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as will be seen as we proceed. Throughout the problems that will be given in this part of our subject the original object whose projections are required is to b3 always understood as resting on a horizontal plane, and the view first given of it, is its pl:in or horizontal projection, such a A, No. 1 (Fig. 157), which shows all that would be seen of such an object when looked at from above. Now, to find the projection of a point situate in a plane which is inclined to the VP and HP, we must know the angle made by the plane in which the point lies with either the VP or HP. Our first problem is Problem 56. Given the plan of a point, it is required to find its elevation when the plane in which it lies is inclined to tJie HP at an angle of 30. Let a (Fig. 158) be the plan of the given point, IL being the inter- secting line. At O draw a line OP, making with the IL an angle of 30. Through the point a draw a projector into the VP. Set off on OP from O, a distance O^', equal to that which point a is from the IL. Through a draw a line parallel to the IL, cutting the projector from a in plan in point a" ; then a" is the required elevation. In the same way, the elevation of a line lying on a plane inclined to the VP and HP is found, for we have only to find the vertical projection of its ends, and join them by a right line, and the desired elevation is obtained. 122 FIRST PRINCIPLES OF Problem 57. Given the plan of a line, to find its elevation when the plane in which it lies is equally inclined to the VP and HP.