SECT. IV. PLANE TRIGONOMETRY. 1033. Plane Trigonometry is that branch of mathematics whose object is the investigation and calculation of the sides and angles of plane triangles. It is of the greatest importance to the architect in almost every part of his practice; but the elements will be sufficient for his use, without pursuing it into those more abstruse subdivisions which are essential in the more abstract relations which connect it with geodisic operations. 1034. We have already observed that every circle is supposed to be divided into 360 equal parts, called degrees, and that each degree is subdivided into 60 minutes, these minutes each into 60 seconds, and so on. Hence a semicircle contains 180 degrees, and a quadrant 90 degrees. 1035. The measure of an angle is that arc of a circle contained between those two lines which form the angle, the angular point being the centre, and such angle is estimated by the number of degrees contained in the arc. Thus, a right angle whose measure is a quadrant or quarter of the circle is one of 90 degrees (Prop. 22. Geometry); and the sum of the three angles of every triangle, or two right angles, is equal to 180 degrees. Hence in a right-angled triangle, one of the acute angles being taken from 90 degrees, the other acute angle is known; and the sum of two angles in a triangle taken from 180 degrees leaves the third angle; or either angle taken from 180 degrees leaves the sum of the other two angles. 1036. It is usual to mark the figure which denotes degrees with a small : thus, 60° means 60 degrees; minutes are marked thus': hence, 45′ means 45 minutes; seconds are marked thus", 49" meaning 49 seconds; and an additional comma is superadded for thirds, and so on. Thus, 58° 14′ 25′′ is read 58 degrees, 14 minutes, 25 seconds. 1037. The complement of an arc is the quantity it wants of 90 degrees. Thus, AD (fig. 395.) being a quadrant, BD is the complement of the arc AB, and, reciprocally, AB is the complement of BD. Hence, if an arc AB contain 50 degrees, its complement BD will be 40. 1038. The supplement of an arc is that which it wants of 180 degrees. Thus, ADE being a semicircle, BDE is the supplement E of the arc AB, which arc, reciprocally, is the supplement of BDE. Thus, if AB be an arc of 50 degrees, then its supplement BDE will be 130 degrees. C Fig. 395. AB Н 1039. The line drawn from one extremity of an arc perpendicular to a diameter passing through its other extremity is called a sine or right sine. Thus, BF is the sine of the arc AB, or of the arc BDE. Hence the sine (BF) is half the chord (BG) of the double arc (BAG). 1040. That part of the diameter intercepted between the arc and its sine is called the versed sine of an arc. Thus, AF is the versed sine of the arc AB, and EF the versed sine of the arc EDB. 1041. The tangent of an arc is a line which touches one end of the arc, continued from thence to meet a line drawn from the centre, through the other extremity, which last line is called the secant of the arc. Thus, AH is the tangent and CH the secant of the are АВ. So EI is the tangent and CI the secant of the supplemental arc BDE. The latter tangent and secant are equal to the former; but, from being drawn in a direction opposite or contrary to the former, they are denominated negative. 1042. The cosine of an arc is the right sine of the complement of that arc. Thus BF, the sine of AB, is the cosine of.BD. 1043. The cotangent of an arc is the tangent of that arc's complement. Thus AH, which is the tangent of AB, is the cotangent of BD. 1044. The cosecant of an arc is the secant of its complement. Thus CH, which is the secant of AB, is the cosecant of BD. 1045. From the above definitions follow some remarkable properties. I. That an arc and its supplement have the same sine, tangent, and secant; but the two latter, that is, the tangent and the secant, are accounted negative when the arc exceeds a quaarant, or 90 degrees. II. When the arc is 0, or nothing; but the secant then becomes the radius CA, which is the least it can be. As the arc increases from 0, the sines, tangents, and secants all increase, till the arc becomes a whole quadrant AD, and then the sine is the greatest it can be, being equal to the radius of the circle; under which circumstance the tangent and secant are infinite. III. In every arc AB, the versed sine AF, and the cosine BK or CF, are together equal to the radius of the circle. The radius CA, the tangent AH, and the secant CH, form a right-angled triangle CAH. Again, the radius, sine, and cosine form another right-angled triangle CBF or CBK. So also the radius, cotangent, and cosecant form a right-angled triangle CDL. All these right-angled triangles are similar to each other. 