In the second case, when AC and C are given to find the hypotenuse BC, a solution may also be obtained by help of the secant, for CA: CB:: R: sec. C.; if, therefore, this proportion be made R sec. C:: AC: CB, CB will be found. = In the third case, when the hypotenuse BC and the side AB are given to find AC, this may be done either as directed in the Table, or by the 47th of the first; for since AC2 BC2 - BA2, AC = √BC2 — BA2. This value of AC will be easy to calculate by logarithms, if the quantity BC2-BA2 be separated into two multipliers, which may be done; because (Cor. 5. 2.), BC2—BA2=(BC + BÁ). (BC—BA). Therefore AC = √(BC+BA) (BC—BA). When AC and AB are given, BC may be found from the 47th, as in the preceding instance, for BC= √BA2+AC2. But BA2+AC2 cannot be separated into two multipliers; and therefore, when BA and AC are large numbers, this rule is inconvenient for computation by logarithms. It is best in such cases to seek first for the tangent of C, by the analogy in the Table, AC: AB:: R: tan. C; but if C itself is not required, it is sufficient, having found tan. C by this proportion, to take from the Trigonometric Tables the cosine that corresponds to tan. C, and then to compute CB from the proportion cos. C: R:: AC : CB. PROBLEM II. In an oblique angled triangle, of the three sides and three angles, any three being given, and one of these three being a side, it is required to find the other three. This problem has four cases, in each of which the solution depends on some of the foregoing propositions. CASE I. Two angles A and B, and one side AB, of a triangle ABC, being given, to find the other sides. SOLUTION. Because the angles A and B are given, C is also given, being the sup plement of A+B; and, (2.) Sin. C sin. A :: AB: BC; also, : Two sides AB and AC, and the angle B opposite to one of them, being given, to find the other angles A and C, and also the other side BC. SOLUTION. The angle C is found from this proportion, AC : AB :: sin. B: sin. C. Also, A=180°-B-C; and then, sin. B: sin. A :: AC: CB, by Case 1. In this case, the angle C may have two values; for its sine being found by the proportion above, the angle belonging to that sine may either be that which is found in the tables, or it may be the supplement of it (Cor. def. 4.). This ambiguity, however, does not arise from any defect in the solution, but from a circumstance essential to the problem, viz. that whenever AC is less than AB, there are two triangles which have the sides AB, AC, and the angle at B of the same magnitude in each, but which are nevertheless unequal, the angle opposite to AB in the one, being the supplement of that which is opposite to it in the other. The truth of this appears by describing from the centre A with the radius AC, an arc intersecting BC in C 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'Ă 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'=AC'C, and therefore, AC'B, which is the supplement of AC'C, 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 (25. 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' (25. 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 (C-B) are both given, B and C may be found. For B (C+B)+ (C-B), and C= (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 angle B and C ; for BC AB2-2 cos. AXAB.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. B ДА C D SOLUTION II. Let D be the difference of the sides AB, AC. Then (Cor. 7.) 2 √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 arc corresponding is between 89°, and 89° 1'; but it cannot be found true to seconds, because the sines of 89° and of 890 1', differ only by 50 (in the two last places,) whereas the arcs themselves differ by 60 seconds. Two arcs, 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 arc 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. The 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 arc in the tables. 1. If ADB be a circle, of which the centre is C, DB, any arc of that circle, and the are DBE double of DB; and if the chords DE, DB be drawn, also the perpendiculars to them from C, viz. CF, CG, it has been demonstrated (8. 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 cosine of the arc BD, and CG the cosine of the half of BD; whence the cosine of the half of any arc 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 arc, cos. A is a mean proportional |