But, because cos B'-COS A' 2 sin (A'+B'). Sin (A-B') = ¦ (art. 25 ch. iii.), and consequently cos (bc) cos a = 2 sin .sin a+b we have obviously, sin A= = a+b-c sin i̟la ↳ h—c), sin f ́a+c—b) Whence, making s=a+b+c, there results sin (48—a). •in (4s—b) And, sin c = ✔ silla sig b The expressions for the tangents of the half angles might have been deduced with equal facility; and we should have obtained, for example, tan A = √ sin (şa—¿). sin (18—c) (iii.) Thus again, the expressions for the cosine and cotangent of half one of the angles, are sin † • . sin 4 (s—n) COS A sin 4 s. sin †(s—a) cot Asin (†s—b), sm (†s—c)° The three latter flowing naturally from the former, by means of the values tan == cot Cor. 1. When two of the sides, as b and c, become equal, then the expression for sin Cor. 2. When all the three sides are equal, or a = b = c, Cor. 3. In this case, if a = b = c = 90°; then sin a = = sin 45°; and A = B = C = 90°. = v3 Cor. 4. If a=bc60°: then sin a = = {√3 = sin 35 15 51": and A=B=c=70°31′42": the same as the angle between two contiguous planes of a tetraedron. Cor. 5. If a b c were assumed = 120° ; then sin 1 = sin 60® 13 = 1; and AB = c = 180°; which shows that no such triangle can be constructed (conformably to th. 2); but that the three sides would, in such case, form three continued arcs completing a great circle of the sphere. PROBLEM III. Given the three angles of a spherical triangle, to find If from the first and third of the equations marked 1 (prob. 1), cos c be exterminated, there will result COS A. sin c + cos c. sin a. But, it follows from th. 7, that sin c for sin c this value of it, and for cos b = cos a. sin b. cot a. sin b. sin b sill B sin b = cos a . = cos a. sin a sin cot A, cot a, we shall have, cot A. sin c+cos c. cos b = Now, cot a. sin B= (th. 7). So that the preceding equation at length becomes, COS A sin c cost. sin B- sin A. cos c. cos b. COS B. sin c = cos b. sin A sin B. cos c. cos a. Exterminating cos b from these, there results (IV.) cos A=cos a. sin B. sin c-cos B. cos c. So like-cos B=cos b. sin a. sin c--cos a. COS C. wise cos c = cos c. sin a. sin B-Cos A. COS B. This system of equations is manifestly analogous to equation 1; and if they be reduced in the manner adopted in the last problem, they will give The expression for the tangent of half a side is (V.) The values of the cosines and cotangents are omitted, to save room; but are easily deduced by the student. Cor. 1. When two of the angles, as B and C, become equal, then the value of cos a becomes cos ja= Cor. 2. When A=BC; then cos a Cor. 3. When A=B=c=90', then ab=c=90°. Cor. 4. If A=B=C60°; then cos ļa sin GO = 1. sin 69 So that a = b = c = 0. Consequently no such triangle can be constructed: conformably to th. 3. Cor. 5. If A=B=c=120°; then cos a 12 C = 3=cos 54°44'9". Hence a=b=c=109°28′ 8′′. Schol. If, in the preceding values of sin ja, sin b, &c. the quantities under the radical were negative in reality, as they are in appearance, it would obviously be impossible to determine the value of sin fa, &c. But this value is in fact always real. For, in general, sin (x-10)=cos: therefore, sin ( ' ! ' ° — ¦ 0 ) = — COS †(A+B+C); a quantity which is always positive, because, as a +B+C is necessarily comprised between 0 and, we have (A + B + c) — ÷0 greater than nothing, and less than Further, any one side of a spherical triangle being smaller than the sum of the other two, we have, by the property of the polar triangle (theorem 4), 10-A less than 10 − B + ! ~ — C ; whence (B+C-A) is less than 0; and of course its cosine is positive. 2 PROBLEM. IV. Given two sides of a spherical triangle, and the included angle; to obtain expressions for the other angles. 1. In the investigation of the last problem, we had cos A. sin c = cos a. sin bcos c. sin a and by a simple permutation of letters, we have • cos b': COS B. sin c = cos b. sin a COS C sin b. cos a: adding together these two equations, and reducing, we have sin c (cos A+ COS B) = (1 cos c) sin (a + b). Now, we have from theor. 