The angle C is the angle contained by the two lines PM and Mp; so that to find C it is only necessary to construct the right-angled triangle P,Mp, having the hypotenuse MP=MP1=MP.

The angle A is found similarly by constructing the rightangled triangle P ̧Np, having P ̧Ñ='NP2=NP.

The angle B is the angle contained by PQ and PR, which are at right angles to PS. But these two lines are equal respectively to PQ and PR, at right angles to SP, and SP

Let the triangle QPR be constructed having the side QPQP, and RP ̧= RP,

The angle QP ̧R= B.

2°

It is obvious that pPpP,; also when the triangle QP ̧R is constructed P, should be on the line Sp.

5

PROBLEM II. Fig. 73.

Given two faces of a trihedral angle and the dihedral angle contained by them, to find the third face and the other dihedral angles.

Let a and ẞ be the two given faces and C their inclination to one another; it is required to find y, A and B.

Having laid down the two angles a and B as in the last problem, from the point P, draw Pp at right angles to SM, and at the point M make the angle pMP, the given angle C; also make MP=MP1, and from P, draw Pop at right angles to Mp.

2

=

Sp is the projection of the third edge of the solid angle; for p is the projection of the point P in the line SP when the two faces are inclined to one another at the angle C.

To find the third face; draw pN at right angles to SN, and from the centre S describe a circle with the radius SP,, cutting PN produced in the point P. NSP, is then the rabatment of the angle whose projection is NSp: therefore NSP1 = y.

A and B may now be found as in the last problem.

PROBLEM III. Fig. 74.

Given two faces of a trihedral angle and the dihedral angle opposite one of them, to find the third face and the other two angles.

Let a and ẞ be the two given faces and A the given dihedral angle,

Having drawn the angles a and ß in the horizontal plane, take a vertical plane at right angles to their common arm SM; that is, take PF at right angles to SM for a ground line. Now SF being the horizontal trace of a plane, find the vertical trace of the plane when it is inclined at the angle A. That is draw eD at right angles to SF, make eD1 = eD, and the angle eDE=A. SFE is a plane inclined at the angle A to the horizontal face B. (See Prob. XIX. Chap. II.)

If the face a revolve about SM till SP, lies in the plane. SFE, it is evident a trihedral angle would be formed having the faces 2 and B and the angle A.

But as the triangle SMP, revolves about SM, P, describes a circle on the vertical plane, since it is at right angles to SM. Therefore the point P where that circle meets FE is the vertical trace of the third edge of the trihedral angle.

Sp is the horizontal projection of that edge; and P ̧ found as in the last problem. P,SN=7.

is

As the circle P,PQ cuts FE in two points there are two solutions to the problem; that is, two trihedral angles may be formed from the given data. In the second case Q,SF is the third face.

If the circle touched EF there would be but one solution, and if the line and circle did not intersect the solution would be impossible-no trihedral angle could be formed from the given data. This problem corresponds with what is known as the ambiguous case in the solution of plane triangles.

THEOREM. Fig. 75.

If from any point P within the trihedral angle S, perpendiculars Pp, Pq, Pt are drawn to the faces, a trihedral angle is formed at P such that its faces are the supplements of the dihedral angles of S and its dihedral angles the supplements of the faces of S.

Proof. The plane PpNt which contains the perpendiculars Pp and Pt is perpendicular to the two faces which intersect in SN...(Theor. IV.); and therefore at right angles to SN...(Theor. VI.).

Therefore the dihedral angle between the two faces whose common section is SN is measured by the angle pNt. But in the quadrilateral PpNt, the angles at p and t are right angles; therefore pPt and pNt are supplementary.

In the same way it may be proved that the dihedral angles at M and R are the supplements of the angles pPq and tPq, respectively.

Again, as the edges SN, SM, SR are respectively perpendicular to the three faces of the trihedral angles at P, from what has just been proved the dihedral angles of P are the supplements of the faces of S.

The angles S and P are called supplementary.

By means of the principle established in this Theorem the last three cases of the solution of the trihedral angle may be reduced to the first three. When the three angles A, B and C are given, 180A, 180 - B, 180-C are the faces of the supplementary angle. Having found as in the first case the inclinations of the faces of this supplementary angle which may be called A', B, C; then 180-A', 180-B, 180 C'are the faces of the trihedral angle whose inclinations are A, B and C.

Similarly, the fifth case resolves itself into the second and the sixth into the third.