Note. If an angle be required to contain more than 90 degrees, this number must be applied on the arc F G produced, and the remaining number of degrees added. For example, if the angle is to contain 170 degrees, 90 degrees must first be applied on the arc, and then 80 degrees more; each quantity to be taken from the same line of chords as the 60 degrees were, with which the arc F G was described. The angles GA B and ECD, in Prob. II. contain 25 degrees each. The angle B A C in Prob. I. contains 72 degrees. EXAMPLES. 1. Make an angle to contain 50 degrees. 2. Make an angle to contain 95 degrees. 3. Make an angle to contain 104 degrees, and bisect it by Prob. I. PROBLEM IV. Any angle being given, to find the number of degrees it contains. 1. Let G AB as in Prob. II. be the given angle. 2. From A as a centre, and with 60 degrees taken from any line of chords as a radius, describe an arc G F, cutting AG, and A B (produced if necessary) in G and F. 3. Apply the distance from G to F on the same line of chords, and the number of degrees thus measured will be the content of the angle G A B. 4. If the arc G F contain more than 90 degrees refer to the note in Prob. III. EXAMPLES. 1. Draw two right lines making any acute angle, and determine the number of degrees it contains. 2. Make any obtuse angle, and decide its contents in degrees. 3. Draw three lines making any triangle; measure the three angles, and see whether the sum is equal to 180 degrees. 4. Extend either side of the above triangle, and see if the exterior angle be equal to the sum of the two interior and opposite angles. PROBLEM V. To make an angle with the Protractor, which shall contain any given number of degrees. 1. Draw a line as CD in Prob. II. 2. Place the protractor on CD, its lower edge coinciding with CD, and its centre with C. 3. Look round the edge of the protractor for the number of degrees required. Mark the paper at that point as E. 4. Join E C, and E C D will be the angle required. The nature of the protractor may be learned in a few minutes, by making any angle by Prob. III. as ECD in Prob. II., and having determined its contents, place the lower edge of the protractor on the line CD, till its centre touches C, and the line CF (produced if necessary) will be seen to coincide with the exact number of degrees (marked on the upper edge or ends of the protractor) which the angle E CD is known to contain. EXAMPLES. 1. Make an angle that shall contain 90 degrees. 2. Make an angle to contain 45 degrees. 3. Make an angle to contain 135 degrees. PROBLEM VI. From a given point, to draw a right line that shall be parallel to a given right line. 1. Let A be the given point, and BC the given right line. 2. From the point A draw a right line AD, cutting BC at any angle with it. 3. Draw FA E, making the angle FAD equal to the angle ADB (Prob. II.) and É A F shall be parallel to B D C. E A B Many cases occur in which it is far more correct to employ this or the following problem, for drawing one line parallel to another, than to use a parallel ruler, or any other instrument. PROBLEM VII. To draw a line parallel to a given line at any given distance. 1. Let B C in Prob. VI. be the given line. 2. From B and C as centres, with the given distance as a radius, describe two circles. 3. Draw a straight line, (either above or below B C) touching both the circles; and it shall be parallel to BC as was required. When this problem is understood, parts of circles only will be necessary. The line EF, in Prob. VI. is nearly half an inch distant from BC. EXAMPLES. 1. Draw a line parallel to and above another line, at 2 inches distance. 2. Draw a line parallel to and below another line at 3 inches distance. 3. Draw two lines, each parallel to another line; the first three quarters of an inch distant from the given line, and the second half an inch from the first, and a quarter of an inch from the given line. PROBLEM VIII. To make a Triangle, whose sides shall be equal to any three given right lines, if either two of them be greater than the third. 1. Let DE, FG, and HI, be the given lines. 2. Draw a line as A B equal to DE. 3. From the point A as a centre, with a radius equal to F G, describe an arc as at C. 4. From the point B as a centre, with HI as a radius, describe another arc, cutting the first arc in the point C. 5. Draw the lines A C, CB; and ACB will be the triangle required. EXAMPLES. Make triangles, whose sides shall be equal to the following dimensions given :— 3. 3 inches, 1 inch, and 3 inches. 4. Base 4 inches, and the other sides, 2 inches and 4 inches. 5. 2 inches, 2 inches, and 2 inches. 6. Make a triangle, each of its sides being 14 inches. The first two of the above examples will be scalene, the next two, isosceles, and the last two, equilateral triangles. PROBLEM IX. Through a given right line to draw a perpendicular that shall bisect it. 1. Let A B be the given line. 2. From A and B as centres, with any one radius greater than half the distance from A to B, describe arcs cutting each other in C and D. 3. Draw the line CD, which shall bisect AB in E, and be also perpendicular to it. Such angles as AEC and BEC, are called Adjacent. The angles A EC and BE D, are Vertical or Opposite. By the above process any given arc may be bisected; for by conceiving A and B as the extremities of the arc, the line CD would cut it into two equal parts; as may be further seen, on reference to the diagrams of Probs. XII. and XXIV. PROBLEM X. At a point in a right line to erect a perpendicular, when the given point is near the middle of the line. 1. Let A B be the given line and C the given point. с 26 2. From C as a centre, with any one radius, describe arcs cutting A B in D and E. 3. From D and E with any one radius, greater than half DE, describe arcs cutting each other in F. 4. Draw the line FC, which shall be perpendicular to A B. PROBLEM XI. At a point in a right line to erect a perpendicular, when the given point is at or near the end of the line. 1. Let C be the given point on the line A B. 2. Take any point above the given line, as D, and from it as a centre, with the radius D C, describe a circle, or arc, cutting A'B in E. 3. Draw a straight line from E, through the centre, cutting the arc in F. 4. Draw the line F C, which will be perpendicular to AB as required. ANOTHER METHOD. AE CB 1. From the point C in the line AB make CE equal to 3 equal parts, taken from any scale, (say a scale of inches or half-inches on a common foot rule.) 2. From C as a centre, with 4 equal parts as a radius, describe an arc as at F. 3. From E as a centre, with 5 equal parts as a radius, describe an arc, cutting the first arc in F. 4. Join FC, which will be perpendicular to A B. By this problem, using a line and peg, or a 20 feet rod, a perpendicular can be laid down upon ground. Timbers also can be put together at right angles. The proportions measured on each simply requiring that they should be multiples of 3, 4, and 5; the latter number being the hypothenuse of a right-angled triangle, of which the former two are the base and perpendicular. |