2. In ABC, A = 50°, B = 75°, c = 60 in. Solve the triangle. 3. In ABC, A = 131° 35', B = 30°, b = 5} ft. 4. In ABC, B = 70° 30', C = 78° 10', a = 102. 5. In ABC, B = 98° 22', C = 41° 1′, a = 5.42. Find a. Solve the triangle. 56. Case II. Given two sides and an angle opposite to one of them. In the triangle ABC let a, b, A be known, and C, B, c be required. The triangle will first be constructed with the given elements. At any point A of a straight line LM, unlimited in length, make angle MAC equal to angle A, and cut off AC equal to b. About C as a centre, and with a radius equal to a, describe a circle. This circle will either: (1) Not reach to LM, as in Fig. 50. (2) Just reach to LM, thus having LM for a tangent, as in Fig. 51. (3) Intersect LM in two points, as in Figs. 52, 53. Each of these possible cases must be considered. In each figure, from C draw CD at right angles to AM; then CD=b sin A. In case (1), Fig. 50, CB < CD, and there is no triangle which can have the given elements. Hence, the triangle is impossible when ab sin A. = In case (2), Fig. 51, CB CD. Hence, the triangle which has elements equal to the given elements is right-angled when a = b sin A. In case (3), Figs. 52, 53, CB> CD; that is, a > b sin A. If a> b, then the points B, B1, in which the circle intersects LM, are on opposite sides of A, as in Fig. 52, and there is one triangle which has three elements equal to the given elements, namely, ABC. If a <b, then the points of intersection B, B1, are on the same side of A, as in Fig. 53, and there are two triangles which have elements equal to the given elements, namely, ABC, AB,C. For, in ABC, angle BAC=A, AC = b, BC=a; in ABC, angle B1AC=A, AC b, B1C = a. Both triangles must be solved. In this case, Fig. 53, the given angle is opposite to the smaller of the two given sides. Hence, there may be two solutions when the given angle is opposite to the smaller of the two given sides. The words "may be" are used, for in cases (1), (2), the given angle is opposite to the smaller of the two given sides. Case II. is sometimes called the ambiguous case in the solution of triangles. = The ambiguity in Case II. is also apparent in the trigonometric solution. The angle B is found by means of the relation, The angle B is thus determined from its sine. Now there is always an ambiguity when an angle of a triangle is determined from its sine alone, for sin x = sin (180° - x). Figure 53 shows the two angles which have the same sine, namely, ABC, AB,C. In Fig. 52, the given condition, namely, that b<a, shows that B<A; accordingly, only the acute angle corresponding to sin B can be taken. If, in equation (1), b sin A> a, then sin B>1, and, accordingly, B is impossible and there is no solution. If, in equation (1), b sin A = a, then sin B = 1, and B = 90°. The consideration of the trigonometric equation (1) leads, therefore, to the same results as the preceding geometrical investigation. Checks: A+B+C=180°, and, as in Case I. Other checks will be found later. EXAMPLES. 1. Solve the triangle STV, given: ST = 15, VT = 12, S = 52°. FIG. 54. Both values of V must be taken, since the given angle is opposite to the smaller of the given sides. The two triangles corresponding to the two values of Vare STV, STV1, Fig. 54, in which In the ambiguous case, care must be taken that the calculated sides and angles are combined properly. 2. Solve ABC, given: a = 29 ft., b = 34 ft., A = 30° 20'. 57. Case III. Given two sides and their included angle. In the triangle ABC, a, b, C are known, and it is required to find A, B, c. In this case, c can be determined from the relation c2 a2+b2 - 2 ab cos C, Art. 54; angle = A can be determined from the relation sin A sin C. C Checks: a2= b2+c2 - 2 be cos A, b2a2+ c2 2 ac cos A, the result in Ex. 1, Art. 54 a; other checks will be found later. EXAMPLES. 1. In triangle PQR, p = 8 ft., r = 10 ft., Q = 47°. Find q, P, R. 58. Case IV. Three sides given. If the sides a, b, c are known in the triangle ABC, then the angles A, B, C can be found by means of the relations (3), Art. 54. Checks: Relations (1), Art. 54: A+B+C=180°. Other checks will be shown later. A =7 C-10 FIG. 57. B Angle C is in the second quadrant since its cosine is negative. Check: 18° 12' +33° 7′30′′+128° 40′52′′-180°0'22". The discrepancy is due to the fact that four-place tables were used in the computation. Had five-place tables been used, the discrepancy would have been less. Find P, Q, R. Find R, S, T. 2. In PQR, p = 9, q = 24, r = 27. 3. In RST, r = 21, = s 24, t = 27. Find A, B, C. b = 26, c = 74. 6. Solve Ex. 1, using five-place tables. Find A, B, C. 59. The aid of logarithms in the solution of triangles. It was pointed out in Art. 6 that an expression is adapted for logarithmic computation when, and only when, it is decomposed into factors. In Cases I., II., Arts. 55, 56, the expressions used in solving the triangle can be computed with the help of logarithms. On the other hand, the side opposite to the given angle in Case III., Art. 57, and the angles in Case IV., Art. 58, are found by evaluating expressions which are not adapted to the use of logarithms. Other relations between the sides and angles of a triangle will be found in Arts. 61, 62. By these relations the computations in Cases III., IV., can be made both without and with logarithms. These relations are useful not merely for purposes of computation; they are important in themselves, and valuable because many important properties of triangles can be deduced from them. The explanations given in Arts. 55-57 are presupposed in Arts. 60-62. The general directions to be observed in working the problems are as follows: |