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Draw the line AB of any convenient length; from a scale of chords take 60 degrees with the dividers, and with one foot in B describe an arc from the line AB; from the same scale take the given number of degrees, 36, and lay it on the arc, from C to D; draw a line from B through D, and the angle at B will be an angle of 36 degrees.

PROBLEM VII. To make an obtuse E angle, suppose of 110 degrees. Fig. 28.

Fig. 28.

110o

B

Take a chord of 60 degrees as before, and describe an arc greater than a quadrant; set off 90 degrees from B to C, and from C to E set off the excess above 90, which is 20; draw a line from G through E, and the angle will contain 110 degrees.

[It is best, however, in making obtuse angles, to take from the scale the chord of half the angle, and set it off twice. This will save taking two separate chords.]

NOTE. In a similar manner angles may be measured; that is, with a chord of 60 degrees describe an arc on the angular point, and on a scale of chords measure the arc intercepted by the lines forming the angle.

A more convenient method of making and measuring angles is to use a protractor instead of a scale and dividers.

Fig. 29.

B

L

PROBLEM VIII. To make a triangle of three given lines, as BO, BL, LO. Fig. 29, any two of which are greater than the third.

B
L

B

L

Draw the line BL from B to L; from B, with the length of the line BO, describe an arc as at 0; from L, with the length of the line LO, describe another arc to intersect the former; from O draw the lines OB and OL, and BOL will

Fig. 30.

PROBLEM IX. To make a right an

gled triangle, the hypothenuse and angles being given. Fig. 30.

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Suppose the hypothenuse CA 25 rods or chains, the angle at C 35° 30' and consequently the angle at A 54° 30'. See note after the 39th Geometrical Definition.

NOTE. When degrees and minutes are expressed, they are distinguished from each other by a small cypher at the right hand of the degrees, and a dash at the right hand of the minutes; thus, 35° 30′ is 35 degrees and 30 mi

nutes.

Draw the line CB an indefinite length; at C make an angle 35° 30'; through where that number of Degrees cuts the arc draw the line ČA 25 rods, which must be taken from some scale of equal parts; drop a perpendiculur from A to B, and the triangle will be completed.

[A scale of equal parts may be found on one side of Gunter's scale, occupying half its length. It will be known by slanting lines which cross it at each end. The length of the scale, not occupied by these oblique lines, is equally divided into several larger divisions, numbered on one side, and likewise twice as many smaller, numbered on the other. In taking distances from the scale, each of these divisions, (either the larger or the smaller, as is most convenient,) must be considered 1, 10, 100, &c. rods, chains, or other dimensions of length. If each division be called 1, it will be easy to take off the required number. But the scale is not usually long enough for this. When each division is called 10, as many divisions must be taken, as there are tens in the given number. For the excess of tens, in this case, the little scale with the oblique lines, is used. Each side of this little scale is divided into 10 equal parts, and each of these parts is, of course, 1. Then, to take off the hypothenuse, 25, above, we should take in the dividers 2 of the divisions of the large scale, and 5 of those of the small one. There is one of these little scales for the greater, and one for the smaller divisions of the large scale. When each division of the large scale is called 100, each of those of the small one becomes 10, and

are drawn across parallel lines, running the whole length of the scale, from each division on one side, to the next higher on the other. The parallel lines divide the width of the scale into 10 equal parts. Since each oblique line, then, in crossing the scale passes over one division of length, it is evident, that, in passing one tenth across, (that is, to the first parallel line,) it will pass over one tenth of a division of length; in passing two tenths across, (that is, to the second parallel line,) it will pass over two tenths of a division of length, and so on. The parallel lines are numbered at the end of the scale. To take off a distance, containing hundreds, then, as 234, we must place one foot of the dividers on the second division of the larger scale, and on the parallel line marked 4, and extend the other foot to the third oblique line. Decimals may evidently be taken off in a similar manner; the divisions of the larger scale being made units, and those of the smaller, tenths and hundredths.]

NOTE. The length of the two legs may be found by measuring them upon the same scale of equal parts from which the hypothenuse was taken.

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Suppose the angle at C 33° 15', and the leg AC 285. Draw the leg AC making it in length 285; at A erect a perpendicular an indefinite length; at C make an angle of 33° 15'; through where that number of degrees cuts the are, draw a line till it meets the perpendicular at B.

NOTE. If the given line CA should not be so long as the chord of 60°, it may be continued beyond A, for the pose of making the angle.

pur

PROBLEM XI. To make a right angled triangle, the hypothenuse and one leg being given. Fig. 32.

Fig. 32.

40

B

28

Suppose the hypothenuse AC 40, and the leg AB 28. Draw the leg AB in length 28; from B erect a perpendicular an indefinite length; take 40 in the dividers, and setting one foot in A, wherever the other foot strikes the perpendicular will be the point C.

NOTE. When the triangle is constructed, the angles may be measured by a protractor, or by a scale of chords.

PROBLEM XII. To make a right angled triangle, the two legs being given. Fig. 33.

Fig. 33.

A

38

46

Suppose the leg AB 38, and the leg BC 46.

Draw the leg AB in length 38; from B erect a perpendic

ular to C in length 46; and draw a line from A to C.

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Suppose the side BC 98; the angle at B 45° 15', the angle at D 108° 30', consequently the other angle 26° 15'.

Draw the side BC in length 98; on the point B make an angle of 45° 15'; on the point C make an angle of 26° 15', and draw the lines BD and CD.

PROBLEM XIV. To make an oblique angled triangle, two sides and an angle opposite to one of them being given. Fig. 35.

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Suppose the side BC 160, the side BD 79, and the angle at C 29° 9'.

Draw the side BC in length 160; at C make an angle of 29° 9', and draw an indefinite line through where the degrees cut the arc; take 79 in the dividers, and with one foot in B lay the other on the line CD; the point D will be the other angle of the triangle.

PROBLEM XV. To make an D oblique angled triangle, two sides and their contained angle being given. Fig. 36.

Fig. 36.

B

209

Suppose the side BC 109, the side BD 76, and the angle at B 101° 30'.

Draw the side BC in length 109; at B make an angle of 101° 30', and draw the side BD in length 76; draw a line from D to C and it is done.

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Draw the line AB the length of the proposed Square; from B erect a perpendicular to C and make it of the same length as AB; from A and C, with the same distance in the dividers, describe arcs intersecting each other at D, and draw the lines AD and DC.

PROBLEM XVII. To make a rec

tangle. Fig. 38.

Fig. 38.

D

Α'

Draw the line AB equal to the longest side of the rectangle; on B erect a perpendicular the length of the shortest side to C; from C, with the longest side, and from A, with the shortest side, describe arcs intersecting each other at

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