(494) CASE IX. Given two angles, and the side adjacent to both of them, to find a side opposite to one of the given angles. The LA 51° 30′ } Required the side BC. The side AC= 80 19 DETERMINATION OF THE SPECIES. 1. The ACD is acute or obtuse, according as the A is of the same, or of different species with the side Ac. 2. The ing as the A A is of the same species as CD; therefore accordA and the BCD are of the same, or of different species, BC is acute or obtuse. SOLUTION. 1. In the triangle ADC, find the ACD. * Thus rad x cos AC=cot LAX cot LACD. LACD=53° 26' acute. * .. cot ▲ ▲ : rad :: cos ac: cot ACD=78° 4′ acute. and BCD= ≤ ACB· 2. In the triangle BDC, rad x cos BCD=tan DC X cot BC. In the triangle ADC, rad × cos ACD=tan DC × cot ac. Hence cos / ACD: COS / BCD :: cot AC: cot BC=63° 50' acute. Because the A is of the same species as the BCD, the side BC is acute. Log cot BY CASE IX. page 205. log tan a+log cos b-10; hence Log cot a log cot b+log cos (c-4)—log cos and the corresponding arc is 63° 49′a or BC. −78° 4′. = +9.69164, ACB = 61 41 Answer. LACD=125° 52′ obtuse, BCD ACD— ▲ ACB= 64° 11' acute, and the side BC=115° 20′ obtuse. (495) CASE X. Given two angles, and the sides adjacent to both of them, to find the other angle. The A = 51° 30' Given The ACB =131 30 The side AC= 80 19 } Required the angle B. 2. The angles A and B are of the same, or of different species, according as the ACD is less or greater than the ▲ Acb. SOLUTION. 1. In the triangle ADC, find the ACD. Thus rad x cos AC=cot LAX Cot LACD. .. cot Arad :: cos AC: cot ACD=78° 4' acute. * 2. In the triangle BDC, rad x cos B=cos DC X sin BCD. In the triangle ADC, rad× cos a=cos DC × sin_ACD. Hence sin / ACD: sin BCD:: cosa: cos / B=59° 16' acute. Because the ACD is less than the ACB, the LB is of the same species as the A, viz. acute. BY CASE X. page 205. Log cot log tan A+ log cos b-10; hence 78° 4'. Log cos Blog cos A+ log sin (c-)-log sin = +9.70844, and the corresponding arc is 59° 16′ = ▲ B. Given EXAMPLE II. 37' The AA=21° 41 Required the B. The side AC= 57 30 Answer. The ACD 125° 52′ obtuse, ▲ BCD = ACD – LACB 64° 11' acute, and B=48° 30′ acute. = (496) CASE XI. Given the three sides, to find the angles. Given The side AC= 80° 19′ Required the angles The side AB=120 47 The sum of the sides AC and BC being less than 180°, the perpendicular falls nearest to the less side BC; therefore if it falls without the triangle, the rule determines ED in fig. 2. tan AB : tan : tan ED tan 60° 23′ 30′′ (AC+BC)=tan 72 4 30 = Ꭱ The segment AD=}AB+ED=74° 40', and DB=46° 7′. In the triangle ADC, the segment AD and the hypothenuse AC are given to find the A=51° 31′, and the ▲ ACD=78° 3′. In the triangle BDC, the segment BD and the hypothenuse BC are given to find the ▲ B=59° 17′, and the ▲ BCD=53° 25′. Lastly, ACB 2 ACD + 2BCD=131° 28′. OR THUS, Let BC be considered as the base. Then the sum of the sides AB and AC being greater than 180°, the perpendicular falls nearest to the greater side AB; therefore, if it falls without the triangle, the rule determines Ed in fig. 2. tan 31° 55′ = (AB+ AC) tan 100 33 = 9.79438 10.72992 32 = A : tan ED or Ed Here the fourth proportional is greater than half the base, therefore the perpendicular falls without the triangle. The segment dв=Ed-1 BC=40° 37′; the segment BD= BC+supplement of Ed=139° 23′. 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′. Answer. LA 126° 36', and B=48° 32'. = (497) CASE XII. Given the three angles to find the sides. [ The LA= 51° 30′ Given The B= 59 16 The c=131 30 Required the sides ac, BC, 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. (A+B) = cot 55° 23′ = cot tan (B-A)=tan 3 53 = :: tan ACB=tan 65 45 = 9.83903 8.83175 10.34634 In the triangle ADC, the A and the ACD are given, to find AC=80° 19′, and AD=74° 41'. In the triangle BDC, the ▲ B and the ▲ BCD are given, to find BC=63° 50′, and DB=46° 7'. Lastly, AB AD+DB = 120° 48′. 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 : tan (c-B)=tan 36 7 9.86312 :: tan BACtan 25 45 : tan LEAd =tan 75 9.68336 1=10.57227 d Then / cad=2 BAC+ LEAd=100° 46′; and hence the CAD, being the supplement,=79° 14′ and BAD=130° 44′. 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'. EXAMPLE II. The LA 48° 31' The c=125 20 Required the sides AC, BC, and AB. Answer. Ac=83° 9′, BC=56° 42′, and AB=114° 29′. CHAP. IX. IV. PRACTICAL RULES FOR SOLVING ALL THE DIFFERENT CASES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES, WITHOUT A PERPENDICULAR, WITH THEIR APPLICATION BY LOGARITHMS. CASE I. (498) 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, As sine of the other given side, Is to sine of its opposite angle. 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 proportion, 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 (428). * 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 |