The base AB is still the middle part, for it is not connected either with AC or the angle c, therefore these parts are the extremes disjunct; the former being separated from AB by the angle A, and the latter by the side BC. Here we apply the second rule, recollecting that the cosine of the complement of an angle, or the hypothenuse, is the sine itself. Hence, rad x sine AB=sine AC X sine C. (S) CASE II. FIRST, Let AB, BC, and the angle c, be the parts under consideration, in the triangle ABC. This is exactly like the first part of the preceding case; the three parts are connected, BC is the middle part, AB and the complement of c are the extremes conjunct. Hence, rad x sine BC=tang AB X cot C. SECONDLY. Let BC, the hypothenuse AC, and the angle, the parts concerned in the triangle ABC. be This is similar to the second part of the preceding case; the perpendicular BC is still the middle part, for the angle c separates it from ac, and the side AB from the angle A: therefore ac and the angle A are extremes disjunct, that is not joined to BC the middle part. Here we must apply the second rule, taking care to remember that the cosine of the complement of the hypothenuse, or an angle, is the sine itself. Hence, rad x sine BC=sine ACX sine A. (T) CASE III. FIRST, Let BC, AC, and the angle c, bè the parts under consideration, in the triangle ABC. of The three parts follow each other, without the intervention any other quantity; therefore the complement of c is the middle part, BC and the complement of ac are the extremes conjunct, that is, joined to the angle c. Hence, rad x cos c=cot ACX tang BC. SECONDLY. Let AB, the angle A, and the angle c, be the parts under consideration, in the triangle ABC. The complement of the angle c is here the middle part, being separated from the angle A, by Ac; and from the side AB, by the perpendicular BC; therefore, AB and the complement of the angle A, are the extremes disjunct, or not joined to c. Hence, rad x cos c=sine AX cos AB. (U) CASE IV. FIRST, Let AB, AC, and the angle A, be the parts under consideration, in the triangle ABC. This is exactly similar to the first part of case 3d. The complement of the angle A is the middle part, AB, and the the complement of AC, are the extremes conjunct, that is, they are joined to the angle A. Hence, rad x cos A=cot AC X tang AB. SECONDLY. Let BC, the angle A, and the angle c, be the parts under consideration, in the triangle ABC. This is exactly of the same nature with the second part of the 3d case. The complement of the angle A is the middle part; BC, and the complement of c, are the extremes disjunct, the former being separated from the middle part A, by the base AB, and the latter by the hypothenuse AC. Hence, rad x cOS A COS BCX sine c. (W) CASE V. FIRST, Let the hypothenuse AC, the angle ▲, and the angle c, be the parts under consideration, in the triangle ABC. The three parts follow each other; therefore the complement of AC is the middle part, the complement of A and the complement of c are the extremes conjunct, that is, they are joined to the middle part AC. Hence, rad x cos AC=cot AX cot c. SECONDLY. Let the hypothenuse AC, the base AB, and the perpendicular BC, be the parts under consideration, in the triangle ABC. The complement of AC is here the middle part, being separated from AB by the angle A, and from BC by the angle c; therefore AB and BC are the extremes disjunct. Hence, rad x COS ACCOS ABX COS BC. SCHOLIUM. The preceding cases include all the varieties that can possibly happen in the practice of right-angled spherical triangles. Any of the equations may be turned into a proportion by putting the required term last, that with which it is connected first, and the other two in the middle in any order. These equations are exactly the same as those already given (O. 164.) and therefore Napier's rules are universally true. 1 GENERAL RULES FOR THE SOLUTIONS OF ALL THE DIFFERENT CASES OF RIGHT-ANGLED SPHERICAL TRIANGLES. Every spherical triangle consists of six parts, three sides, and three angles; any three of which being given, the rest may be found. In a right-angled spherical triangle, two given parts, besides the right angle, are sufficient to determine the rest. The The questions arising from a variation of the given and required parts are 16, but if distinguished by the data, the number of cases is 6. THE GIVEN QUANTITIES ARE, EITHER 1. The hypothenuse and an angle. 6. The two angles. (X) RULE İ. E G Draw a rough figure as in the margin; and let AG, AH; FB, FH; CI, CD; E1, EG, be considered as quadrants, or 90° each; then you have eight right-angled sphericaltriangles, every two of which will have equal angles at their bases: And the triangles CGF and EDF will have their respective sides and angles either equal to those of ABC, the triangle under consideration, or they will be the complements thereof. (L. 147.) Then, B In triangles having equal angles at their bases. The sines of their bases have the same ratio to each other, as the tangents of their perpendiculars. And, The sines of the hypothenuses have the same ratio to each other, as the sines of the perpendiculars (M. 163.) (Y) Illustration. 