PROBLEM XXVII.

Through a given straight line, to draw a plane that shall have a given angle of inclination to a given plane.

The several cases of this problem are:

1. When the given straight line is in the given plane.

The construction in the last problem immediately furnishes the solution of this, which is its converse.

Let aaa' (fig. XXVII. 1) be the given plane, and a, a' the traces of the given line in it. Draw a'b perpendicular to xy, and join ab; then ab is the horizontal projection of the given line joining a and a' in space. From any point h in aa draw he perpendicular to ab, and produce it if necessary. From the centre b, at the distances ba, be describe circles cutting xy in p, q; join a'p, and from q draw qr perpendicular to it: then from what has been stated in the last problem, qr is the length of the line drawn from c perpendicular to the line joining a, a' in space. Take cN equal to qr; join hN; and at the point N draw Ng to meet he, produced if necessary in g, making the angle hNg equal to the given angle of inclination of the planes. Join ag, and produce it to meet ay in B; and join a3 : the plane aßa' is the plane required. For by the last problem the angle hNg is the angle of inclination of the plane aßa' to the plane aaa'; and the angle hNg being made equal to the given angle of inclination, the plane aßa', passing through the given line whose traces are a, a', has the required inclination to the given plane aaa'.

2. When the given straight line is parallel to the given plane.

In this case, straight lines drawn through the traces of the given straight line, parallel to the traces of the given plane, are the traces of a plane passing through the given line, and parallel to the given plane; and a plane drawn, by the first case, through the given line to have the given angle of inclination to this parallel plane, will also have the given angle of inclination to the given plane (Prop. 40).

It is unnecessary here to go through the construction of this case with a figure, but the student is recommended to do so, as an exercise. 3. When one of the traces of the given straight line is in one of the traces of the given plane.

Let the vertical trace of the given straight line be in the vertical trace of the given plane. If from the horizontal trace of the given line, a perpendicular be drawn to the given plane, and also a straight line making the given angle of inclination with that plane, and a circle be described on that plane from the foot of the perpendicular as a centre, at the distance of the intersection of the oblique line; then a plane passing through any tangent to the circle, and the horizontal trace of the given line, will make the given angle of inclination with the given plane (Prop. 16 and Def. 7). If, therefore,

from the vertical trace of the given line a tangent be drawn to the circle, the plane passing through this tangent and the horizontal trace of the given line will pass through this line, and make the given angle with the given plane. This tangent to the circle thus described on the given plane will be the trace on that plane of the plane required, and therefore to determine this trace we have only to effect the above construction on one of the planes of projection.

Let aaa' (fig. XXVII. 3) be the given plane, [bc, b'd] the given straight line, and LMN the given angle of inclination. Draw cm, c'n' perpendiculars to aa, aa', and determine by Problem XV., p and p' the projections of the foot of the perpendicular from the point [c, d] upon the plane aaa'. The length of the perpendicular is the hypothenuse of a right-angled triangle of which the sides are its projections cp and qp'. Drawing therefore pP perpendicular to cm, and equal to qp', joining cP, cP is the length of the perpendicular from c upon the plane aaa'. Join mP. If now the triangle cPm be conceived to turn about the side cm until its plane is perpendicular to the horizontal plane, Pp, being perpendicular to cm, will, in this position, be perpendicular to the horizontal plane; and since pP is equal to qp', P will be the point of which p and p' are the projections, that is, the foot of the perpendicular from the horizontal trace c of the given line, on the plane aad', in its true position: cP will be the perpendicular in its true position; and mP, being a line meeting cP in the plane, will be perpendicular to it, or make the angle mPc a right angle. Draw cR making the angle PcR equal to MLN, and therefore the angle cRP equal to the given angle of inclination LMN. If therefore in this position of the triangle cPm, a line be drawn in the plane aad, through R perpendicular to Pm, it will also be perpendicular to cR (Prop. 16); consequently a plane passing through this perpendicular and Re will have the required angle of inclination LMN to the plane aaa' (Def. 7).

Produce me to meet xy in e, and draw ee' perpendicular to xy, meeting aa' in e'. Restoring now the triangle mPc to its original position in the horizontal plane, conceive the plane aua' also to be brought down upon that plane by turning about the line aa. The point e will describe a circle about a as a centre; the line of which me is the horizontal projection, being at right angles to aa, will be brought down upon mE, perpendicular to aa, or in em produced, and its intersection E with the circle described by e' will be the position of this point on the horizontal plane; and aE will be the vertical trace of the given plane brought down upon the horizontal plane, by the former plane turning about aa. Also the point U, which is the vertical trace of the given line, will describe a circle and be in the point B, making «B equal to ab'.

From the centre m at the distances mP, mR describe circles intersecting mE in Q, V; then, from the centre Q, at the distance QV,

describe the semicircle VDFS; and from B draw BDT, BFW touching this circle in D, F, and meeting mE in T, W.

Conceive the plane maE to be restored to its true position by revolving about am, so that aE coincides with ae', when the plane exe is likewise in its true position, that is, vertical. Also conceive the plane cmG again to become vertical, by revolving about cm. The points Q, T, W will describe circles about the centre m, and be found at the points P, H, G in the plane cmG, mP being equal to mQ, mH equal to mT, and mG equal to mW. In this position cP, that is cQ, being perpendicular to the plane aaa', that is, maE, QD perpendicular to BT, and QF to BW, the line joining c and D in space will also be perpendicular to BT (Prop. 16), and the line joining c and F in space will be perpendicular to BW. Since the lines joining c and D, c and F make the same angles with the plane aaa' which the line joining c and V, or cR does, it follows from what has already been stated that the plane passing through BT and c, and also that through BW and c will have the angle of inclination to the plane aaa' equal to the angle cRP, that is, to the given angle LMN. BT and BW are therefore the traces on the given plane, of planes having the required angle of inclination to it; and it remains only to determine the horizontal traces of these lines of which the vertical trace is b, in order to obtain the traces of the latter planes.

