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1024. Prob. XXVIII. To inscribe a circle in a regular polygon.
Bisect any two sides of the polygon by the perpendiculars GO, FO (fig. 386.), and their intersection ( will be the centre of the inscribed circle, and OG or Of will be the radius.
1025. PROR. 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 circumscribing circle; and OC or OD will be the radius.
1026. PROB. XXX. To make a triangle equal to a given quadrilateral ABCD.
Draw the diagonal AC (fig. 388.), and parallel to it DE, meeting BA produced at E, and join CE; then will the triangle CEB
Fig. 385. be equal to the given quadrilateral ABCD.
1027. PROB. XXXI. To make a triangle equal 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 rectangle 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. Then DF will be the rectangle equal to
Fig. 389. the given triangle ABC.
1029. ProB. XXXIII. 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
А с duced at F, then will BF be the side of the square BFGH equal to the given rectangle BD, as required.
1030. PROB. XXXIV. To draw a catenary, c C d (fig. 392.) A catenary is a curve formed by a flexible
Fig. 392. cord or chain suspended by its two extremities. Let c, d, in the line A B (fig. 392.) be the two points of suspension, and from them let 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. 1032. Prob. XXXVI. To draw u diugonal scale.
Fig 394. Let it be of feet, tenths and hundredth parts of a foot. Set off on AB (fig. 391.) as 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 diagonals with the horizontal lines give hundredth parts of a foot, inasmuch as each tenth is divided by the diago
Fig. 394. nals into ten equal parts in descending.
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 cqual 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 o ; thus, 60% means co degrees; minutes are marked thus': hence, 45' means 45 minutes; seconds are marked thus", 49" ineaning 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 supplemente 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.
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 chiord (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 are. Thus, AF is the versed sine of the arc AB, and EF the versed sine of the arc EDB.
104). 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. 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 are exceeds a quadrant, or 90 degrees. II. When the arc is (, 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 are 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 forin another right-angled triangle CBF or CBK. So also the radius,
20 40 60
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, morcover, 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
Verseul Sines following proportion :
As the number of degrees or minutes in the semicircle
So is 3•14159265, &c. to the length of the said arc. Now the length of the arc being denoted by the letter a, and its sine and cosine by s and c, these two will be expressed by the two following series, viz. —
2.3.4 18.104.22.168.6 a?
al = 1 +
2 22 720 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. : 000290888208665 = the length of an arc of one minute. Hence, in this case,
and fa3 = '000000000004
Also from 1.
take ļa?=0.0000000423079, &c.
leaves c= -9999999577, the cosine of one minute. Example 2. For the sine and cosine of 5 degrees : Here 180° : 50:.3.14159265, &c. : 08726646 =a, the length of 5 degrees.
Hence a= 08726646
-fa3 = .00011076
Th0a5 = '00000004
- 199= 00380771
+ya+= '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 grenter 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=VCB2- BF, or c=v1-5There 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.
1019. Proz. 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. 395.), where BF is the sine, CF or BK thc 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 : 1. 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.
Observation 3. All the cases in trigonometry are comprised in three varieties only; viz. Ist. When a side and its opposite angle are given. 2d. When two sides and the con.
tained angle are given. 3d. When the three sides are given. More than these three varieties there cannot possibly be ; and for cach 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 triangle have the same proportion to each other as the sines of their opposite angles have. That is,
As any one side
To the sine of its opposite angle. For let ABC (fig. 397.) be the proposed triangle, having A B 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.
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
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 supple. ments 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 inay 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,
LA 37° 20':
As the side BC=232
LA= 37 20 37 20
=152 56 101 44
27 04 78 16 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, divi. sion 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'. To find the side AC. As sine LA 370 20
5 27 So sine L B
9.658037 78 16
9.990829 To opp. side AC 174.04
2.573521 Example 2. In the plane triangle ABC,
Let AB = 365 yards,
24 45 Herein two angles are given, whose sum is 81° 57'. Therefore 180°– 810 57'=LC. As sine LC 98° S'
=24911657 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
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.
For let ABC (fig. 399.) be the proposed triangle, having the two given sides AC, BC, including the given anglé c. With the centre C and radius E CA, the less of these two sides, describe a semicircle, meeting the other side of BC produced in E, and the unknown side AB in G. Join AE, CG, and draw DF 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
Fig. 339. 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.