at the bases is less than the sum of the angles at the bases of the triangles whose vertex is in S. Hence, the sum of the angles about the point O is greater than the sum of the angles about the point S. But the sum of the angles about the point O is equal to four right angles (34); therefore the sum of the plane angles, which form a solid angle S, is less than four right angles. 358. Scholium. It is supposed, in this demonstration, that the solid angle is convex, or that the plane of neither of the faces would, by being produced, cut the solid angle; if it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude whatever. THEOREM. 359. If two solid angles are respectively contained by three plane angles which are equal, each to each, the planes of any two of these angles in the one will have the same inclination to each other as the planes of the homologous angles in the other. = Demonstration. Let the angle ASC = DTF (fig. 197), the Fig. 197. angle ASB DTE, and the angle BSC ETF; we say that the two planes ASC, ASB, will have, with respect to each other, an inclination equal to that of the planes DTF, DTE. Take SB of any magnitude, and draw BO perpendicular to the plane ASC; from the point O, where this perpendicular meets the plane, draw OA, OC, perpendicular respectively to SA, SC; join AB, BC. Take also TE = SB; and draw EP perpendicular to the plane DTF; from the point P draw PD, PF, perpendicular respectively to TD, TF; and join ED, EF. The triangle SAB is right-angled at A, and the triangle TDE at D (332); and, since the angle ASB DTE, we have also SBA = TED. Moreover, SB = TE; therefore the triangle SAB = TDE; consequently SA = TD, and AB = DE. It may be shown, in a similar manner, that SC = TF, and BC= EF. This being supposed, the quadrilateral SAOC is equal to the quadrilateral TDPF; for, if we apply the angle ASC to its equal DTF, because SA TD, and SC TF, the point A will fall upon D, and the point C upon F. At the same time AO, perpendicular to SA, will fall upon DP, perpendicular to TD; and, in like manner, OC upon PF; therefore the point O will fall upon the point P, and we shall have AO=DP. But the triangles AOB, DPE, are right-angled at Q, and P, the hypothenuse AB = DE = = and the side AO = DP; consequently the triangles are equal (56); hence OAB = PDE. But the angle OAB is the inclination of the two planes ASB, ASC; and the angle PDE is the inclination of the two planes DTE, DTF; therefore these two inclinations are equal to each other." It should be observed, however, that the angle A of the rightangled triangle OAB is not properly the inclination of the two planes ASB, ASC, except when the perpendicular BO falls, with respect to SA, on the same side as SC; if it should fall on the other side, the angle of the two planes would be obtuse, and, added to the angle A of the triangle OAB, it would make two right angles. But, in the same case, the angle of the two planes TDE, TDF, would be likewise obtuse, and, added to the angle D of the triangle DPE, it would make two right angles; therefore, as the angle A would be always equal to D, we infer, in like manner, that the inclination of the two planes ASB, ASC, is equal to that of the two planes TDE, TDF. 360. Scholium. If two solid angles are respectively contained by three plane angles which are equal, each to each, and if, at the same time, the angles of the one are disposed in the same manner as the homologous angles of the other, these solid angles will be equal, and, being applied the one to the other, will coincide. Indeed, we have already seen that the quadrilateral SAOC may be placed upon its equal TDPF; thus, by placing SA upon TD, SC would fall upon TF, and the point O upon the point P. But, on account of the equality of the triangles AOB, DPE, the line OB perpendicular to the plane ASC is equal to PE perpendicular to the plane TDF; moreover the perpendiculars are directed the same way; therefore the point B will fall upon the point E, the line SB upon TE, and the two solid angles will coincide entirely with each other. This coincidence, however, does not take place except by supposing that the plane angles are disposed in the same manner in each of the two solid angles; for if the plane angles were disposed in a contrary order in the one from what they are in the other; or, which comes to the same thing, if the perpendiculars OB, PE, instead of being directed the same way with respect to the planes ASC, DTF, were directed contrary ways, it would be impossible to make the solid angles coincide with each other. Still it would not be the less true, that, agreeably to the theorem, the planes of the homologous angles would be equally inclined to each other; so that the two solid angles. would be equal in all their constituent parts, without the property however of coinciding, when applied the one to the other. This kind of equality, which is not absolute, or does not admit of superposition, deserves to be distinguished by a particular denomination; we shall call it equality by symmetry. Thus the two solid angles under consideration, which are respectively contained by three plane angles equal, each to each, but disposed in a contrary order in the one from what they are in the other, we shall call angles equal by symmetry, or simply symmetrical angles. The same remark is applicable to solid angles contained by more than three plane angles; thus a solid angle contained by the plane angles A, B, C, D, E, and another solid angle contained by the same angles in the inverse order A, E, D, C, B, may be such that the planes of the homologous angles shall be equally inclined to each other. These two solid angles, which would be equal without admitting of superposition, we shall call solid angles equal by symmetry, or symmetrical solid angles. There is not properly an equality by symmetry among plane figures; all those to which we might give this name, have the property of absolute equality, or equality by superposition. The reason is, that a plane figure may be reversed, and the upper side be taken for the under. It is not so with respect to solids, in which the third dimension may be taken in two different ways. PROBLEM. 361. Three plane angles forming a solid angle being given, to find, by a plane construction, the angle which two of these planes make with each other. Solution. Let S (fig. 198) be the proposed solid angle in Fig. 198. which the three plane angles ASB, ASC, BSC, are known; the angle made by two of these planes with each other, ASB, ASC, for example, is required. The same construction being supposed as in the preceding theorem, the angle OAB would be the angle sought. It is proposed to find the same angle by a plane construction, or by lines raced upon a plane. 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 OA b, in order that the same solution may be adapted to every case without 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 meeting in I and K the circum CS the perpendiculars B'I, EK, 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 EAb 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 EAb 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 6 O, and from the point O let fall upon SC the indefinite perpendicular OC B", Geom. 16 |