In Fig. 49 α, in Fig. 49 b, Also, ADAC cos BAC; ADAC cos BAC (Art. 40). DC2 + AD2=AC2. Hence, in both figures, BC2 = AC2 + AB2 — 2 AC • AB cos A ; that is, a2 = b2 + c2 – 2 bc cos A. This formula also holds for Fig. 49 c; for there, COS A = cos 90° = 0. (3) Similar formulas for b, c, can be derived in like manner, or can be obtained from (3) by symmetry: b2 = c2 + a2 2 ca cos B, c2= a2 + b2. 2 ab cos C. (3') These formulas can be expressed in words: In any triangle, the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these two sides multiplied by the cosine of their included angle. NOTE. In Fig. 49 a, A is acute and cos A is positive; in Fig. 49 b, A is obtuse and cos A is negative. Hence formula (3) shows that in Fig. 49 a, a2 is less than 62+ c2, and that in Fig. 49 b, a2 is greater than b2 + c2. Fig. 49 c, a2 = b2 + c2. Relation (3) may be expressed as follows: In cos A = b2 + c2- a3 (4) and similarly for cos B, cos C. Ex. Derive the formulas for b2 and for c2. Each of the relations (1), (3), (3'), involves four of the six ele. ments of a triangle. If any three of the elements in any one of these relations are known, then the fourth element can be found by solving the equation. Inspection shows that relations (1) are serviceable in the solution of Cases I., II., Art. 53, and that relations (3), (3'), are serviceable in the solution of Cases III., IV., Art. 53. The student is advised to try to work some of the examples in Arts. 55-58 before reading the text of the articles. (See Arts. 20-24, 34.) 54 a. Substitution of sines for sides, and of sides for sines. Since α b с = sin A sin B sin C' the sines of the opposite angles can be sub stituted for the sides of triangles, and vice versa, when they are involved homogeneously in the numerator and denominator of a fraction, or in both members of an equation. for cos(AB). 2 sin (A + B) cos } (A — B) cos (AB) = sin (A + B) = cos | C, since } (A + B) + } C = 90°. 2. Derive two other relations similar to that in Ex. 1. 55. Case I. Given one side and two angles. In triangle ABC, suppose that A, B, ɑ are known; it is required to find C, b, c. In this case (see Fig. 47 a, Art. 54), Ex. 1, Art. 54 a. Other checks will be discovered later, = sin B sinடு EXAMPLES. 1. Solve the triangle PQR, given: PQ = 12 in., Q = 40°, P = 75°. Solution: R = PR= RQ = 2. In ABC, A = 50°, B = 75°, c = 60 in. Solve the triangle. 3. In ABC, A = 131° 35', B = 30°, b = 51 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 a <b 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 AB,C, angle B,AC-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, ABC. 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 V are STV, STV1, Fig. 54, in which In the ambiguous case, care must be taken that the calculated sides and angles are combined properly. |