Cor. When a = b, the first of the above equations be comes 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, cot A = co. a. sin b sm c cot c. cos b. Then, choosing the subsidiary angle o so that tan = tan a. cos c, that is, finding the angle 4, 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 A=cotc(coto.sinb-cosb)=cot(cos.sinb - sino.cosb). And this, since sin (b-4)=cos. sin b - sino.cos b, becomes cot A = cot c sin (b - φ). 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 в * 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 H2 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 = tan ; then will PROELEM VI. 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. sin b. sin A Suppose the given angles are A, and B, and b the given side: then th. 7, gives sin a = ; or, sin a, a fourth proportional to sin B, sin b, and sin A. Scholium. 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 B, there can be only one solution when B = or, when B <○ (a right angle), •• a < + O .... .... .... B< B > .... .... b > a, b>-a, cos(a+b): cos(a-b):: cotc: tan(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 The triangle is susceptible of two forms and solutions when B < a < ᄋ.... b <a, 2. With the data A, B, and b, the triangle can exist but in It is susceptible of two forms, b<○.... A <○.... B > A. when b > .... A > .... B > A, b>....<. B > ....A > ○ B < b<O .... b<O...A.. B < 4, It may here be observed, that all the analogies and formulæ of spherical trigonometry, in which cosines or cotangents are not concerned, may be applied to plane trigonometry; taking care to use only a side instead of the sine or the tangent of a side; or the sum or difference of the sides instead of the sine or tangent of such sum or difference. The reason of this is obvious: for analogies or theorems raised, not only from the consideration of a triangular figure, but the curvature of the sides also, are of consequence more general; and therefore, though the curvature should be deemed evanescent, by reason of a diminution of the surface, yet what depends on the tri angle alone will remain, notwithstanding. We have now deduced all the rules that are essential in the operations of spherical trigonometry; and explained under what limitations ambiguities may exist. That the student, however, may want nothing further to direct his practice in this branch of science, we shall add three tables, in which the several formulæ, already given, are respectively applied to the solution of all the cases of right and oblique-angled spherical triangles, that can possibly occur. I. given TABLE I. For the Solution of all the cases of Right-Angled Spherical Triangles. Angle opposite to the} Given. Values of the terms required. Its sin Required. sin given leg sin hypoth. { Hypothenuse, and one leg. Angle adjacent to the given leg. e} Other leg. } Its cos = II. Hypothenuse. } Its sin = sin given ang. { Idem. { Ambiguous. In working by the logarithms, the student must observe that when the resulting logarithm is the log. of a quotient, 10 must be added to the index; when it is the log. of a product, 10 must be subtracted from the index. Thus when the two angles are given, 10: Log. cos hypothen. = log. cos one angle + log. cos other angle — Log. cos either leg = log. cos opp. angle - log. sin adjac. angle + 10. In a quadrantal triangle, if the quadrantal side be called radius, the supplement of the angle opposite to that side be called hypothenuse, the other sides be called angles, and their opposite angles be called legs: then the solutions of all the cases will be as in this table; merely changing like for unlike in the determinations. |