depends the solution of all questions, that can be proposed concerning the measure of extension, and its parts; and the art of determining all these things from the knowledge of some of them, is reduced to the solution of these two general questions. 1. Knowing three of the six parts, the sides and angles—which constitute a rectilineal triangle; to find the other three. 2. Knowing three of the six parts, which compose a spherical triangle; that is a triangle formed on the surface of a sphere by three arches of circles, which have their centre in the centre of the same sphere-to find the other three. The first question is the object of what is called Plane Trigonometry, because the six parts, considered here, are in the same plane: it is also denominated Rectilineal Trigonometry. The second question belongs to Spherical Trigonometry, wherein the six parts are considered in different planes. But the only object here is to explain The solutions of the former question : viz. PLANE TRIGONOMETRY. Plane Trigonometry is that branch of geometry, which teaches how to determine, or calculate three of the six parts of a rectilineal triangle by having the other three parts given or known. It is usually divided into Right angled and Oblique angled Trigonometry, according as it is applied to the mens suration of Right or Oblique angled Triangles. In every triangle, or case in trigonometry, three of the parts must be given, and one of these parts, at least, must be a side ; because, with the same angles, the sides may be greater or less in any proportion. Right ANGLED PLANE TRIGONOMETRY. P1.5. Fig. 1. 1. In every right-angled plane triangle ABC, if the hypothenuse AC be made the radius, and with it a circle, or an arc of one, be described from each end; it is plain (from def. 20.) that BC is the sine of the angle A, and AB is the sine of the angle C; that is, the legs are the sines of their opposite angles. Fig. 2. If one leg AB be made the radius, and with it, on the point A, an arc be described ; then BC is the tangent, and AC is the secant of the angle A, by def. 22 and 25. Fig. 3. 3. If BC be made the radius, and an arc be described with it on the point C; then is AB the tangent,and AC is the secant of the angle C, as before. Because the sine, tangent, or secant of any given arc, in one circle, is to the sine, tangent, or secant of a like arc (or to one of the like number of degrees) in another circle; as the radius of the one is to the radius of the other ; therefore the sine, tangent, or secant of any arc is proportional to the sine, tangent, or secant of a like arc, as the radius of the given arc is to 10.000000, the radius from whence the logarithmic sines, tangents, and seçants, in most tables, are calculated, that is; If AC be made the radius, the sines of the angle A and C, described by the radius AC, will be proportional to the sines of the like arcs, or angles in the circle, that the tables now mentioned were calculated for. So if BC was required, having the angles and AB given, it will be, Fig. 1. As S.C: AB :: S.A :BC. That is, as the sine of the angle C in the tables, is to the length of AB ;(or sine of the angle C, in a circle whose radius is AC;) so is the sine of the angle A in the tables, to the length of BC. (or şine of the same angle, in the circle, whose radius is AC.) In like manner the tangents and secants represented by making either leg the radius, will be proportional to the tangents and secants of a like arc, as the radius of the given arc is to 10.000000, the radius of the tables aforesaid. Hence it is plain, thi.t if the name of each side of the triangle be placed thereon, a proportion will arise to answer the same end as before : thus if AC be made the radius, let the word radius be written thereon; and as BC and AB, are the bines of their opposite angles ; upon the first let S. A, or sine of the angle A, and on the other let S.C, or sine of the angle C, be written. Then, When a side is required, it may be obtained by this proportion, viz. As the name of the side given is to the side given, So is the name of the side required to the side required. Thus, if the angles A and C, and the hypothenuse AC were given, to find the sides; the proportion will be Fig. 1. 1. R: AC:: S.A: BC. That is, as radius is to AC, so is the sine of the angle A, to BC. And, 2. R: AC:: S.C : AB. That is, as radius is to AC, so is the sine of the angle C to AB. When an angle is required, we use this proportion, viz. As the side that is made the radius, is to radius, So is the other given side, to its name. Thus, if the legs were given to find the angle A, and if AB be made the radius, it will be Fig. 2. AB:R::BC: TA. That is, as AB, is to radius, so is BC, to the tangent of the angle A. After the same manner, the sides or angles of all right angled plane triangles may be found, from their proper data. We here, in plate 4, give all the proportion requisite for the solution of the six cases in rightangled trigonometry; making every side possible the radius. In the following triangles this mark angle denotes it to be known, or the quantity of degrees it contains to be given; and this mark' on a side, denotes its length to be given in feet, yards, perches, or miles, &c. and this mark', either in an angle or on a side, denotes the angle or side to be required. From these proportions it may be observed that to find a side, when the angles and one side are given, any side may be made the radius; and in an to find an angle, one of the given sides must be made the radius. So that in the 1st, 2d, and 3d cases, any side as well required as given may be made the radius, and in the first statings of the 4th, 5th, and 6th cases, a given side only is made the radius. RIGHT ANGLED TRIANGLES. CASE I. The angles and hypothenuse given, to find the base and per pendicular. PL. 5. Fig. 4. In the right angled triangle ABC, suppose the angle A = 46°. 30. and consequently the angle C 43o. 30'.(by cor. 2. theo. 5.); and AC 250 parts, (as feet, yards, miles, &c.) required the sides AB and BC. 1st. BY CONSTRUCTION. Make an angle of 46'. 30', in blank lines, (by prob. 16. geom.) as CAB; lay 250, which is the given hypothenuse, from a scale of equal parts, from A to C; from C, let fall the perpendicular (BC, by prob. 7. geom.) and that will constitute the triangle ABC. Measure the lines BC, and AB, from the same scale of equal parts that AC was taken from; and you have the answer. 2d. BY CALCULATION. 1. Making AC the radius, the required sides are found by these propositions, as in plate 4, case 1. R: AC:: S.A: BC. |