angle G'-30° 13'; the times corresponding in the column of Hours P. M., are S' 4 1 36'; G'4h 1m 44; and if we wish to find the log. sine, cosine, &c., corresponding to any intermediate time, as, for example, 4 1m 39, which differs 3 from the angle S', we must find the tabular logarithm corresponding to S', and apply the correction for 3', given by the table at the bottom of the page, as in the following examples: nave before explained : +3 Secant 10.06338 Nearly the same results are obtained by using the angle G', in the manner we These corrections must be applied by addition or subtraction, according to the directions given above, so as to make the required logarithm fall between those which correspond to the times S' and G'. The inverse process will give the time corresponding to any logarithm. Thus, if the log. sine 9.70167 be given, the difference between this and 9.70159, corresponding to S' = 4 1 36, is 8; seeking this in the column A, in the second line of the table at the bottom of the page, it is found to correspond to 3; adding this to the time S'4 1 36, we get 4" 1" 39" for the required time. We may proceed in the same manner with the logarithms in the columns B, C; using the numbers corresponding, marked B, C, respectively, in the table at the bottom of the page. To find the degrees, minutes, and seconds, corresponding to any given logarithm sine, cosine, &c. by Table XXVII. Find the two nearest numbers to the given log. sine, cosine, &c., in the colurin marked sine, cosine, &c., respectively, one being greater, and the other less, and take their difference, D; take also the difference, d, between the given logarithm and the logarithm corresponding to the smallest number of degrees and minutes; then say, As the first found difference is to the second found difference, so is 60" to a number of seconds to be annexed to the smallest number of degrees and minutes before found. The three columns of differences may also be used, by an inverse operation to that which we have explained in the preceding article. EXAMPLE V. Find the degrees, minutes, and seconds (less than 90°), corresponding to the log. sine 9.61400. Next less log. S'=24° 16′ 9.61382 D=29 Log. of smallest angle S'=24° 16′ is 9.61382 Given log.... 9.61400 d=18 Then say, As 29: 18 :: 60′′: 38′′, nearly; which, annexed to 24° 16′, give 24° 16′ 38′′, answering to log. sine 9.61400. Subtracting 24° 16′ 38′′ from 180°, there remain 155° 43′ 22′′, the log, sine of which is also 9.61400. The quantity 38" may also be found by inspection in the side column S' of the page opposite d= 18, in the column of differences between the two columns, A, A. If we use the angle G', we shall have d' equal to 11, the difference of the logarithms 9.61411 and 9.61400, and the corresponding number of seconds in column G', is 37", making 24° 16′ 37′′. To find the arithmetical complement of any logarithm. The arithmetical complement of any logarithm is what it wants of 10.00000, and is used to avoid subtraction. For, when working any proportion by logarithms, you may add the arithmetical complement of the logarithm of the first term, instead of subtracting the logarithm itself, observing to neglect 10 in the index of the sum of the logarithms. The arithmetical complement of any logarithm is thus found-Begin at the index, and write down what each figure wants of 9, except the last significant figure, which take from 10.* Thus, the arithmetical complement of 9.62595 is 0.37405, that of 1.86567 is 8.13433; and that of 10.33133 is 89.66867, or 9.66867. * When the index of the given logarithm is greater than 10, as in some of the numbers of Table XXVII., the left-hand figure of it must be neglected; and when there are any ciphers to the right hand of the last significant figure, you may place the same number of ciphers to the right hand of the other figures of the arithmetical complement. PLANE TRIGONOMETRY. PLANE TRIGONOMETRY is the science which shows how to find the measures of the sides and angles of plane triangles, some of them being already known. It is divided into two parts, right-angled and of lique-angled; in the former case, one of the angles is a right angle, or 90°; in the latter, they are all oblique. In every plane triangle there are six parts, viz. three sides and three angles; any three of which being given (except the three angles), the other three may be found by various methods, viz. by Gunter's scale, by the sliding rule, by the sector, by geometrical construction, or by arithmetical calculation. We shall explain each of these methods; but the latter is by far the most accurate; it is performed by the help of a few theorems, and a trigonometrical canon, exhibiting the natural or the logarithmic sines, tangents, and secants, to every degree and minute of the quadrant. The theorems alluded to are the following: THEOREM 1. any right-angled triangle, if the hypotenuse be made radius, one side will be the sine of the opposite angle, and the other its cosine; but if either of the legs be made radius, the other leg will be the tangent of the opposite angle, and the hypotenuse will be the secant of the same angle. In 1st. If, in the right-angled plane triangle ACB (fig. 1), we make the hypotenuse AB radius, and upon the centre, A, describe the arc BE, to meet AC produced in E, then it is evident that BC is the sine of the arc BE (or the sine of the angle BAC), and that AC is the cosine of the same angle; and if the arc AD be described about the centre B (fig. 2), AC will be the sine of the angle ABC, and BC its cosine.. 