or a = 31 54' 46' and a = 63 49' 32". To verify this result, and to apply the formulæ (8) and (9) of Art. 9, the side a may now be found independently of the angles C and B. Let a', b', c', be the sides, and A', B', C', the angles, of the polar triangle, then 128 30', A B' 180-b Whence there are given of the polar triangle A' B'C', the two sides a' and c', and the included angle B', from which by the preceding method, its remaining parts may be found, and thence the remaining parts of the primitive triangle. To find the angles A' and C' by the formulæ, = 9.7344660 180 C' 120` 48' 9". 180 - b'. = The third side b' is now found in the same way as a in the last example, and thence B by the relation B This example may also be solved Analogies, given in Årt 10. by means of Napier's Second EXERCISES. 1. Given a 84° 14' 29", b find the angles A and B. = = 36' 45' 28", to 44° 13' 45", and C Ans. A = 130' 5' 22'', B = 32° 26' 6". 39° 23', B = 33° 45' 3", and c = 68° 46′ 2′′, to find the Ans. a 43 37' 36", b = 37° 10', C=120 59' 46". 3. Given A = 31° 34' 26", B = 30° 28′ 12", and c = 70° 2' 3", to find C. Ans. C 130° 3′ 50′′. 2. Let A = other parts. = CASE III. (1.) Given the three sides (a, b, c) of a spherical triangle, to find the angles: (2.) Or given the three angles (A, B, C) to find the sides. When the three sides are given, the angles are found at once from (4), (5), or (6), of Art. 7. When the three angles are given, the same formula may be applied to the polar triangle. This, however, never occurs in any applications of Spherical Trigonometry. Professor De Morgan remarks (in his Elements of Spherical Trigonometry), "Delambre, who probably calculated more spherical triangles than any man of his day, says, he never met with this case (the three angles given to find the sides) but once, and then he could have done without it." EXAMPLE. 1. Given a = 63' 50', b A, B, C. The formula (6) of Art. 7, is, employed in preference to the others, for the reason given in the analogous case of plane triangles (Art. 23). To find A:To find B: 8 = 132° 28' cosec = 10.1321377 sα = 68° 38' cosec = 10.0309254 cosec = 10.1321377 sin = 9.9690746 8- b = 52° 9' sin = 2. Given a = 68° 46' 2'', b = 43° 37′ 38", and c = 37° 10', to find the angles. Ans. A = 120° 59' 46", B = 39° 23', C = 33' 45' 2". 3. The three sides of a spherical triangle are 48 24' 16", 59° 38' 27", and 70' 23' 42", what are the angles? Ans. 52° 32' 54", 66 20′ 40′′, and 90'. EW 4. Given a = A, B, C. = 40 0' 10", b 50° 10′ 30′′, c = 76° 35' 36", to find 121° 36' 20". Ans. A = 31° 15' 2", B = 42° 15′ 13" 26, C = THE AREA OF A SPHERICAL TRIANGLE. 13. To find the area of a spherical triangle. N D Let the planes of the great circles CAD, CND, intersect in the diameter CD, and let A Ñ be the arc of another great circle whose poles are C and D; then O being the centre of the sphere whose radius OA or O D is r, we have by the Mensuration, Art. 13, and Euc. vi. 33, Lune ACND: surface of sphere :: arc A N : 2 wr By means of this expression for the lune, the area of a spherical triangle may be found in terms of its angles and the radius of the sphere. Let ACB be a spherical triangle on the surface of the sphere whose radius is r. Produce the sides A C, BC, till they meet again in c in the opposite hemi- is the A (A+B+C) -22, Hence, since the surface of the hemisphere is 2 r, the area of a spherical triangle is to the surface of the hemisphere as the excess of its three angles above two right angles is to four right angles. The expression A+B+C 180° is called the spherical excess, being the excess of the angles above two right angles. Cor. If A be a spherical triangle, and E the spherical excess, then 1. The angles of a spherical triangle, measured on the surface of the earth, are 83 10' 57", 66° 15′ 16′′, and 30' 33' 48": find the spherical excess, and the area of the triangle, the earth's diameter being 7957.5 miles, and its form being taken as spherical. = Ans. E 1"; area = 115 1223 miles. 2. The area of a spherical equilateral triangle is one-fourth of the surface of the sphere; what are its angles? 3. Prove that if E be the spherical excess, sina sin b sin C. 5. Deduce from Exercises 3 and 4 the area of a spherical triangle in terms of two sides and the included angle. MENSURATION OF PLANES. THE term mensuration is applied to those methods or formulæ by which the lengths of curve lines, and the areas and volumes of superficial and solid figures, are determined. PARALLELOGRAM AND TRIANGLE. C D F C 1. To find the area of a square, a rectangle, or any parallelogram. First let A B, A D, be two adjacent sides of a rectangle, and A E, A G, each equal to the unit with which A B and AD are compared; then the square A E F G, described on A E or A G, will be the square unit with which the area A B C D is compared. And since the parallelograms ABCD, AEF G, are equiangular, they are to one another (Euc. vi. 23) in a ratio compounded of the ratios of their sides; that is, they are to one another as A B.AD to A E.A G. But A E F G = A E.AG1; hence A B C D AB.A D. Consequently, if the adjacent sides of the rectangle be denoted by a and b, = area ab.* A E Cor. The area of a square is found by squaring one of its sides. C F Next, let A B D C he any parallelogram. Upon A B describe the rectangle A B F E, so that this rectangle and the parallelogram A B D C may be on the same base, and between the same parallels. Then, by Euc. i. 35, and the preceding expression for the area of a rectangle, ABDC=ABFE = AB.AE=AB.CG. C B But (Plane Trig., Art. 20), G C A C sin CAB. Hence denoting the sides A C, A B, by b and c, and the included angle by A, we have for the area of the parallelogram ABDC, area = c p b c sin A, p being the altitude of the parallelogram. 2. To find the area of a triangle. Since the triangle A B C (fig. to last Art.), is half the parallelogram ABDC (Euc. i. 41), we have, or, ABC: A = cpbc sin A, in which b, c, are two sides of the triangle, A their included angle, p the perpendicular on c from the opposite angle C, and A the area of the triangle. *This proof, it will be noticed, applies only to cases in which the two sides of the rectangle can be exactly measured by a common linear unit. It is easily extended, however, to those cases in which the sides are incommensurable with the linear unit. |