1046. The sine, tangent, or secant of an angle is the sine, tangent, or secant of the arc by which the angle is measured, or of the degrees, &c. in the same arc or angle. The method of constructing the scales of chords, sines, tangents, and secants engraved on mathematical instruments is shown in the annexed figure. 1047. A trigonometrical canon (fig. 396.) is a table wherein is given the length of the sine, tangent, and secant to every degree and minute of the quadrant, compared with the radius, which is expressed by unity or 1 with any number of ciphers. The logarithms, moreover, of these sines, tangents, and secants, are tabulated, so that trigonometrial calculations are performed by only addition and subtraction. (See 632. et seq.) Tables of this sort are published separately, and we suppose the reader to be provided with such. 1048. PROBLEM I. To compute the natural sine and cosine of a given arc. The semiperiphery of a circle whose radius is 1 is known to be 3.141592653589793, &c.: we have then the following proportion: Ás the number of degrees or minutes in the semicircle Sines 70 60 50 Secants Tangents 70 40 30 50 90 60 40 10 30 88 2882 10 Now the length of the arc being denoted by the letter a, and its sine and cosine by s these two will be expressed by the two following series, viz.. 10 Example 1. Let it be required to find the sine and cosine of one minute. of minutes in 180 degrees being 10800, it will be, first, as 10800 1:3.14159265, &c. : *000290888208665 = the length of an arc of one minute. Hence, in this case, take la2=0·000Q000423079, &c. leaves c= 9999999577, the cosine of one minute. Example 2. For the sine and cosine of 5 degrees : Here 180°: 5°:.3.14159265, &c.: 08726646=a, the length of 5 degrees. Hence a 08726646 +5=00000004 These collected give s=08715574, the sine of 5 degrees. And for the cosine 1=1· -fa2= 00380771 +4= '00000241 These collected give c = 99619470, the cosine of 5 degrees. In the same way we find the sines and cosines of other arcs may be computed. The greater the arc the slower the series will converge; so that more terms must be taken to make the calculation exact. Having, however, the sine, the cosine may be found from it by the property of the right-angled triangle CBF, viz. the cosine CF= √✓CBo— BFo, or c=√1-5%. There are other methods of constructing tables, but we think it unnecessary to mention them; our sole object being here merely to give a notion of the mode by which such tables are formed. 1049. PROB. II. To compute the tangents and secants. Having, by the foregoing problem, found the sines and cosines, the tangents and secants are easily found from the principles of similar triangles. In the arc AB (fig. 895.), where 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: I. CF: FB::CA: AH, by which we find that the tangent is a fourth proportional to the cosine, sine, and radius. II. CF CB::CA: CH, by which we find that the secant is a third proportional to the cosine and radius. III. BF: FC::CD: DL, by which we find that the cotangent is a fourth proportional to the sine, cosine, and radius. IV. BF: BC:: CD: CL, by which we find that the cosecant is a third proportional to the sine and radius. Observation 1. There are therefore three methods of resolving triangles, or the cases of trigonometry; viz. geometrical construction, arithmetical computation, and instrumental operation. The method of carrying out the first and the last does not need explanation: the method is obvious. The second method, from its superior accuracy in practice, is that whereof we propose to treat in this place. Observation 2. Every triangle has six parts, viz. three sides and three angles. And in all cases of trigonometry, three parts must be given to find the other three. And of the three parts so given, one at least must be a side; because, with the same angles, the sides may be greater or less in any proportion. viz. Observation 3. All the cases in trigonometry are comprised in three varieties only; 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. More than these three varieties there cannot possibly be; and for each of them we shall give a separate theorem. 