XIII. Two sides and the included angle being given to find the other two angles. Take the sum and difference of the two given sides, and the half sum and half difference, and also half the given angle. Under heads (1) and (2) put down the following logarithms. Under (1) and (2) log. cot. of half the given angle. Add together the logarithms under (1) and (2), and (rejecting 20 from each index) the results will be the log. tangents of half the difference and half the sum (a) of the required angles respectively. The sum of the two arcs thus found will be the angle opposite to the greater of the given sides: and their difference will be the angle opposite to the less of the given sides. (a) NOTE. If half the sum of the given sides be greater than 90°, half the sum of the required angles will also be greater than 90°: in this case, the arc found as above under (2) must be subtracted from 180° to get half the sum of the required angles. EXAMPLE. (178). In the spherical triangle ABC given a = 124° 10′, 6 89° 0′ 15′′, C=112° 1' 30": required the other two angles A and B. a=124° 10' 0" b= 89 0 15 a+b..213 10 15 (1) (2) a-b.. 35 9 45 cot.C....9.828783 cot. C ..9.828783 (a+b)..106 35 7 cosec(a+b)0.018455 sec.(a+b)0.544480 (a—b).. 17 34 52 sin. (a—b) 9 480090 cos. (a—6)9.979225 tan. (4-B) 9.327328 tan. (arc.)10.352488 tan. (A-B).. 11° 59′ 45′′ arc. 66° 3 0" Examples to this and the following rule are easily formed out of those already given. XIV. (See Ex. 169, &c.) Given two angles and the included side, to find the other two sides. Take the sum and difference of the two given : angles and the half sum and half difference: and also half the given side. Under heads (1) and (2) put down the following logarithms. Under (1) and (2) log. tan. of half the given side. } of half the sum of given angles. Under (1) log. cosec. Under (2) log. sec. Under (1) log. sin. Under (2) log. cos. Add together logarithms under (1) and (2), and (rejecting 20 in each index) the results will be log. tangents of the difference and the sum (a) of required sides respectively. The sum of the two arcs will be the side opposite the greater of the given angles: and their difference will be the side opposite to the less of the given angles. of half the difference of given angles. (a) NOTE. If half the sum of the given angles exceed 90°, the arc taken out under (2) must be subtracted from 180°, to give half the sum of the required sides. EXAMPLE. (180) In the spherical triangle ABC given 4=113° 33′ 30′′, B=51° 30′ 20′′, and c=60° 18 4": required the other two sides. A-B 62 3 10 tan.c....9.764070 tan. & c....9.764070 (A+B) 82 31 55 cosec. (A+B) 0.003700 sec.(A+B) 0.886220 (A—B) 31 1 35 .... 30 9 2 sin.(A-B) 9.711935 cos. (A-B) 9.932952 tan.(a-b) 9.479725 tan.(a+b) 10.583242 (a-b).. 16° 47′ 45′′ (a+b) ..75° 22′ 15′′ (a-b) ..16 47 45 a....92 10 0 b....58 34 30 XV. Right angled spherical triangles. In a right angle spherical triangle ABC (fig. 10), A being the right angle, the two sides b and c, and the complements* of the three other parts, namely, co. a, co. B and co. C are called the five circular parts. By the following rules, called from the name of their author, Napier's Rules, any two of these parts being given, the other three may be found. In applying the rules, we must select one of the three parts concerned, such, that the other two may either be both adjacent to it, or both separate from it. The part so selected is called the middle part. Rule I. will apply to the former case; Rule II. to the latter. *The complement of an angle is what it wants of 90°: thus in fig. Art. 5, the angle P is the complement of the angle A. It may be easily proved that the sin. tan. sec. cos. cot. and cosec. of an angle are the cos. cot. cosec. sin. tan. and sec. of its complement respectively. Thus let the two right angled triangles CPN C,P,N, (fig. 11,) be equal to each other in every respect: that is C=C, PP, and N-N,, then P, is the complement of A or co. A. Hence in applying the above rules, sin. A is substituted for cos. co. A; tan. A for cot. co. A, &c. P,N, = C,N, RULE I. The sine of the middle part is equal to the product of the tangents of the two parts adjacent to it. RULE II. The sine of the middle part is equal to the product of the cosines of the two parts opposite to, or separated from it. Having written down the equation according to the case, make a dash under the part required, and determine its magnitude by applying the proper signs (or) to each term according to the rule in Art. 34. EXAMPLES. (1). In the right angled spherical triangle ABC (fig. 12) given b=74° 19′ 30′′, c=38° 56', and 4= 90° : required the other parts. To find B (fig. 12'), the right angle A not being considered as a part, c will be the middle part, and complement B and 6 are the adjacent parts. ... By Rule I. sin. ctan. co. B tan. b Calculation. log. sin. c+10..19.798247 log. tan. b...... 10.551887 see note page 61) and determining the sign of B by art. 34 we have log. cot. B.. 9 246360 |