perpendicular to the plane BAD. See GEOMETRY, SPHERICAL. Therefore as the two planes, EBF and DBF, are perpendicular to the same plane AB D, their line of common section BF is also perpendicular to the plane ABD, and therefore perpendicular to B D and B E, which it meets in that plane. BF is therefore the tangent of the arc B C, B E the sine of B E, and the angle BEF, the angle made by the planes CDA and BDA, is the same as the spherical angle BAC. Now, in the right angled plane triangle EBF, BE rad. = BF, cot. BEF; or rad. sin. AB= tan. BC cot. BA C. PROP. II.-In any right angled spherical triangle, the rectangle of radius and the sine of either of the sides containing the right angle is equal to the rectangle of the sine of the angle opposite that side, and the sine of the side opposite the rectangle. From C (fig. 20,) in the plane of CDA, draw CG perpendicular to AD, and from G in the plane BDA draw GH perpendicular to DA meeting D B in H, and join CH. Then it may be shown, as in the last proportion, that C H G is a right angled plane triangle, CG the sine of CA, CH the sine of B C, and the angle C G H is equal to the spherical angle BAC. Now in the plane triangle CG H, C H, rad. = Rad. sin. CE tan, E F Rad. sin. CE sin. C F Rad. sin. EF = tan. EC Rad. sin. E F = sin. F C = The first four of these equations have been cot. E CF; or rad. cos. A C = cot, AC B cot. A; fifth equation. By taking C as the pole of a great circle, and producing BC and AC and completing the figure as above, the ninth and tenth equations may be deduced in the same manner as the seventh and eighth have been. Napier, the inventor of logarithms, devised a very simple expedient for recollecting these ten equations. He called A B, BC, the complement of A, the complement of C, and the complement of AC, the five circular parts of a spherical triangle, the right angle not being considered. In every case two of these parts will be given to find a third, and of these three parts that which has either one of the others immediately adjoining to it, on each side, or is separated from both of them, by one of the two remaining parts, is called the middle part; and the other two are called adjoining or opposite extremes, according as they are adjacent to or separated from the middle part. Thus when complement A is the middle part, A B and complement AC are adjoining extremes, and BC and complement C opposite extremes; when BC is the middle part, A B and complement C are adjoining, and complement AC and complement A opposite extremes; when complement AC is the middle part, complement A and complement C are adjoining, and A B and B C opposite extremes. With these explanations Napier's rules are 1. The rectangle of radius and the sine of the middle part is equal to the rectangle of the tangents of the adjoining extremes. 2. The rectangle of radius and the sine of the middle part is equal to the rectangle of the courses of the opposite extreme. Suppose, for example, that A B, BC, and ▲ A, were the three parts concerned in the formation of an equation. Here A B is the middle part, and BC and the complement of A are adjoining extremes, and by rule first Rad. sin. A B = tan. B C tan. comp. A = tan. BC cot. A, which agrees with the first equation. Again, let AC, AB, and B C, be the three parts concerned, two of them given and one required. Here the complement of AC will be the middle part, and AB, and BC, which are separated from it, one by A and the other by C, will be the opposite extremes. And by the second rule rad. sin. complement AC = cos. AB cos. BC; or rad. cos. AC cos. AB. cos. BC, which agrees with the sixth equation. = PROP. IV. In any spherical triangle, as ABC, (figs. 22 and 23), the following proportions obtain, A D and D B being the segments of the base made by a perpendicular from the vertical angle. 