In order to this, make upon a plane the angles B'SA, ASC, B"SC, equal to the angles BSA, ASC, BSC, in the solid figure; take B'S, B'S, each equal to BS in the solid figure; from the points B' and B" let fall B'A and B"C perpendicularly upon SA and SC, which will meet in a point O. From the point A, as a centre, and with the radius AB describe the semicircumference B'bE; at the point O erect upon B'E the perpendicular Ob meeting the circumference in b; join A b, and the angle EAb will be the inclination sought of the two planes ASC, ASB, in the solid angle.

We have only to show that the triangle AOb of the plane figure is equal to the triangle AOB of the solid figure. Now the two triangles B'SA, BSA, are right-angled at A, and the angles at S are equal, consequently the angles at B and B' are also equal. But the hypothenuse SB' is equal to the hypothenuse SB; therefore the triangles are equal; hence SA in the plane figure is equal to SA in the solid figure, also AB' or its equal Ab in the plane figure, is equal to AB in the solid figure. It may be shown, in the same manner, that SC in one figure is equal to SC in the other; whence it follows that the quadrilateral SAOC in one figure is equal to SAOC in the other, and that thus AO in the plane figure is equal to AO in the solid figure; consequently the right-angled triangles AO b, AOB, have their hypothenuses equal and one side of the one equal to one side of the other; they are therefore equal, and the angle EAb, found by the plane construction, is equal to the inclination of the planes SAB, SAC, in the solid angle.

When the point O falls between A and B' in the plane figure, the angle EAb becomes obtuse, and always measures the true inclination of the planes. It is on this account that we have designated the required inclination by EA b, and not by O.A b, in order that the same solution may be adapted to every case with.out exception.

362. Scholium. It may be asked, if, any three plane angles, taken at pleasure, can be made to form a solid angle.

In the first place, it is necessary that the sum of the three given angles should be less than four right angles, otherwise the solid angle could not be formed (357); it is necessary, moreover, that, after having taken two of the angles at pleasure B'SA, ASC, the third CSB" should be such that the perpendicular

BC to the side SC shall meet the diameter B'E between its extremities B' and E. Thus the limits of the magnitude of the angle CSB" are such as require the perpendicular B" C to terminate at the points B' and E. From these points let fall upon CS the perpendiculars B'I, EK, meeting in I and K the circumference described upon the radius SB", and the limits of the angle CSB" will be CSI and CSK.

But, in the isosceles triangle B'SI, the line CS produced being perpendicular to the base B'I,

And, in the isosceles triangle ESK the line SC being perpendicular to EK, the angle CSK CSE. Moreover, on account of the equal triangles ASE, ASB', the angle ASE ASB'; therefore CSE or CSK ASC-ASB'.

Hence we infer that the problem will be possible, while the third angle is less than the sum of the two others ASC, ASB', and greater than their difference, a condition which accords with the theorem art. 356; for, by this theorem, we must have CSB" <ASC+ASB', also ASC< CSB" + ASB', or

CSB" ASC-ASB'.

PROBLEM.

363. Two of the three plane angles, which form a solid angle, being given together with the angle which their planes make with each other, to find the third plane angle.

Solution. Let ASC, ASB' (fig. 198), be the two given plane Fig. 198. angles, and let us suppose, for the present, that CSB" is the third angle sought; then, by constructing the figure as in the preceding problem, the angle contained by the planes of the two first would be EA b. Now, as we determine the angle EA b by means of CSB", the two others being given; so we can determine CSB" by means of EA b, and thus solve the proposed problem.

Having taken SB' of any magnitude at pleasure, let fall upon SA the indefinite perpendicular B'E, make the angle EA b equal to the angle of the two given planes; from the point b, where the side Ab meets the circumference described with the centre A and the radius AB', let fall upon AE the perpendicular b O, and from the point O let fall upon SC the indefinite perpendicular OCB", Geom.

16

which terminate in B" making SB" = SB'; the angle CSB" will be the third plane required.

For, if a solid angle be formed of the three planes B'SA, ASC, * CSB", the inclination of the planes containing the given angles ASB', ASC, will be equal to the given angle EA b.

364. Scholium. If a solid angle is quadruple, or formed by Fig. 199. four plane angles ASB, BSC, CSD, DSA (fig. 199), we cannot, by knowing these angles, determine the mutual inclination of their planes; for with the same plane angles any number of solid angles may be formed. But, if a condition be added, if, for example, the inclination of the two planes ASB, BSC, be given, then the solid angle is entirely determinate, and the inclination of any two of the planes may be found. Suppose a triple solid angle formed by the plane angles ASB, BSC, ASC; the two first angles are given as well as the inclination of their planes; we can then, by the problem just resolved, determine the third angle ASC. Afterward, if we consider the triple solid angle formed by the plane angles ASC, ASD, DSC, these three angles are known; thus the solid angle is entirely determinate. But the quadruple solid angle is formed by the union of the two triple solid angles of which we have been speaking; therefore, since these partial angles are known and determinate, the whole angle will be known and determinate.

