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must

should be considered that in general the arc

3

have the same cosine as some one of the three arcs,

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for the number

m

must be either of these forms n, n + 1, or

3

n+, where n is an integer. If it have the form n, that is, if 3 measures m, then

2m

2m

•±}s=2n+}a; therefore

cos ( rija)=cos (2n+A)=cos A.

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"If it have the form n+};

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2m

3

~±ja=2n+}; therefore

(2±A)=cos (2nx+f«±‡a)=cos j(2«±ˆ). "If it have the form n+};

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2m

COS (A)=cos (2n +1A)=cos (4% ±a).

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2mx A), whatever be

"And hence it follows that the cos 3

the value of m, must be equal to one or other of the quanti. ties.

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Cos. (4-A),

which correspond to the three roots of the cubic equation already found."

Ex. 2. Given the side of a cube, to find the side of another of double capacity.

Let the side of the given cube be a, and that of a double one y, then 2a3=y', or, by putting 2a=b, it will be a2b=y3 : there are therefore to be found two mean proportionals between the side of the cube and twice that side, and the first of those mean proportionals will be the side of the double cube. Now these may be readily found by means of two parabolas: thus:

Let the right lines AR, AS, be joined at right angles; and a parabola AмH be described about the axis AR, with the parameter a; and another parabola AMI about the axis As,with the Then AP, PM=Y, parameter b; cutting the former in M.

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are the two mean proportionals, of which y is the side of the double cube required.

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For, in the parabola AMH the equationis yax, and in the parabola AMI P it is xby. Consequently ay::y :x, and y:x:: x: b. Whence yx = ab; or, by substitution, yby=ab, or, by squaring, y3b: ab; or lastly, y=a2b2a3, as it ought to be.

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GENERAL SCHOLIUM.

On the Construction of Geometrical Problems.

Problems in Plane Geometry are solved either by means of the modern or algebraical analysis, or of the ancient or geometrical analysis. Of the former, some specimens are given in the Application of Algebra to Geometry, in the first volume of this Course. Of the latter, we here present a few examples, premising a brief account of this kind of analysis.

Geometrical analysis is the way by which we proceed from the thing demanded, granted for the moment, till we have connected it by a series of consequences with something an. teriorly, known, or placed it among the number of principles known to be true.

Analysis may be distinguished into two kinds. In the one, which is named by Pappus contemplative, it is proposed to ascertain the truth or the falsehood of a proposition advanced; the other is referred to the solution of problems, or to the investigation of unknown truths. In the first we assume as true, or as previously existing, the subject of the proposition advanced, and proceed by the consequences of the hypothesis to something known; and if the result be thus found true, the proposition advanced is likewise true. The direct de. monstration is afterwards formed, by taking up again, in an inverted order, the several parts of the analysis. If the consequence at which we arrive in the last place is found false, we thence conclude that the proposition analysed is also false. When a problem is under consideration, we first suppose it resolved, and then pursue the consequences thence derived till we come to something known. If the ultimate result thus obtained be comprised in what the geometers call data, the question proposed may be resolved: the demonstration (or rather the construction) is also constituted by taking the

parts of the analysis in an inverted order. The impossibility of the last result of the analysis will prove evidently, in this case as well as in the former, that of the thing required.

In illustration of those remarks take the following examples.

Ex. 1. It is required to draw, in a given segment of a circle, from the extremes of the base A and B, two lines AC, BC, meeting at a point c in the circumference, such that they shall have to each other a given ratio, viz. that of м to N.

Analysis. Suppose that the thing is af fected, that is to say, that AC: CB:: M: N, and let the base AB of the segment be cut A in the same ratio in the point E. Then EC,

being drawn, will bisect the angle ACB (by th. 83 Geom.); consequently, if the circle be completed, and CE be produced to

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meet it in F, the remaining circumference will also be bisected. in F, or have FAFB, because those arcs are the double measures of equal angles; therefore the point F, as well as E, being given, the point c is also given.

Construction. Let the given base of the segment AB be cut in the point E in the assigned ratio of м to N, and complete the circle; bisect the remaining circumference in F; join FE, and produce it till it meet the circumference in c: then drawing CA, CB, the thing is done.

Demonstration. Since the arc FA = the arc FB, the angle ACF= angle BCF, by theor. 49 Geom.; therefore AC: CB :: AE EB, by th. 83. But AE: EB: MN, by construction; therefore AC: CB: M: N. Q. E. D.

Ex. 2. From a given circle to cut off an arc, such that the sum of m times the sine, and n times the versed sine, may be equal to a given line.