Since the lines BT, BW are in the plane aaa', their horizontal traces will be in aa, the horizontal trace of that plane; consequently, the plane aua' being again turned down upon the horizontal plane, as aaE, producing BT and WB to meet aa in 7 and o, these are the horizontal traces of the lines BT and BW in space, and are therefore points in the horizontal traces of the planes passing through these lines. Drawing therefore lcß, oyc through c, the horizontal trace of the given line, these are the horizontal traces of the planes having the given angle of inclination LMN to the given plane aaa', the vertical traces of these planes being b'ß, b'y.

4. When the given straight line is not parallel to the given plane, and neither of its traces is in a trace of that plane.

In this case, if a plane be drawn parallel to the given plane, and having its vertical trace passing through that of the given line, then a plane drawn, by the last case, passing through the given line and having the given angle of inclination to this parallel plane, will also have the given inclination to the given plane (Prop. 40).

The construction of this case is left as an exercise for the student. Remarks.-1. The traces of the planes in the third case may be determined by first determining the projections of the lines BT, BW, and thus obtaining their horizontal traces, but the method which has been given is much more simple. The student may, however, adopt that method as an exercise, and also as a verification of the other.

2. The distinction between the two planes which thus make a given dihedral angle with the given plane is this, that the lower dihedral angle which is turned towards the horizontal trace of the given plane is in the one case obtuse, and in the other acute.

3. When PR is equal to QB, the circle described upon the plane aaa', from the centre Q or P, at the distance PR, passes through the point B or b'; the tangents to it drawn from that point are in one straight line, to which QB or Pb' is at right angles; the given line, joining c and b' in space, is itself the perpendicular from c on this tangent to the circle; and therefore the two planes having the required angle of inclination to the plane aaa' coincide. Also, since the given line makes a less angle with the plane aaa', than any straight line drawn from c to a point in that plane nearer to the perpendicular on it than B or b', the angle formed with the plane aaa' by the plane passing through the given line and the line which is at right-angles to QB or Pb', in the plane aaa', is less than the angle formed with that plane by any other plane passing through the given line. The limit of the problem therefore is, that the given angle of inclination shall not be less than that of the given straight line to the given plane.

PROBLEM XXVIII.

To reduce to the horizontal plane, the angle contained by two straight lines.

When, in the survey of a country, the angles which are observed between objects whose positions are to be determined, are not horizontal angles, in order to lay down the positions of those objects in a plan, it is necessary that the observed angles should be reduced to the horizontal plane; that is, that from the observed angle contained by two straight lines drawn from any station to two objects, the angle contained by the intersections of the horizontal plane with two vertical planes passing through the two straight lines, or the angle formed by their horizontal projections, should be determined. In order to do this, it is necessary that, besides the angle contained by the two lines, the angles which these lines make with the vertical should be observed.

If we conceive vertical planes to pass through the lines containing the given angle, their intersection will be perpendicular to a horizontal plane (Prop. 32). The height of the angular point of the given angle above this horizontal plane being assumed, and the angles which the lines containing that angle make with the vertical line passing through the angular point being known, the horizontal distances from this vertical at which these lines meet the horizontal plane will be determined, since they are the bases of right-angled triangles of which the perpendicular and vertical angles are given;

and the lengths of the lines containing the given angle, from the angular point to the horizontal plane, will also be determined, since they are the hypothenuses of the same right-angled triangles. The latter two lines and the given angle contained by them will determine a triangle of which the base is horizontal; and this base with the former two horizontal lines will determine a horizontal triangle, of which the angle opposite to this base is the given angle reduced to the horizontal plane, for it is the angle contained by the intersections of that plane with the vertical planes passing through the lines containing the given angle. On these principles we have the following construction.

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To render the references in this construction clearer, we will call the given angle A; the two lines containing that angle, from the angular point to their intersection with the horizontal plane, L and L'; the angles which these lines make with the vertical, V and V', a the angle A reduced to the horizon; and 7, l' the horizontal lines containing that angle, from the intersection of the vertical through the angular point with the horizon, to the intersection of L and L' with the same plane. Let nm (fig. XXVIII.) be perpendicular to xy, and n be the angular point of the given angle. Make the angle mnp equal to V, and mnq=V'; then np and nq, being the hypothenuses of right-angled triangles of which the perpendicular is the height of n above the horizontal plane, and the vertical angles are V, V', are equal to L and L' respectively; and mp and mq, being the bases of the same right-angled triangles, are equal to 7 and l', the sides which contain the horizontal angle a. At n make the angle pnr equal to A; and from n as a centre, at the distance nq describe a circle cutting nr in r; and join pr: pr is equal to the base of the triangle whose sides are 7 and l', and their included angle is a. To construct this triangle, we have only to describe a circle from m as a centre, at the distance mq, and another from p as a centre, at the distance pr, intersecting the former in s; joining ms, ps; mps is evidently the triangle of which the sides are 1, l', and pr; and pms is the required horizontal angle a included by 7 and l'.

MISCELLANEOUS PROBLEMS.

PROBLEM XXIX.

The position of a point on a given plane being given, to determine its horizontal and vertical projections.

In order that the point may be represented in its true position with reference to the traces of the given plane, it is necessary that this plane should be turned down upon one of the planes of projection, the horizontal for example. Its position on the given plane being then determined by a parallel to the vertical trace thus