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those perpendiculars lie in the same direction; therefore the point B will fall upon the point E, the line SB upon TE, and the two solid angles will wholly coincide.
This coincidence, however, takes place only when we suppose that the equal plane angles are arranged in the same manner in the two solid angles; for if they were arranged in an inverse order, or, what is the same, if the perpendiculars OB, PE, instead of lying in the same direction with regard to the planes ASC, DTF, lay in opposite directions, then it would be impossible to make these solid angles coincide with one another. It would not, however, on this account, be less true, as our Theorem states, that the planes containing the equal angles must still be equally inclined to each other; so that the two solid angles would be equal in all their constituent parts, without, however, admitting of superposition. This sort of equality, which is not absolute, or such as admits of superposition, deserves to be distinguished by a particular name: we shall call it equality by symmetry.
Thus those two solid angles, which are formed by three plane angles respectively equal to each other but disposed in an inverse order, will be called angles equal by symmetry, or simply symmetrical angles.
The same remark is applicable to solid angles, which are formed by more than three plane angles: thus a solid angle, formed by the plane angles A, B, C, D, E, and another solid angle, formed by the same angles in an inverse order A, E, D, C, B, may be such that the planes which contain the equal angles are equally inclined to each other. Those two solid angles, likewise equal without being capable of superposition, would be called solid angles equal by symmetry, or symmetrical solid angles.
Among plane figures, equality by symmetry does not properly exist, all figures which might take this name being absolutely equal, or equal by superposition; the reason of which is, that a plane figure cannot be inverted, and the upper part taken indiscriminately for the under. This is not the case with solids; in which the third dimension may be taken in two different directions
PROPOSITION XXIV. PROBLEM.
The three angles which form a solid angle being given, to find by a plane construction the angle contained between two of these planes.
Let S be the proposed solid angle, in which the three plane angles ASB, ASC, BSC, are known, it is required to find
the angle contained by two of these planes, such as ASB, ASC.
Conceive the same construction to be made as in the preceding Theorem; the angle OAB would be the angle sought. It is required to find the same angle by a plane construction, or one performed on a plane.
On a plane, therefore, make the angles B'SA, ASC, B❝SC equal to the angles BSA, ASC, BSC, in the solid figure; take B'S and B'S each equal to BS in the solid figure; from the points B' and B", at right angles to SA and SC draw B/A and B/C, which will intersect each other at the point O. From A as a centre, with the radius AB', describe the semicircle B'bE; at the point O, erect Ob perpendicular to B'E, and meeting the circumference in b; join Ab: the angle EAb will be the required inclination of the two planes ASC, ASB in the solid angle.
All we have to prove is, 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; the angles at S are equal: hence the angles at B and B are also equal. But the hypotenuse SB' is farther equal to the hypotenuse SB; hence these triangles are equal; therefore SA of the plane figure is equal to SA of the solid figure, and likewise AB', or its equal Ab, in the former to AB in the latter. In the same way, it might be shewn that SC is equal in both; hence it follows, that the quadrilateral SAOC must be equal in both, and consequently AO of the plane figure is equal to AO in the solid. Thus, in both figures the right-angled triangles AOb, AOB have each the hypotenuse and a side respectively equal; hence they are themselves equal; and the angle EAB, found by the plane construction, is equal to the inclination of the two 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 still measures the true inclination of the planes. It is for this reason that EAb, not OAb, has been employed to designate the required inclination, that so the same solution might suit every possible case.
Scholium. A question may arise, whether, if any three angles be assumed at pleasure, a solid angle can be formed with them.
Now, first, the sum of the three given angles must be less than four right angles, otherwise (22. V.) no solid angle can be formed; and farther, two of these angles B'SA, ASC being assumed at pleasure, the third CSB" must be such that B/C, perpendicular to the side SC, shall meet the diameter B'E between its extremities B' and E. Thus the limits to the magni tude of the angle CSB" are such as would make the perpendicular B/C terminate in the points B' and E. From these points, draw BI and EK at right angles to CS, and meeting the circumference described with the radius SB" in I and K; the limits of the angle CSB" will be CSI and CSK.
