ZP 39° 04', the Co-latitude; the An gle P 30° 00'; and the Side C P (= ACAP20° 34′ +90) = 110° 34'; to find the Side C Z, the Co-altitude required; which is done by H Cafe 4, thus. The Segment LP is found as in the foregoing Cafe of this Problem to be 35° 06′; hence CL will be 75° 28'. Then fay, Whofe Complement to a Quadrant is BC = 130 46', and is the Altitude of the Sun, for the given Time and Declination. Cafe 3. Suppofe the Declination North 20° 34 as before, but the Time 5 in the Morning, or 7 at Night; I demand the Altitude of the Sun for those Moments in the Latitude of Chichester 50% 561? By By thefe Data, conftruct the Scheme, and you will have formed the Oblique Triangle C P Z, wherein there will be given the Side ZP= 39° 04'; the Side CP 69° 26; and the Angle included ZPC 105° 00'; to find the Side ZC, the Complement of the Altitude required. Let fall the Perpendicular Z L on the Side CP continued out; and then to find PL, fay by Case 4, and Analogy to Variety 1. As Radius = Is to the Co-fine of P 105° 00' To the Tangent of P L = To which add The Sum is CP= 39° 047 II 52 CL 81° 18' Now fay again; As the Co-fine of PL=11° 52' Co. Ar. 0.0093810 Is to the Co-fine P Z = 39° 04′ So is the Co-fine of C L = 81° 18' To the Co-fine of C Z = 83° 07' Or the Sine of BC= 69 53' Altitude of the Sun required. PRO PROBLEM VI. The Latitude of the Place, Declination and Altitude of the Sun being given, to find the Hour of the Day. CZ P, in which there is given all the three Sides; for the Side C Z = 38° 25' 38° 25' the Co-altitude; the the Co-latitude; and the Side Z P = 39° 04′ Side C P 69° 26' = the Co-declination; to find the Angle at P= the Hour from Noon when the Obfervation was made. This is done by Cafe 5, or Theorem 39, thus. Here it is CPZP AM = 30° 22' = Then Then (the Sine of C P = 69° 261 Co. Ar. 0.0286016 the Sine of Z P = 39° 04′ Co. Ar. 0.2005049 CZ+AM = 34°24′ 2 9.7520231 1 2 4° 01' 8.8453874 18.8265170 The of which is the Sine of P = 15o 9.4132585 The Double whereof is 30° P, the Angle of the Hour fought. This 30° reduced into Time gives juft two Hours, fo that the Obfervation 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 Cafe 2 of the last Problem; and alfo if the Time be before fix in the Morning, or after fix at Night, as in Cafe 3, of the foregoing Problem; fuppofing all the three Sides given there as here. PPOBLEM XVII. Given the Latitude, Declination and Altitude of the Sun, to find his Azimuth. 38° 25', and Z P = 39° 04′; Whence we find the Angle at Z the Azimuth, as the Angle P was found in the laft Problem; but for Variety's fake, I fhall here ufe the Method by a Perpendicular, according to Prob. 5, of Chap. 18. Making therefore Z P the Bafe, the Perpendicular let fall thereon is CI; and let LZ = Ì M; then say, As the Tang. of ZP=19° 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 47° 02′ 10.0309388 Hence P M 94° 04'; wherefore P M-Z P= ZM 55° 00'; and the half of Z M is ZL=27° 30′ Therefore in the Right-angled Triangle L Z C there's given the Side C Z = 38° 25', and LZ= 27° 30', to find the Angle LZ C = the Azimuth from the South. Thus As the Tangent of C Z = 38° 251 To the Tangent of LZ 27° 30 So is Radius = To the Co-fine of LZC = 48° 59′ 9.8993682 9.7164767 10.0000000 - 9.8171685 Wherefore the Angle CZ P=131° 01'; and confequently the Azimuth of the Sun, or Point of the Compafs he is then upon, is 3 Points nearly from 1/1/10 the Eaft towards the South; that is, on the Point SE and by E in the Morning; or SW and by Win the Afternoon. Thus alfo may the Azimuth be found for a South Declination, and for any Time of the Day. This Problem is very ufeful to a Diallift, in finding the Plane's Declination; and to the Navigator in |