2dly. If the leg AC (fig. 3) be made radius, and the arc CD be described about the centre A, CB will be the tangent of that arc, or the tangent of the angle CAB; and AB will be its secant. 3dly. If the leg BC (fig. 4) be made radius, and the arc CD be described about the centre B, CA will be the tangent of that arc, or the tangent of the angle B, and AB will be its secant. And Now, it has been already demonstrated (in Art. 55, Geometry) that the sine, tangent, secant, &c. of any arc in one circle is to the sine, tangent, secant, &c. of a similar arc in another circle as the radius of the former circle to the radius of the latter. since in any right-angled triangle there are given either two sides, or the angles and one side, to find the rest, we may, if we wish to find a side, make any side radius; then say, As the tabular number of the same name as the given side is to the given side of the triangle, so is the tabular number of the same name as the required side, to the required side of the triangle. If we wish to find an angle, one of the given sides must be made radius; then say, As the side of the triangle made radius is to the tabular It will not be necessary to add any further description of the uses of the sector or sliding rule; for what we have already given will be sufficient for any one tolerably well versed in the use of Gunter's scale. radius, so is the other given side to the tabular sine, tangent, secant, &c. by it represented; which, being sought for in the table of sines, &c., will correspond to the degrees and minutes of the required angle. THEOREM II. In all plane triangles, the sides are in direct proportion to the sines of their opposite angies (by Art. 58, Geometry). Hence, to find a side, we must say, As the sine of an angle is to its opposite side, so is the sine of either of the other angles to the side opposite thereto. But if we wish to find an angle, we must say, As any given side is to the sine of its opposite angle, so is either of the other sides to the sine of its opposite angle. THEOREM III. In every plane triangle, it will be, as the sum of any two sides is to their difference, so is the tangent of half the sum of the two opposite angles to the tangent of half their difference (by Art. 59, Geometry). THEOREM IV. As the base of any plane triangle is to the sum of the two sides, so is the difference of the two sides to twice the distance of a perpendicular (let fall upon the base from the opposite angle) from the middle of the base (by Art. 60, Geometry). THEOREM V. In any plane triangle, as the rectangle contained by any two sides including a sought angle, is to the rectangle contained by the half sum of the three sides and the same haif sum decreased by the other side, so is the square of radius to the square of the cosine of half the sought angle (by Art. 61, Geometry). In addition to these theorems, it will not be amiss for the learner to recall to mind the following articles: 1. In every triangle, the greatest side is opposite to the greatest angle, and the greatest angle opposite to the greatest side. 2. In every triangle equal sides subtend equal angles. (Art. 39, Geometry.) 3. The three angles of any plane triangle are equal to 180°. (Art. 35, Geometry.) 4. If one angle of a triangle be obtuse, the rest are acute; and if one angle be right, the other two together make a right angle, or 90°; therefore, if one of the acute angles of a right-angled triangle be known, the other is found by subtracting the known angle from 90°. If one angle of any triangle be known, the sum of the other two is found by subtracting the given angle from 180°; and if two of the angles be known, the third is found by subtracting their sums from 180°. 5. The complement of an angle is what it wants of 90°, and the supplement of an angle is what it wants of 180°. In the two following tables we have collected all the rules necessary for solving the various cases of Right-angled and Oblique-angled Trigonometry. Angles C Side BC. Subtract half the given angle, A, from 90°; the remainder is half the sum of the other augles. Then say, As the sum of the sides, AC, AB, is to their difference, so is the tangent of the half sum of the other angles to the tangent of half their difference; which added to and subtracted from the half sum, will give the two angles B and C; the greatest angle being opposite to the greatest side. Sine B side AC: sine A: side BC. Let fall a perpendicular, BD, opposite to the required angle; then, as AC sum of AB, BC:: their difference: twice DG, the distance of the perpendicular from the middle of the All the angles. base; hence, AD, CD, are