Now Æ Z is known ; therefore to find the other Part 2 0, there is given in the Triangle O NB, or rather in the small one Z HO, the Side ZH= 180 30' the Reclination ; and the Angle at Z = the Declination; Therefore to find ( Z or ON, say thus ; 35° CO', As Radius 10.0000000 To the Co-sine of Declin. Z = 350 00! 9.9133645 So is Co-tang. of Reclin. H 2 = 18° 30' 10.4754801 To the Co-tangent of O Z = 22° 13' 10.3888446 To which add Æ Z= 51° 32' The Sum is ÆO = 73° 45', the new Latitude fought. To find the New Declination F Q. The Declination is always an Arch of a Circle contained between the Meridian of the Place and Pole of the Reclining Plane, and such is the Arch F Q.; to find which in the Right-angled Triangle F QZ there is given Q2 = 71° 30' the Co-Reclination, and Angle at Z = 35° oo'; Therefore fay, As Radius 719 30 10.0000000 9.9769566 9.7585913 9.7355479 Variety Variety 2. Suppose the foregoing Plane recline just so far as to pass thro' the Pole ; then will the Hour-Lines be all Parallel to one another as on a direct East or West Dial. And the Æ Anlogies for finding the Reclination S ZH, and Declination Æ 2. are as follows. E 1. For the Reclination Z H. As Co-fine of Z = 35 00 38 28 9.1133645 10.0000000 10.0999135 10.1865490 2. For the new Declination Æ Q. As Radius 10.0000000 To the Sine of ZQ= 56° 56' 9.9232628 So is the Sine of Z = 35° 00' 9.7585913 To the Sine of Æ Q= 28° 43' 9.6818541 Variety 3. Suppose the fame Plane recline from the Zenith 53° 30'; Quere the new Latitude and Declia nation ? NC D In this case we must find PO, .N the Complement WZ H Η of the new Latitude ; for which P Purpose we must find either ON A B by the Triangle FY ONB; or O Z by the Triangle OHZ. Now this latter ( by the A S nalogy in Var. 1.) will be found 58° 47', from which take ZP = 380 28', there will remain 20° 19' = PO, whose Complement is 69° 41', the Distance of O from the Northern Intersection of the Equinoctial with the Meridian ; and is the new Latitude fought. Also the foregoing Analogy will find FQ = 190 57' the new Declination. PROBLEM VIII. Pratice. с Plane declining to N the West 35° oo', recline from the Zenith 18° 30'; Quere the Latitude in which it will be an A erect Plane, and its Declination as fuch? H This being just the reverse of the last Problem, requires no fresh Calcula Æ tion ; tion ; the Triangles O H Z and Z FQ, with the Parts known in each, being the same, but only in a Position reverse to the former in the Schemes of the foregoing Problem ; thus O Z will be found here 22@ 13', then Æ Z-OZ = Æ O = 29° 19' the new Latitude fought, by a Process the Reverse of the foregoing. Alfo F 2 = 32° 57' the new Declination, the same as before in the South Declining Recliner. Variety 2. Also here if the Plane recline so as to pass through Æ, it will then be an Equinoctial Plane ; and its Reclination, and new Declination are found as in the second Variety of the last Problem. Variety 3. If the fame Plane recline so far as to lie between the Intersection of the Meridian and Equinoctial Æ, and the Horizon S, as the Plane AbB; then we must find o Z, by the Triangle o Zh; then oZ - ÆZ= 0Æ, the new Latitude ; and a f will be the new Declination. Thus you may see the same Canons for Calculation and the new Latitude and Declination both for Norch and South Declining Reclining Planes. PROBLEM IX. To find the new Latitude and Declination of Direct East and West Reclining Planes. PraEtice. In either of these kinds of Recliners, the Complement of the given Latitude is the new Latitude ; and the Complement of the Reclinacion is the new Declination. For N For let NS be a direct East Plane reclining from the Zenith 400 00'; draw W Æ E for the Equinoctial; then because the H H E Arch of the Meridian comprehended between the Equi Æ noctial and the Plane is Æ S; therefore that is the S new Latitude, viz. 38° 28', the Complement of the given Latitude Æ Z. Also the Distance of the Pole of the Plane X, from the Meridian is Z Q, that must be the new Declination, viz. 50oo', the Complement of the Reclination given. PROBLEM. X. To discover whether the North or South Pole be elevated above any given Plane. PraEtice. From the Schemes of the foregoing Problems this is easily known, as also from the following general Rules arising 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 |