Through the point [m, m'] draw the plane stv parallel to the plane pqr (Prob. IX.): stv is the plane required; for parallel planes have the same inclination to a plane which cuts them (Prop. 40).

PROBLEM XXIII.

Through a given straight line, to draw a plane having a given angle of inclination to the horizontal plane.

When the given line cuts both planes of projection, its traces on these planes will be points in the traces of the plane required; and if another point in either trace can be found, both traces will be determined.

Let [ab, a'b'] (fig. XXIII.) be the given straight line, and LMN the given angle. From a and b' draw aa' and b'b at right angles to xy, so that band a' are the traces of the given line (Prob. I.). Draw LN at right angles to MN; and at the point a' make the angle aa'm' equal to the angle NLM, so that the angle a'm'a is equal to the given angle LMN. From the centre a, at the distance am', describe the circle m'pq; and from b draw bpt touching the circle m'pq, and meeting xy in t. Join a't; a't, tb are the traces of the plane required.

Since aa' is perpendicular to the plane xby, and ap is at right angles to bt (III. 16), the line joining a' and p, in space, is perpendicular to the trace tb (Prop. 16). Consequently the angle formed by this line with pa is the inclination of the plane a'tb to the horizontal plane. The triangle formed by this line with pa and aa' is right-angled at a, as is the triangle a'am'; and pa is equal to m'a, and aa' common to this triangle and the triangle m'aa': therefore the angle formed at p with pa, by the line joining a and p in space, that is, the inclination of the plane a'tb, is equal to the angle a'm'a, that is, to the given angle LMN.

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The angle formed by the line joining a' and p in space, with pa, also be shown to be equal to the angle am'a', by conceiving the plane paa to turn about aa' until it coincides with the vertical plane, when this angle will coincide with a'm'a.

The given angle LMN may be such that am' becomes equal to bb'. In this case, the horizontal trace of the plane required is parallel to the line of level xy, and therefore the vertical trace is also parallel to xy (Prop. 23).

Other particular cases are:

1. When the given line is parallel to the horizontal plane. In this case, its vertical projection a'b' (fig. XXIII. 1) is parallel to xy (35. 2), and the horizontal trace of the plane required will be parallel to the horizontal projection ab (Props. 23, 19) of the line. The only difference in this case is, that the horizontal trace of the plane is found by drawing ut parallel to ab, to touch the circle m'pq. The vertical trace is found as before by joining t and the vertical trace a' of the given line.

2. When the given line is parallel to the vertical plane. In this case, its horizontal projection ab (fig. XXIII. 2) is parallel to xy, and the vertical trace of the plane required will be parallel to the vertical projection a'b' of the given line (Props. 23, 19).

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At any point n', in b'y, draw n'l, making the angle b'n'l equal to the given angle NML, and meeting b'l, perpendicular to xy, in l and through 7 draw is parallel to a'b', meeting xy in s. From as a centre, at the distance b'n', describe the circle n'rq: and from s draw sru touching the circle, and meeting b'b in u. Drawing uw parallel to xy, uw and b'a' will be the horizontal and vertical projections of a line parallel to the given line, and whose horizontal trace u; and from what has been stated, ls and su will be the traces of a plane passing through that line and making an angle with the horizontal plane equal to In'b', that is, to the given angle LN. Therefore drawing bt parallel to us, and through t, tv parallel to ls or a'b'; bt and to are the traces of a plane parallel to the former plane (Prob. IX.), and therefore making an angle with the horizontal plane equal to the given angle, and also passing through the given line whose horizontal trace is b.

3. When the given line is parallel to both planes of projection. In this case, both projections of the given line are parallel to the line of level (35. 2), and both traces of the plane required will also be parallel to this line (Props. 23, 19).

