therefore if it falls without the triangle, the rule determines Ed in fig. 2. tang BC tang 31°.55'. tang A (AB+AC)=tang 100°.33'= tang (AB-AC)=tang 20°.14'= tang ED or Ed=tang 72°.32'= Here the fourth proportional is greater than half the base, therefore the perpendicular falls without the triangle. B DE The segment dв Ed-BC-40°.37'; the segment BD= BC+supplement of Ed=139°.23'. SECONDLY. In the triangle Adв, or ADB, the B may be found=59°.17'. And in the triangle ACD, or Acd, the ▲ c may be found=131°.28′. EXAMPLE II. The side AC 57°.30′ Given The side BC=115°.20′ The side AB 82°.28' Required the angles Answer. LA=126°.34', and B=48°.31'. (K) CASE XII. Given the three angles to find the sides. The LA 51°.30') Given The B 59° 16' Required the sides AC, BC, The c=131°.30' and AB. Let the angles A and B, which are of the same species, be considered as base angles, then the perpendicular will fall within the triangle. And since the sum of the two angles A and B is less than 180°, the sum of the sides AC and BC is less than 180°, therefore the perpendicular falls nearest to B, the greater angle. Cot (A+B)=cot 55°.23'= tang (B-A)=tang 3°.53'= 9.83903 8.83175 ::tang ềZACB=tang 65.45 = - 1034634 tang ECD tang 12°.19'= In the triangle ADC, the ▲ A and the ▲ ACD are given, to find AC 80°.19′, and AD=74°.40′. In the triangle BDC, the B and the ▲ BCD are given, to find BC 63°.50', and DB=46°.7'. Lastly, AD+DB=AB=120 ̊.47'. OR THUS, Let BC be considered as the base. The base angles being here of different species, the perpendicular falls without the triangle. And because the sum of the two angles B and C is greater than 180°, the perpendicular falls nearest to the less angle B, or greatest side AB, consequently the rule determines the LEAd. Cot (B+c)=cot 95°.23' 8.97421 d. tang (c-B)=tang 36°.7' 9.86312 = tang BACtang 25°.45'-9.68336 : tang/EAd=tang 75°.1'= 10-57227 B E Then BAC+ LEAd=cAd=100°.46'; and hence the CAD, being the supplement,=79°.14' and Z BAD=130°.44'. SECONDLY. In the triangle Adc, or ADC, the side AC will be found=80°.19', and the side dc or DC may likewise be found. In the triangle adв, or ADB, the side AB will be found= 120°.48', and the side dв, or DB, may likewise be found. Lastly, 180°-(DC+dB), or dc-dв, or DB-DC=BC=63°.50'. Answer. Ac=83°.8'.32", BC-56°.41'.40", and AB=114°.28'.45". CHAP. IX. PRACTICAL RULES FOR SOLVING ALL THE DIFFERENT CASES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES, WITHOUT a PERPENDICULAR, WITH THEIR APPLICATION BY LOGA RITHMS. CASE I. (L) When two sides and an angle opposite to one of them are given, to find the rest. RULE. 1. To find the other opposite angle. Sine of the side opposite to the given angle, Is to sine of the given angle; As sine of the other given side, To the angle found by this proportion, and its supplement, add the given angle. Then, if each of these sums be of the same species with respect to 180°, as the sum of the given sides, the problem is ambiguous; that is, the angle thus found may be either acute or obtuse. But, if only one of these sums be of the same species with the sum of the sides, that value of the angle, found by this pro'portion, must be taken, whether it be acute or obtuse, which when added to the given angle agrees with the sum of the sides. In this case the problem is not ambiguous. 2. To find the angle contained between the given sides. Find the angle opposite to the other given side, by the first part of the rule, and note whether it be acute, or obtuse, or ambiguous. Then, Sine of half the difference between the two given sides, Is to sine of half their sum; As tangent of half the difference between their opposite angles, Is to cotangent of half the angle* contained between the given sides. (M. 188.) 3. To find the third side. Find the angle opposite to the other given side, by the first part of the rule, and note whether it be acute, obtuse, or ambiguous. Then, Sine of half the difference between the two angles, Is to sine of half their sum; As tangent of half the difference between the two given sides, Is to tangent of half the required side. (N. 189.) CASE II. (M) When two angles, of an oblique-angled spherical triangle, and a side opposite to one of them are given, to find the rest. * Since a side, or an angle, of any spherical triangle is always less than 180°; the half of any side or angle must always be acute. The ambiguity therefore ascribed to Case I. and II. arises from the first proportion in each case; if the angle, or side, found by these proportions be ambiguous, the remaining parts of the triangle will necessarily be ambiguous, but if the angle, or side, found by these proportions be determinate, the remaining parts of the triangle will also be determinate. The ambiguous parts derived from the first proportion, in Case I. or II. are always supplements of each other; but the remaining parts of the triangle, when ambiguous, are not supplements of each other, as is obvious both from the constructions and calculations following. RULE. 1. To find the other opposite side. Sine of the angle opposite to the given side, As the sine of the other given angle, Is to the sine of its opposite side. To the side found by this proportion, and its supplement, add the given side. Then if each of these sums be of the same species with respect to 180°, as the sum of the given angles, the problem is ambiguous; that is, the side thus found may be either acute or obtuse. But, if only one of these sums be of the same species as the sum of the given angles, that value of the side, found by this proportion, must be taken, which when added to the given side agrees with the sum of the angles. In this case the problem is not ambiguous. 2. To find the side adjacent to the two given angles. Find the side opposite to the other given angle, by the first part of the rule, and note whether it be acute, obtuse, or ambiguous. Then, Sine of half the difference between the two given angles, Is to sine of half their sum; As angent of half the difference between the two sides, Is to tangent of half the third side. (N. 189.) 3. To find the third angle. Find the side opposite to the other given angle, by the first part of the rule, and note whether it be acute, obtuse, or ambiguous. Then, Sine of half the difference between the two sides containing the required angle, Is to sine of half their sum; As tangent of half the difference between the other two angles, Is to cotangent of half the required angle. (M.188.) CASE III. (N) When two sides and the included angle, of an obliqueangled spherical triangle, are given, to find the rest. RULE. 1. To find the other two angles. Cosine of half the sum of the two given sides, As cotangent of half the included angle, Is to tangent of half the sum of the other two angles. Half the sum of these two angles must be of the same species as half the sum of the given sides. Secondly, Sine of half the sum of the two given sides, Is to sine of half their difference; As cotangent of half the included angle, Is to tangent of half the difference between the other two angles. (M. 188.) Half the difference between these angles is always acute. Lastly, Half the sum of the two angles increased by half their difference, gives the angle opposite to the greater side, and diminished by the same, leaves the angle opposite to the less side. (C. 35.) 2. To find the third side. Find the two required angles by the first part of the rule. Then, Sine of half the difference between these angles, Is to sine of half their sum; As tangent of half the difference between the given sides, Is to tangent of half the third side. (N. 189.) OR, without finding the other two angles. To the sum of the logarithmical sines of the given sides, add double the logarithmical sine of half the contained angle, and reject 30 from the index. Look for the remainder in the table of logarithmical sines, and take the degrees and minutes answering to it. Then take the difference between twice the natural sine of those degrees, and the natural cosine of the difference between the given sides; the remainder will be the natural cosine of the side required. This side is acute or obtuse, according as the double natural sine is less, or greater, than the natural cosine of the difference between the given sides. (X. 192.) |