= article, except that PN' is not now necessarily PN. Then, in the triangle PZZ, PZ is the co-latitude of the given place, and PZ' that of the required place; angle Z is the direction of the inclination, and ZZ' its quantity; and ZPZ' is the difference of longitude; hence, in this triangle, there are given PZ, ZZ', and angle PZZ', to find PZ and angle P. Hence, if PZc, ZZ'i, angle Za, and angle Pd, and PZ' = c', then 1. To find the difference of longitude d (361), Rad: cos atan i : tan 6, and sin (c―0): sin 0=tan a : tand. 2. To find the co-latitude c' (361), Cos : cos (c) = cos i : cos c'. Then the latitude of the required place is '90-c′; and if h, h', and t, represent the same parts as in last article, then, as in it, cot h': rad sin l': tan t'. GEODETIC SURVEYING. 552. The method of geodetical surveying is employed when a large portion of the earth's surface, extending several degrees, is to be accurately measured. The method consists in forming a series of large triangles, by supposing various conspicuous objects, as the summits of high edifices and mountains, to be connected, so that the sides of any one triangle serve as bases for three contiguous triangles. Such a series of triangles is called a system of triangulation; and all the angles of the system being accurately measured, and only a single side, all the other sides can be computed on the principles of spherical trigonometry, or more simply by means of methods deduced from these principles. The object of the survey may be either the construction of an accurate map, or the determination of the elements of the figure of the earth (see articles 557 and 602.) 553. When the angles at the stations, subtended by any other two stations, are taken by means of a sextant, a repeating circle, or any other instrument by which the inclined angles in the planes of the objects are measured, the angles at the different stations must be reduced to the corresponding horizontal angles by computation; but when a theodolite is used in the survey, this reduction is unnecessary, as by means of it the horizontal angle is directly measured. 554. PROBLEM I.-Given the zenith distances of two stations observed at a third station, and the inclined angle at the latter, subtended by the distance between the former, to find the corresponding horizontal angle. Consider the two zenith distances and the inclined angle to be the three sides of a spherical triangle; and find the spherical angle contained by the two former, and it will be the required horizontal angle." = Let OAB be the three stations, OC, OD, horizontal lines at the station O, and OZ a vertical line; also let MNP be a spherical triangle, whose centre is O. Also, let MNz = the zenith distance of A, MP == the zenith distance of B, PONo, the inclined angle at O, COD = 0, the horizontal angle at 0; then, if s = half the sum of the sides, by article 349, · sin (s) sin (s—') Sin2 10 = Cos2 10 = Or, Or logarithmically, M P ... N A [1] [2] 2 L. sin 10 = L. sin (s)+L. sin (s-z)+L. cosec z+L. cosec z'—20, and 2 L. cos 0= L. sin s+ L. sin (so) + L. cosec z + L. cosec z - 20. 555. When the zenith distances differ by only 2 or 3 degrees in excess or defect from 90 degrees, that is, when two objects are elevated or depressed by only 2 or 3 degrees, the following formulas may be employed. In this case, let e, e' the elevations or depressions of A and B, and d0-o; then e 90-z, é: 90', and O= o+d; also, if d, e, e', are expressed in parts of the radius, d=1(e+e') tano-(e-e')2 coto [3] Or 4 dm – n, if m and n denote the two terms, of which is the coefficient. 