Then, (1.) CA: CD:: R: cof C; and BA:BD:: R: cof B; wherefore C and B are given, and confequently A. 44 SOLUTION II. Let D be the difference of the fides AB, AC. Then (Cor.7.) 2√AB.AC: √ (BC+D)(BC—D) : : R ; fin ¦ BAC. 14 Let S be the fum of the fides BA and AC. Then (1 Cor. 8.) 24/AB.AC: √√(s+BC)(S—BC) :: R. cof1⁄2 BAC. SOLUTION IV. S and D retaining the fignifications above, (2 Cor. 8.) √√ (S+BC)(S—BC) : ^/(BC+D)(BC—D) :: R: tan BAC. It may be obferved of these four folutions, that the first has the advantage of being easily remembered, but that the others are rather more expeditious in calculation. The fecond folution is preferable to the third, when the angle fought is less than a right angle; on the other hand, the third is preferable to the fecond, when the angle fought is greater than a right angle; and in extreme cafes, that is, when the angle fought is very acute or very obtuse, this diftinction is very material to be confidered. The reafon is, that the fines of ángles, angles, which are nearly 90°, or the co-fines of angles, which are nearlyo, vary very little for a confiderable variation in the corresponding angles, as may be feen from looking into the tables of fines and co fines. The confequence of this is, that when the fine or co-fine of fuch an angle is given, (that is, a fine or co-fine nearly equal to the radius), the angle itself cannot be very accurately found. If, for inftance, the natural fine .9998500 is given, it will be immediately perceived from the tables, that the arch correfponding is between 899 and 89°, 1'; but it cannot be found true to feconds, because the fines of 89° and of 89°, 1', differ only by 50 (in the two laft places), whereas the arches themselves differ by 60 feconds. Two arches, therefore, that differ by 1", or even by more than 1", have the same fine in the tables, if they fall in the laft degree of the quadrant. The fourth folution, which finds the angle from its tangent, is not liable to this objection; nevertheless, when an arch approaches very near to 90, the variations of the tangents become exceffive, and are too irregular to allow the proportional parts to be found with exactness, so that when the angle fought is extremely obtufe, and its half of confequence very near to 90, the third folution is the best. It may always be known, whether the angle fought is greater or lefs than a right angle by the fquare of the fide oppofite to it being greater or lefs than the fquares of the other two fides. SEC SECTION III. CONSTRUCTION OF TRIGONOMETRICAL In all the calculations performed by the preceding rules, tables of fines and tangents are neceffarily employed, the conftruction of which remains to be explained. Thefe tables ufually contain the fines, &c. to every minute of the quadrant from 1 to 90°, and the first thing required to be done is to compute the fine of 1', or of the least arch in the tables. D G 1. If ADB be a circle, of which the centre is C, DB any arch of that circle, and the arch DBE double of DB; and if the chords DE, DB be drawn, and also the perpendiculars to them from C, viz. CF, CG, it has been demonstrated, (S. 1. Sup.) that CG is a mean proportional between AH, half the radius,and AF, the line made up of the radius and the perpendicular CF. Now CF is the co-fine of th arch BD, and CG the co-fine of the the half of BD; whence the co-fine of the half of any arch BD,of a circle of which the radius, is a mean B H C E proportional between and 1+ cof BD. Or for the greater generality, generality, fuppofing Aany arch, cof A is a mean proportional between and 1+ cof A, and therefore (cofA)2= (1 + cof A) or cof A= (1+ cof A). 2. From this theorem, (which is the fame that is demonstrated 8. 1. Sup. only that it is here expreffed trigonometrically), it is evident, that if the co-fine of any arch be given, the co-fine of half that arch may be found. Let BD, therefore, be equal to 60°, fo that the chord BD radius, then the cofine, or perpendicular CF was fhewn (9. 1. Sup.) to be, and therefore cof BD, or cof 30°= N = (1 + 1 ) = √ 2 = 13. In the fame manner, ١٢/ به cof 15° = √(1+cof 30°),and cof 7°, 30 ́= √ 1⁄2 (1+cof 15o), &c. In this way the co-fine of 3o, 45′, of 1o, 52′ 30′′, and fo on, will be computed, till after twelve bifections of the arch of 60°, the cofine of 52". 44"". 03''''. 45. is found. But from the co-fine of an arch its fine may be found, for, if from the square of the radius, that is, from 1, the fquare of the cofine be taken away, the remainder is the fquare of the fine, and its fquare root is the fine itself. Thus, the fine of 52". 44 ́ ́ ́. 03 ́ ́ ́ ́. 45". is found, " 3. But it is manifeft, that the fines of very fmall arches are to one another, nearly as the arches themfelves. For it has been shewn, that the number of the fides of an equilateral polygon infcribed in a circle may be fo great, that the perimeter of the polygon and the circumference of the circle may differ by a line lefs than any given line, or which is the fame, may be nearly to one another in the ratio of equality. Therefore their like parts will alfo be nearly in the ratio of equality, fo that the fide of the polygon will be to the arch which it fubtends nearly in the ratio of equality; and therefore, half the fide of the polygon to half the arch fubtended by it, that is to fay, the fine of any very fmall arch will be to the arch itself, nearly in the ratio of equality. Therefore, if two arches are both very small, the first will be to the fecond as the fine of the firft to the fine of the fecond. Hence, from the fine of 52" 44". 03. 45. being found, the fine of 1' becomes known; for, as 52". 44 ́ ́ ́. 03 ́ ́ ́ ́. 45'. to I, fo is the fine of the former arch to to the fine of the latter. Thus, the fine of 1' is found 0.0002908882. 4. The fine of 1' being thus found, the fines of 2', of 3', or of any number of minutes, are found by the following propo fition. THEOREM. Let AB, AC, AD be three fuch arches, that BC the difference of the first and second is equal to CD the difference of the second and third; the radius is to the co-fine of the common difference BC as the fine of AC, the middle arch, to half the fum of the fines of AB and AD, the extreme arches. D Draw CE to the centre; let BF, CG, and DH perpendicular to AE, be the fines of the arches AB, AC, AD. Join BD, and let it meet CE in I; draw IK perpendicular to AE, alfo BL and IM perpendicular to DH. Then, because the arch BD is bifected in C, EC will be at right angles to BD, and will bifect it in I, and BI will be the fine, and EI the co-fine of BC or CD. And, fince BD is bifected in I and IM is parallel to BL, (2.6.) LD is alfo bifected in M. Now BF is equal to HL, therefore, BF+DH DH+HL= DL+2LH2LM+2LH2MH, or 2KI; and therefore, IK is half the fum of BF and DH. But because the triangles CGE, IKE B A F GK M IL are equiangular, CE: EI:: CG: IK, and it has been fhewn, that EI cof BC, and IK (BF+DH); therefore R : cof BC: fin AC : 1⁄2 (fin AB+ fin AD). Q. E. D. · : |