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This instrument may be made either of wood or metal. Fig. 1 represents it applied to, and trisecting the angle н C в, and fig. 2 represents it shut up. The pieces H C, E C, F C, and GC, are all of the same length, and moveable on the joint c. The joints a, b, c, and d, are all equally distant from c. The connecting pieces a e, e b, bh, hc, c i, and i d, are all equal; and the pieces ef, fh, h g, and g i, are equal to each other, but longer than the preceding pieces. Two sockets, ƒ and g, fit, and move up or down on the pieces E C and F C. The pieces are all connected by pivots at the joints, represented by the small letters a, b, c, d, &c., and the connecting pieces fit in between the other when the apparatus is shut.

In applying this instrument it is only necessary to lay the centre c on the vertex of the given angle, c &, on one of the sides forming it, and to move н c till it coincides with the other; then each of the angles, H C E, E C F, and F C G, will be a third of the angle H C G. For it is manifest, that the angle H C E cannot be increased without increasing the angle a e b, and that a e b cannot be increased without diminishing the angle bef and the distance fb. But because be is equal to bh, fe to fh, and fb common to the two triangles ƒ e b and f h b; the angle fh b must be always equal to the angle fe b, and consequently b hc to a e b; therefore н CE must in all positions of the apparatus, continue equal to E C F. In the same manner it might be shown that the angles E C F and F C G will always continue equal. Hence the angle H C B has been trisected by the straight lines E C and F C. If the instrument had been applied to the angle D C B, it would have taken the position represented by the dotted lines.

Note. It is evident that instruments may be made on the same principle to divide an angle into any other number of equal parts.

PROB. 12. To cut off from a given line ▲ B, supposed to be very short, any proportional part.

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From the ends A From A to D set off

Suppose, for example, it were required to find the &c. of the line A в in the first figure below. and B draw A D B C perpendicular to A B. any opening of the compasses 12 times, and the same from B to c. Through the divisions 1, 2, 3, &c. draw lines 1 f, 2 g, &c. parallel to A B. Draw the diagonal A c, and 1 d will.be the of A B; 2 c,, and so on. The same method is applica

ble to any other part of a given line.

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PROB 13. To make a diagonal scale, say of feet, inches, and tenths of an inch.

Draw an indefinite line A B, on which set off from A to в the given length for one foot, any required number of times. From the divisions A, C, H, в, draw A D, C E, &c. perpendicular to a B. On A D and B F set off any length ten times; through these divisions draw lines parallel to A B. Divide A c and DE into 12 equal parts, each of which will be one inch. Draw the lines A 1, G 2, &c. and they will form the scale required; viz. each of the larger divisions from E to 1, 1 to 2, &c. will represent a foot; each of the twelve divisions between D and E an inch; and the several perpendiculars parallel to R C in the triangle E C R, To, ToT, &c. of an inch.

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Note. If the scale be meant to represent feet, or any other unit, and tenths and hundredths, then D E must be divided into ten instead of twelve equal parts.

PROB. 14. Given the side of a regular polygon of any number of sides, to find the radius of the circle in which it may be inscribed.

Multiply the given side of the polygon by the number which stands opposite the given number of sides in the column entitled radius of circum. circle; the product will be the radius required.

Thus, suppose the polygon was to be an octagon, and each side 12, then 1.3065628 x 12=15.6687536 would be the radius sought. Take 15.67 as a radius from a diagonal scale, describe a circle, and from the same scale, taking off 12, it may be applied as the side of an octagon in that circle.

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PROB. 15. Given the radius of a circle to find the side of regular polygon (sides not exceeding 12) inscribed in it. Multiply the given radius by the number in the column entitled factors for sides, standing opposite the number of the proposed polygon; the product is the side required.

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Thus, suppose the radius of the circle to be 5, then 5×1.732051=8.66025, will be the side of the inscribed equilateral triangle.

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PROB. 16. To reduce a rectilinear figure of 6, 7, or more sides, to a triangle of equal area.

This is a very useful problem, as it saves much labour in computation.