7, and sin b sin B == sin c Freeing these equations from their denominators, and respectively adding and subtracting them, there results sin c (sin a + sin x) = sin c (sin a + sin b), S and sin c (sin a Dividing each of these two equations by the preceding, there will be obtained Comparing these with the equations in arts. 25, 26, 27, ch. iii., there will at length result cns (a--b) tan (A+B) = cot c. • cvs $(a+b) tan (AB): =cotic. sina-by sin a+b) Cor. When a=b, the first of the above equations becomes tan a = tan B = cot c. sec a. And in this case it will be, as rad: sin c :: sin a or sin b: sin c. And, as rad: cos a or cos B :: tan a or tan b : tan c. 2. The preceding values of tan (A+B) tan !(A — B) are very well fitted for logarithmic computation: it may, notwithstanding, be proper to investigate a theorem which will at once lead to one of the angles, by means of a subsidiary angle. In order to this, we deduce immediately from the second equation in the investigation of prob. 3, tan ❤ tan a. cos C, that is, finding the angle o, whose tangent is equal to the product tan a. cos c, which is equivalent to dividing the origin. al triangle into two right-angled triangles, the preceding equation will become cot a cot c (cot p. sin b-cos b)=' cos b). cot c (cos p. sin b sin . And this, since sin(b−q)=cos q. sin b- sin. cos b, becomes cot c cot A = . sin (b-). Which is a very simple and convenient expression. PROBLEM V. Given two angles of a spherical triangle, and the side comprehended between them; to find expressions for the other two sides. 1. Here, a similar analysis to that employed in the preceding problem, being pursued with respect to the equations iv, in prob. 3, will produce the following formulæ : sin a+ sin b sin c sin A+ sin B 2. If it be wished to obtain a side at once, by means of a subsidiary angle; then, find o so that cot A COS C tan p; then will Given two sides of a spherical triangle, and an angle oppo, site to one of them; to find the other opposite angle. Suppose the sides given are a, b, and the given angle в : then from theor. 7, we have sin a = ; or, sin a, a sin a. sin B sin b fourth proportional to sin b, sin в, and sin a. PROBLEM VII. Given two angles of a spherical triangle, and a side opposite to one of them; to find the side opposite to the other. Suppose the given angles are A, and B, and the given side: then th. 7, gives sin a proportional to sin B, sin b, and sin a. = sin b. sin A sin B Scholium. ; or, sin a, a fourth In problems 2 and 3, if the circumstances of the question leave any doubt, whether the arcs or the angles sought are greater or less than a quadrant, or than a right angle, the difficulty will be entirely removed by means of the table of mutations of signs of trigonometrical quantities, in different quadrants, marked vir in chap. 3. In the 6th and 7th problems, the question proposed will often be susceptible of two solutions by means of the subjoined table the student may always tell when this will or will not be the case. : * The formulæ marked vi, and vII, converted into analogies, by making the denominator of the second member the first term, the other two factors the second and third terms, and the first member of the equation the fourth term of the proposition, as cos (a+b): cos (ab) :: cot c: tan (A+B), sin (a+b); sin (a-b):: cot do: tan (A-B), &c. &c. are called the Analogies of Napier, being invented by that celebrated geometer. He likewise invented other rules for spherical trigonometry, known by the name of Napier's Rules for the circular parts: but these, notwithstanding their ingenuity, are not inserted here; because they are too artificial to be applied by a young computist, to every case that may occur without considerable danger of misapprehension and error, VOL. II. 8 |