;:., 



F N 



Fig. 34. 



and along D B aet off D Q, eqnal to B. Join o F ; tho triangle 



n tho requirement* aet fortli in 



alao the triangle K o H, obtained by nutting off o n along D T 

 equal to B, and D K along D K equal to c. 



Tho triangloa o D F, K D a are symmetrical and equal in every 

 reapeot; but if tho position of the given angle had been required 

 to be opporite to one of the given aides, instead of being in- 

 diiil.-<l Ix'twoen thorn, a Tory different result would have been 

 obtained. 



ill suppose, firstly, that it ia required to place tho anglo 

 opposite tho shorter of tho two given sides. At tho point L in tho 



straight lino of inde- 

 finite length, x Y, 

 make the anglo x L it 

 eqnal to tho given 

 angle A, and as this 

 angle is to bo opposite 

 to tho shorter aide, 

 Bet off along L x the 

 straight line L N, 

 equal to c ; and from 

 N as a centre, with 

 a radius equal to B, 

 describe the ore o p, cutting the straight line L M in tho points 

 o, p. Join o N and P N. Either of tho triangles o N L, p N L, 

 will satisfy the requirements of the data, for in the triangle 

 O N L the sides O N, N L are equal to B and c respectively, 

 while the angle o L N is opposite to the shorter side o N ; and 

 in the triangle p N L, tho sides P N, N L are equal to B and c 

 respectively, while the angle p L N ia opposite to tho shorter 

 side P N. 



If it bo required to place the angle opposite to the longer of 

 the two given sides, it is manifest that we must set off L Q along 

 ii X equal to B ; and from Q as centre, with a radius equal to c, 

 describe on arc cutting tho straight line L M in R. By joining 

 B Q, we get a triangle, R Q i>, that satisfies the requirements of 

 the data, the sides L Q, Q R being equal to B and c respectively, 

 and the angle Q L R, which is equal to tho angle A, opposite to 

 the longer side R q. 



The learner may make an endless variety of practical exercises 

 on this problem, by varying tho length of the given sides and 

 the opening of the given angle. Practice of this kind will bo 

 found to ensure neatness and accuracy in geometrical or 

 mechanical drawing, and will tend to render tho draughtsman 

 skilful in the use of his compasses and parallel ruler. 



PROBLEM XXVI. To draw a triangle of which one side and 

 two of the angles are given. 



Let A represent the length of tho given side of tho required 

 triangle, and B and c tho given angles, and first let both of tho 



given angles be ad- 

 jacent to the given 

 side, or in other 

 words, let them be 

 at its opposite ex- 

 tremities, on tho 

 same side of it. 



Draw any straight 

 line, x Y, of indefinite 

 length, and in it take 

 D E equal to A. At 

 tho point D make the 

 anglo E D F equal to 



the angle B, and at the point E make the angle i> E F equal to c. 

 Let the sides D F, E F meet in the point P ; tho triangle F D E 

 satisfies the requirements of the data; as will also the trianglo 

 ODE, constructed by making the angle a D E equal to c, and 

 the angle o E D equal to B. 



Next, let one of tho given angles be opposite to the given 

 ride, as, for example, when the anglo eqnal to tho larger angle B 

 is required to bo in this position. Take H K, in the straight 

 line of indefinite length, x Y, and at the point H make the angle 

 K H L equal to the angle c. Through K draw K M parallel to 

 H L, and at tho point K in the straight line M K make tho anglo 

 M K N equal to the anglo B, and let tho straight lino K N meet 

 the straight lino H L in N. The triangle N H K has the anglo 

 K H N equal to c, and tho angle H N K equal to B (for it is equal 

 to its alternate angle N K ar, which was made equal to B), and tho 

 larger angle H N K ia opposite to tho sido u K, which is equal 



to A. If it be required to havo the (mailer angle opposite to 

 the given ido, the angle K H o mnat be made equal to the laifet 

 angle B, and the amo method of construction followed M indi- 

 cate! by the dotted line* in the figure. 



PKOULKM XX VII. To draw a triangle of wktck on* tide, itt 

 A / 



D H 



M* It 



altitude, and one of ilt angle* adjacent to the yiven rids, an 

 given. 



Let A represent tho length of the given side of the required 

 triangle, B its altitude, and c the given angle. Draw any 

 straight line, x Y, of indefinite length, and, by Problem JL 

 (page 192), draw the straight line D E, also of indefinite length, 

 parallel to it, at a distance from it equal to B. Set off r a ia 

 x Y equal to A, and at the point F in the straight line a w make 

 the angle o F H equal vo the given angle c. Let F H meet D K 

 in H. Join o H. The triangle FOB answers the requirement* 

 of the data, for it has a side r a equal to A, an angle o r H 

 equal to o, and it is of the altitude H K, which is equal to the 

 given altitude B. A triangle equal to tho triangle o F H in 

 every respect, and symmetrical with it, may be obtained by 

 making an angle at o, in the straight line F o, equal to c, and 

 following the same process of construction. 



If the given angle be an obtuse angle, as c, the line which 

 represents the altitude of the triangle required will fall on a 

 point in x Y without the line which is set off upon it eqnal t> 

 the given side. If it be an acute angle, as the angle z, the 

 line representing tho altitude of tho triangle may fall between 

 the extremities of the line set off equal to the given aide, a* 

 N o in tho triangle N L M, which is drawn having the aide L x 

 equal to A, and the angle M L N equal to the given angle z ; 

 but whether this be the cose or not depends entirely on the 

 size of the angle and the relative proportions of the altitude and 

 given side. 



In the construction of right-angled triangles, as one angle U 

 always necessarily known, less data are required than in the 

 construction of obtuse-angled and acute-angled triangles ; thu* 

 any right-angled triangle may oe constructed if we know 



1. The length of either of the sides containing the right angle (a* 

 A B and A c in Fig. 37). 



2. The length of either of the sides containing the right angle, and 

 the side which subtends the right angle (a* A B and B c, or A c and B c, 

 in Fig. 37). 



3. The side which subtends the right angle, and the perpendicular 

 let fall on it from the right angle (as A D aiid B c in Fig. 37). 



Thus, if the sides that contain the right angle be eqnal to r 

 and R, draw at right angles to each other A B and AC, and 

 make A B equal to P, and A c equal to K, and join B c : ABO 

 will be the triangle required. 



Again, if one of tho sides containing the right angle be given 



equal to P, and tho side that sub- P .^ 



tends the right angle equal to 8, 



draw B c equal to 8 ; bisect it in 

 E, and from E as centre, with the 

 distance E B or EC, describe the 

 semicircle BAG. Then from B as 

 centre, with a radius equal to P, 

 draw an arc cutting the semicircle 

 BAG in A. Join A B, AC; tho 

 triangle ABC will bo the trianglo 

 required. 



If tho sido which subtends the 

 right angle be given equal to s, and the perpendicular let fall 

 on it from the right angle equal to Q, dw B c equal to a, bisect 

 it in E, and draw tho semicircle B Ac as before; through x draw 

 E P perpendicular to B c, and along it aet off B o eqnal to Q. 

 Through o draw o A parallel to B c, cutting the circumference 

 in A, and from A draw A B, AC, to the points B and C, The 

 altitude, A D, of the triangle A B c is equal to q. 



IV. J7. 



