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 řethod, 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 has 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: a 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 ; Demonstr. A Demonstr. For, let ABC be the proposed triangle, having as 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 : : AD 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 sinę 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 he 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 aß = 345, from some convenient scale of equal parts.—Make the angle A = 37°15.-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, structed, which gives two triangles, and showing that the case is ambiguous. Then, the sides ac measured by the scale of equal parts, ant the angles B and c measured by the line of chords, or other instrument, will be found to be nearly as below; viz. AC 174 <c115o4. or 64 LB 27° Вс 2. Arithmetically. log. 2:365488 9.782796 2.537819 152 56 or 101 44 taken from 180 00 180 00 leaves LB 27 04 or 78 16 AB the sum 3. Instrumentally. In the first proportion. -Extend the compasses from 232 to 345 on the line of numbers; then that extent reaches, on the sines, from 3704 to 640, the angle c. In the second proportion.—Extend the compasses from 37°; to 27° or 78°, on the sines ; then that extent reaches, on the line of numbers, from 232 to 174 or 374į, the two yalues of the side AC. When two Sides and their Contained Angle are given. FIRST add the two given sides together, to get their sum, and subtract them, to get their difference. Next subtract the given angle from 180°, or two right angles, and the remainder will be the sum of the two other angles; then divide that by 2, which will give the half sum of the said unknown angles. Then say, As the sum of the two given sides, To the tang. of half the diff. of the same angles. Then add the half difference of the angles, so found, to their half sum, and it will give the greater angle, and subtracting the same will leave the less angle: because the half sum of any two quantities, increased by their half difference, gives the greater, and diminished by it gives the less. Then all the angles being now known, the unknown side will be found by the former theorem. Note. Instead of the tangent of the half sum of the un-known angles, in the third term of the proportion, may be used the cotangent of half the given angle, which is the same thing Demonst. Let ABC be the proposed E triangle, having the two given sides AC, BC, including the given angle c. With the centre c, and radius ca, the less of these two sides, describe a semicircle, meeting the other side BC produced in D, E, and the unknown side AE in A, G. AD, CG, and draw of parallel to AE. Then be is the sum of the two given sides AC, CB, or of IC, CB; and ed is the difference of the same two given sides AC, Join AE, AC, BC, or of CD, CB. Also, the external angle ACE, is equal to the given sum of the two internal angles CAB, CBA; but the angle ADE, at the circumference, is equal to half the angle Ace at the centre; therefore the same angle AdE is equal to half the given sum of the angles CAB, CBA. Also, the external angle AGC, of the triangle BCG, is equal to the sum of the two internal angles GCB, GBC, or the angle GCB is equal to the difference of the two angles AGC, GBC; but the angle cab is equal to the said angle AGC, these being opposite to the equal sides AC, CG; and the angle DAB, at the circumference, is equal to half the angle DCG at the centre; therefore the angle DAB is equal to half the difference of the two angles CAB, CBA ; of which it has been shown that ADE or CDA is the half sum. Now the angle DAE, in a semicircle, is a right angle, or AE is perpendicular to AD; and DF, parallel to As, is also perpendicular to AD: consequently As is the tangent of cda the half sum, and of the tangent of DAB the half difference of the angles, to the same radius AD, by the definition of a tan gent. But the tangents AE, DF being parallel, it will be, as BE : BD :: AE : DF; that is, as the sum of the sides is to the difference of the sides, so is the tangent of half the sum of the opposite angles, to the tangent of half their difference. EXAMPLE I. In the plane triangle ABC, AB 345 yards Given AC 174.07 yards C { ZA 37° 20' Required the other parts. 1. Geometrically. Draw AB = 345 from a scale of equal parts. Make the angle A = 37° 20'. Set off Ac = 174 by the scale of equal parts. Join BC, and it is done. , Then the other parts being measured, they are found to be nearly as follows viz. the side BC 232 yards, the angle B 270, and the angle c 11504. 2. Arithmetically. The side AB 345 From 180° 00' the side AC 174:07 take 2 A 37 20 their sum 519:07 sum of c and B 142 40 sheir differ. 170.93 half sum of do. 71 20 As |