sin CQ sin AQF similarly sin AQ sin CQFi therefore sin BQ = sin CQ; therefore BQ+CQ=. Hence if D be the middle point of BC DQ=1 (BQ +CQ) = 1 T. 188. If three arcs be drawn from the angles of a spherical triangle through any point to meet the opposite sides, the products of the sines of the alternate segments of the sides are equal. E D Let P be any point, and let arcs be drawn from the angles A, B, C passing through P and meeting the opposite sides at D, E, F. Then sin BD sin BPD sin CD sin CPD therefore sin BD sin BPD sin BP sin BF; sin CE sin AF Similar expressions may be found for and sin DE and hence it follows obviously that sin BD sin CE sin AF -1; therefore sin BD sin CE sin AF =sin CD sin AE sin BF. 189. Conversely, when the points D, E, F in the sides of a spherical triangle are such that the relation given in the preceding Article holds, the arcs which join these points with the opposite angles respectively pass through a common point. Hence the following propositions may be established : the perpendiculars from the angles of a spherical triangle on the opposite sides meet at a point; the arcs which bisect the angles of a spherical triangle meet at a point; the arcs which join the angles of a spherical triangle with the middle points of the opposite sides meet at a point; the arcs which join the angles of a spherical triangle with the points where the inscribed circle touches the opposite sides respectively meet at a point. Another mode of establishing such propositions has been exemplified in Arts. 139 and 140. 190. If AB and A'B' be any two equal arcs, and the arcs AA' and BB' be bisected at right angles by arcs meeting at P, then AB and A'B' subtend equal angles at P. = ز For PA = PA' and PB = PB'; hence the sides of the triangle PAB are respectively equal to those of PA'B'; therefore the angle APB = the angle A'PB'. This simple proposition has an important application to the motion of a rigid body of which one point is fixed. For conceive a sphere capable of motion round its centre which is fixed; then it appears from this proposition that any two fixed points on the sphere, as A and B, can be brought into any other positions, as A' and B', by rotation round an axis passing through the centre of the sphere and a certain point P. Hence it may be inferred that any change of position in a rigid body, of which one point is fixed, may be effected by rotation round some axis through the fixed point. (De Morgan's Differential and Integral Calculus, page 489.) 191. Let P denote any point within any plane angle AOB, and from P draw perpendiculars on the straight lines 04 and then it is evident that these perpendiculars include an angle which is the supplement of the angle AOB. The corresponding fact with respect to a solid angle is worthy of notice. Let there be a solid angle formed by three plane angles, meeting at a point 0. From any point P within the solid angle, draw perpendiculars PL, PM, PN on the three planes which form the solid angle; then the spherical triangle which corresponds to the three planes LPM, MPN, NPL is the polar triangle of the spherical triangle which corresponds to the solid angle at 0. This remark is due to Professor De Morgan. OB; a 192. Suppose three straight lines to meet at a point and form a solid angle; let a, b, and y denote the angles contained by these three straight lines taken in pairs : then it has been proposed to call the expression J(1 - cos'a - cos*ß - cosøy + 2 cos a cos ß cos y), the sine of the solid angle. See Baltzer's Theorie...der Determinanten, 2nd edition, page 177. Adopting this definition it is easy to shew that the sine of a solid angle lies between zero and unity. We know that the area of a plane triangle is half the product of two sides into the sine of the included angle: by Art. 156 we have the following analogous proposition; the volume of a tetrahedron is one sixth of the product of three edges into the sine of the solid angle which they form. Again, we know in mechanics that if three forces acting at a point are in equilibrium, each force is as the sine of the angle between the directions of the other two: the following proposition is analogous; if four forces acting at a point are in equilibrium each force is as the sine of the solid angle formed by the directions of the other three. See Statics, Chapter II. 193. Let a sphere be described about a regular polyhedron; let perpendiculars be drawn from the centre of the sphere on the faces of the polyhedron, and produced to meet the surface of the sphere : then it is obvious from symmetry that the points of intersection must be the angular points of another regular polyhedron. This may be verified. It will be found on examination that if S be the number of solid angles, and F the number of faces of one regular polyhedron, then another regular polyhedron exists which has S faces and F solid angles. See Art. 151. 194. Polyhedrons. The result in Art. 150 was first obtained by Euler; the demonstration which is there given is due to Legendre. The demonstration shews that the result is true in many cases in which the polyhedron has re-entrant solid angles ; for all that is necessary for the demonstration is, that it shall be possible to take a point within the polyhedron as the centre of a sphere, so that the polygons, formed as in Art. 150, shall not have any coincident portions. The result, however, is generally true, even in cases in which the condition required by the demonstration of Art. 150 is not satisfied. We shall accordingly give another demonstration, and shall then deduce some important consequences from the result. We begin with a theorem which is due to Cauchy. 195. Let there be any network of rectilineal figures, not necesa sarily in one plane, but not forming a closed surface ; let E be the number of edges, F the number of figures, and S the number of corner points : then F+S = E + 1. This theorem is obviously true in the case of a single plane figure; for then F=1, and S=E. It can be shewn to be generally true by induction. For assume the theorem to be true for a network of F figures; and suppose that a rectilineal figure of n sides is added to this network, so that the network and the additional figure have m sides coincident, and therefore m + 1 = corner points coincident. And with respect to the new network which is thus formed, let E', F', S' denote the same things as E, F, S with respect to the old network. Then E' = E +n-m, F" =F+1, S'= S+n-(m + 1); therefore F + S'- E' = F +S- E. But F+S=E+1, by hypothesis; therefore F" +S' = E' +1. 196. To demonstrate Euler's theorem we suppose one face of a polyhedron removed, and we thus obtain a network of rectilineal figures to which Cauchy's theorem is applicable. Thus F-1+S=E+1; therefore F+S = E +2. = 197. In any polyhedron the number of faces with an odd number of sides is even, and the number of solid angles formed with an odd number of plane angles is even. Let a, b, c, d,....., denote respectively the numbers of faces which are triangles, quadrilaterals, pentagons, hexagons,..... Let a, b, 7, 8, ...... denote respectively the numbers of the solid angles which are formed with three, four, five, six, ...... plane angles. Then, each edge belongs to two faces, and terminates at two solid angles; therefore 2E = 3a + 4b + 50 + 6d + ......, 2E = 3a +43 + 5y + 68 + ...... From these relations it follows that a+c+e+ and aty+€+ are even numbers. 198. With the notation of the preceding Article we have F= a + b +c+d+ 7 2E - 3F=b+ 2c + 3d + 2E – 3S = B+ 27 +38+ Thus 2E cannot be less than 3F, or less than 3S. = ...... + ......) |