From the data, 2S A+B+C 296°, or S = 148°; = SA = 78°, SB = 16° 50′, S — C= 53° 10'. Note 1. Since cos S is (Art. 37), while the cosines of S-A, SB, and S-C are +, the quantities under the radical signs are essentially positive, and hence no attention need be paid to the negative signs in the formula. b = 137° 11.8', and c = 115° 55.8'. The values of a, b, and c may also be obtained by aid of the sine or cosine formulæ of Art. 159. If all the sides are to be computed, the tangent formulæ are the most convenient, as only four different logarithms are required. If but one angle is required, the sine formula will be found to involve the least work. Note 2. The triangle is always possible for any values of the given elements, provided S is between 90° and 270°, and each of the quantities S-A, S-B, and S - C between 90° and — 90° (Art. 37). CASE V. 171. Given two sides and the angle opposite to one of them. 1. Given a 58°, b = 137° 20', = B = = 131° 20'; find A, C, To find C and c, we have by Arts. 160 and 162, Also, † (B+A)=100° 39', † (B — A)= 30° 41′. From which we obtain C = 94° 41.6', and c = 115° 53.6'. Using the second value of A, we have † (B+A) = 120° 41′, † (B — A) = 10° 39'. From which we obtain C = 147° 26.4', and c = 150° 56.8'. Thus the two solutions are: 1. A= 69° 58', C= 94° 41.6', c = 115° 53.6'. 2. A 110° 2′, C=147° 26.4', c = 150° 56.8'. = As in the corresponding case in the solution of plane oblique triangles (compare Arts. 124 to 126), there may sometimes be two solutions, sometimes only one, and sometimes none, in an example under Case V. After the two values of A have been obtained, the number of solutions may be readily determined by inspection; for by Art. 136, (b), if a is <b, A must be <B, and if a is >b, A must be > B. That is, only those values of A can be retained which are greater or less than B according as a is greater or less than b. Thus, in Ex. 1, a is given <b; and since both values of A, 69° 58' and 110° 2', are < B, we have two solutions. Also if the data are such as to make log sin A positive, there will be no solution corresponding. or, 2. Given a = 58°, c=116°, C= 94° 50'; find A, B, and b. Since a is given <c, only values of A which are < C can be retained; hence there is but one solution, corresponding to the acute value of A. To find B and b, we have by Arts. 160 and 162, From which we obtain B= 131° 21.8', and b = 137° 23.6'. Thus the only solution is A=70° 5', B = 131° 21.8', b = 137° 23.6'. 3. Given b=126°, c = 70°, B= 56°; find C. Since both values of C are>B, while c is given<b, there is no solution. CASE VI. 172. Given two angles and the side opposite to one of them. 1. Given A= 110°, B= 131° 20′, b = 137° 20'; find a, c, To find c and C, we have by Arts. 160 and 162, Using the first value of a, we have c=150° 53.8', and C 147° 23'; In Case VI., as well as in Case V., there are sometimes two solutions, sometimes only one, and sometimes none; and it may be shown, exactly as in Art. 171, that only those values of a can be retained which are greater or less than b according as A is greater or less than B. |