is to 6, or as 2 is to 3, which is the first branch of the proposition. In the next place, since the base of the circumscribed cylinder is equal to a great circle, and its altitude to the diameter, the volume of the cylinder (T. II.) will be equal to a great circle multiplied by its diameter. But (T. XII.) the volume of the sphere is equal to four great circles multiplied by a third of the radius; in other terms, to one great circle multiplied by 3 of the radius, or by of the diameter. Hence the sphere is to the cir cumscribed cylinder as 2 to 3, and consequently the volumes of these two bodies are as their surfaces. EIGHTH BOOK. SPHERICAL GEOMETRY. DEFINITIONS. I. A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles. Those arcs, named the sides of the triangle, are always supposed to be each less than a semi-circumference; the angles, which their planes form with each other, are the angles of the triangle. II. A spherical triangle takes the name of right-angled, isosceles, equilateral, in the same cases as a rectilineal triangle. III. A spherical polygon is a portion of the surface of a sphere, terminated by several arcs of great circles. IV. A lune is that portion of the surface of a sphere, which is included between two great semicircles meeting in a common diameter. V. A spherical wedge, or ungula, is that portion of the solid sphere which is included between the same great semicircles, and has the lune for its base. VI. A spherical pyramid is a portion of the solid sphere, included between the planes of a polyedral angle whose vertex is the centre; the base of the pyramid is the spherical polygon intercepted by the same planes. VII. The pole of a circle is a point on the surface of the sphere equally distant from all the points of the circumference. THEOREM I. In every spherical triangle, any side is less than the sum of the other two. Let O be the centre of the sphere; and draw the radii OA, OB, OC. Imagine the planes AOB, AOC, COB; those planes will form a polyedral angle at the point 0; and the angles AOB, AOC, COB will be measured by AB, AC, BC, the sides of the spherical triangle. But (B. V., T. XIX.) each of the three plane triangles composing a polyedral angle is less than the sum of the other two; hence, any side of the triangle ABC is less than the sum of the other two. THEOREM II. The sum of all the three sides of a spherical triangle is less than the circumference of a great circle. Let ABC be any spherical triangle; produce the sides AB, AC till they meet again in D. The arcs ABD, ACD will be semi-circumferencessince (B. VII., T. VI., C. II.) two great circles always bisect each other. But in the triangle BCD we have (T. I.) the side BCBD + CD: add AB+ AC D C B A E to each; we shall have AB+ AC+BC < ABD + ACD, that is to say, less than a circumference. THEOREM III. The sum of all the sides of any spherical polygon is less than the circumference of a great circle. Let us take, for example, the pentagon ABCDE. Produce the sides AB, DC till they meet in F; then, since BC is less B F C D G than BF + CF, the perimeter of the pentagon ABCDE will be less than that of the quadrilateral AEDF. Again, produce the sides AE, FD till they meet in G; we shall have ED < EG + DG: hence the perimeter of the quadrilateral AEDF is less than that of the triangle AFG, which last is itself less than the circumference of a great circle (T. II.): hence the perimeter of the polygon ABCDE is less than this same circumference. A E THEOREM IV. If a diameter be drawn perpendicular to the plane of a great circle, its extremities will be the poles of that circle, and also of all small circles parallel to it. B I D F H P P A M M S For, DC being perpendicular to the plane AMB, is perpendicular to all the straight lines CA, CM, CB, etc., drawn through its foot in this plane; hence all the arcs DA, DM, DB, etc., are quarters of the circumference. So likewise are all the arcs EA, EM, EB, etc.; hence the points D and E are each equally distant from all the points of the circumference AMB; therefore (D. VII.) they are the poles of that circumference. E Again, the radius DC, perpendicular to the plane AMB, is perpendicular to its parallel FNG; hence (B. VII., T. VI., C. IV.) it passes through O, the centre of the circle FNG; therefore, if the oblique lines DF, DN, DG be drawn, they will be equal (B. V., T. V.); but, the chords being equal, the arcs are equal: hence the point D is the pole of the small circle FNG; and, for like reasons, the point E is the other pole. Cor. I. Every arc DM drawn from a point in the arc of a great circle AMB to its pole, is a quarter of the circumference, which, for the sake of brevity, is usually named a quadrant; and this quadrant, at the same time, makes a right angle with the arc AM. For (B. V., T. XVI.), the line DC being perpen dicular to the plane AMC, every plane DMC passing through the line DC is perpendicular to the plane AMC; hence the an gles of these planes, or the angle AMD, is a right angle. Cor. II. To find the pole of a given arc AM, draw the indefinite arc MD perpendicular to AM; take MD equal to a quadrant: the point D will be one of the poles of the arc AMD. Or thus: at the two points A and M, draw the arcs AD and MD perpendicular to AM; their point of intersection, D, will be the pole required. Cor. III. Conversely, if the distance of the point D from each of the points A and M be equal to a quadrant, the point D will be the pole of the arc AM; and also the angles DAM, DMA will be right angles. For, let C be the centre of the sphere, and draw the radii CA, CD, CM. Since the angles ACD, MCD are right, the line CD is perpendicular to the two straight lines CA, CM; it is, therefore, perpendicular to their plane: hence the point D is the pole of the arc AM, and consequently the angles DAM, DMA are right. Scholium. The properties of these poles enable us to describe arcs of a circle on the surface of a sphere with the same facility as on a plane surface. It is evident, for instance, that by turning the arc DF, or any other line extending to the same distance, round the point D, the extremity F will describe the small circle FNG; and by turning the quadrant DFA round the point D, its extremity A will describe the arc of the great circle AM. If the arc AM were required to be produced, and nothing were given but the points A and M through which it was to pass, we should first have to determine the pole D, by the intersection of two arcs described from the points A and M as centres, with a distance equal to a quadrant; the pole D being found, we might describe the arc AM and its prolongation from D as a centre, and with the same distance as before. Lastly, if it be required from a given point P to draw a perpendicular on the given arc AM, produce this arc to S till the distance PS be equal to a quadrant; then from the pole S, and with the same distance, describe the arc PM, which will be the perpendicular required. |