SOLUTION OF PROBLEMS REQUIRING THE USE OF THE INSTRUMENTS THAT HAVE BEEN DESCRIBED. PROBLEM I. At a given point in a given straight line, to erect a perpendicular to the line. 27. Let A be the given point, and BC the given line. From A lay off any two distances AB and AC equal to each other. Then, from the points B and C, as centres, with a radius greater than BA, describe two arcs intersecting each other in D: B draw AD, and it will be the perpendicular required. PROBLEM II. A From a given point without a straight line, to let fall a perpendicular on the line. 28. Let A be the given point and BD the given line. From the point A as a centre, with a radius sufficiently great, describe an arc cutting the line BD in the two points B and D: then mark a point E, equally distant from the points B and D, and draw ᏁᎬ : ᏁᎬ will be the perpendicular required. PROBLEM III. At a point, in a given line, to make an angle equal to a given angle. 29. Let A be the given point, AE the given line, and IKL the given angle. From the vertex K, as a centre, E with any radius, describe the arc IL, terminating in the two sides of the angle. From the point A as a centre, with a distance AE equal to KI, describe the arc ED; then take the chord LI, with which, from the point E as a centre, describe an arc cutting the indefinite arc DE, in D; draw AD, and the angle EAD will be equal to the given angle K. PROBLEM IV. To divide a given angle, or a given arc, into two equal parts. 30. Let C be the given angle, and AEB the arc which measures it. From the points A and B as centres, de- A\ scribe with the same radius two arcs cutting each other in D: through D and the centre C draw CD: the angle ACE will be equal to the angle ECB, and the arc AE to the arc EB. PROBLEM V. E Through a given point to draw a parallel to a given line. 31. Let A be the given point, and BC the given line. From A as a centre, with a radius greater than the shortest distance from B A to BC, describe the indefinite arc ED: from the point E as a centre, with the same radius, describe the arc AF; make ED=AF, and draw AD: then will AD be the parallel required. PROBLEM VI. Two angles of a triangle being given, to find the third. CEH equal to the other: the remaining angle HEF will be the third angle required. PROBLEM VII. To lay down, on paper, a line of a given length, so that any number of its parts shall correspond to the unit of the scale. 33. Suppose that the given line were 75 feet in length, and it were required to draw it on paper, on a scale of 25 feet to The length of the line 75 feet, being divided by 25, will give 3, the number of inches which will represent the line on paper Therefore, draw the indefinite line AB, on which lay off a L A C B distance AC equal to 3 inches: AC will represent the given line of 75 feet drawn to the required scale. REMARK I. This problem explains the manner of laying down a line upon paper, in such a manner that a given number of parts shall correspond to the unit of the scale, whether that unit be an inch or any part of an inch. When the length of the line to be laid down is given, and it has been determined how many parts of it are to be represented on the paper by a distance equal to the unit of the scale, we find the length which is to be taken from the scale by the following RULE. Divide the length of the line by the number of parts which is to be represented by the unit of the scale: the quotient will show the number of parts which is to be taken from the scale. EXAMPLES. 1. If a line of 640 feet in length is to be laid down on paper, on a scale of 40 feet to the inch; what length must be taken from the scale? 40)640(16 inches. 2. If a line of 357 feet is to be laid down on a scale of 68 feet to the unit of the scale, (which we will suppose half an inch), how many parts are to be taken? REMARK II. When the length of a line is given on the paper, and it is required to find the true length of the line which it represents, take the line in the dividers and apply it unit to which it is equal. Then multiply this number by the number of parts which the unit of the scale represents, and the product will be the length of the line. For example, suppose the length of a line drawn on the paper was found to be 3.56 inches, the scale being 40 feet to the inch then, 3.55 X 40 142 feet, the length of the line. PROBLEM VIII. Having given two sides and the included angle of a triangle, to describe the triangle. 34. Let the line B=150 feet, and B C=120 feet, be the given sides; and A=30 degrees, the given angle: to describe the triangle on a scale of 200 feet to the inch. Draw the indefinite line DG, and 120 H at the point D, make the angle GDH equal to 30 degrees; then lay off DG equal to 150, equal to three quarters of an inch, and DH equal to 120, equal to six tenths of an inch, and draw GH: DGH will be the required triangle. PROBLEM IX. The three sides of a triangle being given, to describe the triangle. 35. Let A, B and C, be the sides. Draw DE equal to the side A. From the point D as a centre, with a radius equal to the second side B, describe an arc from E as a centre, with a radius equal to the third side C, describe another arc intersecting the former in D AH BH CH F; draw DF and EF, and DEF will be the triangle PROBLEM X. Having given two sides of a triangle and an angle opposite one of them, to describe the triangle. point H, as a centre, with a radius equal to the other given side B, describe an arc cutting DF in F; draw HF: then will DHF be the required triangle. If the angle C is acute, and A the side B less than A, then the Br arc described from the centre E with the radius EF = B will cut the side DF in two points, F and D G, lying on the same side of D: hence there will be two triangles, DEF, and DEG, either of which will satisfy all the conditions of the problem. PROBLEM XI. The adjacent sides of a parallelogram, with the angle which they contain, being given, to describe the parallelogram. 37. Let A and B be the given sides, and C the given angle. Draw the line DE=A; at the point D D, make the angle EDF=C; take AH F from F, as a centre, with a radius FG=DE, the other from E, as a centre, with a radius_EG=DF; through the point G, where these arcs intersect each other, draw FG, EG; DEGF |