Adding 1 to each member of (5), we have Denoting half the sum of the three sides by s, we have b+c+a=28; b+c-a=2s-2 a 2 (s-a). Hence equation (6) will become 1 + cos. A 2 s (s — a) By equation (8), § 15, we have 1+cos. A = 2 (cos. A)2. Therefore equation (7) will become 2 (cos. A)2 = (6.) (7.) 2 s (s — α) By using logarithms, equations (8), (9), and (10) become log. cos. A= [ar. co. log. b+ar. co.log.c+log.s+log. (s—a)] log. cos. B= [ar. co. log. a+ar. co.log.c+log.s+log. (s—b)] ((A) log. cos. C=[ar. co. log. a+ar. co.log.b+log.s+log. (s—c)]} 2 Applying equations (A) to the case already given, we have a=50.25; b=60.5; c=68·4; consequently, ↓ (a+b+c) = 889-575, and s-a=39.325; s-b=29′075 ; 2 Hence A=22° 41′ 17′′; 1⁄2 B= 29° 29′ 9′′; C=37° 49′ 34′′, and A=45° 22′ 34′′; B = 58° 58′ 18′′; C = 75° 39′ 8′′. When the sides of a triangle are expressed in small numbers, equation (5) may be advantageously employed: this equation, when the letters are properly permuted so as to apply to each angle, gives Nat. cos. A b2 + c2 - a2 Nat. cos. B = (B.) 2 ac Nat. cos. C = As an example, suppose the sides to be 5, 7, and 9. Re Hence, A = 33° 33′ 27′′; B = 50° 42′ 13′′; C = 95° 44′ 21′′. NOTE.-In taking the logarithm of -7, we took it as though it had been +7, and at the right annexed the letter n to indicate that the number is negative. So in the value of cos. C we write the n, thus indicating that the natural cosine of this angle is negative. We then seek in the tables for the angle as though its cosine was positive, and find 84° 15′ 39", which taken from 180° gives the obtuse angle 95° 44' 21" for the true value of C. § 51. Additional Examples of Oblique Triangles. 1. Given, two angles of a triangle equal 47° 13' and 63° 36', and the side opposite the first angle equal 27-5, to solve the = 69° 11'. triangle. The other angle The other sides Ans. 33.562. 35.024. 2. Given, two sides of a triangle equal 47·13 and 63-36, and the angle opposite the first side equal 27° 50', to solve the tri The other side €92.719. Second Ans. The other angles = The other angles = {141° 7′ 13′′. 11° 2′ 47′′. 3. Given, two sides of a triangle equal 49-5 and 101-5, and the angle opposite the first side equal 30° 25', to solve the triangle. Ans. The question is impossible. 4. Given, two sides of a triangle equal 630 and 800, and the angle opposite the first side equal 100°, to solve the triangle. Ans. The question is impossible. 5. Given, two sides of a triangle, 77.5 and 90-25, and the included angle equal 83° 38', to solve the triangle. The other side =112.25. Ans. The other angles = { 43° 19' 38". 53° 2′22′′. 6. Given, the three sides of a triangle, 40, 50, and 60, to find the angles. 41° 24′ 35′′. Ans. The angles 55° 46′ 16′′. 82° 49' 9". 7. Given, two sides of a triangle, 100 and 90, and the angle opposite the first side, 111° 15', to solve the triangle. Ans. The other side =21.823. 11° 44' 7". 57° The other angles = {1700' 53". 8. In an obtuse-angled triangle one of the acute angles is 29° 15′, the other acute angle is 2° 45'; the side opposite the first angle is 6-3. What are the other parts? 9. Given, two sides of a triangle, 332-21, 237-61, and their contained angle equal to 72° 29′ 48′′, to find the other parts. Ans. The other angles = { 40° 59' 35". 66° 30′ 37′′. The other side = 345·46. 10. The three sides of a triangle are 10, 15, and 20. What are the angles? 28° 57' 18". Ans. 46° 34' 3". 104° 28' 39". 11. The three sides of a triangle are 123-48, 135∙61, 140·91. What are the angles? 53° 0' 10". Ans. 61° 17′ 50′′. 65° 42' 0. 12. Two sides of a triangle are 59-34 and 12-13, and their included angle is 150° 38'. What are the other parts? $52. SPHERICAL TRIGONOMETRY treats of the methods of computing the unknown parts of a spherical triangle, when certain parts are given. But before proceeding to the investigation of the relation of the different parts of a spherical triangle, we will give some GENERAL TRIGONOMETRIC FORMULAS. $53. Under $15 and §16 we have already given several general and useful formulas which were needed in Plane Trigonometry. We now propose to continue these formulas to a greater extent, giving such additional ones as are needed in Spherical Trigonometry, as well as others which will be found convenient in the reduction and simplification of trigonometric expressions in general. Dividing (10) by (11), § 15, we have sin. a cos. a √1-cos. 2a =tan. a √1+cos. 2a (1.) Multiplying both the numerator and denominator of the fraction constituting the right-hand member of (1) by the numerator, we have tan. a = 1- cos. 2a √1-cos.22a 1-cos. 2α sin. 2a (2.) If we multiply both the numerator and denominator of the right-hand member of (1) by the denominator, we shall obtain √1-cos.2 2a sin. 2a 1+ cos. 2a (3.) |