and these latter angles are spherical angles. For, if we recur to the preceding method, the angle reduced was AZB, which is equal to the inclination of the planes OZ a, OZb. Hence, if from A and B, as stations, the angles subtended by BO, AO, respectively, be observed, and reduced, by the preceding method, to horizontal angles, the three stations will be projected on the surface of a sphere, and the sum of the three angles at O, a, and b, ought to exceed 180°. (Prop. 9, p. 131). If we could determine, independently of observation, the quantity by which the sum of the three angles ought to exceed 180°, we should then be able to judge of the accuracy of observations; which it is desirable to do, since observations made on objects situated nearly in the horizon, are liable to some uncertainty. Now, the Theorem concerning the area of a spherical triangle, (Prop. 13, page 134), enables us to determine the quantity of the excess For, the area is equal to the difference of the sum of the three angles and 180o, and consequently, since the sides of the spherical triangle described on the Earth's surface are nearly rectilinear, they may, with scarcely any error, be considered as rectilinear in the computation of the area. This beautiful application of Albert Girard's Theorem was first made by General Roy, in the Philosophical Transactions for 1790, page 171, where he gave the following rule* for computing, what he terms, the spherical excess. "From the logarithm of the area of the triangle, taken as a plane one, in feet, subtract the constant logarithm 9.3267737, and the remainder is the logarithm of the excess above 180°, in seconds nearly." The rule may be thus investigated, a+b+c 180° = 180o 4 x x; or, if a, b, c, are reduced to seconds, the π p2 * M. Delambre finds, very expeditiously, the spherical excess by means of a chart of the triangles, a scale, and a Table of two pages. See Mem. Inst. 1806, pages 166, 179, 180. Disc. Prel. excess in seconds = 180.60.60 x; now, on the Earth's surface, 1o (taking a mean measurement) corresponds to (60859.1) x 6 feet; .. since an arc (radius) = 360 2 π = log. x — { 2 (log. 6+ log. 60859.1) — log. 2. 10 } As an instance of this rule, take the Example given by General Roy, Phil. Trans. 1790, p. 172. Hence, making the distance from (a) to (c) the base of the triangle, the perpendicular on that base will be equal to sin. 42° 2′32′′, and therefore the area of the triangle= =24704.7 × 19230.56 × sin. 42° 2′ 32′′. 38461.12 x base x perp'. 2 Again, as a second instance, take that given in the succeeding Third Instance, from the Phil. Trans. 1803. page 428, * 4.8468947 5.2636266 20.0276492 (2) from (h)=183496.2 10.0276492 9.3267737 .7008755 log. 5".022; or spherical excess = 5".022. There appears to be an error in the Philosophical Trans. p. 172. where the spherical excess is put down .29, which is nearly equal to 2.(14992). The cause of the error seems to be, the neglect of the divisor 2 in estimating the area of the triangle; for if we add the log. 2 1.1758591, the resulting number expresses the logarithm of .29. to In this manner the spherical excess computed *, will enable the observer to examine the accuracy of his observations, and in some degree to correct them. He may then proceed to calculate the sides of the triangles by the rules of Spherical Trigonometry. But these rules, although they must give exact results, will not give results very expeditiously. It is, however, very desirable to obtain results expeditiously, when several hundred triangles are to be solved; and, this desideratum, M. Legendre has put us in possession of, by combining sufficient exactness with conciseness in the following Theorem: "A spherical triangle being proposed, of which the sides are very small relatively to the radius of the sphere, if from each of its angles one-third of the excess of the sum of its three angles above two right angles be subtracted, the angles so diminished may be taken for the angles of a rectilinear triangle, the sides of which are equal in length to those of the proposed spherical triangle." The demonstration of this rule will be given in the Appendix, but its utility may be here shewn. Suppose in a spherical triangle, two sides and the included angle to have been determined, respectively, equal to 53° 58′ 35′′.75, 71934.2, 79211.22 feet; compute the spherical excess, which is equal to 1".08 nearly (see the second Example, p. 190), then (1′′.08)=.36, and 53° 58′ 35′′.75—.36= 53° 58′ 35′′.39: with this, as an included angle, and the given values of the two sides, resolve the triangle as a rectilinear triangle, by the method † given in page 83, Plane Trigonometry. ૐ * In two of the instances that have been given, the sum of three observed angles is less than 180°; .. in order to correct that sum, we must add the defect plus the spherical excess; that is, in the first instance, we must add 1"+.15; in the second, .75 +1.08. If all the observed angles were made under similar circumstances, and with equal care, we should have angles better suited to calculation than the observed angles, by adding to each of the latter, one-third of the preceding corrections 1".15 and 1".83. But if two of the observations should have been made more correctly than the third, then to this third is the correction chiefly to be applied. General Roy, probably, acted thus; for his corrections, as they now stand, appear rather arbitrary. + This method has been used, in preference to the common method, because the logarithms of the two sides were already taken out. See page 83. The Side c, opposite the included Angle, computed. We must now add to A and B one-third of the spherical excess, and we shall have the spherical triangle completely resolved; that is, * 3 Angles of the Spherical Triangle. C' = 53° 58′ 35′′.75 A' 68 24 44.995, B' 57 36 40.345 180 0 1.09 3 Sides of the Spherical Triangle. c = 68896.9 a = 79211.22. b = 71934.2 * In this instance solved by this method of Legendre, the conditions of the triangle were, two sides and an included angle; but it is plain, that by the same method, a triangle with any three other conditious may be solved. |
