559. If the angle of the triangle ABC at the right station B be denoted by R, a more concise expression can be found, с sin (R — d) sin O sin l' r sin R namely, C-0= [2] where CO is again expressed in minutes. But this formula, which is only approximative, requires that the correction of the angle C should be small, or that PB should be nearly BC, supposing a circle ABPC to be described through A, B, and C. = When O falls within the circle, but not within the triangle ABC, then dR, and sin (R. - d) becomes negative. When O lies to the left of C, take BOC CAB R, and CAr, supposing the letters A, B, to remain unaltered. d, = 1 The value to be substituted for is 3437-75, the logasin l' rithm of which is 3.53627. The values of r and 7 must be known, but approximate values are sufficient, as, for example, such as would be obtained by solving a spherical triangle ABC of the magnitude common in geodetical operations, as a plane triangle. 560. When O lies to the left of C, the letters A and B may be interchanged, and then the formulas [1] and [2] apply without alteration. EXAMPLE. Given c = 12, 7 = 4581·8, r = 5000, O= 74° 32', and d = 139° 39′, to find C, O being to the left of C. L. sin 65° 7' L. 5000 ... sin 65° 7' 5000 r + =9-95769 L. cos 49° 39' = 9.81121 = 3.66103 L. m 00018143 = 4·25872 L. n 00014131 = 4·15018 L. (mn)·00004012 L. 12 = = L. 1'-655 = 0.21881 Or, CO 1' 39"3, and C 74° 30′ 20′′"7. = = L. 12 = To calculate C-O by the second method, there must be given angle R, that is, ABC (supposing A and B interchanged, as O is to the left of C), and also angle d and r. By observation, let R = 56° 7′ 45′′, then here r = 4581·8 (the above), and 0 = 74° 32′ as above; but the value of d is the excess of the d given above over O, or d = 65° 7′. c sin (Rd) sin O go sin l'sin R ι C-0= 5.60336 sin 8° 59′ 15′′ · sin 74° 32′ • 12 = 1.07918 L. r d) = 9-19373 L. R 23.79316 = 3.66103 9.91923 13.58026 23.79316 L. 1.6327 = •21290 = Or, C—0—1′ 37′′-96, which is 1"-3 less than the value found by the former method, which is the more correct, and requires only three more logarithms than the latter. The first formula is easily proved by means of two analogies obtained from the triangles ACO, BCO, namely, с r: c = sin (0+d): sin CBO, and sin CBO = sin (0+ d) l:c=sind:sin CAO, and sin CAO = But CBO, CAO, are small angles, and are therefore nearly proportional to their sines; and therefore the number of minutes in CBO and CAO respectively will be very nearly sin CBO sin l' and sin CAO sin 1 Also, angle CAO+C=AQB = sin d. CBO + 0, and C — O — CBO — CAO sin (0+ d) __ sin d) r Were the central distance c considerable, it would then be necessary to find the angles CBO, CAO, by means of the analogies in the preceding paragraph, and then their difference would give C — O. The side AB being known in the triangle ABC, and its angles at A and B being also known by observation, the sides AC, BC, can be calculated with sufficient accuracy for determining angle C by the method in this problem, although these two angles are spherical angles; for the angles given in the problem are supposed to be reduced to horizontal planes at these angles. By describing a circle about A, B, C, cutting AO in P, the second formula can be proved by means of the triangles CPO, PBO, provided PB be very nearly equal to BC or r. EXERCISES. 1. The place of observation being to the right of the centre of the station, and O = 60°, d = 58° 41′ 0′′-6, 7 = 19000, r = 20000, and c = 10, to find C. 2. The data being as in the that 7 and rare ten times less; = Ans. C 59° 59' 57"-7. preceding exercise, except what is C? Ans. C 59° 54′ 13′′-32. 3. The angles in this example being expressed according to the centesimal, or the new division of the circle, namely, 0 = 44° 25′ 92′′-6, d = 38° 14′ 81′′·5, and r = 4596·27 metres, 74041·89, and c = 41.69; find C, supposing O to be situated on the right of the station C. = Ans. C 44° 44' 44"-4. 4. Given O, d, c, and r, as in the preceding example, and R = 66° 66′ 66′′7; find C by the formula [2.] Ans. C44° 44′ 42′′-6. The answer to the third example is the correct one, and it therefore appears, that, with a distance c equal to about of r or l, the error of the second method is 1′′-84, or 066 of the sexagesimal division. II. Ꮓ 5. Given O 45° 42', d = 50° 18′, r=32656, 7 = 31052, and c=46 feet; find C, the position of O being still to the right of C. Ans. C 45° 42′ 54′′. = 561. Before proceeding with the computations of geodetical measurements, it is necessary that the mean diameter of the earth should be known with considerable accuracy. A mean value, however, is sufficient for most of these purposes. If a base, for instance AB (next fig.), is measured at a level above that of the sea, and is to be reduced to its length ab, if measured at the mean level of the sea, then, supposing this base to be measured at a height of 3000 feet, and to be 40 miles long, the error on the correction of the base, arising from taking the mean diameter, cannot exceed at its greatest of a foot, and will be much less if the place be situated in a medium latitude (see article 602.) 562. PROBLEM III.-Given the length of a base, and its height above the sea, to reduce it to the level of the sea. с в A Multiply together the base and height, and divide the product by the radius of the earth, and the quotient will be the correction to be deducted from the base.' Let B, b = the measured and reduced bases in feet; r, h the earth's radius, and the c = B− b ; Bh and then and C. b=BOr, taking r = 20900394 feet, its logarithm is = 7.3201545, and its arithmetical complement 8-6798455, which may be added for the term Lr, then EXAMPLE.-Let B62546 feet, and h=624, to find b. L(B-6)=4796199+2.795182+8-679845=0.271226. Hence, B-b1·8674, and b = 622.1326. c = or Lc = LB + Lh — Lr, > r a If a, b, denote the polar and equatorial diameters of the earth, and r the radius of a sphere of equal volume, and r the equidifferent mean between a and b, then a = 41706000 feet 7898.87 miles ; b = 41845000 = 7925.19 41798600 = 7916-416 ... 2r = 7912-03 ... ... and The value of r is obtained thus :-Let v, v', be the volumes of the earth and of the equivalent sphere, then •5236 al2 = v = v′ = ·5236 (2 r)3; hence, (2r)3 = ab2, and 3 L2r= La +2 Lb; and as a and b are supposed to be known, r can be found. ... ... EXERCISES. 1. A base, measured at the height of 1245 feet, was 25086 feet long; required the reduced base. Ans. 25084.506. 2. Find the reduced base corresponding to a base of 12543 feet, measured on a plane elevated 996 feet. Ans. 12542-402. 563. When the three angles of any of the spherical triangles of a system of triangulation have been observed, they can be verified by means of the spherical excess-that is, the excess of the three angles of a spherical triangle above two right angles. This excess can be calculated when the area of the triangle is known, and it can be found when any three parts of the triangle are given. Conversely, the area of a spherical triangle is easily and accurately found when the spherical excess is known. 564. PROBLEM IV.-Given the spherical excess, to find the area of the spherical triangle. 'As 180 is to the spherical excess in degrees, so is of the surface of the sphere to the area of the triangle.' Let E the spherical excess, S, T the surfaces of the sphere and of the triangle; then 180: ES: T, or To ES. EXAMPLE.-The three angles of a spherical triangle are |