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of them can be obtained by simple geometrical construction, which has fewer chances of error than even measuring from a scale. A few of such angles are 15, 30, 45, 60, 75, 120, 135, etc., and are thus obtained : For 30, bisect 60 ; for 15, bisect 30 ; for 45, bisect 90 ; for 60, use radius ; for 75, add 15 to 60 ; for 120, mark off radius twice ; for 135, take 45 from a semi-circle. With these simple con- structions committed to memory, and the use of a scale of chords for any angle not easily obtained otherwise, the student will be able to lay down any angle that may be required. We may now proceed with the construction of plane figures, taking first Problem 10 (Fig. 53). To construct an equilateral triangle on a given base. (Note : The base of any triangle is that side of it on which it stands ; the vertex, the point immediately over the base ; and the altitude the height of the vertex from the base.) With the given base AB as a radius, and from A and B as centres, describe arcs cutting each other in C, the vertex, join AC and BC, and the triangle is constructed. If the altitude only be given as CD (Fig. 54) : Then, as the sum of the angles of any triangle are together equal to two right angles, or 180, and as the triangle required is equi-angular, the angle at its vertex will be one-third of 180, or 60. To construct it, draw EF, GH through C and D at right angles to CD, and from C, with any convenient radius, describe a semi-circle cutting EF in a and c ; with the same radius, and from a and c as centres, cut the semi-circle in d and e, draw lines through Cd and Ce, and produce them to meet GH in g and h, then gCh is an equilateral triangle having an altitude CD. 28