256 



THE POPULAB EDUCATOB. 



altitude, height, because it shows the height of the top or vertex 

 of the triangle from its base. In Fig. 24 (page 209), c E is the alti- 

 tude of the triangle ABC, and D E the altitude of the triangle 

 A B D. If, then, we have to determine the altitude of an equilateral 



. triangle already drawn, as in 



~ " H c the triangle A B c in the same 



figure, it is manifest that we 

 have only to draw a straight line 

 from the point c perpendicular 

 to the base A B ; or, what is the 

 same thing, bisect the base A B, 

 and join the point of bisection 

 and the point c, which is the top, 

 vertex, or apex of the triangle. 



Fig. 26. 



But to proceed with the problem under consideration. 



Let the straight line A represent the altitude of the equi- 

 lateral triangle required. Draw any straight line B c of indefinite 

 length, and from q, point D, taken as nearly as possible in the 

 centre of the line, draw D E at right angles to B C. Then, along 

 the straight line D E set off D F equal to A. and from D as a 

 centre with the distance D F, describe the semicircle H P G, 

 cutting the straight line B C in the points G and H. Then from 

 G as centre with the distance G D, describe the arc D i, cutting 

 the semicircle H P G in L, and from H as centre with the distance 

 H D, describe the arc D K, cutting the semicircle H p G in M. 

 Through p draw the straight line i K parallel to B c, or, what is 

 the same thing, touching the arcs D I, D K (see Problem X., 

 page 192), and through the points L and M, draw the straight 

 lines D N, D o, meeting I K in N and o. The triangle D N o is 

 an equilateral triangle, having its altitude D p equal to the given 

 altitude A. 



If we join L p, the triangle D L p is an isosceles triangle, 

 having the side D L equal to the side D I 1 As the sides 

 D L, D F are equal, the angles which they subtend, namely, 

 the angles D L p, r> p L, are equal to one another. Now, the 

 third angle, LDP, of the triangle D L F, is an angle of 30 

 degrees, and each of the angles D L F, D F i is therefore equal to 

 180-30 divided by 2, or 150-^ 2 = 75 degrees. 



Again, in the triangle L N p, the angle L N p is an angle of 

 60 degrees ; the angle F L N is equal to 180 75, or 105 degrees, 

 since the angles F L N, p L D are together equal to two right 

 angles, and of these the angle p L D has been shown to be an 

 angle of 75 degrees ; and the angle N F L is equal to 90 75, 

 or 15 degrees, since the angle N F D is a right angle, and L F D 

 an angle of 75 degrees. Its value can also be found by sub- 

 tracting the value of the angles FNL, NLP from 180, thus: 

 180 - (60+ 105) = 180 - 165 = 15 degrees. 



PROBLEM XX. To draw an angle which sJiall contain a given 

 number of degrees. 



Although it is plain, from the preceding problems, that it is 

 possible to draw many angles containing a given number of 

 degrees without the aid of any instruments, except a pair of 

 compasses and a ruler, it is necessary to resort to the protractor 

 or scale of chords in drawing the great majority of angles when 

 the extent of their opening is stated. The protractor has been 

 described already (page 113). The scale of chords will bo found 

 on any "Plane Scale" of boxwood or ivory, sold by mathe- 

 matical instrument makers, and consists of a line graduated 

 ^ or divided in such a manner 



as to show the opening of any 

 angle from 1 degree to 90, in 

 degrees only. The method of 

 using the scale of chords is as 

 follows : 



On any straight line, x T, set 

 off a portion, A B, equal to the 

 opening of an angle of 60 de- 

 grees, as marked on the scale of 

 chords. From the point A as 

 a centre, with A B as radius, 



Fig. 27. 



describe the arc B z. Then, supposing it be required to draw 

 an angle of 40 degrees, apply the compasses to the scale of 

 chords, and open the legs to the extent of 40 degrees, as marked 

 on the scale. From B as a centre, with the radius thus ob- 

 tained, draw an arc, cutting the arc B z in the point c. Join A c ; 

 the angle B A c is an angle of 40 degrees. 



