60°, 65°, and 85°, on a sphere whose diameter is 20; required the area of the triangle. E60+65 +85-18030, Sr2 = 314-16; and TE 183 x 314-16 52.36 feet. 30 EXERCISES. 1. Find the area of a spherical triangle, described on a sphere of 15 feet radius, its angles being 75°, 50°, and 85°. Ans. 117.81 feet. 2. The diameter of a sphere is 50, and the angles of a spherical triangle described on it are 75° 15', 82° 12′, and 35° 3'; find the area of the triangle. Ans. 136-3. 565. In other cases the area of a spherical triangle, by means of which the spherical excess is found when the accurate values of the angles are unknown, cannot be accurately found unless by methods that are tedious; but as the spherical excess of triangles, whose sides are very small compared with the radius of the sphere, such as those formed in a geodetical survey, amounts only to a few seconds, from 2 to 10 generally, and at most to about 30"; it will be determined with sufficient accuracy by considering the spherical triangle—of which at least three parts must be given, one of them being a side except when the radius of the earth is known as a plane triangle, whose given parts are equal to those of the spherical triangle. An error of 1" on an angle would produce an error of little more than 1 foot on the opposite side of a triangle whose sides are 20 miles long. When the two sides a, b, and the contained angle C are given, the area will be obtained nearly by the formula Tab sin C. When the side c and the angles A and B are given, sin A sin B sin Csin (A+B), a=c b = c sin (A+B)' sin (A+B)' sin Asin B and hence, T=122 sin (A+B) 566. PROBLEM V.-Given three parts of a spherical triangle, one of which is a side, to find the spherical excess. 'From the logarithm of the area of the triangle in feet subtract the number 9.3267737, and the remainder will be the logarithm of the spherical excess in seconds.' L. EL.T-9.3267737. EXAMPLE. In a spherical triangle LHS, the stations of which are Leith Hill, Hanger Hill, and Severndroog Castle, the observed angle at L is 35° 23′ 14′′, and the containing sides LS and LH are 144760 and 127660. Let the sides opposite to the angles L, H, S, be denoted by l, h, s, then L. Ths sin L. L. + h 72380 L. sin L 35° 23′ 14′′ = 9.76275 -4-85962 L. T- 9.72842 •40165 L. E= Hence, E2·522 2′′·52. 567. In calculating the spherical excess, it is generally sufficient to carry the logarithms to the fifth decimal place inclusive, and therefore an error of a few seconds in the given angles, or of several feet in the sides, if they do not much exceed the usual limits, will not produce an error on the spherical excess amounting to 100 part of 1". EXERCISES. 1. In a spherical triangle WSL, the sides LW, SW, are 158840 and 75014, and angle W = 65° 26′ 48′′; what is the spherical excess? Ans. 2"-55. 2. In a triangle CSW, CS = 63489, CW = 133640, and angle C = 16° 35' 2"; what is the spherical excess? Ans. 0"-97.. 3. In a triangle HLW, the side HL 127660, angle H = 84° 59′ 57′′, and angle L = 17° 42′ 37′′; what is the spherical excess ? Ans. 1"-2. In the preceding exercise, the side HW or LW is first to be found, considering the triangle as plane, then this side and HL and the included angle are given, as in the preceding example. Or use the formula (565), T = }} c2· sin A⚫ sin B sin (A+B) 568. The sum of the observed angles of a triangle above 180°, may be called the observed spherical excess. 569. PROBLEM VI.-Having given the three observed angles and the spherical excess, to find the true values of the angles. 'Subtract the observed from the true spherical excess, and add one-third of the remainder to each of the observed angles, and the sums will be the true spherical angles." When the observed exceeds the true spherical excess, the remainder is negative, that is, the third of the difference is to be subtracted from each observed angle. When the observed spherical excess is negative, that is, when the sum of the observed angles is less than 180, the remainder will be the sum of the two spherical excesses. The rule applies only to the case in which there is reason to consider that the three angles have been observed under equally favourable circumstances; when one of them is considered to be perfectly correct, the correction must be distributed equally between the other two angles; and if two of them are quite correct, the whole correction must be applied to the third angle. When the angles, or any two of them, have not been observed under equally favourable circumstances, the correction must be distributed among them, according to the degrees of probability of accuracy, or what is called the weight of each observation, as determined by the principles of probability, as afterwards explained. Let E and E' denote the true and observed spherical excess, and E, the error or whole correction, then 1 = E-E'. E1 Let A1, B1, C1, be the observed angles, and a, ß, y, their corrections respectively, and A, B, C, the true angles; then a+B+y=E2, and when the corrections are to be equally distributed, a=1E,,B=3E,, 7=E1. = 1 1 1 And AA,+E1, B=B2+}E1, C=C2+3E.. Or generally, A = A2+α, B=B2+ß, C=C1 +7. In the exercises the correction is understood to be equally distributed, unless otherwise expressed. EXAMPLE. The observed angles A1, B1, C1, are 65° 40' 10"-82, 40° 30′ 45′′-94, and 73° 49′ 5′′-69; find the true angles, the true spherical excess being 6"-42. A1 = 65° 40′ 10′′-82 40 30 45.94 C1 73 49 5-69 Here B, = 180 0 2.45 Hence, E,E-E'-6"-42-2"453"-97. } E1 = 1′′-32. And Therefore, 1 A = A2+ α = 65° 40′ 12"-14 C 180 0 6.41 EXERCISES. 1. The three observed angles of a triangle are 28° 40' 32"-26, 54° 28′ 17′′-35, and 96° 51′ 12′′-95; required the true angles, the spherical excess being 4"-3. Ans. 28° 40′ 32′′·84, 54° 28′ 17′′-93, and 96° 51′ 13′′-53. In the two following examples, A1, B1, C1, and E, are given to find A, B, and C; angle A, in the last being correct. 2. A1 = 50° 42′ 36′′-8, E = 2′′·13, A = 50° 42′ 38′′-43. =79 49 32-25, B 79 49 3388. C49 27 48-19, = B1 C49 27 49.82. 3. A, 39° 40′ 17′′-24, E1"-04, A = 39° 40′ 17′′-24. B=68 14 31.10. 5 13.72, 1 570. When three parts of a spherical triangle in a geodetic survey are given, the other parts can be calculated by means of the rules of spherical trigonometry; but there are more simple and expeditious methods of accomplishing this -the method of Legendre, by an equal-sided plane triangle;· and that of Delambre, by means of the triangle of the chords or the chordal triangle. Of these, Legendre's is the most simple. If from each of the angles of a spherical triangle one-third of the spherical excess is deducted, the remainders are the angles of a plane triangle, the lengths of whose sides are equal to those of the spherical triangle. This is Legendre's theorem, on which his method is founded, which is adopted in the following problems. 571. PROBLEM VII.-Given the three angles of a spherical triangle, to find the angles of the equal-sided plane triangle. 'From each of the angles of the spherical triangle deduct one-third of the spherical excess, and the remainders are the required angles." Let A, B, C, be the angles of the spherical triangle; A', B', C', those of the plane triangle, and E the spherical excess; then A' = A — } E, B'=B—3E, and C'=C -} E. EXAMPLE. The three angles of a spherical triangle are 48° 12' 30"-02, 55° 17′ 36′′-31, and 76° 29′ 55′′-8; find the angles of the equal-sided plane triangle. EXERCISES. In the following exercises, the angles A, B, C, of the spherical triangle are given to find the angles A', B', C', of the equal-sided plane triangle. Given. 1. A = 76° 13′ 4′′-32, B48 15 13.8, C55 31 45.04, Answers. = A' 76° 13′ 3′′-27. = |