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21. There are no adjacent angles. They are all vertical, because the lines forming each are produced to form an opposite angle.

22. The pupil is not yet ready to do this in the geometrical way. The teacher should be satisfied if he adds 650 and 25°, and uses the protractor to make an angle of 90°; etc.

24. If the pupil examines a clock, he will see that the number of degrees between 12 and 1 is ta of 360°, or 30°. He has learned already that the length of the sides has nothing to do with the magnitude of the angle.

25. The minute hand goes 90° in a quarter of an hour. The hour hand goes 30° in an hour; 15° in i hr.; 71° in hr.

To ascertain the angle at 12:15, the pupils should draw a clock face, locating the hands properly. Some will place the hour hand at 12, forgetting that it has gone 71° in hour; and will give the answer as 90° instead of the correct one of 90° — 71°, or 82° 30'. At 6:30, the minute hand is at 6, and the hour hand is half way between 6 and 7, or 1 of 30° = 15°. At 8:20, the minute hand is at 4, and the hour hand of the way between 8 and 9— the number of hour spaces being 41, corresponding in degrees to 30° x 41 = 130°.

Pupils should understand that while the angle at 4 o'clock is 30° x 4, or 120°, and while the angle at 5 o'clock is 30° x 5, or 150°, the angle at 7 o'clock is not 30° x 7, or 210°. By making a drawing, they will see that in the last case the angle should be measured on the left, which will make it 30° x 5, or 150°.

NOTE. — Angles of 1800, 2100, etc., may be left for more advanced work.

1260. From 26 should be learned that two lines perpendicular to a third line are parallel to each other; and from 27, that two lines running in the same direction and making the same angle with a third line, are parallel to each other.

29. DE will be drawn parallel to BC by means of the protractor, an angle of 58° being made at the intersection of AB and

DE. Six of the twelve angles will contain 58° each ; the remaining six will each measure 180° — 58° = 122°. The pupils should be permitted to ascertain this for themselves.

The card suggested in the note is to be used in schools in which small wooden triangles are not obtainable.

32. To draw from P, a line parallel to the oblique line AB, place the perpendicular of the triangle on the line AB, Fig. 4, and place the ruler XY against the base of the triangle. Holding the ruler in position, slide the triangle along it until the perpendicular passes through the point P. A line PE drawn along this side of the triangle will be parallel to AB.

Fig. 4. Time should be given the pupils to discover this or a similar method of drawing by means of a ruler and a triangle a line parallel to another line. The method may be made the subject of a composition.

33. Q R and UV are drawn parallel by means of the ruler and the triangle. They may lie in any position, care only being taken to cut them by a line making angles of 50° and 130° with one of them. Three of the remaining six angles will measure 50° each; and the others, 130° each.

1261. 35. If the work is done as it should be, the angle at C will measure 80°.

The line AB does not need to be horizontal ; nor should all the pupils draw AB of the same length.

36. The third angle will measure 60°.

37. There are 68° in a, and 57° in c. In 6, there are 180° - (68° + 57°), or 55°. In d, which is vertical to b, there are 55o.

38. The angle e should measure 28°, and f 120°. There are 180° in e (28°) +9+f (120°).

39. PRQ = 180° — (70° + 60°) = 50°. PRS = 180° – 50° = 130°. PRS is therefore equal to the sum of the angles P and Q.

40. 180°. Ans.
41. 180° — (36° + 65°).

45. In measuring the side of a triangle, use the smallest fraction marked on the ruler. When the ruler in use has the denominations of the metric system on one face, that face should be used, and the length of the line given in millimeters. This will not require any teaching of the metric system beyond showing pupils how to read their rulers, and it will do away with the need of using fractions.

46. The length of each side should be marked. If two of them are not found exactly equal, the construction is faulty.

47. The three sides should be equal.
50. Each of the oblique angles will contain 45°.
52. Angles 2 and 3, 90° each ; 4, 50°; 5 and 6, 40° each.

53. Angles 1 and 4, 671° each ; 2 and 3, 90° each ; 5 and 6, 221° each.

54. The angle p contains 120°; m, 30°; n, 30°.

1266. 56. Construct this parallelogram by drawing two lines of the given lengths, meeting at an angle of 60°. By means of the ruler and the triangle, construct the other two sides.

If the work is properly done, these two sides will measure 2 in. and 3 in., respectively; and the remaining angles will measure 120°, 60°, and 120°, respectively. From this exercise, the pupils should learn that the opposite sides and the opposite angles of a parallelogram are equal, and that the sum of the four angles is 360°. The angles being oblique, the figure is a rhomboid.

The work of the scholars should show the variety suggested in previous exercises. It is not essential that the longer of the two

given sides should be taken as the base, nor that the base should be parallel to the lower edge of the paper.

57. Different pupils will construct this trapezoid in different ways. Some, seeing that one angle is a right angle, will use the triangle to draw the second side, 3 in.; and at the extremity of this side, will draw, by the same means, an indefinite perpendicular line to form the third side, which is parallel to the base. The fourth side is drawn to make an angle of 60° with the base.

The remaining angles will measure 90° and 120°, respectively; and the sides will measure 5 in., 3 in., nearly 34 in., and nearly 31 in.

58. The triangle cut off will form a rhombus when opened out, unless the base and the perpendicular are equal. In this case, the paper, when opened, will form a square.

Making one angle of the triangle 30o (or 40°) will give a rhombus containing 60° (or 80°).

60. When the three sides of a triangle are equal, its three angles are equal; but the rhombus has four equal sides without having equal angles.

61. A triangle that contains three equal angles, has its sides equal; but an oblong has four angles of 90° each, with unequal sides.

62. To construct the rhomboid, draw a base of 21 in. Two inches above, draw a parallel line 21 in. long, with the extremities of the latter on the right or the left of the extremities of the base. If the two remaining sides are exactly equal to each other, the work is correctly done.

It is not necessary to draw the altitude, though a broken line may be used. While different methods may be employed, the use of an incorrect one should not be permitted. The work done on the board by a pupil, should be criticised by the class if it be faulty; or a better way may be suggested, if the one employed by the pupil at the board require too much time or unnecessary work. The lengths of the two remaining sides should be measured, and marked on the papers. Those of different pupils should be different, there being no limit except the size of the paper: they must, however, be longer than 2 in. each; and they should not be just 21 in., which would make the figure a rhombus.

63. If the line that shows the altitude of the rhomboid can be drawn within the figure, a right-angled triangle can be cut from one side and transferred to the other, making the figure a rectangle. Arithmetic, Art. 929, 5, last parallelogram. .

In the case of a rhomboid whose altitude does not fall

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Fig. 5.

within the figure, several cuts will be necessary to change it into a rectangle. See Fig. 5.

65. The areas will be equal, because each rhomboid is equal in area to a rectangle 3 in. by 2 in.

66. The three rhomboids in Fig. 6 have their respective sides equal each to each, but their angles are unequal; hence their altitudes and their areas are unequal.

To construct these rhomboids, draw base lines of three inches,

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and inclined to each, at any angle except one of 90°, a two-inch line. Use the ruler and the triangle to complete the figures.

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