nc the comet or star's co-lat. and xnc the diff. between the star's long, at c and Sirius's long. nc=67°.8'.26" the star's co-latitude, hence its latitude oc= 22°.51′.34", and xnc=22°.16′.42", and since the star's longitude has been shewn to be greater than the comet's; from Sirius's longitude 101°.13.44" take 22°.16′.42′′, the remainder 78°.57.2" is the unknown star's longitude. EXAMPLE II. Suppose the distance of an unknown star at x was measured to be 65°.47.42", from a known star (Capella) at c having 22°. 51'.57′′ North latitude, and 2$.18°.57.57′′ of longitude; and its distance from another known star at T (Procyon)=25°.42′.10′′, this star having 15°.58'.14" South latitude, and 35.22°.55′.42′′ of longitude; required the latitude and longitude of the unknown star? First. In the triangle cmT, we shall have cm=mo+oc=112°.51'.57′′ Capella's distance from the pole of the ecliptic nearest the comet, or unknown star. = CMT=33°.57.45" the diff. long. between Capella and Procyon. TMm = 74°.1′.46′′- Procyon's distance from the pole of the ecliptic. To find TC 51°.6′.56", the distance between the known stars; and the angle mcT=43°.37.41". Secondly. In the triangle xCT, we have given xc=65°.47′.42′′ XT=25°.42.10" To find the angle xcr at the star c. By any of the methods (W. 231. or P. 248.) xCT will be found 24°.58'; this angle being less than mcr shews the longitude of the star at x to be greater than that at c. Hence mcr XCT 18°.39.41"-xcm. Thirdly. In the triangle xcm, we have cm=112°.51′.57" To find xm the unknown star's co cx= 65°.47′.42′′ latitude, and xmc the difference be4xcm= 18°.39.41"tween the star's longitude at x, and Capella's longitude at c. xm=50°.26′ the star's co-lat., hence its latitude vx=39°.34', and xmc=22°.15'.6", and since the given star's longitude has been shewn to be less than the required star's longitude, this difference must be added to Capella's longitude, viz. to 25.18°. 57′.57′′ add 22°.15'.6", the sum is 3.11°.13.3" the longitude required. |