To construct a scale of chtirds, a quadrant of a circle is drawn, 

 and the arc of, the quadrant is divided into ninety equal parts, 



corresponding to the number of degrees in a right angle. Straight 



lines are then drawn from one extremity of the arc to each of 



the points of division, and the length of each line in succession, 



from that which is drawn to the point nearest to the extremity 



of the arc from which the lines are drawn, to that which is 



drawn to the other extremity, is transferred to the scale. The 



radius of any circle, whether large or small, is the chord of an 



angle of 60 degrees ; but the 



learner must bear in mind 



that no chord of an angle of 



60 degrees, except that which 



is marked on his scale, will 



suffice for the length of the 



line A B, as the proportions 



of the chords of the other 



angles of the scale have been 



determined by the aid of 



the quadrant of a circle 



whose radius is equal to the 



SO 45 60 75 00 



Fig. 28. 



chord of an angle of 60 degrees of the length laid down on 

 the scale. 



To render this perfectly intelligible, in Fig. 28 B A c is a 

 quadrant of a circle, and th'e angle B A C is an angle of 90 

 degrees. As it would require an arc of considerable size to 

 divide it clearly into 90 portions of equal size, let us be 

 satisfied with dividing the arc B c into six equal parts in the 

 points D, E, F, G, H. The straight lines drawn from A to each 

 of these points divide the angle BAG into six equal angles of 

 15 degrees each, and the angles BAD, B A E, B A F, BAG, 

 BAH, are respectively angles of 15, 30, 45, 60, and 75 degrees. 

 Draw the lines B D, B E, B F, B G, B H, B c, from the extremity 

 B of the arc B c through the points D, E, F, G, H, c. These 

 lines represent the chords of the angles BAD, BAB, B A F, 

 B A G, B A H, and B A c respectively, or chords of angles of 15, 

 30, 45, 60, 75, and 90 degrees, and by setting off the length of 

 each in due order along any straight line, we construct a scale 

 of chords for angles having these openings, based on the quad- 

 rant of a circle whose radius is equal in extent to the length of 

 the chord of an angle of 60 degrees, as marked on the scale. 



To make an angle greater than 90 degrees by means of the 

 scale of chords, it is only necessary to draw a semicircle instead 

 of a quadrant of a circle, and having set off 90 degrees on the 

 arc, to set off in addition the chord of the number of degrees 

 by which the given angle exceeds 90. Thus, in Fig. 28, to 

 draw an angle of 120 degrees, first draw the semicircle B x, with 

 a radius equal to the chord of an angle of 60 degrees, as marked 

 on the scale. Open the compasses to the whole extent of the 

 scale, and setting one foot on B, with the other draw a small 

 arc, cutting the arc B X in c. Then reduce the opening of the 

 compasses to the extent of the chord of an angle of 30 degrees, 

 and setting one foot on c, with the other cut the arc c x in K. 

 Join A K ; the angle B A K is an angle of 120 degrees, being 

 formed by the angles B A C, G A K, 

 the former of which is an angle of 

 90 degrees, while the latter is one of 

 30 degrees. 



A scale of chords can be readily 

 constructed without drawing lines 

 from one extremity of the arc of the 

 quadrant to every point of section in 

 succession between the extremity from 

 which the chords are drawn and the 

 other extremity. The method which 

 we are now going to bring under the 

 reader's notice has the advantage of 



Fig. 29. 



simplicity ; but in Fig. 28 the actual chords of the angles from 

 15 to 90 degrees are shown in succession, and the angles them- 

 selves that the chords subtend are also shown by straight lines 

 drawn from the point B to the different points of section of the 

 arc. In Fig. 29, having drawn a quadrant of a circle, ABC, 

 as before, join A B, the chord of the right angle A c B, and divide 

 the arc A B into nine (or ninety, if it be large enough) equal 

 parts in the points a, fc, c, d, e, /, g, h. Then, setting one foot of 

 the cpmpasses at A, draw arcs through the points a, b, c, etc., in 

 succession, cutting the straight line A B in the points numbered 

 10, 20, 30, etc. The distances along A B intercepted between the 

 extremity A and each arc in succession are respectively chorda 

 of angles of 10, 20, 30, 40, 50, 60, 70, 80, and 90 degrees, 



