The Scheme being constructed, there is P formed the Oblique Triangle C ZP, in which there is given ZP = 39° 04', the Co-latitude ; the An. gle P = 30° oo'; H and the Side CP(= AC + AP=20° 34' +90) = 110° 34'; to find the Side CZ, the Co-altitude required ; which is done by Cafe 4, thus. The Segment L P is found as in the foregoing Case of this Problem to be 35° 06'; hence CL will be 75° 28'. Then fay, As the Co-sine of LP=35° 06' Co. Ar. c.0871672 To the Co-fine of Z P = 399 04' 9.8900929 So is the Co-sine of CL= 75° 28' 9.3995754 To the Co-sine of Z C = 760 14' 9.3768355 Whose Complement to a Quadrant is BC= 132 46', and is the Altitude of the Sun, for the given Time and Declination. Case 3. . Suppose the Declination North 200 34' as before, but the Time 5 in the Morning, or z at Night ; I demand the Altitude of the Sun for those Moments in the Latitude of Chichester 50, 56' ? By B В La P 0 given the Side ZP= H N Let fall the Perpendicular Z L on the Side CP continued out ; and then to find P L, fay by Case 4, and Analogy to Variety 1. As Radius 10.0000000 Is to the Co-fine of P = 1050 00 9.4129962 So is the Tangent of ZP = 399 04? - 9.9094022 To the Tangent of P L= IIS 52 9.3223984 To which add CP 6926 The Sum is 81 18' CL = Now say again ; 9.1797265 To the Co-fine of CZ = 83° 07' 9.0792004 Or the Sine of BC = 62 53' = Altitude of the Sun required. PRO PROBLEM VI. A The Latitude of the Place, Declination and Altitude of the Sun being given, to find the Hour of the Day. Practice. Suppose I observe the Declination of P the Sun to be 200 Æ 34' North, and his Altitude to be 51° 35' in the Latitude B of Chichester 50° 56', I demand what Hour H of the Day the Obfervation was made ? Having prepared the Diagram, there S will be formed the N This is done by Case 5, or Theorem 39, thus. Here it is CPZP = AM = 30° 22'. 2 2 Then (the Sine of CP= 69° 26' Co. Ar. 0.0286016 the Sine of ZP= 39°04' Co. Ar, 0.2005049 { = 349 24' 9.7520231 the Sine of CZTAM 40 01 8.8453874 The Sum of all these is 18.8265170 The of which is the Sine of P = 159 9.4132585 The Double whereof is 30o = P, the Angle of the Hour fought. This 30° reduced into Time gives just two Hours, so that the Observation was made either at 10 aClock in the Morning, or at 2 in the Afternoon. After this Manner the Angle P of the Hour may be found when the Sun hath South Declination, as in the Triangle to Case 2 of the last Problem ; and also if the Time be before fix in the Morning, or after six at Night, as in Cafe 3, of the foregoing Problem ; supposing all the three sides given there as here. PPOBLEM XVII. Practice. N 38° 25', and Z P= 390 04'; Whence we find the Angle at 2 = the Azimuth, as the Angle P was found in the last Problem ; but for Variety's fake, I shall here use the Method by a Perpendicular, according to Prob. 5, of Chap. 18. Making therefore Z P the Base, the Perpendicular let fall thereon is CI; and let LZ = L M; then fay, As the Tang. of IZP=190 32 Co. Ar. 0.4500489 To the Tang of CZ+CP=53° 55' 10.1374113 So is the Tang. of CZ-CP=15° 31' 9.4434786 To the Tangent of PM 470 oz' 10.0309388 Hence P M=94°04'; wherefore P M-Z P= ZM=550 001; and the half of Z M is ZL=270 30'. Therefore in the Right-angled Triangle LZG there's given the Side Č Ž = 38025', and L Ż 27° 30', to find the Angle LZC= the Azimuth from the South. Thus As the Tangent of CZ = 38° 25' 9.8993082 To the Tangent of L2 = 279 302 9.7164767 So is Radius To the Co-line of LZC = 489 59' 9.8171685 10.0000000 Wherefore the Angle C Z P=131° 01'; and consequently the Azimuth of the Sun, or Point of the Compass he is then upon, is 3 Points nearly from the East towards the South ; that is, on the Point S E and by E in the Morning ; or SW and by W in the Afternoon. Thus alfo may the Azimuth be found for a South Declination, and for any Time of the Day. This Problem is very useful to a Diallist, in finding the Plane's Declination; and to the Navigator in |