P

A

E

D

Analysis. Suppose it done, and that AEE'B is the given circle, BE'E the required arc, ED its sine, BD its versed sine; in DA (produced if necessary) take BP and nth part of the given sum; join PE, and produce it to meet BF to AB or to ED, in the point F. Then, since M. ED+N. BD=n. BP = n . PD † n . BD; consequently m. ED=n. PD; hence PD: ED :: mn. But PD: ED: (by sim. tri.) PB PB: BF: m : n. Now PB is given, therefore BF is given in magnitude, and, being at right angles to PB, is also given in position; therefore the point r is given and consequently PF given in position; and therefore the point E, its intersection

E.

B

FM

BF; therefore

with the circumference of the circle AEE'B, or the arc BE is given. Hence the following

Construction. From B, the extremity of any diameter AB of the given circle, draw BM at right angles to AB; in AB (produced if necessary) take BP an nth part of the given sum; and on Bм take BF So that BF: BP: n: m. Join PF, meeting the circumference of the circle in E and E', and BE or BE' is the arc required.

Demonstration.

From the points E and E', draw ED and E'D' at right angles to AP. Then, since BF: BP ::n: m, and (by sim. tri.) BF: Bp :: de: DP; therefore DE : DP ::n : m. Hence m . DE=n. DP; add to each n . BD, then will m . DE +n. BD=n. BD + a. DP = n. PB, or the given sum.

Ex. 3. In the given triangle ABH, to inscribe another triangle abc, similar to a given one, having one of its sides parallel to a line man given by position, and the angular points a, b, c, situate in the sides AB, BH, AH, of the triangle ABB respectively.

Analysis. Suppose the thing done, and that abc is inscribed as required. Through any point c in вн draw CD parallel to men or to ab, and cutting AB in D; draw CE parallel to be, and DE to ac, intersecting each other in E. The triangles DEC, acb, are similar, and DC: ab

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C

E

B

A

a

D

CE: bc; also BDC, Bab, are similar, and Dc : ab :: BC: Bb. Therefore BC: CE :: Bb: bc; and they are about equal angles, consequently B, E, C, are in a right line. Construction. From any point c in вH, draw CD parallel to nm; on CD constitute a triangle CDE similar to the given one; and through its angles E draw BE, which produce till it cuts AH in c through c draw ca parallel to ED and cb paral. lel to Ec; join ab, then abc is the triangle required, having its side ab parallel to mn, and being similar to the given triangle.

Demonstration. For, because of the parallel lines ac, DE, and cb, Ec, the quadrilaterals BDEC and Bacb, are similar; and therefore the proportional lines DC, ab, cutting off equal angles BDC, Bab; BCD, вba; must make the angles EDC, ECD, respectively equal to the angles cab, cba; while ab is parallel to DC, which is parallel to men, by construction.

Ex. 4. Given, in a plane triangle, the vertical angle, the perpendicular, and the rectangle of the segments of the base made by that perpendicular; to construct the triangle.

E

G

D

B

Analysis. Suppose ABC the triangle required, BD the given perpendicular to the base ac, produce it to meet the periphery of the circumscribing circle ABCH, whose centre is o, in H; then, by th. 61 Geom. the rectangle BD. DH=AD . DC, the given rectangle : hence, since BD is given, DH and вн are given; therefore BIHI is given: as also ID=OE: and the angle Eoc is ABC the given one, because EOC is measured by the arc KC, and ABC by half the arc AKC or by KC. Consequently EC and AC = 2Ec are given. Whence this

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K

H

Construction. Find DH such, that DB DH = the given rectangle, or find DH =

AD DC

BD

; then on any right line GF

take FE the given perpendicular, and EGDH; bisect FG in o, and make EOC the given vertical angle; then will oc cut EC, drawn perpendicular to OE, in c. With centre o and radius oc, describe a circle, cutting ce produced in a : through F parallel to AC draw FB, to cut the circle in в ; join AB, CB, and ABC is the triangle required.

Remark. In a similar manner we may proceed, when it is required to divide a given angle into two parts, the rectangle of whose tangents may be of a given magnitude. See prob. 40, Simpson's Select Exercises.

Note. For other exercises, the student may construct all the problems except the 24th, in the Application of Algebra to Geometry, at page 371, vol. i. And that he may be the better able to trace the relative advantages of the ancient and the modern analysis, it will be advisable that he solve those problems both geometrically and algebraically.

MECHANICS.

Definitions and preliminary Notions.

1. Mechanics is the science of equilibrium and of motion. 2. Every cause which moves, or tends to move a body, is called a force.

3. When the forces that are applied simultaneously to a body, destroy or annihilate each other's effects, then there is equilibrium.

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