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Suppose A B C D E F G to be the proposed space to be reduced to a triangle. Lay a parallel ruler from A to c, and move it until it pass through B, marking the point 1 in which it cuts A G continued. Then lay the ruler through 1 and D, and move it until it pass through c, and mark the point 2 where it cuts AG. Next lay the ruler through 2 and F, move it up till it pass through D, marking the point 3 where it cuts A G continued. Again, lay the ruler through 3 and F, move it up until it pass through E, the point of intersection with G A produced.

1

A

2.

G

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C

F

E

and mark 4, Lastly, draw

the right line 4 F; so shall the triangle 4 F G be equal in area to the irregular polygon A B C D E F G. Here B1 is parallel to a c; so that if c 1 were drawn, the triangle a 1 c would be equal to A C B and by the mechanical process this reduction is effected. In like manner, the other triangles are referred, one by one, to equal triangles, having their bases on G A or its prolongation. Hence the principle of the reduction is obvious.

PROB. 17. To reduce a simple rectilinear figure to a similar one upon either a smaller or a larger scale.

Pitch upon a point p any where about the given figure A B C D E, either within it, or without it, or in one side or angle; but near the middle is best. From that point p draw lines through all the angles; upon one of which take P a to PA in the proposed proportion of the

a

B

b

E

D

scales, or linear dimensions; then draw a b parallel to a в, b c to в c, &c.; so shall a b c d e be the reduced figures sought, either greater or smaller than the original. (Hutton's Mens.)

Otherwise to reduce a Figure by a Scale.-Measure all the sides and diagonals of the figure, as A B C D E, by a scale; and lay down the same measures respectively from another scale, in the proportion required.

To reduce a Map, Design, or Figure, by Squares.-Divide the original into a number of little squares, and divide a fresh paper, of the dimensions required, into the same number of other squares, either greater or smaller, as required. This done, in every square of the second figure, draw what is found in the corresponding square of the first or original figure..

The cross lines forming these squares may be drawn with a pencil, and rubbed out again after the work is finished. But a more ready and convenient way, especially when such reductions are often wanted, would be to keep always at hand frames of squares ready made, of several sizes; for by only just laying them down upon the papers, the corresponding parts may be readily copied. These frames may be made of four stiff or inflexible bars, strung across with horse hairs, or fine catgut.

When figures are rather complex, the reduction to a different scale will be best accomplished by means of such an instrument as Professor Wallace's Eidograph, or by means of a Pantograph, an instrument which is now considerably improved by simply changing the place of the fulcrum. See the Mechanics' Oracle, part II. page 33.

CHAPTER IV.

TRIGONOMETRY.

SECTION I-Plane Trigonometry.

1. Plane Trigonometry is that branch of mathematics by which we learn how to determine or compute three of the six parts of a plane, or rectilinear triangle, from the other three, when that is possible.

The determination of the mutual relation of the sines, tangents, secants, &c. of the sums, differences, multiples, &c. of arcs or angles; or the investigation of the connected formulæ, is, also, usually classed under plane trigonometry.

2. Let A C B be a rectilinear angle: if about c as a centre, with any radius, c A, a circle be described, intersecting c A, C B, in A, B, the arc A B is called the measure of the angle A c B. (See the next figure.)

3. The circumference of a circle is supposed to be divided or to be divisible into 360 equal parts, called degrees; each degree into 60 equal parts, called minutes; each of these into 60 equal parts, called seconds; and so on to the minutest possible subdivisions. Of these, the first is indicated by a small circle, the second by a single accent, the third by a double accent, &c. Thus, 47° 18' 34" 45"", denotes 47 degrees, 18 minutes, 34 seconds, and 45 thirds. So many degrees, minutes, seconds, &c. as are contained in any arc, of so many degrees, minutes, seconds, &c. is the angle of which that arc is the measure said to be. Thus, since a quadrant, or quarter of a circle, contains 90 degrees, and a quadrantal arc is the measure of a right angle, a right angle is said to be one of 90 degrees.

4. The complement of an arc is its difference from a quadrant; and the complement of an angle is its difference from a right angle.

5. The supplement of an arc is its difference from a semicircle, and the supplement of an angle is its difference from two right angles.

6. The sine of an arc is a perpendicular let fall from one extremity upon a diameter passing through the other.

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