CHAP. V. On the Resolution of the Cases of Rectilinear Triangles-1st, When the Triangles are Right-angled.--2d, When Oblique. Reasons for introducing different Solutions of the same Case. Examples, &c. In a triangle there are 3 sides and 3 angles: any three of these being given, the remaining may be obtained. There is one exception to this, which takes place when the three quantities given are the 3 angles. The reason of the exception is this: take any triangle, then, externally or internally, other triangles may be formed with sides parallel to the sides of the proposed triangle, which triangles shall have the same angles, but greater sides or less sides: the magnitudes of the sides therefore are independent of the angles, and consequently cannot be determined from them. We will begin with the solutions of the cases of right-angled triangles. 1st Case of right-angled triangles, in which two sides are given. Here, besides the right angle C, a, b, are given, and c, A, B are 2d Case, in which the hypothenuse and one of the acute angles are mined exactly as b has been; or thus, from b, a = √(c2-b2) = √{ (c + b) (c-b) }. 1 2 In Logm. log. a = { log. (c+b)+log. (c - b) } . 3d Case, in which a side and the acute angle (which is not opposite to Example. b=31.76 A=17° 12′ 51′′ ... B =72°47′ 9". Computation for a. log. b or log. 31.76 = 1.5018805 log. tan. 17° 12′ 51′′ = 9.4911132 log. a + 10 = 10.9929937 ... a = 9.8399 Computation for c. log. 101° = 10. 10+log. b. = 11.5018805 log. cos. 17° 12′51′′ = 9.9800967 log. c = 1.5217838 ..c 33.249. We now proceed to the Cases of Oblique-angled Triangles. First Case, in which two angles and a side opposite to one of the angles are given. Here, A, B, a, are given, and b, c, C are required. Solution. Example. C determined. A = A+B+C=180°.•.C=180°-(A+B) 41°13′22′′, B=71° 19′ 5′′, a = 55 Second Case of oblique-angled triangles, in which, two sides and an angle opposite to one of the sides are given. Here, a, b, B are given, and A, C, care required. Example 1. (ambiguous). a=178.3, b=145, B=41°10′ Computation for A. sin. B log. sin. B In logarithms, or log. sin. 41°10′=9.8183919 log.sin. A=log.sin. B+log.a-log.b log. a or log. 178.3=2,2511513 This case may be ambiguous, or 12.0695432 will admit of two solutions, when a > b, and B is acute: for, let MN =a, MP=b, ∠ MNP=B, take Mn =MN,then MP(<Mn) falls between log. b or log. 145=2.1613680 log. sin. A =9.9081752 .. N, n, N, n, and another line MQ, also between N, n, can be taken equal to it;... the triangle may be MNP, or MNQ, and the angle A may be either MP2 or its supplement M2N3; but, if bor MP be > a or MN, MP would fall beyond N and n, as MP' does, and no other line equal to it can be drawn between P' and N: in this case, A has one value only. If B be obtuse, A cannot; therefore here also the case is not ambiguous. A and B being known, C = 180° - (A + B) is known. .:. A = 54° 2'22" and C 84 47 38 or A=125 57 38 and C = 12 52 22 Example 2. (not ambiguous) a=145, b=178.3, B=41° 10′ log. sin. 41° 10'=9.8183919 log. 145 =2.1613680 11.9797599 log. 178.3= 2.2511513 log. sin. A= 9.7286086 .. A 32° 21′ 54′′. In this instance the supplement of A cannot belong to the case proposed. Computation of c in 1st Example. log. 145=2.1613680 log. sin. 84° 47′38′′ 9.9982047 12.1595727 log. sin. 41° 10'=9.8183919 log.c=2.3411808 ... c=219.37 Third Case, in which two sides and the included angle are given, Here, a, b, Care given, and A, B, c are required. The above is a complete solution of the case, in which two sides and the included angle are the quantities given. But, the analytic art is required to furnish, besides merely adequate solutions, commodious and concise ones. And, of this latter character are the solutions which have been given of the third case by Dr. Maskelyne, in the Introduction to Taylor's Logarithms, and by Legendre, in his Trigonometry, p. 369, 4th edit. These solutions we now proceed to explain. |