PROBLEMS.

PROBLEM I.

To bisect a given rectilineal angle, that is, to divide it into two equal parts.

Let two right lines be drawn, of any length, containing the given angle. Print (with the lead pencil) any letters at the extremities of these lines, (as BAC in the diagram annexed,) simply that the lines may be easily referred to.

D

1. Having fixed the lead pencil leg in the compasses, place the steel leg very accurately on the vertex A. Gently open the compasses any distance less than the length of the line A B, or A C. With this distance as radius describe an arc, as DE, (with the pencil leg,) cutting the lines A B and A C in the points D and E. Print the letters DE.

B

2. Next place the steel leg carefully on the point D, and with the same radius as before, (or any other greater than half the distance from D to E,) describe an arc as at F. Remove the steel leg to the point E; and with the radius last used, describe another arc, cutting the first arc in the point F.

3. Draw the straight line A F, and it will bisect the angle B A C, as was required to be done; for the angle BAF is equal to the angle CAF.

A similar process to that described in this problem will, with equal accuracy, bisect a curvilineal angle, provided its legs are curved with equal radii.

When the diagrams are drawn in ink, the pencil lines must be rubbed out. The letters of reference may be rubbed out, or inked in, at pleasure.

EXAMPLES FOR PRACTICE.

1. Draw an acute angle, with its vertex downwards, and bisect it.

2. Make any obtuse angle, and bisect it.

3. Draw any two straight lines crossing each other in any direction, and bisect each of the four angles thus formed. If the operations be correct, the lines of bisection of the opposite angles will be in the same straight line.

PROBLEM II.

At a given point in a given straight line, to make an angle equal to a given rectilineal angle.

Let ECD be the given angle; A B the given right line; and A the given point. It is required from A to draw a line which, with B A, shall contain an angle equal to the angle ECD.

1. From the vertex C (in the given angle) as a centre, and with any radius, describe an arc, cutting the legs of the given angle in the points E and D.

C

A

E

FB

2. From the point A, (in the given line,) as a centre, and with the radius CE, describe an arc GF, cutting A B in the point F.

3. Take DE as a radius, and from the point F as a centre, describe an arc, cutting F G in the point G.

4. Draw the right line A G, when the angle GAB will be equal to the angle E CD.

EXAMPLES.

1. Draw an obtuse angle, with its vertex upwards, and make another angle equal to it.

2. Draw an acute angle, and upon each of its sides make another equal to it in opposite directions.

PROBLEM III.

To make an angle that shall contain any number of degrees.

1. Draw a line, as A B in Prob. II.

2. Take 60 degrees in the compasses from any line or scale of chords, (marked CHO, CH, or C,) which may be found on the protractor scale, and from A as a centre, describe the arc GF.

3. Take the number of degrees which the required. angle is to contain, from the same line of chords, and from F as a centre, describe an arc, cutting F G in G.

4. Draw A G, and G A B will be the angle required.

Note. If an angle be required to contain more than_90 degrees, this number must be first applied on the arc FG (produced if necessary), and the remaining number of degrees added; for example, if the angle is to contain 170 degrees, 90 degrees must first be applied to 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 GAB and ECD in Prob. II. contain 25 degrees each.

By means of the scale of chords, the number of degrees in a given angle may be ascertained.

Let ECD (Prob. II.) be the given angle.

1. From the point C, with a radius of 60 degrees, describe an arc, cutting the legs of the angle in D and E.

2. Take the distance from E to D in the compasses, and apply it to the scale, when the number of degrees contained in the given angle will be shown.

Note. If the angle be obtuse, 90 degrees must first be applied to the arc, and the number of degrees in the remainder ascertained; which number added to 90, will give the number of degrees in the given angle.

EXAMPLES.

1. Make an angle to contain 99 degrees, (99°.) 2. Make an angle to contain 105°.

3. Draw two straight lines meeting each other, and determine the number of degrees in the contained angle.

4. Make an angle to contain 80 degrees, and bisect it by Prob. I.

5. Make an angle of 154°, and upon one of its legs make an angle of 90°: show that the remainder contains 64°.

PROBLEM IV.

To make an angle with the protractor which shall contain any given number of degrees.

1. Draw a line, as A. B, and take the point O in it. 2. Place the protractor across the line A B, so that the centre mark (on its blank edge CT) shall coincide with the point O, and the number of degrees required (marked on its graduated edge), reckoned from C, shall meet the line AB.

3. Draw a line along the edge CT, when the angle COA will be the angle sought.

In the figure, the angles CO A, CO B are each right angles. The dotted protractor ct, represented in another position, gives the angle c OA equal to 45 degrees, and its supplement COB 135 degrees. Another method of using the protractor is to place its edge CT on the line A B, its centre on O, and then to look round its graduated edge for the number of degrees required, and mark the paper at that number: a line drawn from this point to the centre O will give the required angle this method is, in general, not so expeditious as that explained before.

Note. Some little difficulty may be experienced (at first) in reading the divisions on the graduated edge. The student is therefore recommended to make several angles with the protractor, containing different numbers of degrees, and to verify them by the second part of Prob. III.

EXAMPLES.

1. Make an angle of

95°.

2. Make an angle of 49°. 3. Make an angle of 135°.

Verify these by the
Scale of Chords.

4. Make contiguous angles of 15°, 20°, 25o, 30°, and show, by the scale of chords, that the total forms a right angle.

PROBLEM V.

From a given point to draw a right line that shall be parallel to a given right line.

Let A be the given point, and BC the given right line.

1. In BC take any point D, and draw the straight line AD.

2. Make the angle FAD equal to the angle ADB (Prob. II.), and produce the line FA towards E: the line EF will be parallel to BC.

E A

F

B

D C

Many cases occur in which it is far more correct to employ this, or the following problem, than to use a parallel ruler.

Note. When a line, as AD, meets two other lines, as BC, EF, such angles as FAD, BD A, or EAD, A DC, are called alternate angles.

PROBLEM VI.

To draw a line parallel to a given line at any given distance.

Let A B be the given straight line.

1. From A and B as centres, with the given distance as radius, describe arcs as at C and D.

2. Draw the line CD touching the arcs, and it shall be parallel to A B.

A