Therefore, cot c sin a=cos a cos B+cot C sin B. Hence, we may write the three symmetrical equations, cot a sin b=cos b cos C+cot A sin C cot b sin c=cos c cos A+cot B sin A (9.) cot c sin a cos a cos B+cot C sin B That is, in every spherical triangle, the cotangent of one of the sides into the sine of a second side, is equal to the cosine of the se cond side into the cosine of the included angle, plus the cotangent of the angle opposite the first side into the sine of the included angle. IX. We shall terminate these formulas by demonstrating Napier's Analogies, which serve to simplify several cases in the solution of spherical triangles. If from the first and third of equations (2.), cos c be eliminated, there will result, after a little reduction, R cos A sin c-R cos a sin b-cos C sin a cos b. By a simple permutation, this gives R cos B sin c-R cos b sin a-cos C sin b cos a. Hence by adding these two equations, and reducing, we shall have sin c (cos A+cos B)=(R—cos C) sin (a+b) sin c sin a sin b But since sin C sin A sin B' we shall have 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 these two equations successively by the preceding one; we shall have And reducing these by the formulas in Articles XXIII. and XXIV., there will result Hence, two sides a and b with the included angle C being given, the two other angles A and B may be found by the analogies, cos (a+b): cos(a-b):: cot C: tang (A+B) sin(a+b): sin (a—b) : : cot C: tang (A—B). If these same analogies are applied to the polar triangle of ABC, we shall have to put 180°-A', 180°-B', 180°-a', 180°-b', 180°-c', instead of a, b, A, B, C, respectively; and for the result, we shall have after omitting the ', these two analogies, cos (A+B): cos (A—B) :: tang c: tang (a+b) sin (A+B) sin (A—B) :: tang c: tang (ab) by means of which, when a side c and the two adjacent angles A and B are given, we are enabled to find the two other sides a and b. These four proportions are known by the name of Napier's Analogies. X. In the case in which there are given two sides and an angle opposite one of them, there will in general be two solutions corresponding to the two results in Case II. of rectilineal triangles. It is also plain that this ambiguity will extend itself to the corresponding case of the polar triangle, that is, to the case in which there are given two angles and a side opposite one of them. In every case we shall avoid all false solutions by recollecting, 1st. That every angle, and every side of a spherical triangle is less than 180°. 2d. That the greater angle lies opposite the greater side, and the least angle opposite the least side, and reciprocally. NAPIER'S CIRCULAR PARTS. XI. Besides the analogies of Napier already demonstrated, that Geometer also invented rules for the solution of all the cases of right angled spherical triangles. The circular parts, as they are called, are the two sides c and b, about the right angle, the complements of the oblique angles B and C, and the complement of the hypothenuse a. Hence there are five circular parts. The right angle A not being a circular part, is supposed not to separate the circular parts c and b, so that these parts are considered as adjacent to each other. If any two parts of the triangle be given, their corresponding circular parts will also be known, and these together with a required part, will make three parts under consideration. Now, these three parts will all lie together, or one of them will be separated from both of the others. For example, if B and c were given, and a required, the three parts considered would lie together. But if B and C were given, and b required, the parts would not lie together; for, B would be separated from C by the part a, and from b by the part c. In either case B is the middle part. Hence, when there are three of the circular parts under consideration, the middle part is that one of them to which both of the others are adjacent, or from which both of them are separated. In the former case the parts are said to be adjacent, and in the latter case the parts are said to be opposite. This being premised, we are now to prove the following rules for the solution of right angled spherical triangles, which it must be remembered apply to the circular parts, as already defined. 1st. Radius into the sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. 2d. Radius into the sine of the middle part is equal to the rectangle of the cosines of the opposite parts. These rules are proved by assuming each of the five circular parts, in succession, as the middle part, and by taking the extremes first opposite, then adjacent. Having thus fixed the three parts which are to be considered, take that one of the general equations for oblique angled triangles, which shall contain the three corresponding parts of the triangle, together with the right angle: then make A=90°, and after making the reductions corresponding to this supposition, the resulting equation will prove the rule for that particular case. For example, let comp. a be the middle part and the extremes opposite. The equation to be applied in this case must contain a, b, c, and A. The first of equations (2.) contains these four quantities: hence R2 cos a= =R cos b cos c+ sin b sin c cos A. If A=90° cos A=0; hence R cos a=cos b cos c; that is, radius into the sine of the middle part, (which is the complement of a,) is equal to the rectangle of the cosines of the opposite parts. Suppose now that the complement of a were the middle part and the extremes adjacent. The equation to be applied must contain the four quantities a, B, C, and A. It is the first of equations (8.). B R2 cos A sin B sin C cos ɑ- -R cos B cos C. Making A=90°, we have sin B sin C cos a=R cos B cos C, or R cos a=cot B cot C; A that is, radius into the sine of the middle part is equal to the rectangle of the tangent of the complement of B into the tangent of the complement of C, that is, to the rectangle of the tangents of the adjacent circular parts. Let us now take the comp. B, for the middle part and the extremes opposite. The two other parts under consideration will then be the perpendicular b and the angle C. The equation to be applied must contain the four parts A, B, C, and b : it is the second of equations (8.), R2 cos B sin A sin C cos b-R cos A cos C. = Making A 90°, we have, after dividing by R, == R cos B-sin C cos b. Let comp. B be still the middle part and the extremes adjaThe equation to be applied must then contain the four four parts a, B, c, and A. It is similar to equations (9.). cent. cot a sin c=cos c cos B+cot A sin B But if A=90°, cot A=0; hence, cot a sin c-cos c cos B ; or R cos B cot a tang c. And by pursuing the same method of demonstration when each circular part is made the middle part, we obtain the five following equations, which embrace all the cases. R cos a=cos b cos c=cot B cot C R cos B-cos b sin C=cot a tang c R cos C=cos c sin B=cot a tang b (10.) R sin c=sin a sin C-tang b cot BJ We see from these equations that, if the middle part is required we must begin the proportion with radius; and when one of the extremes is required we must begin the proportion with the other extreme. We also conclude, from the first of the equations, that when the hypothenuse is less than 90°, the sides b and c will be of the same species, and also that the angles B and C will likewise be of the same species. When a is greater than 90°, the sides b and c will be of different species, and the same will be true of the angles B and C. We also see from the two last equations that a side and its opposite angle will always be of the same species. These properties are proved by considering the algebraic signs which have been attributed to the trigonometrical lines, and by remembering that the two members of an equation must always have the same algebraic sign. SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES BY LOGARITHMS. It is to be observed, that when any element is discovered in the form of its sine only, there may be two values for this element, and consequently two triangles that will satisfy the question; because, the same sine which corresponds to an angle or an arc, corresponds likewise to its supplement. This will not take place, when the unknown quantity is determined by means of its cosine, its tangent, or cotangent. In all these cases, the sign will enable us to decide whether the element in question is less or greater than 90°; the element will be less than 90°, if its cosine, tangent, or cotangent, has the sign + ; it will be greater if one of these quantities has the sign In order to discover the species of the required element of the triangle, we shall annex the minus sign to the logarithms of all the elements whose cosines, tangents, or cotangents, are negative. Then by recollecting that the product of the two |