A A B 1005. PROB. IX. To find a fourth proportional to three lines AB, AC, AD. Let two of the lines AB, AC (fig. S67.), be so placed as to form any angle at A, and set out AD on AB; join BC, and parallel to it draw DE; then AE will be the fourth proportional required. 1006. PROB. X. To find a mean proportional between two Ines AB, BC. Place AB, BC (fig. 368.) A joined together in one straight line AC, which bisect in the point 0; then with the centre 0 and radius DA or OC describe the semicircle ADC, to meet which erect the perpendicular BD, which will be the mean proportional be tween AB and BC sought. 1007. PROB. XI. To find the centre of a circle. Draw any chord AB (fig. 569.), and bisect perpendicularly with the line CD, which bisected in O will be the centre required. 1009. PROB. XII. To describe the circumference of a circle through three points A, B, C. From the middle point B (fig. 370.) draw the chords BA, BC to the two other points, and bisect these chords perpendicularly by lines meeting in O, which will be the centre; from the centre 0, with the distance of any one of the points, as OA, describe a circle, and it will pass through the two other points B which, as a diameter, describe a semicircle cutting the given circumference in D, through which draw BADC, which will be the tangent required. 374.) describe an are; with the centre B and distance BC describe another arc cutting the former in C; draw AC, BC, and ABC will be the triangle required. lines AD, BD (fig. 376.); from the intersection D, which will be the centre of the circle, draw the perpendiculars DE, DF, DG, and they will be the radii of the circle re quired. 1014 PROB. XVIII. To describe a circle about a given triangle ABC. Bisect any two sides with two of the perpendiculars DE, DF, DG (fig. 377.), and I will be the centre of the circle. Through the centre C draw any diameter AB (fig. 378.); from the point B as a centre, with the radius BC of the given circle, describe an arc DCE; join AD, AE, DE, and ADE is the equilateral triangle sought. 1016. PROB. XX. "To inscribe a square in a given circle. (Half AB, BC, &c. forms an octagon.) Draw two diameters AC, BD (fig. 379.) crossing at right angles in the centre E; then join the four extremities A, B, C, D with right lines, and these will form the inscribed square ABCD. 1017. PROB. XXI. To describe a square about a given circle. Draw two diameters AC, BD crossing at right angles in the centre E (fig. 380.); then through the four extremities of these draw FG, IH parallel to AC, and FI, GH parallel to BD, and they will form the square FGHI. Fig. 379. E Fig. 380 G 1018. PROB. XXII. To inscribe a circle in a given square. Bisect the two sides FG, FI in the points B and A (see fig. 380.); then through these two points draw AC parallel to FG or IH, and BD parallel to FI or GH. Then the point of intersection E will be the centre, and the four lines EA, EB, EC, ED radii of the inscribed circle. 1019. PROB. XXIII. To cut a given line in extreme and mean ratio. Let AB be the given line to be divided in extreme and mean ratio (fig. 381.); that is, so that the whole line may be to the greater part as the greater part is to the less part. Draw BC perpendicular to AB, and equal to half AB; join AC, and with the centre C and distance CB describe the circle BDF; then with A as a centre and distance AD describe the arc DE. Then AB will be divided in E in extreme and mean ratio, or so that AB is to AE as AE is to EB. 1020. PROB. XXIV. To inscribe an isosceles triangle in a given circle that shall have each of the angles at the base double the angle at the vertex. Draw any diameter AB of the given circle (fig. 382.), and divide the radius CB in the point D in extreme and mean ratio (by the last problem); from the point B apply the chords BE, BF, each equal to the greater part CD; then join AE, AF, EF; and AEF will be triangle required. 1021. PROB. XXV. To inscribe a regular pen- D tugon in a given circle. (Half AD, &c. is a decagon.) Inscribe the isosceles triangle AB (fig. 383.) having each of the angles ABC, ACB double the angle BAC (Prob. 24.); then bisect the two arcs ADB, AEC, in the points D, E; and draw the chords AD, DB, AE, EC; then ADBCE will be the inscribed regular pentagon required. Fig. 383. B E D Fig. 384. 1022. PROB. XXVI. To inscribe a regular hexagon in a circle. (Holf AB, &c. forms a dodecagon.) Apply the radius of the given circle AO as a chord (fig. 384.) quite round the circumference, and it will form the points thereon of the regular hexagon ABCDEF. 1023. PROB. XXVII. To describe a regular pentagon or hexagon about a circle. In the given circle inscribe a regular polygon of the same name or number of sides as ABCDE (fig. 385.) by one of the foregoing problems; then to all its angular points draw (Prob. 13.) tangents, and these will by their intersections form the circumscribing polygon required D 1024. PROB. XXVIII. To inscribe a circle in a regular polygon. Bisect any two sides of the polygon by the perpendiculars GO, FO (fig. 396.), and their intersection O will be the centre of the inscribed circle, and OG or OF will be the radius. 