Draw any line pr' (fig. XXIII. 3), perpendicular to xy, and therefore to the projections a'b', ab of the given line, cutting them in a', a and xy in m. In ab take an equal to ma', and in a'b', a'n' equal to ma. At n, draw np, meeting r'p in p, and making the angle anp equal to the angle MLN, and therefore the angle apn equal to the angle LMN; and at n' draw n'r' meeting pr' in r', and making the angle a'n'r' equal to the angle LMN. Through p and r' draw pq, r's parallel to xy: these are the traces of the plane required.

For the plane xa'y being in its vertical position, conceive the triangle a'n'r' to turn about a'r', and at the same time the triangle apn to turn about ap until both these triangles are in the vertical plane passing through ama'. Then an being equal to ma', and a'n' equal to ma, the points n and n' will coincide in the point A of the given line, of which a and a' are the projections; the lines pn, n'r! will form one straight line in the plane whose traces are pq and r's', because the angles on one side of pn and n'r' at the point of meeting are together equal to two right angles; and this plane will pass through the given straight line. And since mp, and the continued line formed by pn and n'r' are both at right angles to the horizontal trace pq, the angle which the plane whose traces are pq, r's' makes with the horizontal plane is the angle apn, that is, is equal to the given angle LMN.

PROBLEM XXIV.

To find the angle made by two given straight lines; that is, the projections of two straight lines being given, to construct the angle formed by these lines in space.

If the horizontal and vertical projections of the two given straight lines do not meet in the same perpendicular to xy, the lines in space do not meet, and the angle which they make with each other is then considered to be the angle formed by one of them with a parallel to the other, meeting the first. It is, therefore, the angle formed by the two given straight lines, when they meet, or the angle formed by one of them and a parallel to the other meeting the first, that we have to construct.

When both the intersecting straight lines meet the horizontal plane, the angle formed by them is the vertical angle of a triangle of which the base is the distance between the horizontal traces of the lines, and the sides are the distances of these traces from the point where the lines intersect in space. It is therefore this angle which we have

to construct.

Let ab and a'b' (fig. XXIV.), ac and a'd' be the projections of the two given lines, or of one of them and the parallel to the other, which meet in the point of which a and a' are the projections. Drawing b'b and c'e perpendicular to xy, b and c are the horizontal traces of the lines; and joining bc, this line with the two given lines in space will form a triangle of which the angle opposite to be is the required angle, formed in space by the two given straight lines at their point of intersection A, of which a and a' are the projections; that is, the required angle is the angle at the point [a, a'] in the triangle b [a, a'] c. Drawing ad perpendicular to bc, the line joining d and the point [aa] or A, in space, will be perpendicular to be (Prop. 16), and will be the hypothenuse of a right-angled triangle of which the base is ad, and perpendicular ma'. Taking, therefore, md" equal to ad, a'd" is the distance from d to the point [a, a'], or the length of the line d [a, a']. Since the plane da [a, a'] is perpendicular to be (Prop. 4), if the triangle b [a, a'] c turn about be until it coincide with the horizontal plane, the line d [a, a'] will come down upon da, the point [aa] coinciding with a point A, in da produced, so that dA, is equal to d"a'; and the angle bAc will be the required angle b [a, ac thus brought down upon the horizontal plane.

In the construction of the triangle of which the vertical angle is required, we have made use of the base and the perpendicular: we might have employed the sides. Since ba, ca are the horizontal projections of the sides of the triangle, and ma' is the height of their point of intersection above the horizontal plane, the sides of the triangle are the hypothenuses of right-angled triangles of which ba, ca are the bases, and ma' is the perpendicular. Taking, therefore,

mb" equal to ab, and me" equal to ac, a'b" and a'c" are the required sides; and if from b and c, as centres, and at distances equal a'b" and a'd' respectively, arcs are described, their intersection A, will be the vertex of the triangle, and the angle bA,c the angle required.

