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position of A on the IL ; at C draw Ct, at an angle of 45 with the IL and equal to ac, or the projected length of Ab in the HP ; then Ct will be a plan of Ab when at 45 to the VP, and at the angle, b Al) with the HP. To obtain its elevation, draw from, t a projector perpendicular to the IL, and from b another parallel to it, to cut the one from t inp, join C and p, and the line Cp will be the elevation of the original line AB, inclined to both planes of projection. Here it will be noticed that the original line AB, in addition to its having been moved on A as a joint from B to b, has also, while making the angle bAd with the HP, been swung round on A through 45. Now to make this matter of the projection of inclined lines still more clear, as much depends on the student having a thorough grasp of this first part of the subject. We will assume that the two projections, Cp in the VP, and Ct in the HP, are given, and it is required to find the real length of the line of which they are the projections. Here the line Ct is the plan or horizontal projection of the line Cp, 40 FIRST PRINCIPLES OF the latter being a line having one of its ends C, in the HP, and the other end p a given distance above that plane. C/> is also the projected length of the hypothenuse (or longest side) of a right-angled triangle, having Ct for its base, and a line equal to the vertical height of p from the HP for its perpendicular. With these two sides given, we can find the third side, or the actual line of which Cp is the projection. Therefore, at t in the line Ct,, and perpendicular to it, draw a line indefinitely, and from it cut off in h, a length tli equal to the height that p in the line Cp is above the HP or IL, join C and h ; then Ch is the real length of the original