Note ; The Declination of this Culminating Point is North, when the Right Ascension of Medium Cæli, 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 Add the Declination of Medium Cæli The Sum is the Altitude of Medium Cæli 54 40 15 36 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 Case. PROBLEM XXVIII. Given the Altitude of Mid-heaven, and the Meridian Angle, to find the Altitude of the Nonagesima Degree, or the Angle the Ecliptic makes with the Horizon. PraEtice. In the Scheme to Prob. 24, and in the Rectangle Triangle E HM, right-angled at H, there is given the Side HE=548 40', the Altitude of Mid-heaven, as found by the last ; and the Meridian Angle HEM = 729 14', as found per Prob. 25 ; to find the An gle L ᄂ gle H M E, the Altitude of the Nonagesima, or goth Degree distant from M in the Ecliptic. To do which this is The Proportion. As Radius 10,0000000 To the Sine of E = 724 14' 9.9787770 So is the Co-line of HE = 54° 40' 9.762 1775 To the Co. sine of HME = 56° 35' 9.7409545 the Altitude of the Nonagesima Degree, as required. PROBLEM XXIX. Given the Altitude of Mid-heaven, and the Meridian Angle, to find the Place of the Nonagesima Degree. Practice. In the aforesaid Triangle, there are the same things given to find the Side M E, whose Complement to 90°, is the Arch, which deducted from the Place of Mid-heaven, fatisfies the Problem. As Radius 10.0000000 To the Co-fine of H E= 72° 14' 9.4846630 So is the Co-tangent of HE = 54° 40' 9.850593 I To the Co-tangent of ME = 77° 49' 9.3342561 whose Complement to 90° is 12° 11'; this subducted from the Place of Mid-heaven, or Point in the Ecliptic culminating, which lies in & 17° 32', ( as per Prob. 24, ) and there will remain the Place of the Nonagesima Degree ; viz. in Leo 2 059 21' as was required. VOL. II. L 1 2 Note. Note. When the Places im H 8 Il add ? of Mid-heaven is in { co m TI, fub. S 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 Nonagesima Degree. See Leadbetter's Astronomy, Page 167. PROBLEM XXX. Given the same Things, as in the two last, to find the Cufp of the Ascendant, or Horoscope, or Minute of the Ecliptic ascending the Horizon. Practice. The Place of Medium Cæli, or the Point E, as S 21 found by Prob. 24, is distant from Aries 4 17 32 To which add the Arch E M = 2 17 49 The Sum is the Place of the Ascendant 7 05 That is in the Sign Scorpio m 5° 21'. Or thus ; S m The Sum is the Place of the Asc. 7 05 21, as before, The Dif. is the Place of the Desc. I 05 21 PROBLEM XXXI. Given the Sun's Altitude and Distance from the Afcendant or Descendant, to find the Parallactic Angle. PraEtice. Practice. Adniit the Sun's Place be in my 149 E P 59; and his Altitude Æ 400 oo'; then in the Scheme adjoin'd, and in the Right-angled H Triangle BCM, there B M is given B C = 40° 00', the Altitude ; and CM = 50° 22', the Distance of m 14° æ 59' from the Ascen S dant m 05° 21'; to find the Angle BCM made by the Ecliptic, and Circle of Altitude. To which say thus ; 9.9181627 As Radius 10.0000000 Is to Co-tangent of CM = 50° 22' So is the Tangent of B C = 40° oo' 9.9238135 To the Co-fine of BCM = 45° 59' 9.8419762 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 Nonagesima Degree, 'cís plain if that be given, the Problem may be answered with only that Difference. PROBLEM XXXII. Given the Latitude of the Place, and Oblique Afcension of a Star or Planet ; to find the Time of its cosmica! Rising PraEtice, Pratice. Admit the Oblique Afcenfion of Fomabant be 29° 54', AC P ( as found by the foregoing Problems, E I demand the Time of his Cofmical Ri H A 0 sing at Chichester, Lat. 509 56' ? According to this, let the Scheme be constructed ; and S there will be formed the Oblique Triangle YBA, in which there is given the Angle at 1 = 23° 29', the greatest Declination of the Sun; The Angle Y B A = 140° 56', the Supplement of the Co-Latitude to 180°; and the Side included 1 B 29° 54', the Oblique Ascension ; to find the Side » A, the Point A being that Point in the Ecliptic which riseth with the Star Fomabant. In order to this, let fall the Perpendicular Y D, then find the Angle DY B, by this Proportion ; As Radius 10.0000000 Is to the Co-fine of YB = 29° 54' 9.9379674 So is the Tangent of P BI= 39°04' 9.9094022 To the Co-tangent of D 1 B = 549 52' 9.8473696 Then |