556. But if d, e, e', denote the number of minutes in these arcs, then d=13751{(e+e')2 tano-(e-e')2 coto} [4] Or, denoting the terms within the parentheses respectively by m and n, d=13751 (m—n), or Ld = L (m. -n n) — 4.13833. In formulas [3] and [4] when logarithms are used, they require to be carried only to 5 places in the decimal. = : When e is expressed in minutes, then, since l'=00029888 when radius=1, therefore the reciprocal of 1' 3437-75; and hence, if in the former expression [3] d, e, e', are divided by this number, the formula will be changed into the latter expression, for 4 × 3437·75 = 13751, as in [4.] In the formula, e and e' are considered to be elevations; when either of them is an angle of depression, its sign must be changed. When either of the formulas [1] or [2] is to be employed, the solution is exactly the same as in article 350; it will therefore be necessary only to give an example of the application of the last formula. Here e 60' 182, e' e-e68.008, and by [4], 2 L (e+e') L. tano L. m 1632.7 ... EXAMPLE. Given e an angle of elevation = 1° 0' 10"-92, e' an angle of depression=7′ 49′′-54, and o = 61° 33′ 20"-59, to find O. ... = = - 7'-826, e + e' = 52·356, 10.22505 =3-43794 2L (e-e') = 3.66512 3.89017 = L (mn) 6132-8 L. 0.4460 = 3.78766 1.64933 Hence, d=0'44626"-76, which is negative, because mn; and Oo―d — 61° 33′ 20′′·59 — 26"-76 =61° 32′ 53"-83. = 2L (e+e') L. tano = The same example may be solved in the following manner as an illustration of formula [3]: The length of 1° = 01745 when radius = 1; and hence the lengths of e and e' can be found by means of the preceding or by article 283, Part I.; or, more readily, by means of a table of the lengths of circular arcs. It is thus found that e = '01750, e' =— - ·00228, e + e′ = ·01522, e- ·e' = 01978. = 4·36482 2 L (e—e') -= 4. Given e == 41′ 50′′-68, 0 = 56° 38′ 33′′-34, to find O. L. m 00013797 = 4·13977 L. n ·00065691 = 4.81751 And m 00012973: n = 26"-75. = 4.59246 = 10.22505 557. The horizontal angle found by this problem is just the spherical angle at the station at which the angle is formed, contained by arcs of two great circles of the earth, considered as a sphere, passing through that point. EXERCISES. = 1. Given z = 88° 12′, ≈′ = 88° 39′, and o = 63°, to find O by [1.] Ans. 063° 1' 30"-35. 2. Given e 25′ 47′′-2, e' =- 1", and o 66° 30' 38"-9, to find O by [4.] Ans. 0 66° 30′ 36′′-89. 1° 6', and o= 97° 36', to Ans. O 97° 34′ 30′′. e' = — 57′ 31′′-91, and Ans. 0 56° 38′ 54′′-78. 3. Given e 1° 30', e' —— find O by [2] or [4.] = 5. Given e 25′ 11′′·9, e′ = = 1° 15′ 42′′·97, and o = 61° 48' 10"-61, to find O. Ans. 0 61° 48′ 18"-38. = 558. PROBLEM II.-To reduce angles taken out of the centre of a station to their corresponding angles at the centre. Let ABC be three stations, the centre of an object at C being chosen for the centre of that station, at which it is impossible to A place an angular instrument for determining the angle ACB; and let an angle AOB be taken at a point near to this station. The angle ACB is called the central angle, AOB the observed angle, AOC the angle of direction, OC the central distance, and BC, AC, the right and left distances, and AB the base. Let ACBC, AOB = 0, AOC=d, OC= c, BC=r, and AC, then [1] с sin (0+ d) C-0= r sin l' in which the correction CO is expressed in minutes. When O is to the left of C, and still without the triangle ACB, d will be negative, as the right and left sides OA, OC, of this angle then become respectively the left and right sides. In this case O+d becomes O d, and as then d>0, sine of (0 — d) is negative, and sin d becomes sin (d), and is also negative, so that sin (0 ~ d) + sin d ) C-0= r ι с sin 1' ( с sin (0+ d) r sin l ́ ( When O lies within the triangle ABC, O+ d is 180, and its sine is negative; and in this case C-O: sin d ι C sin (0 — d) r sin l' B + sind) When C lies within the triangle AOB, d again becomes negative, and sin (0+ d) becomes sin (0- d), and sin d becomes sin (d) — — sin d, so that C-0= + sin d) ι |