Let A B C, be a spherical triangle, (one of those on the common base A B, =c); and the given surface A + B + C → π S. Let IPK be an indefinite perpendicular on the middle of AB, having taken IP = a quadrant, P will be the pole of the arc AB, and the arc CD, drawn through the points P C, will be perpendicular to A B. Let ID p, CD angled triangles, ACD, BCD, in which AC AD = p + c, B D = p cos q, cos (pc), cos b = cos q cos (p + found, page 48, that 9, the rightb, BC = a, give cos a = But it was c, will c). Substituting in this formula the values cos a + cos b = 2 cos q cos p cosc, 1 cos c = 2 cos2c, sin b sin C = sin c sin B = 2 sin c cosc sin B; there results, cots A. cos c + cos p cos q sin a sin c sin B Again, from the right-angled triangle BCD, sin a sin B = sin q; cos c. This is the or, cos p cos q = cots sin c sin q relation between P and 9 which will determine the locus of all the points C. Produce IP to K, let PK = 20. Join KC, and let KC = y; in the triangle PKC where we have PC= -q, the angle KPC- p, the side KC will be found by the formula cos KC = cos KPC sin PK sin PC + cos PK cos PC, or Substituting this instead of cos p cos q the value cos y = sin x cos c + sin q (cos x-sin x cot S sin c), or In which, if we take cos x sin a cots sin c = 0, cots sinc cota, there will result cos y = sin x cos c, and thus a constant value of y is determined. Therefore, if after having drawn the arc IP perpendicular to the middle of the base AB, and beyond the pole the part PK such that cot PK cotS sinc, all the vertices of the triangles on the same base c, and of the same surface S, will be situated on the small circle described from K as a pole at the distance KC, such that cos KC = sin P K cosc. This is Lexell's theorem. 48. Given the three sides, BC= a, AC = =b, AB = c, to find the position of the point I, the pole of the circle circumscribing the triangle ABC. Let the angle ACI = a, and the arc AICI = BI=Q; in the triangles CAI, CBI, we shall have by the equation Substituting, in this equation, the values of cos C and sin C, in terms of the sides a, b, c, and putting for the sake of abridgcos2b-cos2c+ 2 cos a cos b cos c), ment, M = √(1 — cos2 a we have tan x = 1 + cos b the angle ACI. From the isosceles triangles ACI, ABI, BCI, we have ACI = (CAB); and, in the same manner, BCI (B+C-A); BAI (A + B-C). = = To which we may add that which gives cot S, and which can be put under the form tan (A+B+C)= cosa - cosb COSC From the value of the tangent of x, already found, we have 49. The surface of a spherical polygon is measured by the sum of the angles, minus the product of two right angles, and the number of sides of the polygon, minus 2. From A draw the arcs AC and AD, the angles of the polygon; it will then be divided into as many triangles, minus two, as the figure has sides; but the surface of each triangle is measured by the sum of the angles, minus two right angles; and it is clear, that the sum of all the angles of the triangles is equal to the sum of all the angles of the polygon. 50. If s be the number of solid angles of a polyhedron, H the number of faces, A the number of its edges, then S+ H = A + 2. Take a point within the polyhedron, and from which draw lines to all the angular points of the polyhedron; imagine from this point, as a centre, we describe a spherical surface which meets all these lines in as many points, then join these points by arcs of great circles, in such a manner as to form, upon the surface of the sphere, the same number of polygons as there are faces of the polyhedron. Let ABCDE be one of these polygons, and n the number of its sides, its surface by the last article will be $-2n +4; S being the sum of the angles A, B, C, D, E. Similarly if we find the value of each of the other spherical polygons, and add them all together, we conclude that their sum or the surface of the sphere which is represented by 8, is equal to the sum of all the angles of the polygons, less twice the number of their sides, plus four times the number of faces. Now as all the angles that meet at the same point, A, is equal to four right angles, the sum of all the angles of the polygons is equal to four times the number of solid angles, it is therefore equal to 4 S. Then double the number of sides AB, BC, CD, &c., is equal to four times the number of edges, or equal to 4A, since the same edge serves for two faces; .. 84S4A +4H; or 2=s AH; or S+ H= A + 2. [See Legendre's Geometry, pp. 228, 229.] Cor. It follows that the sum of all the plane angles which form the solid angles of a regular polyhedron, is equal to as many times four right angles as there are units in S 2, s being the number of solid angles of the polyhedron. The plane angles = sum of all the interior angles of each face, which Prop. 32, of the 1st book of Euclid 51. There can be only five regular polyhedrons. Since every face has n plane angles, the number of plane angles which compose all the solid angles = n H=Sm=2A, and by the last article S + H = A + 2 ; m ... H S, and A= ; substituting these, п m n mS 2 ms 2 8 + S= + 2, Now this must be a positive whole number, and in order that it may be so 2 (m + n) must be greater than mn; and, consequently, since n must be an integer and cannot be less than 3, it can only be 3, 4, or 5. In the same manner m cannot be less than 3, therefore the values of m can only be 3, 4, or 5. |