POLYHEDRAL ANGLES. DEFINITION. A polyhedral angle, improperly called a solid angle, is the angular space contained between several planes which meet in the same point. This point is called the vertex. Three planes at least are required to form a polyhedral angle. A polyhedral angle is called a trihedral, tetrahedral, &c., angle, according as it is formed by three, four, &c., plane angles. A polyhedral angle is named from the letter at its vertex. A polyhedral angle is called regular when all its plane angles are equal and all its diedral angles equal. A trihedral is called birectangular, trirectangular, when two or three of its diedral angles are right angles. When two of the diedral angles are equal, it is called isohedral. PROP. I. If a polyhedral angle be contained by three plane angles, the sum of any two of these angles will be greater than the third. It is unnecessary to demonstrate this proposition, except in the case where the plane angle, which is compared with the two others, is greater than either of them. Let A be a polyhedral angle contained by the three plane angles BAC, CAD, DAB, and let BAC be the greatest of these angles. Then CAD + DAB > BAC. E D Make, also, AE AD, and through E draw any straight line BEC, cutting AB, AC in the points B, C; join D, B; D, C ; = Then, because AD AE, and AB is common to the two triangles DAB, BAE, and the angle DAB = angle BAE, by construction, .. BD BE. But, in the triangle BDC, = BD+DC > BE + EC (ax. 13, cor.), Again, •* AD = AE, and AC is common to the two triangles DAC, EAC, but the base DC > base EC, But ... angle DAC > angle EAC (th. 32). angle DAB = angle BAE; .. angle CAD + angle DAB> angle BAE + angle' The sum of the plane angles which form a polyhedral angle is always less than four right angles. Let P be a polyhedral angle contained by any number of plane angles APB, BPC, CPD, DPE, EPA. Let the polyhedral angle P be cut by any plane ABCDE. Take any point O in this plane; join A, O; B, O; C, O; D, 0; E, 0. Then, since the sum of all the angles of every triangle is always equal E P A B to two right angles, the sum of all the angles of the triangles APB, BPC,..... about the point P, will be equal to the sum of all the angles of the equal number of triangles AOB, BOC,.................... about the point O. Again, by the last Prop., angle ABC angle ABP +angle CBP; in like manner, angle BCD < angle BCP+ DCP, and so for all the angles of the polygon ABCDE. *This sign. signifies "because." Hence the sum of the angles at the bases of the triangles whose vertex is O, is less than the sum of the angles at the bases of the triangles whose vertex is P. .. The sum of the angles about the point O must be greater than the sum of the angles about the point P. But the sum of the angles about the point O is four right angles. .. The sum of the angles about the point P is less than four right angles. PROP. III. If two trihedral angles be formed by three plane angles which are equal, each to each, the planes in which these angles lie will be equally inclined to each other. Let P, Q be two trihedral angles; Let angle APC = angle DQF, angle APB == angle DQE, and angle BPC = angle EQF. Then the inclination of the P Q F A B DE Z planes APC, APB will be equal to the inclination of the planes DQF, DQE. Take any point B in the intersection of the planes APB, CPB. From B draw BY perpendicular to the plane APC, meeting the plane in Y. From Y draw YA, YC, perpendiculars on PA, PC; join A, B; B, C. = Again, take QE PB; from E draw EZ perpendicular to the plane DQF, meeting the plane in Z; from Z draw ZD, ZF, perpendiculars on QD, QF; join D, E; E, F. BA is perpendicular to PA (Geom. of Planes, Prop. 8), and the triangle PAB is right-angled at A, and the triangle QDE is right-angled at D. Also, the angle APB = angle DQE, by hyp. Moreover, the side PB-side QE (by construction); .. the two triangles APB, DQE are identical (cor. 8, th. 15). ... PA=QD, and AB= DE. In like manner, we can prove that PCQF, and BC=EF. Let now the angle APC be placed upon the equal angle DQF, then the point A will fall upon the point D, and the point C on the point F, because PA=QD, and PC=QF. At the same time, AY, which is perpendicular to PA, will fall upon DZ, which is perpendicular to QD; and, in like manner, CY will fall upon FZ. Hence the point Y will fall on the point Z, and we shall have AY=DZ, and CY = FZ. But the triangles AYB, DZE are right-angled in Y and Z, the hypothenuse AB = hypothenuse DE, and the side AY side DZ; hence these two triangles are equal (th. 26, cor. 2). = .. angle YAB= angle ZDE. The angle YAB is the inclination of the planes APC, APB (Geom. of Planes, def. 6); and The angle ZDE is the inclination of the planes DQF, DQË. .. These planes are equally inclined to each other. In the same manner, we prove the angle YCB= angle ZFE, and, consequently, the inclination of the planes APC, BPC is equal to the inclination of the planes DQF, EQF. We must, however, observe that the angle A, of the right-angled triangle YAB, is not, properly speaking, the inclination of the two planes APC, APB, except when the perpendicular BY falls upon the same side of PA as PC does; if it fall upon the other side, then the angle between the two planes will be obtuse, and, added to the angle A of the triangle YAB, will make up two right angles. But, in this case, the angle between the two planes DQF, DQE will also be obtuse, and, added to the angle D of the triangle ZDE, will make up two right angles. |