4 and C'; then, if AC and AC' be drawn, it is evident that the triangles ABC, ABC' have the side AB and the angle at B common, and the sides AC and AC' equal, but have not the remaining side of the one equal to the remaining side of the other, that is, BC to BC', nor their other angles equal, viz. BC'A to BCA, nor BAC' to BAC. But in these triangles the, angles ACB, AC'B are the supplements of one another. For the triangle CAC' is isosceles, and the angle ACC'= the AC'C, and therefore, AC'B, which is the supplement of ACC, is also the supplement of ACC' or ACB; and these two angles, ACB, AC'B are the angles found by the computation above. From these two angles, the two angles BAC, BAC' will be found: the angle BAC is the supplement of the two angles ACB, ABC, (32. 1.), and therefore its sine is the same with the sine of the sum of ABC and ACB. But BAC' is the difference of the angles ACB,' ABC; for it is the difference of the angles AC'C and ABC, (because AC'C, that is ACC' is equal to the sum of the angles ABC, BAC', (32. 1.)). Therefore to find BC, having found C, make sin C: sin (C+B): AB: BC; and again, sin C: sin (C-B) : : AB : BC'. Thus, when AB is greater than AC, and C consequently greater than B, there are two triangles which satisfy the conditions of the question. But when AC is greater than AB, the intersections C and C fall on opposite sides of B, so that the two triangles have not the same angle at B common to them, and the solution ceases to be ambiguous, the angle required being necessarily less than B, and therefore an acute angle. CASE III. Two sides AB and AC, and the angle A, between them, being given, to find the other angles B and C, and also the side BC. SOLUTION. First, make AB+AC: AB-AC :: tan (C+B): tan (C-B.). Then, since (C+B) and 1⁄2 (C-B) are both given, B and C may be found. For B1 (C+B)+ } (C−B), and C=1(C+B)-(C−B.). (Lem. 2.). To find BC. Having found B, make sin B : sin A :: AC: BC. But BC may also be found without seeking for the angles B and C' ; for BC: ✓ AB2-2 cos A×AB.AC + AC”, Prop. 6. This method of finding BC is extremely useful in many geometrical investigations, but it is not very well adapted for computation by logarithms, because the quantity under the radical sign cannot be separated into simple multipliers. Therefore, when AB and AC are expressed by large numbers, the other solution, by finding the angles, and then computing BC, is preferable. CASE IV. The three sides AB, BC, AC, being given, to find the angles A, B, C. SOLUTION I. Take F such that BC: BA+AC: BA-AC: F, then F is either the sum or the difference of BD, DC, the segments of the base, (5.). If F be greater than BC, F is the sum, and BC the difference of BD, DC; but, if F be less than BC, BC is the sum, and F the difference of BD and DC. In either case, the sum of BD and DC, and their difference being given, BD and DC are found. (Lem. 2.) Then, (1.) CA: CD: R: cos. C; and BA: BD:: R: cos. B; wherefore C ́and B are given, and consequently A. 44 SOLUTION II. D Let D be the difference of the sides AB, AC. Then (Cor. 7.) ✔AB.AC: ✔ (BC+D)(BC—D) :: R : sin BAC. SOLUTION III. Let S be the sum of the sides BA and AC. Then (1 Cor. 8.) 2✔AB.AC: ✔(S+BC) (S—BC) :: R; cos BAC. SOLUTION IV. S and D retaining the significations above, (2 Cor. 8.) ✔ (S+BC) (S—BC): √ (BC+D) (BC—D) : : R ; tan ↓ BAC. It may be observed of these four solutions, that the first has the advantage of being easily remembered, but that the others are rather more expeditious in calculation. The second solution is preferable to the third, when the angle sought is less than a right angle; on the other hand, the third is preferable to the second, when the angle sought is greater than a right angle; and in extreme cases, that is, when the angle sought is very acute or very obtuse, this distinction is very material to be considered. The reason is, that the sines of angles, which are nearly 90°, or the cosines of angles, which are nearly-0, vary very little for a considerable variation in the corresponding angles, as may be seen from looking into the tables of sines and cosines. The consequence of this is, that when the sine or cosine of such an angle is given, (that is, a sine or cosine nearly equal to the radius,) the angle itself cannot be very accurately found. If, for instance, the natural sine .9998500 is given, it will be immediately perceived from the tables, that the arch corresponding is between 89°, and 89°, 1'; but it cannot be found true to seconds, because the sines of 89° and of 89°, 1', differ only by 50 (in the two last places), whereas the arches themselves differ by 60 seconds. Two arches, therefore, that differ by 1", or even by more than 1", have the same sine' in the tables, if they fall in the last degree of the quadrant. The fourth solution, 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 