In the Triangle ABC, we have this Analogy; As Radius To the Co-tangent of B = 50° 56' So is the Tangent of AC 10.0000000 20° 34' 9.9094022 - 9.5742761 9.4836783 To the Sine of B A = 17° 43′ Now the Motion of 17° 43′ is performed in 1 Hour 10' 56"; Whence the Sun is due Eaft in the Morning at 7 Hour 10'56", and at 4 Hours 49'04” in the Afternoon, he will be due Weft. In the Triangle C P Z, we have this Analogy; As t CP: R::t Z Pcs P = 72° 17'. This reduced to Time gives 4 Hours 49' 04", the Time in the Afternoon; and this fubducted from 12 Hours, leaves the Time 7 Hours 10' 56" in the Morning; both as before. PROBLEM XII. The Latitude of the Place, and the Sun's Declination, given; to find his Altitude when due Eaft or Weft. Practice. In the two Triangles A B C, or CP Z (fuppofing the Time and Place, the fame as in the laft Problem) there is given in the first Triangle ABC, the Side AC = 20° 34' the Declination, the Angle B = 50° 56' the Latitude of Chichefter; to find the Side B C, the Altitude required. VOL. II. Hh Ana To the Sine of the Altitude BC=26° 54′ 9.8900929. 9 5456745 10.0000000 9.6555816 In the Triangle C P Z, there is given C P, the CoDeclination, and Z P the Co-Latitude, to find the Side Z C, the Co-altitude required. For which you have this Analogy viz. c s ZP: R::cs CP: csCZ =63o 06. PROBLEM XIII. The Latitude of the Place, and Declination of the Sun given, to find the Sun's Azimuth at the Hour of Six. Suppofe the Latitude that of Chichester 50° 56', and the Sun's Declination 20° 34' as. it will be on the 12th of May, A. D. 1735. Then, the Diagram being prepared, you will have therein formed, the two Rightangled Triangles ABC, and CZ P, by either of which this Problem is fatisfied. Practice. For in the Right Triangle ABC, there is the Side BC= 20° 34' the Declination, and the Angle at B 50° 56' the Latitude of Chichester, both given; to to find the Side AB, the Azimuth of the Sun from the Eaft or Weft Points of the Horizon. The Analogy is As Radius 10.0000000 9.5742761 9.7994951 To the Tangent of BC= 20° 34' To the Tang. of the Azim. AB=13° 18′ 9.3737712 Wherefore the Azimuth of the Sun at fix in the Morning is about E by N and ; and at fix in the Evening, W by N and CP Alfo in the Triangle C Z P, there is given the Side the Co-Declination, and ZP = the Co-Latitude; to find the Angle at Z = the Azimuth from the North = AO. The Analogy for this is, SP Z:R::t CP:t Z= Or varied thus, 768 42' tCP: R::s PZ:ct Z = 13° 18'. Or thus, R:ctCP::sPZ: tZ. PROBLEM XIV. The Latitude and Declination given, to find the Altitude of the Sun at the Hour of fix. Practice. In the foregoing Diagram, and in the Triangles ABC and CPZ there are given the fame Things as in the laft Problem, to find the Side AC in the firft, and C.Z in the latter; either of which answers the Problem. To find the Altitude AC, this is the Analogy. VOL. II. Hh 2 As As Radius Is to the Sine of B C = 20° 34' 10.0000000 9.5456745 9.8900929 To the Sine of the Al. at fix AC=15° 49′=9.4357675 In the Triangle CP Z to find CZ = Co-altitude, this is the Analogy; R:cs PZ::cs CP:cs C Z= 74° 11'. Note, The Altitude in Summer, is equal to the Depreffion in Winter. PROBLEM XV. Given the Latitude of the Place, the Declination, and Hour of the Day, to find the Sun's Altitude. Cafe 1. Let the Declination of the Sun be North 20° 34' at the Hour of 10 in the Morning, or 2 in the Afternoon (on the 12th of May, 1735) and in the Latitude of Chichester 50° 56', I demand the Height of the Sun at that Time? According to thefe Data, let the Scheme be conftructed as hath been heretofore taught; then there will be formed the Oblique Spherical Triangle CZP. In this Triangle there is given the Side Z P = 39° 04', the Co-latitude; and the Side C P = : 69° 26' the Co-declination; and the included Angle at P = 30° 00', the Time to Noon; to find the Side C Z, the Co-altitude required. This is found by Cafe 4, in the Synopfis, thus ; As the Radius To the Co-fine of 10.0000000 P = 30° 00' 9.9375306 9.9094022 So is the Tangent of Z P = 39° 04' To the Tangent of PL = 35° 06′ There will remain CL 34° 20′ Wherefore fay again; 9.8469328 As the Co-fine of LP 35° 06! Co. Ar. 0.0871672 Is to the Co-fine of ZP=39° 04' So is the Co-fine of LC=34° 201 To the Co-fine of CZ=38° 25' 9.8900929 9.9168593 9.8941194 The Complement of which to 90° = Z B, is CB 51° 35', which therefore is the Altitude of the Sun the Time fpecified. Safe 2. Let the Declination of the Sun be South 20° 34'; the Latitude, and Hours as above, to find the Altitude? The |