is equal to the sum of the altitude and polar distance, and is of the same name as the latter." That is, LZ ± D, or L= A'+P, where A' is the altitude for the lower culmination. EXAMPLE. On the 2d of May 1841, in longitude 50° 15′ W., the observed meridian altitude of the sun's lower limb was 70° 31' 18" S., the height of the eye being 15 feet; what is the latitude? The principle of the rule is easily explained by a reference to the figure in article 462. Let B be the body, then the zenith distance ZB, and the declination BE, are of the same name, whether P be the south or the north pole; and the latitude EZ is their sum, and of the same name. If B" be the body, then the zenith distance B"Z, and declination B'E, are of different names, and the latitude EZ = B′′Z — B′′E. So when B' is the body, Z and D are of different names, and EZ EB'-ZB'. Let b' be the body at the lower culmination, A' 'N, and Pb'P; then LA'+P. = EXERCISES. 1. If the true altitude of Aldebaran, at a place in longitude 48° 30′ W., on the 20th of May 1841, was 54° 20' 35′′ S.; required the latitude, the star's declination being 16° 11′ 9′′ N. Ans. 51° 50′ 34′′ N. 2. If the meridian altitude of the moon's centre, on the 2d May 1841, in longitude 40° 45′ W., was 25° 13′ 46′′-2 S., when its declination was 8° 26' 3"-8 S.; what was the latitude of the place? Ans. 56° 20′ 10′′ N. 477. PROBLEM XXXIII.-Given the sun's declination and altitude, and the hour of the day, to find the latitude of the place. Let the parts of the triangle PZS, in article 466, represent the same quantities as in that problem; then the polar distance P PS, the zenith distance Z = ZS, and the horary angle H = SPZ are given, to find the colatitude C = PZ. . The side PZ can be obtained by the method in art. 357, or, more concisely, by that in art. 365. In it the quantities a, b, A, and c, are respectively the same as Z, P, H, and C, in this problem. Hence, and And then R: cos Htan P : tan 0, When the perpendicular falls without the triangle between Z and E, then c, and c is negative, that is, c = 0 — (0 —c). When it falls between P and Q, then is negative, and c becomes c+0; and then c = (c + 0) — 0. The algebraic signs of the terms indicate these conditions. EXAMPLE. On the 8th May 1813, at 5h 33m 338-4 apparent time, the altitude of the sun's lower limb was observed to be 15° 40' 57", the longitude of the place being 80° 39′ 45′′ W.; what was the latitude? As in the example in art. 428, the declination of the sun at the time of observation is found to have been 17° 10′ 42′′, and the true altitude of his centre 15° 53' 32"; hence P=72° 49′ 18′′, Z = 74° 6′ 28′′, and H = 83° 23′ 21′′·4. 1. At a given place, in south latitude, when the sun's declination was 15° 8'′ 10′′ S., his true altitude was 39° 5′ 28′′ at 2h 56m 428.7 A.M.; required the latitude of the place. Ans. 52° 12′ 42′′. 2. At a place, in north latitude, when the sun's declination was 6° 47′ 50′′ S., his true altitude was 33° 20′ at 8h 46m A.M.; required the latitude. Ans. 24° 30'. 478. PROBLEM XXXIV.—Given two altitudes of the sun or of a star, and the interval of time between the observations; or the altitudes of two known stars, taken at the same instant, to find the latitude of the place. E B Ꮓ Let P be the pole, Z the zenith, and B, B', the body in two different positions, or two different bodies. Then PB, PB', are the polar distances, and ZB', ZB, the zenith distances; these four quantities being given. Also, when B and B' are the sun in two different positions on the same day, or of a star on the same night, or of two different stars at the same instant, the angle BPB', which measures the elapsed time, or the diffe H N Q R rence of right ascensions of two different stars, is known. But, in the second case, when B, B', are two different stars, and the elapsed time between the observations is measured in mean time, it must be reduced to sidereal time. Hence, the latitude may be found thus : Let P, P' and the polar distances PB, PB', the zenith distances BZ, B'Z, H = the angle BPB', E the side BB', L, C = the latitude and colatitude ZE and ZP. By 355, 1. To find angle B', in triangle PBB'. (B~ B') cotH: tan (B+B') sin (P+P): sin (P~ P') cot H: tan and cos(P+P'): cos(P~ P') From which B and B' can be found. 2. To find E, in triangle PBB'. Sin (B~ B'): sin 1 (B+B') = tan 1 (P ~ P') : tan E. By 349, 3. To find angle B', in triangle BZB'. 2 L cos B' Lsin S+L sin (S-Z) +L cosec E + L cosec Z'. By 361, and 6. Then 5. To find C, in triangle PB'Z. R: cos B'tan Z' : tan 0, When B and B' are the same star at the times of the two observations, P is = P'; and when they represent the sun, if its declination for the middle time between the observations be taken, PB and PB' may then be considered equal to this declination P. The solution may then be simplified, for PBB' will then be an isosceles triangle, and a perpendicular from P on BB' will bisect it, and will form two equal right-angled triangles. Hence, instead of the preceding formulas at No. 1 and 2, take these two: 1. To find B' in one of the right-angled triangles. CotH: R= cos P: : cot B'. 2. To find E in the same triangle. EXAMPLE. If in the forenoon, when the sun's declination was 19° 39′ N., at the middle time between two observations of his altitude, these altitudes corrected were 38° 19′ and 50° 25′; what was the latitude, the place of observation being north, and the interval between the observations one hour and a half? Here P70° 21', Z = = 51° 41′, and Z' — 39° 35'. H = 1h 30m, and 1 H = 11° 15′, and PZ = C. 1. To find angle B' in PBB'. L. Cot H = 10-701338 2. To find E in PBB'. L. Rad = 10. L. Rad = 10. L. Sin P = 9.973942 = 9.825355 And B' 86° 10′ 24′′. L. Sin E = 9.264178 And E 21° 10′ 26′′. 3. To find angle B' in BZB' and PB'Z. |