excessive, and are too irregular to allow the proportional parts to be found with exactness, so that when the angle sought is extremely obtuse, and its half of consequence very near to 90, the third solution is the best. It may always be known, whether the angle sought is greater or less than a right angle by the square of the side opposite to it being greater or less than the squares of the other two sides. SECTION III. CONSTRUCTION OF TRIGONOMETRICAL TABLES. In all the calculations performed by the preceding rules, tables of sines and tangents are necessarily employed, the construction of which remains to be explained. These tables usually contain the sines, &c. to every minute of the quadrant from 1' to 90°, and the first thing required to be done is to compute the sine of 1', or of the least arch in the tables. 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, (8. 1. Sup.) that CG is a mean propor tional between AH, half the radius, and AF, the line made up of the radius and the perpendicular CF. Now CF is the cosine of the arch BD, and CG the cosine of the half of BD; whence the cosine of the half of any arch BD, of a circle of which the radius =1, is a mean proportional between and 1+cos BD. Or for the greater generality, supposing A = any arch, cos ↓ A is a mean proportional between ¦ and 1+cos A, and therefore (cos A)2 = (1+cos A) or cos A = ✓ (1+cos A). 2. From this, theorem, (which is the same that is demonstrated (8. 1. Sup., only that it is here expressed trigonometrically), it is evident, that if the cosine of any arch be given, the cosine of half that arch may be found. Let BĎ, therefore, be equal to 60°, so that the chord BD=radius, then the cosine or perpendicular CF was shewn (9. 1. Sup.) to be=1, and therefore cos BD, or cos 30°= √3 2* In the same manner, cos 15°— ✓(1+cos 30°), and cos 7°, 30=(1+ cos 15°), &c. In this way the cosine of 3o, 45′, of 1o, 52', 30", and so on, will be computed, till after twelve bisections of the arch of 60°, the cosine of 52". 44′′. 03"". 45v. is found. But from the cosine of an arch its sine may be found, for if from the square of the radius, that is, from 1, the square of the cosine be taken away, the remainder is the of the sine, and its square root is the sine itself. Thus, the sine of 52'. 44". 03"". 45v. is found. square 3. But it is manifest, that the sines of very small arches are to one another nearly as the arches themselves. For it has been shewn, that the number of the sides of an equilateral polygon inscribed in a circle may be so great, that the perimeter of the polygon and the cir cumference of the circle may differ by a line less than any given line, or which is the same, may be nearly to one another in the ratio of equality. Therefore their like parts will also be nearly in the ratio of equality, so that the side of the polygon will be to the arch which it subtends nearly in the ratio of equality; and therefore, half the side of the polygon to half the arch subtended by it, that is to say, the sine of any very small 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 second as the sine of the first to the sine of the second. Hence, from the sine of 52′′. 44′′. 03"'. 45. being found, the siue of 1' becomes known; for, as 52". 44". 03""'. 45. to 1, so is the sine of the former arch to the sine of the latter. Thus the sine of 1' is found 0.0002908882. 4. The sine 1' being thus found, the sines of 2', of 3', or of any number of minutes, found by the following proposition. THEOREM. Let AB, AC, AD be three such 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 cosine of the common difference BC as the sine of AC, the middle arch, to half the sum of the sines of AB and AD, the extreme arches. D M Draw CE to the centre; let BF, CG, and DH perpendicular to AE, be the sines of the arches AB, AC, AD. Join BD, and let it meet CE in I; draw IK perpendicular to AE, also BL and IM perpendicular to DH. Then, because the arch BD is bisected in C, EC is at right angles to BD, and bisects it in I; also BI is the sine, and EI the cosine of BC or CD. And, since BD is bisected in I, and IM is parallel to BL, (2. 6.), LD is also bisected in M. Now BF is equal to HL, therefore, BF+DH=DH+HL= DL + 2LH = 2LM + 2LH = 2MH or 2KI; and therefore IK is half the sum of BF and DH. But because the triangles CGE, IKE are equiangular, CE: EI :: CG: IK, and it has been shewn that EI=cos BC, and IK={ (BF+DH); therefore R cos BC :: sin AC: (sin AB+sin AD).Q. E. D. B AF GK H E COR. Hence, if the point B coincide with A, R: cos. BC sin. BC: of :: sin. BD, that is the radius is to the cosine any arch, as the sine of the arch is to half the sine of twice the arch; or if any arch=A, sin. AXcos. A. sin. 2A=sin. AXcos. A, or sin. 2A = 2 Therefore also, sin 2'-2' sin 1'x cos 1'; so that from the sine and cosine of one minute the sine of 2' is found. Hh |