The angle of the two planes ASD, DSC, may be found immediately by means of the second partial solid angle. As to the angle of the two planes BSC, CSD, it is necessary in one of the partial solid angles to find the angle comprehended between the two planes ASC, DSC, and in the other the angle comprehended between the two planes ASC, BSC; the sum of these angles will be the angle comprehended between the two planes BSC, DSC. It will be found, in the same manner, that, in order to determine a quintuple solid angle, it is necessary to know, beside the five plane angles which compose it, two of the mutual inclinations of their planes; in a sextuple solid angle it is necessary to know three of these inclinations, and so on.

SECTION SECOND.

Of Polyedrons.

DEFINITIONS.

365. EVERY solid terminated by planes or plane faces is called à solid polyedron, or simply a polyedron. These planes are themselves necessarily terminated by straight lines.

A solid of four faces is called a tetraedron, one of six a hexaedron, one of eight an octaedron, one of twelve a dodecaedron, one of twenty an icosaedron, &c.

The tetraedron is the most simple of polyedrons; for it requires at least three planes to form a solid angle, and these three planes would leave an opening, to close which a fourth plane is neces

sary.

366. The common intersection of two adjacent faces of a polyedron is called a side or edge of the polyedron.

367. A regular polyedron is one, all whose faces are equal regular polygons, and all whose solid angles are equal to each other. There are five polyedrons of this kind.

368. A prism is a solid comprehended under several parallelograms terminated by two equal and parallel polygons.

To construct this solid let ABCDE (fig. 200) be any polygon, Fig. 200. if, in a plane parallel to ABC we draw the lines FG, GH, HI, &c.. equal and parallel to the sides AB, BC, CD, &c., we shall form the polygon FGHIK equal to ABCDE; if now we connect the vertices of the homologous angles by the straight lines AF, BG, CH, &c., the faces ABGF, BCHG, &c., will be parallelograms, and the solid thus formed ABCDEFGHIK will be a prism.

369. The equal and parallel polygons ABCDE, FGHIK, are called the bases of the prism. The other planes taken together, constitute the lateral or convex surface of the prism. The equal straight lines AF, BG, CH, &c., are called the sides of the prism.

370. The altitude of a prism is the distance between its bases, or the perpendicular let fall from a point in the superior base upon the plane of the inferior.

371. A right prism is one whose sides AF, BG, &c., are perpendicular to the planes of the bases; in this case, each of the

Fig. 206.

sides is equal to the altitude of the prism. In every other case the prism is oblique, and the altitude is less than the side.

372. A prism is triangular, quadrangular, pentagonal, hexagonal, &c., according as the base is a triangle, a quadrilateral, a pentagon, a hexagon, &c.

373. A prism whose base is a parallelogram (fig. 206), has all its faces parallelograms, and is called a parallelopiped.

A parallelopiped is rectangular, when all its faces are rectangles.

374. Among rectangular parallelopipeds is distinguished the cube or regular hexaedron comprehended under six equal squares.

375. A pyramid is a solid formed by several triangular planes proceeding from the same point and terminating in the sides of a Fig. 196. polygon ABCDE (fig. 196).

Fig. 202.

The polygon ABCDE is called the base of the pyramid, the point S its vertex, and the triangles ASB, BSC, &c., compose the lateral or convex surface of the pyramid.

376. The altitude of a pyramid is the perpendicular let fall from the vertex upon the plane of the base, produced if necessary. 377. A pyramid is triangular, quadrangular, &c., according as the base is a triangle, a quadrilateral, &c.

378. A pyramid is regular, when the base is a regular polygon, and the perpendicular, let fall from the vertex to the plane of the base, passes through the centre of this base. This line is called the axis of the pyramid.

379. The diagonal of a polyedron is a straight line which joins the vertices of two solid angles not adjacent.

380. I shall call symmetrical polyedrons two polyedrons which, having a common base, are similarly constructed, the one above the plane of this base and other below it, with this condition, that the vertices of the homologous solid angles be situated at equal distances from the plane of the base, in the same straight line perpendicular to this plane.

If the straight line ST (fig. 202), for example, is perpendicular to the plane ABC, and is bisected at the point O, where it meets this plane, the two pyramids SABC, TABC, which have the common base ABC, are two symmetrical polyedrons.

381. Two triangular pyramids are similar when they have two faces similar, each to each, similarly placed, and equally inclined to each other.