called the centre of the protractor. This point is marked by a notch in the diameter. 62. Another instrument, called a square, (fig. 56,) is also used for drawing and measuring right angles. This consists of two rulers fixed together at right angles with one another. Sometimes the ends of these two are connected by a third ruler, thus forming a right-angled triangle. We may ascertain if a square is accurately made in this very simple manner, (fig. 57.) Draw a straight line AB; divide it into 2 parts at the point O. Apply one edge of the square to the part AO, placing the vertex of the right angle at O; then, by drawing a line along the other edge, make the angle AOC. Reverse the square, keeping the vertex of the right angle at O, but applying one edge to the other part of the line AB; viz. to OB; draw a line along the other edge, and the angle BOD will be made. If OC and OD do not coincide, the sides of the square do not form exactly a right angle. The angle COD will be twice the difference between the angle made by the 2 sides and a right angle. 63. Having examined the construction of these instruments, we will now proceed to use them. To make a right angle. Solution 1. With the Protractor. It is required to make a right angle at the point O in the straight line AB, (fig. 48.) Apply the diameter of the protractor to the given line so that the centre shall fall exactly on the point O. Mark on the paper the point where the division 90° on the arc of the protractor falls. Suppose at C; draw the straight line CO. This line will be perpendicular to AB, and will make 2 R. A. with it. Sol. 2. With the Square. Draw straight lines along the 2 sides which make the right angle, and at their point of junction a right angle will be made upon the paper. If one of the sides of the required angle be given, apply one side of the square to this given line, and draw a line along the other side. Sol. 3. With the Compasses, (fig. 48.) Draw a straight line; take in it any point O, and make OA= OB. From A and B as centres with equal radii describe 2 arcs which will intersect at C. Draw CO. The angles COA and COB are right angles. In the solutions of this problem we have solved another problem also; namely, that of erecting a perpendicular to a given straight line; for the line CO is perpendicular to AB. 64. To draw a perpendicular from a given point C to a given straight line AB, (fig. 48.) From the point C, with a radius greater than the shortest distance from C to the line AB, describe an arc which will cut AB at two points, K and L. From each of these points, with the same radius, describe an arc. These arcs will intersect each other at some point; suppose at M. A line drawn from C to a AB, passing through M, will be the perpendicular required. 65. To make an angle the size of which is given in degrees. Sol. Draw a straight line. Apply to it the diameter of the protractor. Mark on the paper the point where the centre of the protractor lies, and also the point where the given number of degrees on the arc of the protractor lies. Connect these two points by a straight line, and the required angle will be made. 66. To make an angle equal to a given angle AOB, (fig. 58.) Sol. 1. With the Protractor. Measure the number of degrees in the given angle, and then make an angle of the same number of degrees. Sol. 2. With the Compasses. Draw the straight line CD. From O as a centre, with any radius, describe an arc which shall intersect the sides of the angle AOB. From C as a centre, with an equal radius, describe an arc, which will intersect the line CD at D. From D as a centre, with a radius equal to the chord AB, describe an arc which will intersect the former at the point E. Draw EC. The angle ECD = = AOB. 67. Through a given point C, to draw a line parallel to a given line AB, (fig. 61.) Sol. 1. From the point C let fall upon AB the perpendicular DC. Upon DC at C erect the perpendicular CF, which is the line required. Sol. 2. Draw from C a straight line meeting AB at any point E. At C make an angle ECF ang. CEB. CF is the line required. = 68. To divide a given angle O into 2, 4, 8, 16, &c. equal angles, (fig. 59.) Sol. From O as a centre, with any radius, describe an arc AB. From A and B as centres with equal radii describe 2 arcs which will intersect at C. Draw CO, and the angle AOB will be divided into 2 equal angles, COA and COB. Repeat this operation with each of the angles COA and COB, and the entire angle at O will be divided into 4 equal angles. By continuing the process it may be divided into 8, 16, 32, &c. equal angles. 69. To make an angle which shall be 2, 3, 4, times, or 2, 3, as great as the angle O, (figs. 59 and 60.) Sol. From the point M, at which the vertex of the new angle is to lie, draw a straight line of any length. From M as a centre, with radius equal to OB, describe an arc, which will intersect the straight line at N. Beginning at N measure on this arc 2, 3, or 4 times the length of AB; for example, NP, PS, &c.; draw SM, and we shall have the angle SMN=2 AOB. Or, divide the arc AB into 2, 3, or more equal parts, and measure one or more of these parts on the arc NS; for example, NR AB; draw RM; the angle RMNAOB. = III. FIGURES. CONSTRUCTION OF TRIANGLES. 70. To construct a triangle which shall be equal to another triangle ABC, (fig. 62.) . 1. Let the 3 sides of ABC be given. Draw a straight line EF equal to one of the given sides, for example, BC. From E as a centre, with radius equal to a second side AB, describe an arc; and from F, with a radius equal to the third side CA, describe another arc which will intersect the former at D. Draw the straight lines DE and DF. DEF is the required triangle. 2. Suppose the side BC and the two adjacent angles B and C to be known. Draw a straight line EF = side BC. At the point E make the angle DEF=CBA, at point F make the angle DFE=ACB. Produce the sides of the angles E and F until they intersect at D. DEF is the required triangle. 3. Suppose the sides BC and BA, and the included angle ABC, to be known. Draw a straight line EF= side BC. At point E make an angle DEF=CBA. Take ED BA, and draw DF. DEF is the triangle required. = That the triangles thus constructed are in each case equal to the given triangle will be demonstrated hereafter (126, 127, 137.) By equal figures we mean figures which coincide entirely when one is laid upon the other; therefore equal figures must be similar. 71. To construct triangles in and about circles. 1. An equilateral triangle, (fig. 63.) Divide the circumference of a circle with the compasses into 3 equal parts. Join the division points by straight lines, and we shall thus construct an equilateral triangle in a circle. In an equilateral triangle constructed about a circle the sides are tangents to the circle at the above mentioned division points of the circumference. To obtain these tangents draw radii to the division points; and upon the ends of these radii erect perpendiculars. Produce these perpendiculars both ways until they intersect one another, and we shall have an equilateral triangle constructed about a circle. 2. A scalene triangle. Divide the circumference of a circle into three unequal parts. Join the division points by straight lines, and we shall thus have constructed a scalene triangle in a circle. Draw tangents to the circle at the division points, and produce such tangents in both directions until they intersect one another, and we have a scalene triangle about a circle. |