But, because cos B' COS A' 2 sin (A+B). sin (A'B') (art. 25 ch. iii), and consequently, a+b. cos (b − c) — cos a = 2 sin 4+6-c. sin a+c-b: we have, obviously, sin2 = sin (a+b−c) . sin † (a+c—b) sin b. sin c 2 Whence, making s = a+b+c, there results The expressions for the tangents of the half angles, might have been deduced with equal facility; and we should have obtained, for example, sin (s—b). sin (sc), (iii.) sins.sin (s-a) Thus again, the expressions for the cosine and cotangent of half one of the angles, are sins. sin (s- a) sin b. sin c sin s sin (s-a) sin (sb). sin (s—c)* The three latter flowing naturally from the former, by means of the values tan == sin cos' Cor. 1. When two of the sides, as band c, become equal, then the expression for sin 4 becomes When all the three sides are equal, or a = b=c, Cor. 3. In this case, if a = bc90°; then sin 44 = =√2 sin 45°: and A = B = c = 90°. Cor. 4. If a b c = 60°: then sin A= 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 = bc were assumed = 120°: then sin 4= sin 60° sin 120 3 = 1; and A=B = C = 180°; which shows √3 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 If from the first and third of the equations marked L (prob. 1), cos c be exterminated, there will result, COS A. sin c + cos c. sin a. cos b = cos a. sin b. But, it follows from th. 7, that sin c = sin 7. sin c COS A sin A SIO A cos a Substituttheir equi sin a ing for sin c this value of it, and for (th. 7). So that the preceding equation at length becomes, COS A. sin c = cos a. sin B In like manner, we have, - sin A. cos c. cos b. . COS C. cos a. COS B. sin C = cos b. sin A - sin B Exterminating cos b from these, there results COS A So like-cos B = wise } cos c = cos c. sin A. sin 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 (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 α = COS sin B COS A sin A b = c = 90°. Cor. 2. When A B c; then cos a = 1. Cor. Cor. 5. If A 2 B=C=120°: then cos a = = cos 60° sin 120° √3 3√3 = cos 54°44'9". Hence a = b = c = 109°28′18′′. Schol. If, in the preceding values of sin a, sin b, &c, the quantities under the radical were negative in reality, as they are in appearance, it would obviously be impossible to deter mine the value of sin a, &c. But this value is in fact always real. For, in general, sin (r – †O) = = cos: therefore, sin (A+B+C - 0) = −cos (A + B + c); a quantity which is always positive, because, as A+B+c is necessarily comprised between 0 and 0, we have (A + B + c) − ÷ O 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 40 − B + ÷ O− c; whence (BCA) is less than 40; and of course its cosine is. positive. PROBLEM IV. Given Two Sides of a Spherical Triangle, and the Included 1. In the investigation of the last problem, we had - and by a simple permutation of letters, we have 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, sin b sin c and Freeing these equations from their denominators, and respectively adding and subtracting them, there results and sin c (sin a sin B) -- sin c (sin A+ sin B) 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 be comes tan Atan 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 6 : tan c. 2. The preceding values of tan (A + B), tan (AB) 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, Then, choosing the subsidiary angle o 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 sin cot a=cot c(cot 4. sinb-cosb)=(cos.sinb — sinø.cos b). And this, since sin (b-)=cos. sin b- sin .cos b, becomes sin (b). cot A = cot c sin 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 formula: sin a+ sin b sin c sin A+ sin B The formulæ marked vi, and v11, 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 H 2 2. If 2. If it be wished to obtain a side at once, by means of a subsidiary angle; then, find so that cot a = .COS (B-). cot c Cosp PROELEM VI. cot A tan; then will COS C Given Two Sides of a Spherical Triangle, and an Angle Opposite to one of them; to find the Other Opposite Angle. Suppose the sides given are a, b, and the given angle B : then from theor. 7, we have sin A = sin a. sin B fourth proportional to sin b, sin B, and sin a. PROBLEM VII. ; or, sin A, a 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 b the given side then th. 7, gives sin a = sin b. sin a sin B proportional to sin B, sin b, and sin ▲. 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 VII 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. 1. With the data a, b, and в, there can be only one solution when BO (a right angle), B or, when в ÷ O B<O b ... > a, b > 10-a, cos (a+b): cos (A+B), 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. The |