1st. In the triangles ABC and AHG. 1. Sine AG : sine AC :: sine HG : sine BC. and these are propor2. Sine AH: sine AB:: tang HG: tang BC. ( tional by inversion. 2d. In the triangles FGC and FHB. and inversely, &c. 3. Sine BH: sine CG:: sine FB : sine FC. and inversely, &c. 4th. In the triangles EDF and EIG. 7. Sine EI : sine ED: : tang IG : tang DF, and inversely, &c. The student must remember that ABC is the proper triangle in the preceding proportions, and that sine AG, sine AH, &c. are each radius. BH is the complement of the base AB. In any of these cases, for sine, or tangent, write cosine, or co-tangent. 1G=AC the hypothenuse, and DFBC the perpendicular. Since AB and BC are perpendicular to each other, either of them may be considered as the base, and to avoid a number of different figures it will sometimes be necessary to make a and c change places, as is done in some of the succeeding cases; then, in considering these remarks, where AB occurs read BC, and instead of the angle a read c; and the contrary. (Z) RULE II. BARON NAPIER'S RULES. 1. In every right-angled spherical triangle there are five parts, exclusive of the right-angle, which is not taken into consideration: and these five parts are the hypothenuse, the two sides or legs, and their opposite angles. Now in every case proposed for solution, there are three of these five parts concerned, that is, two given (together with the right-angle) and a third required. 2. If the three quantities under consideration, viz. the two which are given and that which is required, are joined together, or follow each other in a successive order, without the intervention of a side, or angle, not concerned in the question; the middle one is called the middle part, and the other two the extremes conjunct, because they are joined to the middle part. 3. If only two of the three things under consideration are connected, or joined together, they are invariably called extremes disjunct, that is, not joined to the middle part, and the other term which is not joined with them, is called the middle part. Then, 4. Radius x sine of the middle part=rectangle, or product, of the tangents of the extremes when they are conjunct. 5. Radius x sine of the middle part=rectangle, or product, of the cosines of the extremes when they are disjunct. Observe to write cosine or co-tangent, instead of sine or tangent; and sine instead of cosine, &c. when you make use of either of the angles, or the hypothenuse; but not when you use the sides or legs. 6. Having found which is the middle part, and written the terms down as expressed in one or other of the above rules, according as the extremes are conjunct or disjunct, turn the terms into a proportion, thus: Put the required term last, that with which it is connected first, and the remaining two in the midany order. dle in (A) RULE III. 1. Radius x sine of either side sine of the opposite angle x sine of the hypothenuse. 2. Radius x sine of either side=tangent of the other side x cot of its opposite 4. = 3. Radius x cosine of either of the angles tangent of the adjacent side x cot hypothenuse. 4. Radius x cosine of either of the angles angle x cosine of its adjacent side. 5. Radius x cosine of the hypothenuse cot of the other 4. 6. Radius x cosine of the hypothenuse cosine of the other side. sine of the other cot of one angle x cosine of one side x Then put the required term last, that with which it is connected first, and the remaining two in the middle in any order, will have a correct proportion. and you NOTE. This rule is the same as BARON NAPIER'S (O. 164.) OF THE DIFFERENT SPECIES OR AFFECTIONS OF RIGHT-ANGLED SPHERICAL TRIANGLES. (B) I. When the hypothenuse and an angle are given. 1. The side opposite to the given angle, is of the same species with the given angle. 2. The side adjacent to the given angle, is acute or obtuse, according as the hypothenuse is of the same, or of different species with the given angle. 3. The other angle is acute or obtuse, according as the hypothenuse and the given angle are of the same or of different species. (C) II. When the hypothenuse and one side are given. 1. The angle opposite to the given side, is of the same species with the given side. 2. The angle adjacent to the given side, is acute or obtuse, according as the hypothenuse is of the same or of different species with the given side. 3. The other side is acute or obtuse, according as the hypothenuse is of the same, or of different species with the given side. (D) III. When a side and its adjacent angle are given. 1. The other angle is of the same species as the given side. 2. The other side is of the same species as the given angle. 3. The z 2 |