But in the isosceles triangle B'SI, since the line CS produced is perpendicular to the base B'I, we have the angle CSICSB'=ASC+ASB'. And in the isosceles triangle ESK, since the line SC is perpendicular to EK, we have the angle CSK CSE. Also, by reason of the equal triangles ASE, ASB, we have the angle ASE ASB'; hence CSE or CSK ASC-ASB'.
Therefore the problem is always possible, when the third angle CSB" is less than the sum of ASC, ASB' the other two, and greater than their difference; conditions agreeing with Theorem 21.; according to which it was required that we should have CSB" ASC+ASB', and also ASC CSB"+ASB', or CSB">ASC — ASB.'
PROPOSITION XXV. PROBLEM.
Two of the three plane, angles which form a solid angle, and also the inclination of their planes being given, to find the third plane angle.
Let ASC, ASB', (see the last figure) be the two given plane angles; and suppose for a moment that CSB" is the third angle required: then, employing the same construction as in the fore going Problem, the angle included between the planes of the two first, the inclination of those planes, would be EAb. Now, as EAb can be determined by means of CSB" the other two being given, so likewise may CSB" be determined by means of EAb, which is just what the Problem requires.
Having taken SB' at pleasure, upon SA let fall the indefinite perpendicular B'E; make the angle EAb equal to the inclination of the two given planes; from the point b, where the side Ab meets the circle described from the centre A with the radius AB', draw 60 perpendicular to AE; from the point O, at right angles to SC draw the indefinite line OCB"; make SB"-SB': the angle CSB" will be the third plane angle required.
For, if a solid angle is formed with the three plane angles B'SA, ASC, CSB", the inclination of the planes, in which are the given angles ASB', ASC, will be equal to the given angle EAB.
Scholium. If a solid angle is quadruple, or formed by four plane angles ASB, BSC, CSD, DSA, a knowledge of all these angles is not enough for determining the mutual inclinations of their planes; for the same plane angles may serve to form a multitude of solid angles. But if one condition is added, if, for example, the inclination of the two planes ASB, BSC is given, then the solid angle is entirely determined, and the inclination of any other two of its planes may be found, as follows: Conceive a triple solid angle to be formed by the three plane angles ASB, BSC, ASC; the first two angles are given, as well as the inclination of their planes; the third angle ASC may therefore be determined, by the Problem we have just solved. Now, examining the triple solid angle formed by the plane angles ASC, ASD, DSC; those three angles are known; hence this solid angle is entirely determined. But the quadruple solid angle is formed by the junction of the two triple solid angles of which we have now been treating; and both these partial angles being known and determined, the whole angle will be known and determined likewise.
The inclination of the two planes ASD, DSC may be found immediately by means of the second partial solid angle. As for the inclination of the two planes BSC, CSD, to obtain this, the inclination of the two planes ASC, DSC must be found in the one partial angle, that of the two planes ASC, BSC in the other; the sum of these two inclinations will be the angle included between the planes BSC, DSC.
In the same manner, we should find that for determining a quintuple solid angle, not only the five plane angles which comit must be known, but also two of the mutual inclinations of their planes; three in a sextuple solid angle; and so on.
1. THE name solid polyedron, or simply polyedron, is given to every solid terminated by planes or plane faces; which planes, it is evident, will themselves be terminated by straight lines. In particular, the solid which has four faces is named a tetraedron; that which has six, a hexaedron; that which has eight, an octaedron; that which has twelve, a dodecaedron; that which has twenty, an iscocaedron; and so on.
The tetraedron is the simplest of all polyedrons; because at least three planes are required to form a solid angle, and these three planes leave a void, which cannot be closed without at least one other plane.
II. The common intersection of two adjacent faces of a polyedron is called the side or edge of the polyedron.
III. A regular polyedron is one whose faces are all equal regular polygons, and whose solid angles are all equal to each other. There are five such polyedrons. (See the Appendix to Books VI. and VII.)
IV. The prism is a solid bounded by several plane parallelograms, which are terminated at both ends by two plane polygons equal and parallel.
To construct this solid, let ABCDE be any polygon; then if in a plane parallel to ABC, the lines FG, GH, HI, &c. be