From the Angle C, let fall the Perpendicular C L on the Side E P; then in the Right-angled Triangle CI P, find L P thus; As Radius 10.0000000 Is to the Co-fine of L P C = 15° 50′ 9.9832019 So is the Tangent of C P = 44° 161 9.9888816 To the Tangent of L P = 43° 10/ 9.9720835 Then LP PELE 66° 39'; then fay, 44° 16' 9.8547428 9.5980754 67° 08' 9.5898712 As the Co.fine of LP = 43° 10′ C. A. 0.1370530 Is to the Co-fine of C P = So is the Co-fine of L E = To the Co-fine of C E Which Complement is tude North. = CD = 22° 52′ the Lati Secondly; Then to find the Longitude, or Angle at E, fay; As the Sine of C E = 67° 08' Co. Ar. 0.0355463 Is to the Sine of C P E = 164° 10′ 9.4359080 9.8438547 This being the Complement of its Longitude to Cancer, D, therefore fubftract it from 90o, there will remain 78° 03' from the Vernal Equinox or Sign V, that is in 18° 03'. Note; This is 30" more than the Tables give it; but in these Operations, which are intended only for Examples, we do not regard fuch fmall Differences from the Truth of Things; and the fame is to be remembered in other Places. PRO PROBLEM XXIII. Given the Latitude of the Place, the Sun's Right Afcenfion, and the Altitude of a known fixed Star; to find the Hour of the Night. Pralice. Admit the Latitude be 50° 56', the Sun's Right Afcenfion 228° 45′; and the Altitude of the Bull's Eye, or Aldebaran 32° 30′ towards the Eaft of the Meridian; thence to find the Hour of the Night. The Right Afcenfion of Aldebaran is 65° 8', and his Decli Æ will be formed the H Triangle C Z P, in which all the Sides are given; ZP = 39° 04 the Co-Latit. CZ = 57° 30′ = the Co-Altitude; and CP 74° 05' the Co A P S N Declination; to find the Angle APE the Difference between the Right Afcenfion of Medium Cæli & Aldebaran. Now 4 1 Now C P Z P = AM 35° 01'; Then Add (the Sine of Z P= 39° 04' Co. Ar. 0.2005049 the Sine of C P 74° 05' Co. Ar. 0.0169777 CZ+AM 2 toge-the Sine of ther. of wh. is the Sine of CPZ=29° 10' 9.6879520 Wherefore the Double of 29° 10' gives 58 20=CPZ; which fub. from the R. Afcen. of Aldeb. 64 08 there will remain the R. Af. of Med. Cali 6 48 Which reduced to Time, makes 9 Hours 12′ 12′′ the Hour of the Night, as required. Note, When the Angle C P Z is equal to the Stars Right Afcenfion; then the Sun's Right Afcenfion fubducted from 360° ( the Cor Cali then ) leaves the Degrees which reduced, gives the Time of Night required. If the Angle CPZ be greater than the Star's Right Afcenfion, the Difference fubducted from 360°, leaves the Right Afcenfion of Medium Cali, from which take the Sun's Right Afcenfion, the Remainder reduced to Time, gives the Hour fought. PRO PROBLEM XXIV. Given the Obliquity of the Ecliptic, and the Right Afcenfion of Medium Cali, or Mid-Heaven; to find the Culminating Point in the Ecliptic. 10.0000000 To the Co-fine of Æ~ E = 23° 29′ 9.9624527 So is the Co-tangent of Æ= 40° 00′ 10.0761865 To the Co-tangent of E = 42° 28′ That is in the Divifion of the Ecliptic There remains the Distance 10.0386392 417-32 from the firft Point of Aries, viz. a 17° 32' the Point of the Ecliptic culminating, or on the Meridian. PRO PROBLEM XXV. Given the Obliquity of the Ecliptic, and Right Afcenfion of Medium Cæli, to find the Meridian Angle. Practice. In the Scheme to the last Problem, in the Triangle EE, there is given, the Side = 40° 00'; and the Angle at = 23° 29′; to find the Angle EE, made by the Interfection of the Ecliptic and Meridian. Given the fame things as in the two laft, to find the Declination of the Culminating Point. In the aforefaid Triangle Æ E there is as before specified, the Side ; and the Angle at, to find the Side E E, which is the Declination of the Culminating Point. |