1. Sin. AC: sin. BC:: sin. B: sin A. 2. Cos. AD: cos. DB:: cos. A C: cos. BC. 3. Sin. AD: sin. DB:: cot. A cot. B. 4. Tan. A D : tan. DB:: tan. A CD: tan. DCB. 5. Cos. A cos. B:: sin. A CD: sin. BCD 6. Cos. A CD: cos. BCD:: cot. A C: cot. BC. For, by proposition II., rad. sin. CD=sin. A C, sin. A; and rad. sin. C D= sin. BC, sin. B, therefore sin. A C, sin. Asin. BC, sin. B; and consequently sin. AC sin. 5C:: sin. B : sin. A. And, by equation 6th, proposition III., rad. cos. A C=cos. A D, cos. DC, and rad. cos. BC cos. BD, cos. DC; whence rad. cos. AC: rad. cos. BC:: cos. A D cos. DC: cos. BD cos. DC, or cos. A C: cos. BC: cos. AD: cos. BD. By proposition I. rad. sin. AD=tan. DC cot. A, and rad. sin. BD=tan. DC cot. B; whence rad. sin. AD: rad. sin. BD:: tan. DC cot. A tan. DC cot. B; and consequently sin. A D : sin. BD:: cot. A: cot. B. By proposition II. rad. sin. DC=tan. AD cot. AC D, and rad. sin. DC tan. D B cot. DC B; therefore tan. AD cot. A CD= tan, DB cot. DCB; whence tan. AD: tan. DB:: cot. DCB: cot. A CD:: tan. ACD: tan. DCB; see 8th deduction from the definitions in Plane Trigonometry. By equation 8th, proposition III., rad. cos. A =cos. DC sin. ACD, and rad. cos. B= : COS. DC sin. BC D, whence rad. cos. A: rad. cos. B: cos. DC sin. A CD: cos. DC sin. BCD, or cos. A cos. B:: sin. A CD sin. BC D. By equation 7th, proposition III., rad. cos. ACD tan. DC cot. AC, and rad. cos. DCB =tan. DC cot. BC, whence rad. cos. A CD: rad. cos. B CD:: tan. DC cot. A C; tan. DC ⚫ cot. BC; or cos. A CD: cos. BCD:: cot. A C : cot. BC. = PROP. V. In any spherical triangle, as A BC, figs. 22 and 23, if CD be a perpendicular from C upon AB, or AB produced, tan. BD+AD BD AD BC+AC 2 ⚫ tan. 2 • tan. BCAC 2 2 tan. For let D be the centre of the sphere, A E the tangent, and D E the secant of the arc b, A F the tangent, and DF the secant of the arc C; then the plane angle EA F is the spherical angle CA B, and the plane angle EDF is measured by the arc a. Join E F; then, by form. 7th, proposition VIII., plane trigonometry, E FED2+DF22ED DF cos. D, and E F2 E A2 + A F2 — 2 EAAF cos. A; whence E D2+D F2 — 2 ED DF cos. D=EA'+A F2-2 EA AF. cos. A, or E D2+2 F2 — E A2+A F2=2 rad. 2= 2 ED DF cos. D-2 EA AF cos. A; or EA AF cos. A-ED DF cos. D-rad. '; EDDF cos. D For let B C=a, AC=b, B D=m, and A D= n; then by proportion 2d, proposition IV. cos. a : cos. bcos.m: cos. n; whence cos. a+ cos. b; cos. au cos. b:: cos. m + cos. n: cos. m in cos. n; or formula 6th, proposition III. plane trigono- sin. b · sin. c a+b whence cot. and unity = sin. b. sin. c cos. acos. b cos.c sin. b. sin. c COS. a- -(cos. b. cos.c-sin. b · sin. c) sin. b. sin. c cos.a-cos.b+c sin. b sin. c : tan. 2 arb 2 m + n ::cot. : 2 mn a+b tan. 2 ab '; and consequently tan. = But form. 4th, proposition III., plane trigono A 2 By this proposition any oblique angled spherical triangle, whose sides are given, may be divided into two right-angled triangles, in each of a+b+c which the hypothenuses and a side are given; for ms n half the difference A the equation determines of the segments of the given base, in fig. 22; or m+n 2 2 half the sum of the segments, the difference of which is the given base in fig. 23. PROP. VI. In any spherical triangle, as ABC, fig. 24, if radius be unity, cos. A = cos.a-cos. b. cos.c sin. b. sin. c = =2 sin. sin. a = 2 sin S sin. S-a, whence cos.' sin. S sin. Sa 2 sin. b sin. c sin. S sin. S-a · R2 sin. b. sin. c rad.2 sin. S sin. S -a cos. b. cos. c rad.2 cos.bsin. b sin. c PROP. VIII. In any spherical triangle, as ABC, fig. 25, if a be any of the sides, cos a= cos. Acos. B cos. C sın. B sin. C For by proposition V. cos. A'= cos. a'. - ; and, by spherical geometry, if A', B′, C', be the angles, and a', b', c', the sides of the supplemental triangle, then cos. A'cos. 180°—d, cos. a, cos. a'cos. 180°—A= cos. A, cos. b'=cos. 180°-B= cos. B, cos. c=cos.180°-C =cos. C, sin. b' sin. 180°-B=sin. B, and sin. c' sin. 180°— C=sin. C; whencecos. A- -cos. B cos. C cos. A+ cos. B. cos. C sin. B sin. C or cos. a= " sin. B sin. C cos.