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and therefore perpendicular to the HP. To find its elevation, through a and c draw projectors perpendicular to the IL, cutting it in A and C ; through A and C with the 60 set-square draw lines intersecting in B, join A, C, and the figure ABC is the required vertical projection of the triangle of which the line a c is the plan. The projection, it will be seen, is an equilateral triangle, for all the sides of the triangle being by the conditions of the problem parallel to the VP, they are projected on that plane into lines of the same length. Problem 23 (Fig. 74). To find the elevation of the triangle obtained in the jrrevious jyroUem when it is inclined 4& to the VP, its plan being the line a c, as before. Here the IL may be considered as a plan of the VP. If we draw a line (with the 45 set-square) of a length equal to a c, at an angle of 42 MECHANICAL AND ENGINEERING THAWING 44 FIRST PRINCIPLES OF 45 with the IL, that line will be a plan of the triangle ABC when at that angle with the VP. Bisect ac in b, through b draw a projector perpendicular to the IL, and from B, in Fig. 73, another parallel to it, cutting the one drawn from b in b'. Then, from a and c draw projectors to the IL, cutting it in a and c, join a'b', ac, c'b' ; the figure a'b'c is the required elevation of the given triangle inclined 45 to the YP. In this case it will be noticed that the elevation obtained is an isosceles triangle, resulting from the altered position of its original with respect to the YP, the plane of its projection. The line ac is bisected in b to find the plan of the vertex B of the triangle ABC, Fig. 73 ; and the vertical projector through this bisection b determines, by its intersection