Particular cases.-1. The general construction assumes that the two lines intersect the horizontal plane. If one of the lines [ac, a'd'] (fig. XXIV. 1) be parallel to this plane, its vertical projection a'd' will be parallel to xy (35. 2), and the horizontal trace of of the plane containing the angle will be parallel to the horizontal projection ac of the line (Prop. 23, 19). The angle may still be brought down upon the horizontal plane by means of the perpendicular ad, the only difference from the general construction being that the horizontal side will remain parallel to bf or ac, having the position A‚Ç,.

If both the given lines be parallel to the horizontal plane, it is evident (Props. 23, 20) that the angle formed by their horizontal projections will be equal to that formed by the lines themselves.

2. The general construction also assumes that the angular point is out of the horizontal plane. When that point is in the horizontal plane, the horizontal traces of the two lines [ab, a'b'], [ac, a'c'] are in the same point a. In this case, a straight line [by, by'] (fig. XXIV. 2) being drawn through any point [b, b'] in one of the lines [ba, b'a'] parallel to the other [ca, da'], the angle made by the line [ba, b'a'] with the line [by, b'y'], at the point [b, b'], will be equal to the angle made by it at the point [a, a'], with the line [ca, d'a'] (I. 29). Constructing as in the figure, according to the general case, the angle yA,a is equal to the former of these angles, and consequently to the latter, which is required.

PROBLEM XXV.

To construct the angle made by a given straight line with a given plane.

If from any point in the given straight line a perpendicular be let fall upon the given plane, the angle contained by these lines will be the complement of the angle of inclination of the given line to the plane. We have therefore only to construct the angle contained by these two lines.

Let ab, a'b' (fig. XXV.) be the projections of the given straight line, and ma, m'a the traces of the given plane. Take a and a' in the projections of the given line, and in the same perpendicular to xy, so that they are the projections of the same point; and from them draw ac and a'd' perpendicular to ma and m'a respectively: these are the projections of the perpendicular from the point [a, a'] on the plane (Prob. XV.). Drawing b'b and c'c perpendicular to xy, b and c are the horizontal traces of the given straight line and the perpendicular from a point in it on the given plane. Constructing as in the figure, according to Problem XXIV., bA,c is equal to the

VOL. II.

F

angle contained by the given straight line and the perpendicular to the given plane; and drawing Ah at right angles to cA, bAh is equal to the angle of inclination of that straight line to this plane.

PROBLEM XXVI.

To construct the angle of inclination of two given planes. If a plane be drawn perpendicular to the intersection of the two planes, its traces on these planes will be perpendicular to the intersection (Def. 3), and therefore the angle contained by these traces will be the inclination of the planes (Def. 7). This angle being brought down upon one of the planes of projection will be the angle required.

Let aaa', aßa' (fig. XXVI.) be the two given planes; then a and a' are the traces of their intersection; and a'b being perpendicular to xy, ab is the horizontal projection of this intersection. Draw any line gch perpendicular to ab, as the horizontal trace of a plane perpendicular to the intersection of the two planes, that is, to the line joining ad' in space. The traces of this plane on the given planes are perpendicular to that line, and pass, the one through g, the other through h, forming a triangle of which gh is the base; and the vertical angle of this triangle, being contained by these traces, is the angle which it is required to construct. The vertex of this triangle is in the vertical plane aba; but the line gh is perpendicular to this plane (Def. 6); therefore the line joining c and the vertex of the triangle is perpendicular to gh (Def. 3); and consequently the triangle being brought down upon the horizontal plane, by turning upon gh as an axis, this line will be in the direction ca. We have therefore only to find the true length of this line in order to construct the angle.

The plane of the triangle being perpendicular to the line joining a and a in space, the line joining c and the vertex is perpendicular to this line. Making then the vertical plane aba turn about its vertical trace ab, until ba coincides with by, the points a and c will describe arcs of circles ap, cq, about the centre b, and have the positions p and q; and the intersection of the two given planes will be upon d'p. Drawing therefore qr perpendicular to a'p, this will be the true length of the perpendicular from the vertex to the base gh of the triangle. Consequently taking cN equal to qr, the angle gNh is the angle of inclination of the two planes.