known, and the triangle ABC is divided into two right-angled triangles, BCD, BAD; then, by Cases IV. and V. of Right-angled Trigonometry, we may find the angle A or C. Either angle, Either of the angles, as A, may also be found by the following rule. From half the sum of the three sides subtract the side BC opposite to the sought angle; take the logarithms of the half sum and remainder, to which add the arithmetical complements of the logarithms of the sides AB, AC (including the sought angle); half the sum of these four logarithms will be the logarithmic cosine of half the sought angle. In calculating by logarithms by any of the preceding rules, you must remember, that the logarithm of the first term of the analogy is to be subtracted from the sum of the logarithms of the second and third terms; the remainder will be the logarithm of the sought fourth term. When the first term is radius (whose logarithm is 10.00000), you need only reject a unit in the second left-hand figure of the index of the sum of the second and third terms. But when the radius occurs in the second or third term, you must suppose a unit to be added to the second left-hand figure of the index of the other term, and subtract therefrom the logarithm of the first term. RIGHT-ANGLED TRIGONOMETRY. Solution of the six cases in Right-angled Trigonometry. CASE 1. The angles and hypotenuse given, to find the legs. Given the hypotenuse AC 250 leagues, and the angle C, opposite to the sude AB,=35° 30′, to find the base CB, and perpendicular AB. BY PROJECTION. Draw the base CB of any length; with an extent equal to the chord of 60°, and on C as a centre, describe the arc DE; from E to D lay off 35° 30′ taken from the line of chords;* through C and D 250 D 35°30' E * In all projections of this kind, the angles are measured from the line of chords; the radius used for draw the line AC, which make equal to 250; from A let fall the perpendicular AB :0 cut CB in B, and it is done; for CB will be 203.5, and AB equal to 145.2. BY LOGARITHMS. By making the hypotenuse CA radius, it will be, To find the perpendicular AB. To find the base BC. As radius..... 10.00000 2.39794 10.00000 Is to the hypotenuse AC 250.. 2.39794 So is the sine angle C 35° 30′.. 9.76895 To the perpendicular AB 145.2 2.16189 BY GUNTER'S SCALE. In all proportions which are calculated by Gunter's scale, when the first and second terms are of the same kind, the extent from the first term to the second will reach from the third to the fourth. Or, when the first and third terms are of the same kind, The extent from the first term to the third will reach from the second to the fourth; that is, we must set one point of the compasses on the division expressing the firs term, and extend the other point to the division expressing the third term; then without altering the opening of the compasses, we must set one point on the division representing the second term, and the other point will fall on the division showing the fourth term or answer. In the present example the work is as follows: Extend from radius, or 90°, to 54° 30' on the line of sines; that extent will reach from 250, the hypotenuse, to 203.5, the base on the line of numbers; and the extent from radius or 90°, to 35° 30′ on the line of sines, will reach froin 250 to 145.2 on the line of numbers. Observe the same method in all the following examples, except in those proportions where the word secant is mentioned, which cases must be wrought by considering the hypotenuse radius,* there being no line of secants on the common Gunter's scale, although it can easily be marked on the line of sines. Note. The radius, according to the nature of the proportion, may be either of the following quantities:— 8 points on the line of rhumbs. 4 points on the line of tangent rhumbs. 90° on the line of sines. 45° on the line of tangents. CASES II. AND III. The angles and one leg given, to find the hypotenuse and other leg. The angle ACB 33° 15′, the leg BC 163 miles, given, to find the hypotenuse and he other leg. BY PROJECTION. Draw the line BC, which make equal to 163 miles; on B erect the perpendicular BA; on C, as a centre, with the chord of 60°, sweep the are BD, which make equal to 33° 15'; draw CD, and continue it to cut AB in A, and it is done; for AB being measured on the same scale that BC was, will be 106.9, and AC 194.9 miles. BY LOGARITHMS. By making the base BC radius, it will be, To find the perpendicular AB. As radius 45° 10.00000 As radius 90°. 1 33°15′ B To find the hypotenuse AC. 10.00000 Is to the base BC 163 So is tangent angle C 33° 15 To the perpendicular AB 106.9 2.02885 Is to the base BC 163 2.21219 10.07765 2.28984 the triangles are measured by scales of equal parts, as was before observed. Instead of using the line chords, it is much more convenient to set off the angles by means of a protractor, or circular are, on which the degrees are marked. Its construction is so simple that it needs no explanation. *Or by using in the analogy, radius: cosine angle, instead of secant angle: radius ; and rooms angle, instead of rosecant angle: radius. |