80. CASE II. Given A, C and b. Second Solution; when the two remaining sides, or when the three unknown parts are all required. We have, by Napier's Analogies, (40) and (41), added to the half sum gives the greater side, and the half difference subtracted from the half sum gives the less side. If C>A, we may write CA, c-a in the place of A― C, a — c. We may now find B by either of Napier's Analogies, (42) and (43), which give* 1. Given A = 135° 5′ 28′′-6, C= 50° 30′ 8′′-6, b = 69° 34′ 56′′-2 ; find a, c and B. We have Then, by (139), (A + C)=92° 47′48′′-6 (A - α) = 1634° 47'28"-1 = ar co log sin (A+C)+0.0005176 ar co log cos(A+C')-1-3116286 * We may also find B by any one of Gauss's Equations, (44), interchanging B and C, b and c. ar co log sin(ac)+0.3726772 arco log cos(a-c)+0.0430243 81. CASE II. Given A, C and b. Third Solution. When the third angle B is alone required, the computation by (134) is in most cases as convenient as any other, but there are other methods (corresponding to those given in Art. 75 for finding a) which may occasionally be serviceable. By (14) and (15) we have cos B = cos B = cos (A + C) — 2 sin A sin C' sin2 6 b (142) the computation of which is similar to that of (130) and (131). EXAMPLE. (143) For the reasons given in Art. 74, (141) is, in this example, not so accurate = 0.5602162 log 9.7483557 B=65° 33′ 9′′.9 as (140). 82. In Art. 14, several formulæ are given, by which B may be computed. By (21) and (22) we have =sin [(4+ C) + ] sin [} (4 + C) — 4] of which (144) is to be preferred when B < 45°, and (145) when 1⁄2 B > 45o. 83. Case II. might have been reduced to Case I. by means of the polar triangle, Art. 8; for there will be known in the polar triangle, two sides and an angle opposite one of them, being the supplements of the given angles and side of the proposed triangle. The polar triangle being solved, therefore, by Case I., and its two remaining angles and third side found, the supplements of these parts would be the two sides and third angle required in the proposed triangle. It is easily seen, also, that all the formula above given for this case might have been obtained by these considerations. 84. CASE III. Given two sides and an angle opposite one of them; or a, b, and A. Fig. 9. First Solution, in which each required part is deduced directly from fundamental formulæ independently of the other two parts. To find c. We have, by (4),* А Fig. 9. C b B c = Q + Q' This formula has been already employed and adapted for logarithms in Case 1; but, for the sake of clearness, it is repeated. The student will remark that a simple transposition of (122) gives (146). It will also be observed that the given angle and the given side adjacent to it, in each of the first four cases, are denoted by A and b, in order that the auxiliaries and 9 may have the same values throughout. Φ The auxiliary will be fully determined by (m), being taken between 0 and 180°, and always positive (Pl. Trig. Art. 174); but, as the cosine of an angle is also the cosine of the negative of that angle [Pl. Trig. (56)], we may take o' in (m') either with the positive or the negative sign, so that co'. There will thus be two values of c answering to the same data, both of which will be admissible, except when + 'exceeds 180°, in which case the only solution is c = '; and except when 'exceeds (which would make Φ c negative), in which case the only solution is c = 0 + Q'. Therefore, eliminating k, we have for finding c, Φ Q tantan b cos A Φ. cos Csin A cos b + sin C cos Asin A sin b cot a (N) = sin A sin b cot a } (n) Here will be fully determined, while 'found by its cosine may be either positive or negative, so that we shall have in general two values of C9', corresponding respectively to the two values of c; but, as before, values greater than 180°, and negative values, being excluded, there will in certain cases be but one solution. sin A cos b Eliminating h= -, we have, then, for finding C, COS 9 To find B. We have several methods: 1st, directly by (3), (147) (148) which gives two values of B, supplements of each other, corresponding respectively to the two values of c and C. We shall presently see how to determine which are the corresponding values of c, C and B. has the same value as in (146), and therefore put 2d. In (123), ting in (123), c— Q = Φ ', we have sin 'cot A sin (149) which gives two values of B by the positive and negative values of q'. 3d. By (124), cos Btan 'cot a (150) which also gives two values of B by the positive and negative values of q'. 4th. In (134), ♬ has the same value as in (147), and therefore putting in (134), C-99', The formula (149) shows that when q' is positive, cot B and cot A have the same sign, that is, B and A are in the same quadrant; and that, when 'is negative, cot B and cot A have different signs, that is, B and A are in different quadrants. A like result follows from (151), with reference to 9'. Hence, that value of B which is in the same quadrant as A, belongs to the triangle in which c=Q+Q', C = I + I'; and that value of B which is in a different quadrant from A, belongs to the triangle in which co- q', C— 9 — 9'. This precept enables us to employ (148) without ambiguity. In the = |
