Hence, having found a subsidiary angle w, by the equation 2. The given quantities being the same as in last example, to deterinine the angle C. tan. A 39° 23' 0" 9.9143020 cos. A cos. c 68 46 2 9.5588979 sin. w, ar. comp. 9.8881335 0.0183921 cot. 73 26 33 9.4731999 sin. (Bw) 39° 41′30′′ 9.8052682 As (B) is negative, cos. C must be negative; hence C is the supplement of this, viz. 120° 59′ 47′′. 3. Given A=30° 28′ 11′′, B = 130° 3′ 11′′, and c=40°; to determine the other parts. a=38° 30′, b = 70°, and C = 31° 34′ 26′′. 4. Given A 31° 34′ 26′′, B=30° 28′ 12′′, and C 70° 2' 3"; to find the angle C. C130° 3' 11". 5. Given A= 34° 15′ 3′′, B=42° 15′ 13′′, and C=76° 35′ 36′′; to find a and b. a = 40° 0′ 10′′, b=50° 10′30′′. 6. Given A = 51° 30′, B = 131° 30', and c 80° 19′ 12′′; to find C. CASE V. C 59° 15′ 59′′. (60.) Given two sides a, b, and the angle A opposite to a; to find the other parts B, C, c. 1. To find the angle B we have, by (46) the proportion, 2. To find C and c, we have, by Napier's analogies, cos. (a ~ b): cos. 1⁄2 (a + b) :: tan. 1⁄2 (A + B): cot. & C (2) cos. § (A ~B): cos. (A+B) :: tan. § (a + b): tan.c Or after either C or c is found by one of these analogies the other part may be found by the proportion although we shall prefer Napier's analogy to this in order that all ambiguity may be avoided. If only one of the parts C, c, be required, then it will be best to find first the angle B, by the proportion (1), which operation must be regarded entirely as subsidiary to the determination of the required part, by one of the analogies (2). The part determined by the proportion (1) admits of a double value, since two arcs answer to the same sine, it becomes necessary, therefore, for us to inquire under what circumstances both these values are admissible, and how we may know which to choose when but one solution exists. Referring to the fundamental formula (A), we have in which expression we may remark that if cos. b is numerically greater than either cos. a or cos. c, the second member must take the sign of cos. b, consequently, B and b must be of the same species if sin. b < sin. a, or sin. b < sin. c, that is, an angle must be of the same species as its opposite side, if the sine of this side is less than the sine of either of the other sides. But if cos. b is numerically less than cos. a, then whether the righthand member be + or will depend upon the magnitude of cos. c, or cos. c will have two values corresponding to + cos. B, and - cos. B; hence an angle has two values, when the sine of its opposite side is greater than the sine of the other given side. EXAMPLES. 1. Given the side a=63° 50′, the side b= 80° 19′, and the angle A=51° 30'; to determine the other parts. The angle B admits of two values, because sin. b > sin. a, so that there exist two triangles, having the data proposed. We shall, however, take the acute value of B. 2. Given a = 40° 36′ 37′′, b = 91° 3′ 25′′, and A=35° 57′ 15′′; to The angle B admits of two values, because sin. b > sin. a. We shall suppose the particular triangle under consideration to have B obtuse. = 3. Given a 40° 18′ 29′′, b=67° 14′ 28′′, and A= 34° 22′ 17′′; to determine the other parts when B is acute. B=53° 35′ 15′′, C=119° 13′ 31′′, c= 89° 47′ 6′′. 4. Given u=84° 14′ 29′′, b=44° 13′ 45′′, and A=130° 5′ 22′′; to determine the other parts. B=32° 26'6", C=36° 45′ 28′′, c=51° 6′ 12′′. 5. Given a = 97° 18′ 39′′, b=86° 53′ 46′′, and A = 97° 21′ 26′′; to determine c. CASE VI. c=89° 21′ 37′′. (61.) Given two angles A, B, and the side a opposite to one of them, to find the other parts. 1. To find b we have sin. A sin. B:: sin. a: sin. b. 2. And to find C and e we may employ Napier's analogies, which need not be here repeated. The nature of the arc b may be discussed, as in the preceding case. Thus the formula (B), art. (50), gives from which it follows, as in the foregoing case, that if cos. B is numerically greater than cos. A, B and b will be of the same species. If cos. B is numerically less than cos. A, then both the values of b, given by the above proportion, will be admissible, for C may be determined so as to render cos. b positive or negative. Hence any side will be of the same species as its opposite angle, if the sine of this angle be less than the sine of either of the other angles; and the species of the side b will be indeterminate if the sine of its opposite angle B be greater than the sine of the other given angle A. There cannot, therefore, be two solutions unless a and A are of the same species. |