@= PROP. VIII. In any spherical triangle, as ABC, fig. 25, if a be any of the sides, and PROF. IX. In any spherical triangle, as ABC, fig. 25, cos. 2 。? cot. and sin. a+b A+B tan. 2 = C03 For equation 1, proposition IV., sin. A: sin. B:: sin. a: sin. b; whence sin. A± sin. B: sin. sin. b and cos. C= sin. b sin. c the last of these expressions cos. csin. a sin. b. cos. C + cos. a b, which, substituted for cos. c in the expression for cos. A, we have cos. a-cos. b・ sin. b · sin. a cos. C. -cos. a· cos. 2 b cos. A= = sin. b sin. c Cos. a sin. a+b 2 tan. sin. M). (sin. a+b 2 + M) sin. a + b M) for mula 5, proposition III., plane trigonometry; whence sin 2 sin. ("++ M). sin. (" 2 b S 2 sin. (" + M.) Before proceeding to exemplify the use of the preceding formulæ, it may be well to recapitulate the rules demonstrated in spherical geometry for determining whether the parts of right angled spherical triangles are obtuse or acute. 1. The sides about the right angle are of the oblique angles are of the same affection as their same affection as their opposite angles, and the opposite sides. 2. When the sides including the right angle are of the same affection, that is both obtuse, or both acute, the side opposite the right angle is acute; but when the sides including the right angle are of different affections, that is, when one of them is acute and the other obtuse, the side opposite the right angle is obtuse. 3. When the oblique angles are of the same affection, the side opposite the right angle is acute; but, when the oblique angles are of different affections, the side opposite the right angle is obtuse. the same affection, the side opposite the right 4. When a side and its adjoining angle are of angle is acute; but, when a side and its adjoining angle are of different affections, the side opposite the right angle is obtuse. To these we may add a property of oblique angled triangles which has also been proved in spherical geometry. When the perpendicular falls within the PROP. X.—In any triangle, as ABC fig triangle, the angles at the base are of the same af fection; but, when it falls without the triangle, the angles at the base are not of the same affection. Examples.-1. In the spherical triangle ABC (fig. 26), right angled at B, given AB 10° 39′' 40" rad. A 23° 27′ 42" to find the other parts. To find AC we have by equation 7, proposition III., rad. cos. A cot. A C. tan. AB; R. cos. A cot. A B. cos. A whence cot. AC = = tan. A B To find BC we have by equation 1, proposition III., rad. sin. A B = tan. BC. cot. A; whence rad. sin. A B tan. A. sin. A B R rad. C, we have by equation 10, pro = cos. a + b + sin. a. sin. b. position III., rad. · cos. C = cos. A B sin. A; 2 C 1 - 2 sin. 2 cos. A B. sin. A rad. And, as A B is acute, C is acute; and as LA is acute, BC is acute; and, as A B and ZA, are of the same affection, AC is acute. A B, 10° 39′ 40", cot. 10-725267 A B, 10° 39′ 40′′, sin. 9-267171 A B, cos. A, sin. 9-637507 9.992438 9-600031 19.592469 rad. 10 2. In the right angled spherical triangle ABC (fig. 26), given AB 151° 23′ 9′′, and BC 16° 35' 14" to find the other parts. Here C, being of the same affection as A B, is obtuse; and A, being of the same affection as BC, is acute; and, as A B and B C are of different affections, A C is obtuse. To find A C, we have equation 6, prop. III. rad. cos. AC cos. A B cos. BC; whence 8-904678 C 66° 58′ 1′′ cos. 9.592469 A C, 147° 16′ 51′ cos. 9·924967 A 31° 52′ 50′′ cot. 9-206227 C 117° 37′ 21′′ cot. 9-718742 3. In the right angled spherical triangle ABC (fig. 26), given AB 29° 12′ 50′′, and its opposite angle C, 37° 26′ 21′′, to find the other parts. If CA and CB were produced till they met, there would then be two right angled triangles, with a common side A B, corresponding to the given data, and the parts to be determined might be A C, B C, and BA C, or their supplements. To find A C, we have rad. sin. AB sin A C, rad. sin. A B sin. C, whence sin. AC = sin. C 4. In any spherical triangle, as A B C (fig. 22), given AC 80° 19′, A B 120° 47′, and ▲ A 51° 30', to find the other parts. Let CD be a perpendicular from C on A B; then rad. . cos. A C cot. A. cot. A CD, whence rad.. cos. A C tan. A. cos. A C cot. A CD A are acute. and A C D is acute, because AC and the angle cot. A tan. A C. cos. A rad. rad.. cos. A cot. A C 19.325228 rad. 10. 9.794150 20.562085 rad. 10. ACD 78° 3′ 36′′ cot. 9.325228 AD 74° 40′ 17′′ tan. 10:562085 A B 120 47 0 BD 46 6 43 Now, tan. AD: tan. BD:: tan. ACD tan. BC D; whence tan. BCD = tan. BD. tan. A CD cot. A D and cos. AD cos. BD:: cos. AC: cos. BC; whence cos. BC= |