to them, and the angle between these lines is the angle required Let :. a and a', ab and a'b', ac and a'c', be the projections of this point and of the two parallel lines. Find b and c, the intersections of the two lines with the horizontal plane, and join bc; we may consider this line as the base of a A, of which a and a' are the projections of the vertex. Our object is to find the vertical Z, and we must :: construct the A. Draw:: ad 1 to b c, and suppose a line drawn from d to the vertex a a'; this line will be 1 to bc; so that if the A revolve round b c, till it coincides with the horizontal plane, this I will be in the direction of da; and to find its length, we see that it is the hypothenuse of a right-angled A, of which the sides are = a d and a'm; :. on xy, take md' = a d, and a'd' will be this hypothenuse. Then produce da to A, so that da = a'd'; join 6 A, CA, and < b A C will be the < required. We have in the construction of the A used its altitude ; we might have employed its sides. For the horizontal projections of the sides are ab and a c, their vertical are a'b' and a'd'; then make m 6" = ab, mo"= ac; draw the hypothenuse a'l", a'c" ; these are the lengths of the lines. With centres b and c, and with radii equal to a'b”, a'c”, describe two arcs ; their intersection determines the vertex A, and consequently the < b AC. im dc) PROBLEM 19. To bisect the angle between two giren lines. 33. Let ab, a'b', a c, a'c', be the projections of the two lines. (See last fig. but one.) 1o. As in the preceding proposition, construct the bac between the lines, and draw af bisecting this angle. It is manifest that, if the a b ac be raised to its first position, the line af will still bisect the angle: we must :: find the projections of af. . Now, during the motion of the A, the point f, where the line cuts b c, does not alter its position ; it is :. its horizontal projection ; and if ff' be drawn 1 to xy, f' will be its vertical projection. One of the projections :: of the required line should pass through f, and the other through f': but they ought also to pass through a and a'; :: af and a' f' are the projections of the line. If, instead of bisecting the angle, it be required to divide it into two parts, having a given ratio to each other, we must draw af so that 2 bac may be similarly divided. The rest of the construction remains the same. PROBLEM 20. 34. It is obvious that, if from any point in the line we draw a t to the plane, the 2 between these two lines is the complement of the required angle. We have :: merely to construct the < between these two lines. Let Im, im', and ab, a'b', be respectively the traces of the plane and the projections of the line. From a draw aa' 1 to wy; draw ac, a'c' I to Im and i m'; then ac and a'ć' will be the projections of a + from a point in the line on the plane. Then construct the 2 between this - and the given line; and the < bac thus found will be the complement of the angle required ; then draw ah 1 to Ac, and the required Z will be bah. PROBLEM 21. To construct the angle between two planes. 35. Let aaa', aßa' be the two planes; find ab, the horizontal projection of their intersection : suppose them to be cut by a plane + to their intersection, the traces of this plane are :: 1 to the projections of that line (Prob. 14); draw :: gch + ab, gch may then be taken for the horizontal trace of the plane 1 to the intersection. This plane cuts the given planes in two lines, which pass, the one through g and the other through h, and which make with gh a A, in which the L opposite to gh is the < required. For the < between two planes is measured by the z between two lines drawn from any the same point of the common intersection of the two planes, and + to it, one in one plane, the other in the other. The vertex of this A is in the vertical plane aba'; now this plane is +gh : gh is tab; :the line from c to the vertex of the A is + gh; :: by being made to coincide with the horizontal plane, it lies in the direction of cb. • To find its length : since the plane of the A is 1 to the intersection of the given planes, and the line required is in the plane of the A, it is :: 1 to this intersection. Let the vertical plane aba', revolve round its vertical trace ba', until ba coincides with xy. The points a and c will describe the arcs ap and cq, round b, and will come to p and q. Then the intersection of the given planes will be in a'p, and the required distance will be the line qr, 1 to a'p: make cn in a b, therefore, = qr, draw the lines ng, nh, and Zhng is the < required. CURVE SURFACES AND TANGENT PLANES. 36. In what has preceded, the required points have been determined by the intersection of planes, and the questions regarding them have been considered as solved when the positions of these planes are determined. We now proceed to treat of a different class of problems, which are dependent upon curve surfaces; it is therefore necessary to make some remarks upon these surfaces. We know that when a curve is traced upon a plane, it may be considered as a series of points connected together and determined by a law, which furnishes what the geometers call the equation to the curve: thus, the circle furnishes us with an instance of this law, since every point in it is at the same distance from one point, called the centre of the circle. This property gives us both a mechanical method of describing the circle, and a geometrical relation by which its equation may be found. Now, surfaces differ from curves in this respect : that their respective points are in space, and are not limited in position to the same and constant plane, but extend in all directions. Still these points are subject to a law, which particularizes and determines the surface in question. The curves just spoken of being always in the same plane are called plane curves. There are curves called curves of double curvature, which are not in the same plane: such is the helix, of which the corkscrew is a familiar instance. We might trace such a curve by the motion of a point, and thus say generally, that curves are |