other (8. 6). Also, the two right-angled triangles ALC, ABD, having the acute angle LAC common, are similar to each other, and to the two triangles AFB, Dfb. Hence the following propositions are derived 1. The triangles DAB, BAF give the following analogy. DA: BA :: BA : AF. That is, the chord of any arc is a mean proportional between the diameter and the versed sine of that arc. 2. Again, the triangles AFB, DFB give the following analogy. AF: BF:: BF: DF. That is, the sine of any arc is a mean proportional between the versed sine and the coversed sine. 3. From the similar triangles CAL, BAF, we deduce this analogy. CA: AL:: AB or 2 AL: AF. That is, radius is to the sine of any arc AM, as twice that sine is to the versed sine of AMB double that arc. 4. Again, CA CL :: AB or 2 AL: BF. : That is, radius is to the cosine of any arc AM, as twice the sine of that arc is to the sine of AMB double that arc. 47. The following corollaries are deduced from art. 46. Cor. 1. From no. 1, DAXAF=BA2 (16. 6). That is, the rectangle under the diameter and the versed sine of any arc is equal to the square of the chord. Cor. 2. From no. 1, DA: AF :: DA2: BA2* :: CA2 : AL2. That is, the diameter is to the versed sine of any arc, as the square of the radius is to the square of the right sine of half that arc. Cor. 3. From no. 3, 2 AL2=CA×AF. But the radius CA is a constant quantity, therefore AL2 varies as AF, that is, the square of the right sine of any arc is as the versed sine of double that arc. Cor. 4. From no. 4, CLx2 AL-CAXBF, or CLXAL= CAXBF. Hence, if A denote any arc or angle, sine Axcos. A= radius × sine 2 A, that is, the rectangle under the sine and cosine of any angle is equal to the rectangle under half the radius and the sine of twice that angle, or, to half the rectangle under the radius and the sine of twice that angle. * By the analytical properties of proportion. Or thus, DA: AF :: DAXDA or DA2: DAXAF=AB2 (15. 5). The Application of Trigonometry to the Mensuration of the Sides and Angles of Triangles described on a Plane, or Plane Trigonometry properly so called. 48. In any right-angled triangle ABC (fig. page 21), if the hypothenuse AC be made radius, the perpendicular BC becomes the sine, and the base AB the cosine of the angle at the base CAB. But if the base AB be made radius, the perpendicular BC becomes the tangent, and the hypothenuse AC the secant of the angle at the base CAB. This is manifest from the definitions. PROP. I. 49. In any right-angled plane triangle, the hypothenuse is to either of the sides, as the radius to the sine of the angle opposite to that side, or to the cosine of the angle adjacent to that side. C G D Let ABC be a right-angled triangle, of which the hypothenuse is AC. From the centre A, with any radius AD, describe the arc DE; from D draw DF at right angles to AB, and from E draw EG touching the arc in E, and meeting AC in G. Then DF is the sine, and AF the cosine of the arc DE, or angle A; also EG is the tangent, and AG the secant of the arc DE, or angle A. A FE B The two right-angled triangles AFD, ABC, having the angle C common, are equiangular; therefore AC: CB:: AD : DF, that is, AC: CB: the radius of the trigonometrical tables: sine of the angle A. Again, AC AB:: AD: AF, or AC: AB :: R: cos. A (cosine of the angle A). PROP. II. 50. In any right-angled plane triangle, as either of the sides is to the other side, so is the radius to the tangent of the angle opposite to the latter. See last fig. Because EG touches the arc in E, AEG is a right angle, and therefore is equal to the angle ABC. From the equiangular triangles ABC, AEG, AB: BC: AE :: EG, or AB: BC: R tan. A. 51. Scholium to Prop. I, II. If the lengths of AC and CB, or AC and AB, or AB and CB, be known in feet, inches, or any other measure, the an gles may be found by the following proportions. AC: AB:: R :: sine C. Whence the sine of the angle C is known, and consequently the quantity of the angle C may be found by seeking the sine in a table of sines and tangents. Again, AC: CB :: R: sine A, and AB: BC :: R: tan. A, and BC: AB :: R: tan. C, and AC: AB :: R: cos. A. Whence the tangents and sines (and consequently the angles) Tan. C: R: AB: BC, Cos. A: R:: AB : AC. The first three terms of each of these analogies are supposed to be known; hence the fourth term is found. 52. Cor. If these analogies be arithmetically expressed, BC and radius be supposed =1; then sine A=AC cos. A= AB AC 53. In every triangle, if a perpendicular be drawn from any of the angles on the opposite side, the segments of that side will be to each other as the tangents of the parts into which the vertical angle is divided by the perpendicular, or as the cotangents of the angles at the base. gents of the opposite angles CAD, BAD. Hence AD: R:: DC: tan. CAD (50), and AD: R::DB: tan. BAD. A D consequently DC: tan. CAD :: DB : tan. BAD (11. 5), or DC: DB:: tan. CAD : tan. BAD (16. 5), or DC: DB:: cot. C: cot. B, because the tangent of an arc or angle is equal to the cotangent of its complement (23). PROP. IV. 54. The sides of any triangle are to one another as the sines of their opposite angles; and, conversely, the sines of the angles of any triangle are to one another as the sides which are opposite to the angles. In the triangle ABC, the side AB: side AC :: sine ZACB, opposite to the former side AB : sine ZABC, opposite to the latter side AC; and conversely. On the side BA (produced if necessary) take BE=CA, and from the points A, E let fall the perpendiculars AD, EF upon BC; then will AD and EF be the sines of the angles ACB and ABC to the equal radii CA and BE. The right-angled triangles ADB, EFB, having the acute angle at B common, are equiangular; therefore AB: EB or AC: AD EF, that is, AB AC :: sine 4C: sine ZB. Otherwise. See fig. Prop. 3. From any angle A of the triangle ABC draw AD perpendicular to BC. About the centres B and C, with the radii BA and CA, suppose arcs to be described; then AD will be the sine of the angles B and C. In the right-angled triangle ABD, AB: AD: R: s. B (49), and in the right-angled triangle ACD, AC AD: R: s. C. Hence s. BxAB=RXAD=s. CXAC. Therefore AB: AC: s. C: s. B (16. 6). In the same manner, AB : BC :: s. C: s. A. 55. Cor. 1. Hence, in any triangle, if two sides and an angle opposite to one of them be given, the remaining side and angles may be found; or, if two angles and a side opposite to one of them be given, the other angle and sides may be found. Thus, if the sides AC, AB, and the angle C, be given, to find the rest, then AB: s. C: AC: :s. B. Because the sum of the angles B and C is the supplement of A (32. 1), 180° (<C+<B)=ZA. Whence, sine C: AB:: s. A: BC. Again, if the angles A and C, and the side AB, be given, to find the rest, then and 180°-(<A+<C)=2B. Hence, sine C: AB:: s. B: AC. 56. Cor. 2. If all the angles of a triangle be given, the pro-` portion of the sides to one another may be found, but not the real lengths of the sides. For the proportion of the sides to one another is the same as that of the sines of the angles which are opposite to them. LEMMA III. 57. If the semi-sum and semi-difference of any two quantities be added together, the aggregate will be the greater quantity; and if the semi-difference be subtracted from the semi-sum, the remainder will be the less quantity. |