Note; The Declination of this Culminating Point is North, when the Right Afcenfion of Medium Cali, is less then 180°, as here; but when it exceeds 180°, then is the Declination South. PROBLEM XXVII. Given the Latitude of the Place, and Declination of the Culminating Point, to find the Altitude of Medium Cæli, or Culminating Point in the Ecliptic. Practice. Admit the Latitude be 50° 56', and Declination of the Culminating Point 15° 361 North. Then to the Complement of the Latitude 39 04 15 36 54 40 If the Declination were South, the Difference between it and the Complement of the Latitude, is the Altitude fought, viz. 23° 28′ in the present Cafe. PROBLEM XXVIII. Given the Altitude of Mid-heaven, and the Meridian Angle, to find the Altitude of the Nonagefima Degree, or the Angle the Ecliptic makes with the Horizon. Practice In the Scheme to Prob. 24, and in the Rectangle Triangle EH M, right-angled at H, there is given the Side HE=54940, the Altitude of Mid-heaven, as found by the laft; and the Meridian Angle HEM 729 14', as found per Prob. 25; to find the An gle gle H M E, the Altitude of the Nonagefima, or 90th Degree diftant from M in the Ecliptic. To do which this is As Radius The Proportion. To the Sine of E = 72° 14′ So is the Co-fine of HE = 54° 40′ 10.0000000 9.9787770 9.7621775 To the Co.fine of HME = 56° 35′ 9.7409545 the Altitude of the Nonagefima Degree, as required. PROBLEM XXIX. Given the Altitude of Mid-heaven, and the Meridian Angle, to find the Place of the Nonagefima Degree, Practice. In the aforefaid Triangle, there are the fame things given to find the Side M E, whofe Complement to 90°, is the Arch, which deducted from the Place of Mid-heaven, fatisfies the Problem. As Radius To the Co-fine of H E 72° 14′ 10.0000000 9.4846630 9.8505931 9.3342561 To the Co-tangent of ME = 77° 49' whofe Complement to 90° is 12° 11'; this fubducted from the Place of Mid-heaven, or Point in the Ecliptic culminating, which lies in a 17° 321, (as per Prob. 24,) and there will remain the Place of the Nonagefima Degree; viz. in Leo 05° 21' as was required. VOL. II. L12 Note. Note. When the Place SHY 8 I add y of Mid-heaven is in mm 7 fub. the Complement of E M to or from the Place of the Mid-heaven; the Sum, or Difference (as here) is the true Place of the Nonagefima Degree. See Leadbetter's Aftronomy, Page 167. PROBLEM XXX. Given the fame Things, as in the two laft, to find the Cufp of the Afcendant, or Horoscope, or Minute of the Ecliptic afcending the Horizon. Practice. The Place of Medium Cali, or the Point E, as S found by Prob. 24, is distant from Aries 4 To which add the Arch EM = 2 The Sum is the Place of the Afcendant 7 That is in the Sign Scorpio m 5° 21' Or thus ; S O To the Place of the Nonag. Deg. 4 05 21 The Sun is the Place of the Afc. 7 05 21, as before, ୪ 0 The Dif. is the Place of the Defc. 1 05 21 PROBLEM XXXI. Given the Sun's Altitude and Distance from the Afcendant or Defcendant, to find the Parallactic Angle. Practice. find the Angle B C M made by the Ecliptic, and Circle of Altitude. To which fay thus As Radius ; -10.0000000 Is to Co-tangent of CM 50° 22 9.9181627 So is the Tangent of B C 40° 00' 9.9238135 the Quantity of the Parallactic Angle, as required. Note. Because the Co-tangent of C M, is the Tangent of the Sun's Distance from the Nonagefima Degree, 'tis plain if that be given, the Problem may be answered with only that Difference. PROBLEM XXXII. Given the Latitude of the Place, and Oblique Af cenfion of a Star or Planet; to find the Time of its cofmical Rifing. Pralice. there will be formed the Oblique Trian gle YB A, in which there is given the Angle at Y = 23 291, the greatest Declination of the Sun; The Angle YBA 140° 56', the Supplement of the Co-Latitude to 180°; and the Side included ? B= 29° 54′, the Oblique Afcenfion; to find the Side YA, the Point A being that Point in the Ecliptic which rifeth with the Star Fomabant. In order to this, let fall the Perpendicular FD, then find the Angle Dr B, by this Proportion; As Radius Is to the Co-fine of YB = 29° 54′ 10.0000000 9.9379674 39° 04′ 9.9094022 To the Co-tangent of D 2 B = 548 521 9.8473696 Then |