Here, retaining the same notation, since we have sin b sin B tan c sin a= and cos B = -; if for the tangents there be substituted their values in sines and cosines, there will arise, sin b SID B sin c. cos a = cos B . cos c . sin a=cos B . COS C. Then substituting for sin A, and sin c. cos a, their values in the known formula (equ. v chap. iii) viz. in sin (a–c)=sin a. cos c — and recollecting that cos a. sin C1 - COS B sin B =tan B, = sin b. cos c tan B, . it will become, sin (a—c) which is the first part of the theorem: and, if in this result we introduce, instead of cos c, its value cos a cos b be transformed into sin (a–c) = tan b. cos a. is the second part of the theorem. Q. E. D. (th. 8), it will tan B; which Cor. This theorem leads manifestly to an analogous one with regard to rectilinear triangles, which, if h, b, and p denote the hypothenuse, base, and perpendicular, and B, P, the angles respectively opposite to b, h; may be expressed thus: h-bh. tan p..... h―f=b. tan в. These theorems may be found useful in reducing inclined lines to the plane of the horizon. PROBLEM II Given the Three Sides of a Spherical Triangle; it is required to find Expressions for the Determination of the Angles. Retaining the notation of prob. 1, in all its generality, we soon deduce from the equations marked 1 in that problem, the following; viz. I cos acos b. cos c As these equations, however, are not well suited for logarithmic computation; they must be so transformed, that their second members will resolve into factors. In order to this, substitute in the known equation 1 COS A 2 sin A, the preceding value of cos ▲, and there will result Whence, making s = a + b + c, there results sin (-b) sin (c) sin b. sin c sin (-a). sin (s—c) (III.) sin a. sin c sin a. sin b And, sin c = 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 (sb) sin (-e) sins. ein (s—ɑ) (iii.) Thus again, the expressions for the cosine and cotangent of half one of the angles, are cot Asin (8-6). sin (s—c)° The three latter flowing naturally from the former, by means of the values tan == cot = (art. 4 ch. iii.) Cor. 1. When two of the sides, as b and c, become equal, then the expression for sin A becomes When all the three sides are equal, or a = b = c; this case, if a = b = c = 90°; then sin A = = √2 = sin 45° : and A = B = c = 90o. Cor. 4. If a=b=c=60°: then sin 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 = 120o : then sin A sin 60° 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 PROBLEM III. Given the Three Angles of a Spherical Triangle, to find Expressions for the Sides. 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 cos b = cos a. But, it follows from th. 7, that sin c ing for sin c this value of it, and for = sin a. sin c sin b. Substitut sin A cos a their equiva cot A sin ccos c cos b cot a sin b. lents cot A, cot a, we shall have, (th. 7). So that the preceding equation at length becomes, COS A. sin c = cos a. sin B In like manner, we have, COS B. sin c = cos b. sin a sin B. Cos C Exterminating cos 6 from these, there results So like- COS B = wise cos c = . cos a. :} (v.) COS A= cos a. sin в sin G-COS B COS C. cos b. sin a sin c- - COS A. cos c. cos c. sin a sin B-COS A. COS B. This system of equations is manifestly analogous to equation I; 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 COS (A+B+C) cos (B+C-A) tana✔ 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, when the value of cos a becomes cosa = sin 60° So that a = = b c = 0. Consequently no such triangle can be contructed: conformably to th. 3. Cor. Cor. 5. If ABC= 120°: then cos a= cos 60° 3 √3 = cos 54° 44'9". Hence a = b = c = 109° 28′ 18′′. 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 a, &c. But this value is in fact always real. For, in general, sin (x - ; 0) = = cos therefore sin ^++ A+B+C O) COS(A+B+C); a quantity which is always positive, because, as A + B + c is necessarily com. prised between 40 and 40, we have (A+B+C) — 40 greater than nothing, and less than 0. 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), O - A less than B+10 C; whence (B+CA) is less than 40; and of course its cosine is positive. 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 b cos c. sin a. cos b and by a simple permutation of letters, we have COS B sin ccos b. sin a - cos c. sin b. cos a: adding together these two equations, and reducing, we have sin c (cos A+COS B) = ( - cos c) sin (a + b). Now we have from theor. 7, sin A Sin sin c Freeing these equations from their denominators, and respectively adding and subtracting them, there results sin c (sin A+ sin B) = sin c (sin a + sin b), and sin c (sin A sin B) sin c (sin a- sin b). 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, Cor. When ab, the first of the above equations becomes tan▲tan B = cot c. sec a. And in this case it will be, as rad; sin c :: sin a or sin 6: sin c. And, as rad: cos a or cos в :: tan a or tan b; tan c. b: 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, cot A = cot a. sin b sin c cot c. cos b. Then, choosing the subsidiary angle so that tan tan a. cos c, that is, finding the angle, whose tangent is equal to the product tan a. cos c, which is equivalent to dividing the original triangle into two right-angled triangles, the preceding equation will become cot C sin cot A cot c(cot4.sin b—cosʊ)=-- (cos. sino-sin .cosb). And this, since sin(b—9)= cos 4. sin b- sin 4. cos ¿ becomes 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 c sin A+ sin B 1+ cos c'sin (A + B) ' sin B sin c sin A Whence, as in prob. 4, we obtain tan (a+b) COS (A-B) tanic. tan 1(a—b)— tanc. sin (A+B)) The formule 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 proportion, as COS |