NOTE.-Because the apparent position of the heavenly bodies is referred to an imaginary spherical surface whose centre we occupy, the geometry of the surface of the sphere early attracted attention. It cannot be studied to any great extent without a knowledge of Trigonometry, but a few important propositions may be given, which illustrate this branch of Geometry. Proposition 7. Theorem. 698. A spherical angle is measured by the arc of a great circle described with its vertex as a pole and included between its sides, produced if necessary. Hyp. Let ABC, AB'C be two arcs of great Os intersecting at A; let AT, AT' be the tangents to these arcs at A; and let OBB' be a plane through the centre 0 to AC, in tersecting the sphere in the great BB'. To prove that the spherical/ BAB' is measured by the arc BB'. A T 아 B B' 'ז. Proof. Since TA and T'A are respectively in the planes of the arcs BA and B'A, and are to their intersection But the spherical / BAB' is measured by TATʼ. (692) ... it is measured by the arc BB'. Q.E.D. 699. COR. 1. A spherical angle is equal to the diedral angle between the planes of the two circles. 700. COR. 2. If two arcs of great circles cut each other, their vertical angles are equal. 701. COR. 3. The angles of a spherical triangle are equal to the diedral angles between the planes of the sides of the triangle. RELATION OF A SPHERICAL POLYGON TO A POLYEDRAL ANGLE. B 702. Because the planes of all great circles pass through the centre of the sphere, therefore the planes of the sides of a spherical polygon form a polyedral angle at the centre O whose face-angles AOB, BOC, etc., are measured by the sides AB, BC, etc., of the polygon (236), and whose diedral angles OA, OB, etc., are equal to the angles A, B, etc., of the spherical polygon ABC, etc. (699). We may therefore speak of all the parts of a spherical polygon as angles, meaning thereby the face-angles, and the diedral angles between the faces, of the polyedral angle whose vertex is the centre of the sphere, and base the spherical polygon. It follows, therefore, that the properties of a spherical polygon and a polyedral angle are mutually convertible. Hence: 703. From any relation proved between the face, and diedral, angles of a polyedral angle, we may infer the same relation between the sides and angles of a spherical polygon. And conversely: From any relation proved between the sides and angles of a spherical polygon, we may infer the same relation between the face, and diedral, angles of a polyedral angle. Therefore: 704. Each side of a spherical triangle is less than the sum of the other two sides (565). 705. The sum of the sides of a spherical polygon is less than a circumference (566). 706. Two mutually equilateral triangles on the same, or on equal spheres, are mutually equiangular, and are either equal or symmetrical (567). SYMMETRICAL SPHERICAL TRIANGLES. 707. Symmetrical Spherical Triangles are those in which the sides and angles of the one are equal respectively to the sides and angles of the other, but arranged in the reverse order. Thus, the spherical triangles ABC and A'B'C' are symmetrical when the vertices of the one are at the ends of the diameters from the vertices of the other. * The corresponding triedral angles O-ABC and O-A'B'C' are also symmetrical. (567) In the same way we may form two symmetrical polygons of any number of sides. Two symmetrical triangles are mutually equilateral and equiangular; yet in general they cannot be made to coincide. Thus, if in the symmetrical triangles ABC, A'B'C', AB is made to coincide with A'B' to bring the vertex C upon the corresponding vertex C', the two convex surfaces would have to be brought together. The triangles are A Да B B in fact right-handed and left-handed, and, though corresponding and equal in every detail, can no more be conceived as superposed on one another so as to occupy the same space, than the form of the right hand on that of the left hand. * Antipodal. Proposition 8. Theorem. 708. Two symmetrical spherical triangles are equivalent. Hyp. Let ABC, A'B'C' be two symmetrical spherical As with their homologous vertices diametrically opposite each other. (707) To prove area ABC area A'B'C'. celes. Proof. Let BA BC, and B'A' = B'C'. If B be placed on the equal < B' A P C 'A' (702), the convexities of the sphere being on the same side, the side BA will fall on B'C', and BC on B'A'. And since BA = B'C', and BC = B'A', ... A will fall on C', and C on A'. ... the two As coincide throughout and are identically equal. CASE II. When the triangles are not isosceles. Proof. Let P and P' be the poles of the small Os passing through the pts. A, B, C, and A', B', C', respectively.* Draw the great arcs PA, PB, PC, and P'A', P'B', P'C'. PA PB = PC. Then (674) Since two symmetric As are mutually equilateral, (707) PA, P'B' PB, P'C' = PC. = ... P'A' = ... the AS PAC and P'A'C', PCB and P'C'B', PBA and P’B'A' are respectively isosceles symmetric As. .. they are identically equal. (Case I) Because sum of As PAC, PCB, PBA = area ABC; and sum of As P'A'C', P'C'B', P'B'A' ... area ABC area A'B'C'. area A'B'C', Q. E.D. * The circle which passes through the three points A, B, C can only be a small circle of the sphere; for if it were a great circle, the three sides AB, BC, CA, would lie in one plane, and the triangle ABC would be reduced to one of its sides. Proposition 9. Theorem.* 709. Two triangles on the same, or on equal spheres, having two sides and the included angle of one equal respectively to two sides and the included angle of the other, are either equal or equivalent. Hyp. Let ABC, DEF be two As having the side AB DE, A ▲ ABC A DEF. E Proof. The ABC may be placed on the ▲ DEF, as in the corresponding case of plane As, and will coincide with it. (104) CASE II. When the given parts are arranged in inverse order, as in As ABC, DEF'. To prove As ABC and DEF' equivalent. Proof. Let the ▲ DEF be symmetrical with the ▲ DEF', having its sides and s equal respectively to those of DEF'. Then in the As ABC, DEF, we have |