Now EZ is known; therefore to find the other Part Z O, there is given in the Triangle O NB, or rather in the fmall one Z HO, the Side Z H 18° 30' the Reclination; and the Angle at Z= 35° 00', the Declination; Therefore to find O Z or ON, fay thus ; As Radius 10.0000000 To the Co- fine of Declin. Z = 35° 00' 9.9133645 So is Co-tang. of Reclin. HZ = 18° 30′ 10.4754801 = To the Co-tangent of OZ 22° 13' To which add The Sum is titude fought. ÆZ = 51° 327 10.3888446 ÆO = 73° 45', the new La To find the New Declination FQ The Declination is always an Arch of a Circle contained between the Meridian of the Place and Pole of the Reclining Plane, and fuch is the Arch F 2; to find which in the Right-angled Triangle F QZ there is given 2 Z = 71° 30' the Co-Reclination, and Angle at Z 35° oo'; Therefore fay, As Radius Q To Sine of Z = 71° 30' To the Sine of F Q = 32° 57′ which is the new Declination required. 10.0000000 9.9769566 9.7585913 9.7355479 Variety Variety 2. Suppofe the foregoing Plane recline juft fo far as to pass thro' the Pole; then will the Hour-Lines be all Parallel to one another as on a direct Eaft or Weft Dial. And the Anlogies for finding the Reclination Z H, and Declination A 호스 S 2, are as follows. 38 28 So is Co-tangent of Z P = To the Co-tangent of ZH 33 04 Variety 3. Suppose the fame Plane recline from the Zenith 53° 30'; Quere the new Latitude and Declination ? In latter (by the Analogy in Var. 1.) will be found 58° 47', from which take Z P = 38° 28', there will remain 20° 19' = PO, whose Complement is 69° 41', the Distance of O from the Northern Interfection of the Equinoctial with the Meridian; and is the new Latitude fought. FQ Alfo the foregoing Analogy will find F2 = 19° 57' the new Declination. PROBLEM VIII. To find the new Latitude and Declination of North Declining Reclining Planes. Practice. B H E tion; the Triangles OH Z and ZFQ, with the Parts known in each, being the fame, but only in a Pofition reverfe to the former in the Schemes of the foregoing Problem; thus O Z will be found here 22 13', then Æ Z. OZ = Æ0 = 29° 19′ the new Latitude fought, by a Procefs the Reverse of the foregoing. Alfo F Q = 32° 57' the new Declination, the fame as before in the South Declining Recliner. Variety 2. Alfo here if the Plane recline fo as to pafs through E, it will then be an Equinoctial Plane; and its Reclination, and new Declination are found as in the fecond Variety of the laft Problem. Variety 3. If the fame Plane recline fo far as to lie between the Interfection of the Meridian and Equinoctial Æ, and the Horizon S, as the Plane AbB; then we must find o Z, by the Triangle o Z b; then o Z — Æ Z = o Æ, the new Latitude; and qf will be the new Declination. Thus you may fee the fame Canons for Calculation And the new Latitude and Declination both for North and South Declining Reclining Planes. PROBLEM IX. To find the new Latitude and Declination of Direct East and Weft Reclining Planes. Practice. In either of thefe Kinds of Recliners, the Compilement of the given Latitude is the new Latitude; and the Complement of the Reclination is the new Declination. For 處 new Latitude, viz. 38° 28', the Complement of the given Latitude EZ. Alfo the Distance of the Pole of the Plane 2, from the Meridian is Z 2, that must be the new Declination, viz. 50° 00', the Complement of the Reclination given. PROBLEM X. To difcover whether the North or South Pole be elevated above any given Plane. Practice. From the Schemes of the foregoing Problems this is cafily known, as alfo from the following general Rules arifing thence. 1. All Horizontal Planes in North Latitude, have the North Pole elevated; but in South Latitude, the South Pole. 2. Upon all erect Planes whether Direct or Declining, if it be a South Plane, the South Pole is elevated; but if it behold the North, the North Pole is elevated. 3. Upon |