1025. PROB. XXIX. To describe a circle about a regular polygon. Bisect any two of the angles C and D with the lines CO, DO (fig. 387.), then their intersection O will be the centre of the cir equal to the given quadrilateral ABCD. parallel to it DE, meeting BA produced at Fig. 387. Fig. 388. 1027. PROB. XXXI. To make a triangle egual to a given pentagon ABCDE. Draw DA and DB, and also EF, CG parallel to them (fig. 389.), meeting AB produced at F and G; then draw DF and DG, so shall the triangle DFG be equal to the given pentagon ABCDE. 1028. PROB. XXXII. To make a rect- E angle equal to a given triangle ABC. Bisect the base AB in D (fig 390.), then raise DE and BF perpendicular to AB, and meeting CF parallel to AB at E and F. F 1 Then DF will be the rectangle equal to the given triangle ABC. 1029. PROB. XXXIII. Fig. 389. C Fig. 390. To make a square equal to a given rectangle ABCD. Produce one side AB till BE be equal to the other side BC (fig. 391.). On AE as a diameter describe a circle meeting BC pro duced at F, then will BF be the side of A catenary is a curve formed by a flexible cord or chain suspended by its two extremi ties. Let c, d, in the line A B (fig. 392.) be the two points of suspension, and from them the cord or chain be hung so as to touch the point C the given depth. From this the curve may be traced on the paper. 1031. PROB. XXXV. To draw a cycloid. Any points b (fig. 393.) in the circumference of a circle rolled along a right line AB till such point is again in contact with the said line, generate a cycloid. Let BC be the circle. Then AB is equal to the semi-circumference of such circle, and any chords at whose extremities b, lines ab, ab, equal to the lengths of arcs they cut off, drawn parallel to AB, will furnish the necessary points for forming the curve. D 1032. PROB. XXXVI. To draw a diagonal scale. Let it be of feet, tenths and hundredth parts of a foot. many times as necessary, the number of feet by equal distances. Divide AG into ten equal parts. On AB raise the perpendiculars BD, GG, and AC, and set off on AC ten equal divisions of any convenient length, through which draw horizontal lines. Then, from the point G in DC to the first tenth part from G to A in BA draw a diagonal, and parallel thereto the other diagonals required. The intersections of these diago nals with the horizontal lines give hundredth parts of a foot, inasmuch as each tenth is divided by the diago. nals into ten equal parts in descending. B SECT. III. 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 importanc 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 ir 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, thest 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 are 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. 1097. 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. K H D L Fig. 395. 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 AB. 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. the sine of AB, is the cosine of BD. Thus BF, 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 are 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 quadrant, or 90 degrees. II. When the arc is 0, or nothing, 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 are AB, the versed cosine BK or CF, are together equal to the radius of the circle. tangent AH, and the secant CH, form a right-angled triangle CAII. sine, and cosine form another right-angled triangle CBF or CBK sine AF, and the The radius CA, the Again, the radius, 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 are 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. Tables of this sort are published separately, and we suppose the reader to be provided with them. 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: As the number of degrees or minutes in the semicircle Sines 60 682882 10 30 20 80 20 40 60 50 Secants Tangents 70 30 50 Chords Fig. 396. 40 50 20 10 Now the length of the are being denoted by the letter a, and its sine and cosine by s and e, these two will be expressed by the two following series, viz. — Example 1. Let it be required to find the sine and cosine of one minute. The number of minutes in 180 degrees being 10800, it will be, first, as 10800: 1::3-14159265, &c. : 100290888208665 = the length of an arc of one minute. Hence, in this case, 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 These collected give s=08715574, the sine of 5 degrees. And for the cosine 1 = 1. -a200380771 +a+= 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 are 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 = CB2 – BF2 or c=√1-52. 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. |