let it meet the circle again in P; let CO be perpendicular to BN, and let it meet AE in R. It is evident that MN=AB+ AC+BC; and that LN=AB+AC -BC. Now, because BD is bisected in E, (3. 3.) and DN in A, BN is parallel to AE, and is therefore perpendicular to BD, and the triangles DAE, DNB are equiangular; wherefore, since DN=2AD, BN=2AE, and BP=2BO=2RE; also PN=2AR. But because the triangles ARC and AED are equiangular, AC: AD :: AR: AE, and because rectangles of the same altitude are as their bases, (1. 6.), AC.AD; AD2 :: AR.AE : AE2, and alternately AC.AD: AR.AE :: AD2: AE2, and 4AC.AD : 4AR.AE :: AD2: AE. But 4AR. AE = 2ARX 2AE = NP.NB=MN.NL; therefore 4AC.AD: MN.NL :: AD2: AE2. But AD: AE:: R: cos. DAE (1)=cos. (BAC): Wherefore 4AC.AD: MN.NL :: Ra: (cos. BAC)2. Now 4AC.AD is four times the rectangle under the sides AC and AB, (for AD = AB), and MN.NL is the rectangle under the sum of the sides increased by the base, and the sum of the sides diminished by the base. Therefore, &c. Q. E. D. COR. 1. Hence 2 ✓AC.AB:: ✓ MN.NL:: R:cos. BAC. COR. 2. Since by Prop. 7. 4AC.AB: (BC+(AB-AC)) (BC(AB-BC)):: R2: (sin. BAC)2; and as has been now proved, 4AC.AB: (AB+AC+BC) (AB+AC-BC) :: R2: (cos. BAC)2; therefore ex æquo, (AB+AC+BC) (AB+AC-BC): (BC+ (AB -AC)) (ВС-(AB-AC)):: (cos. BAC)2: (sin. + BAC)2. the cosine of any arch is to the sine, as the radius to the tangent of But the same arch; therefore, (AB+AC+BC) (AB+AC-BC): (BC+ (АВ-АС)) (ВС-(AB-AC)) :: R3: (tan BAC)2; and (AB+AC+BC) (AB+AC-BC); (BC+AB-AC) (BC-(AB-AC)) :: R: tan BAC. LEMMA II. If there be two unequal magnitudes, half their difference added to half their sum is equal to the greater; and half their difference taken from half their sum is equal to the less. Let AB and BC be two unequal magnitudes, of which AB is the greater; suppose AC bisected in D, and AE equal to BC. It is manifest, A E D B that AC is the sum, and EB the differ C ence of the magnitudes. And because AC is bisected in D, AD is equal to DC; but AE is also equal to BC, therefore DE is equal to DB, and DE or DB is half the difference of the magnitudes. But AB is equal to BD and DA, that is to half the difference added to half the sum; and BC is equal to the excess of DC, half the sum above DB, half the difference. Therefore, &c. Q. E. D. Cor. Hence, if the sum and the difference of two magnitudes be given, the magnitudes themselves may be found; for to half the sum add half the difference, and it will give the greater; from half the sum subtract half the difference, and it will give the less. SECT. II. OF THE RULES OF TRIGONOMETRICAL THE GENERAL PROBLEM which Trigonometry proposes to resolve is: In any plane triangle, of the three sides and the three angles, any three being given, and one of these three being a side, to find any of the other three. The things here said to be given are understood to be expressed by their numerical values; the angles, in degrees, minutes, &c.; and the sides in feet, or any other known measure. Gg The reason of the restriction in this problem to those cases in which at least one side is given, is evident from this, that by the angles alone being given, the magnitudes of the sides are not determined. Innumerable triangles, equiangular to one another, may exist, without the sides of any one of them being equal to those of any other; though the ratios of their sides to one another will be the same in them all, (4.6.) If, therefore, only the three angles are given, nothing can be determined of the triangle but the ratios of the sides, which may be found by trigonometry, as being the same with the ratios of the sines of the opposite angles. For the conveniency of calculation, it is usual to divide the general problem into two; according as the triangle has, or has not, one of its angles a right angle. PROB. I. In a right angled triangle, of the three sides and three angles, any two being given, besides the right angle, and one of those two being a side, it is required to find the other three. It is evident, that when one of the acute angles of a right angled triangle is given, the other is given, being the complement of the former to a right angle; it is also evident that the sine of any of the acute angles is the cosine of the other. This problem admits of several cases, and the solutions, or rules for / calculation, which all depend on the first Proposition, may be conveniently exhibited in the form of a Table; where the first column contains the things given; the second, the things required; and the third, the rules or proportions by which they are found. the two sides. 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 AC=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+BA) (BC-BA). Therefore AC = √(BC+BA) (BC-ВА). When AC and AB are given, BC may be found from the 47th, as in the preceding instance, for BC=BA+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. CR::AC: СВ. 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, b given, to find the other sides. SOLUTION. Because the angles A and B are given, C is also given, being the supplement 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 arch intersecting BC in C |