tables; and these are most commonly used, as they perform the Description of the Table of Log. Sines and Tangents. In the first column of the table are contained all the arcs, the same To compute the Natural Sine and Cosine of a Given Arc. This problem is resolved after various ways. One of these is as follows, viz. by means of the ratio between the diameter and circumference of a circle, together with the known series for the sine and cosine, hereafter demonstrated. Thus, the a + &c. + &c. semicircumference of the circle, whose radius is 1, being •3•141592653589793 &c, the proportion will therefore be, as the number of degrees or minutes in the simicircle, This length of the arc being denoted by the letter a ; and its sine and cosine by s and c; then will these two be expressed by the two following series, viz. a3 as a? + &c. + a? a4 am 2 2.3.4.5.6 + &c. EXAM. 1. If it be required to find the sine and cosine of 1 minute. Then, the number of minutes in 180° being 10800, it will be first, as 10800 : 1 :: 3:14159265 &c. : *000290838208665 the length of an arc of one minute. Therefore, in this case, a = .0002908882 = 0.0000000423079 &c, leaves c = •9999999577 the cosine of 1 minute. Exam. 2. For the sine and cosine of 5 degrees. Here, as 180° : 5° :: 3:14159265 &c. : *08726646 = a the length of 5 degrees. Hence a = •08726646 - 4a3 . .00011076 these collected give s = .08715574 the sine of 5° And, for the cosine, 1=l' ja? 00380771 + zma4 = •00000241 these collected give c = .99619470 the cosine of 5o. arc may After the same manner, the sine and cosine of any other be computed. But the greater the arc is, the slower the the series will converge, in which case a greater number of terms must be taken, to bring out the conclusion to the same degree of exactness. Or, having found the sine, the cosine will be found from it, by the property of the right-angled triangle CBF, viz. the cosine CF = VCB’ — BF?, or c =Ni-s. There are also other methods of constructing the canon of sines and cosines, which, for brevity's sake, are here omitted. PROBLEM II. To compute the Tangents and Secants. THE sinęs and cosines being known, or found by the foregoing problem; the tangents and secants will be easily found, from the principle of similar triangles, in the following manner: In the first figure, where, of the arc AB, BF is the sine, CF or bk the cosine, Ah the tangent, ch the secant, DL the cotangent, and cl the cosecant, the radius being ca or CB or cd; the three similar triangles CFB, CAH, CDL, give the following proportions : 1st, CF : FB :: CA : AH; whence the tangent is known, being a fourth proportional to the cosine, sine, and radius. 2d, cf : CB :: cĄ : ; whence the secant is known, being a third proportional to the cosine and radius. 3d, BF: FC :: CD : DL; whence the cotangent is known, being a fourth proportional to the sine, cosine, and radius. 4th, bf : BC :: CD : CL; whence the cosecant is known, being a third proportional to the sine and radius. As for the log. sines, tangents, and secants, in the tables, they are only the logarithms of the natural sines, tangents, and secants, calculated as above. HAVING given an idea of the calculation and use of sines, tangents and secants, we may now proceed to resolve the several cases of Trigonometry; previous to whịch, however, it may be proper to add a few preparatory notes and observations, as below. Note 1. There are usually three methods of resolving triangles, or the cases of trigonometry; namely, Geometrical Construction, Arithmetical Computation, and Instrumental Operation, in the First Method, The triangle is constructed, by making the parts of the given magnitudes, namely, the sides from a scale of equal parts, and the angles from a scale of chords, or or by some other instrument. Then measuring the unknown parts by the same scales or instruments, the solution will be obtained near the truth. In the Second Method, Having stated the terms of the proportion according to the proper rule or theorem, resolve it like any other proportion, in which a fourth term is to be found from three given terms, by multiplying the second and third together, and dividing the product by the first, in working with the natural numbers; or, in working with the logarithms, add the logs. of the second and third terms together, and from the sum take the log. of the first term; then the natural number answering to the remainder is the fourth term sought. In the Third Method, Or Instrumentally, as suppose by the log. lines on one side of the common two-foot scales ; Extend the Compasses from the first term, to the second or third, which happens to be of the same kind with it ; then that extent will reach from the other term to the fourth term, as required, taking both extents towards the same end of the scale. Note 2. Every triangle bas six parts, viz. three sides and three angles. And in every triangle, or case in trigonometry, there must be given three of these parts, to find the other three. Also, of the three parts that are given, one of them at least must be a side ; because, with the same angles, the sides may be greater or less in any proportion. Note 3. All the cases in trigonometry, may be comprised in three varieties only; viz. 1st, When a side and its opposite angle are given. 2d, When two sides and the contained angle are given. 3d, When the three sides are given. For there cannot possibly be more than these three varie: ties of cases ; for each of which it will therefore be proper to give a separate theorem, as follows: THEOREM I. When a Side and its Opposite Angle are two of the Given Parts. Then the unknown parts will be found by this theorem; viz. The sides of the triangle have the same proportion to each other, as the sines of their opposite angles have. That is, As any one side, Is to the sine of its opposite angle ; other side, Demonstr. 1 D B Demonstr. For, let ABC be the proposed triangle, having AB the greatest side, and bc the least. Take ad = BC, considering it as a radius; and let fall the perpendiculars DE, CF, which will evidently be the sines of the angles A and B, to the radius ad or BC. Now the triangles ADE, ACF, are equiangular; they therefore have their like sides proportional, namely, AC : CF :: AN or BC : DE; that is, the side ac is to the sine of its opposite angle B, as the side bc is to the sine of its opposite angle A. Note 1. In practice, to find an angle, begin the proportion with a side opposite to a given angle. And to find a side, begin with an angle opposite to a given side. Note 2. An angle found by this rule is ambiguous, or uncertain whether it be acute or obtuse, unless it be a right angle, or unless its magnitude be such as to prevent the ambiguity ; because the sine answers to two angles, which are supplements to each other; and accordingly the geometrical construction forms two triangles with the same parts that are given, as in the example below; and when there is no restriction or limitation included in the question, either of them may be taken. The number of degrees in the table, answering to the sine, is the acute angle; but if the angle be obtuse, subtract those degrees from 180°, and the remainder will be the obtuse angle. When a given angle is obtuse, or a right one, there can be no ambiguity; for then neither of the other angles can be obtuse, and the geometrical construc, tion will form only one triangle. 1. Geometrically. Draw an indefinite line ; on which set off AB = 345, from some convenient scale of equal parts.-Make the angle 3701.-With a radius of 232, taken from the same scale of equal parts, and centre B, cross Ac in the two points C, C.-Lastly, join BC, BC, and the figure is con structed, A = |