1050. THEOREM I. When a side and its opposite angle are two of the given parts. Then the sides of the their opposite angles have. triangle have the same proportion to each other as the sines of That is, For let ABC (fig. 397.) be the proposed triangle, having AB the greatest side, and BC the least. Take AD as a radius equal to BC, 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 sine AC is to the sine of its opposite angle B as the side BC is to the sine of its opposite angle A. Fig. 397. B Note 1. In practice, when an angle is sought, the proportion is to be begun with a side opposite a given angle; and to find a side, we must begin with the angle opposite the given side. Note 2. By the above rule, an angle, when found, is ambiguous; that is, it is not certain whether it be acute or obtuse, unless it come out a right angle, or its magnitude be such as to remove the ambiguity; inasmuch as the sine answers to two angles, which are supplements to each other; and hence the geometrical construction forms two triangles with the same parts, as in an example which will follow: and if there be no restriction or limitation included in the question, either result may be adopted. The degrees in a table answering to the sine is the acute angle; but if the angle be obtuse, the degrees must be subtracted from 180 degrees, and the remainder will be the obtuse angle. When a given angle is obtuse, or is one of 90 degrees, no ambiguity can occur, because neither of the other angles can then be obtuse, and the geometrical construction will only form one triangle. Example 1. In the plane triangle ABC, Let AB be 345 feet, BC 232 feet, 37' 20': Required the other parts. First, to the angles at C and B (fig. 398.) Fig. 398. C It is to be observed here that the second and third logarithms are added (that is, the numbers are multiplied), and from the sum the first logarithm is subtracted (that is, division by the first number), which leaves the remainder 9.955127, which, by the table of sines, is found to be that of the angle 115° 36', or 64° 24'. 1051. THEOREM II. When two sides and their contained angle are given. The given angle is first to be subtracted from 180° or two right angles, and the remainder will be the sum of the other two angles. Divide this remainder by 2, which will give the half sum of the said unknown angles; and using the following ratio, we have. As the sum of the two given sides Is to their difference, So is the tangent of half the sum of their opposite angles To the tangent of half the difference of the same angles. Now the half sum of any two quantities increased by their half difference gives the greater, and diminished by it gives the less. If, therefore, the half difference of the angles above found be added to their half sum, it will give the greater angle, and subtracting it will leave the lesser angle. All the angles thus become known, and the unknown side is then found by the former theorem. A Fig. 399. G F B For let ABC (fig. 399.) be the proposed triangle, having the two given sides AC, BC, including the given angle C. With the centre C and radius E CA, the less of these two sides, describe a semicircle, meeting the other side BC produced in D, E, and the unknown side AB in A, G. Join AE, CG, and draw DE parallel to AE. Now BE is the sum of the given sides AC, CB, or of EC, CB; and BD is the difference of these given sides. The external angle ACE is equal to the sum of the two internal or given angles CAB, CBA; but the angle ADE at the circumference is equal to half the angle ACE at the centre; wherefore the same angle ADE is equal to half the sum of the given angles CAB, CBA. Also the external angle AGC of the triangle BGC 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 given 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 AE, is also perpendicular to AD: therefore AE is the tangent of CDA the half sum; and DF, the tangent of DAB, the half difference of the angles to the same radius AD, by the definition of a tangent. 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. It is to be observed, that in the third term of the proportion the cotangent of half the given angle may be used instead of the tangent of the half sum of the unknown angles. Example. In the plane triangle ABC (fig. 400.), C Fg. 100. 180° 00 The side AC 174.07 37 20 142 40 71 20 Log. 2.715226 2.232818 10:471298 9.988890 Log. 9.955126 2.537819 9.782796 2.365489 1052. THEOREM III. When the three sides of a triangle are given. Let fall a perpendicular from the greatest angle on the opposite side, or base, dividing it into two segments, and the whole triangle into two right-angled triangles, the proportion will be As the base or sum of the segments Is to the sum of the other two sides, So is the difference of those sides To the difference of the segments of the base. Then take half the difference of these segments, and add it to the half sum, or the half base, for the greater segment; and for the lesser segment subtract it. Thus, in each of the two right-angled triangles there will be known two sides and the angle opposite to one of them, whence, by the first theorem, the other angles will be found. For the rectangle under the sum and difference of the two sides is equal to the rectangle under the sum and difference of the two segments. Therefore, forming the sides of these rectangles into a proportion, their sums and differences will be found proportional. Example. In the plane triangle ABC (fig. 401.), |