12 Hours, is 2 Hours, 57' 44", the true Time of the Break of Day. And thus it begins and ends on the 17th of August. In this Manner you calculate the fame Things, when the Sun has South Declination. To find when the fhorteft Twilight happens in all the Year, is a Problem of a diftinct Nature from this above; and is very abftruse and perplexed in its Demonstration, as may be feen by the Inquifitive in the Works of Gregory, Keil, &c. However I fhall give the Analogy for finding it in any Latitude, as fuppofe this of Chichester 50° 561, and it is as follows. As Radius 10.0000000 Is to the Sine of the Latitude 50° 56′ 9.8900929 So the Tang. of half the Depreffion 9° 00′ 9.1997125 To the Sine of the Sun's Declination 7o 3' 9.0898054 Now the Times of the Year, when the Sun is upon that Parallel of Declination are March the 28th, and October the rft, and on those two Days the Twilight is shortest of all. PROBLEM XX. Given the Latitude and Longitude of a Star or Planet, to find its Declination. Practice. be 18° 2' 30", and its Latitude 22° 51 47" North; I demand its Declination? Having conftruct ed the Scheme from what is given, there will be formed the Oblique Triangle CPE; in which there is known the Side P E = 23° 29', conftantly; the Side C E = 67° 08 13", the Co-Latitude; and the included Angle CEP = 11° 57' 30", the Complement of the Star's Longitude from Aries V; to find the Side PC the Complement of its Declination. Let fall the Perpendicular PR; then by Cafe 4 of Oblique Triangles, Variety 1, thus fay; As Radius 10.0000000 Is to the Co-fine of PEC = 11° 57′ 9.9904699 So is the Tangent of PE = 23° 29′ To the Tangent of ER 23° 02′ = 9.6379563 9.6284262 Then C E-ER=CR = 44° 06'; Whence fay; As the Co-fine of R E=23° 02′ Co. Ar. 0.0360313 Is to the Co-fine of R C 44° 06′ 9.8562088 So is the Co-fine of PE = 23° 29′ 9.9624527 To the Co-fine of PC 44° 161 VOL. II. 9.8547428 Whofe Complement is 45° 42′ = AC the Declination North, as was required. Example 2. Let it be required to find the Declination of Femakent; whofe Longitude is 50", and Latitude 21° 4' 54" South. According to thefe Data, let the Scheme be prepared, and there will be formed the Oblique Triangle CEP, in which there is given the Side PE 23° 29′; the Side E C=1119 4′ 54′′; and the Angle included PEC= 120° 00′; to find the Side PC; which is done as before, by Æ H R N 29° 59′ P letting fall the Perpendicular S R, in the Complemental Triangle C S e. For therein is given the Side Se PE 23° 29'; the Side Ce= Complement of E c to 180° = 68° 55' 06"; and the included Angle Se C = 60° oo'; to find the Side S C, or its Complement C A = the Declination South. Thus ; As Radius Is to the Co-fine of Se C 10.0000000 60° 00' 9.6989700 9.6379563 9.3369263 To the Tangent of e R 12° 15′ Then Then Cee RCR = 56° 40'; then fay; As the Co-fine of e R = 12° 15' Co. Ar. 0.0100023 Is to the Co-fine of RC So is the Co-fine of Se = 56° 40' 23° 29′ 9.7399748 9.9624527 9.7124298 To the Sine of CA= 31° 02′ the Declination fought, and is South. PPOBLEM XXI. Given the Longitude and Latitule of a Star or Planet, to find its Right Afcenfion? Practice. The Right Afcenfion is required of the Goat Star Capella, whofe Longitude is 18° 02' 30", and its Latitude 22° 57' 47" North. In the first Scheme to the last Problem, in the Oblique Triangle CP E, there is given (as is there fpecified) the Sides P E and C E, and the included Ângle at E; to find the Angle CPE, or its Complement EPA the Complement of the Right Afcenfion to 90°. This I fhall do by the Method without a Perpendicular, by Prob. 1, and 2. Thus; So is the Co-Tangent of PEC = 5° 58' 10.9809169 VOL. II. Kk 2 Then Then fay again by Prob. 2. As Co-fine of Sum of Sides 45° 18' C. A. o.1528009 Is to the Co-fine of their Differ. 21° 50′ 9.9676741 So is Co-tang. of inclu. Ang. E=5° 58′ 10.9809169 Add the half Difference The Sum is The Remainder is Av 164 10 the Angle CPE. 90 74 10 the Right Af cenfion from Aries, as was required. In this Manner alfo the Right Afcenfion of Fomabant (in the fecond Scheme to the foregoing Problem) will be found 340° 35' from Aries. PROBLEM XXII. Given the Right Afcenfion and Declination of a Star or Planet, to find its Longitude and Latitude. Practice. This Problem is the Converfe of the two laft exactly For admit the Right Afcenfion of the Goat Star Capella be 74 10', and its Declination 45° 427 North; then in the Triangle C P E, there is given the Side PE 23° 29′; the Sides C P = 44° 16′, the Co-Declination; and the included Angle at P = 164° 10 Right Afcenfion from Capricorn, to find the Angle at E = the Longitude; and CE Co-Latitude of the Star. = the From |