PROPOSITION VII. PROBLEM. 704. To pass a circumference of a great circle through any two points on the surface of a sphere. Let A and B be any two points on the surface of a sphere. It is required to pass a circumference of a great circle through A and B. From A as a pole, with an arc equal to a quadrant, strike an arc a b, and from B as a pole, with the same radius, describe an arc cd, intersecting a b at P. Then a circumference described with a quadrant arc, with P as a pole, will pass through A and B and be the circumference of a great circle. Q. E. F. 705. COROLLARY. Through any two points on the surface of a sphere, not at the extremities of the same diameter, only one circumference of a great circle can be made to pass. 706. SCHOLIUM. By means of poles arcs of circles may be drawn on the surface of a sphere with the same facility as upon a plane surface, and, in general, the methods of construction in Spherical Geometry are similar to those of Plane Geometry. Thus we may draw an arc perpendicular to a given spherical arc, bisect a given spherical angle or arc, make a spherical angle equal to a given spherical angle, etc., in the same way that we make analogous constructions in Plane Geometry. PROPOSITION VIII. THEOREM. 707. The shortest distance on the surface of a sphere between any two points on that surface is the arc, not greater than a semi-circumference, of the great circle which joins them. Let A B be the arc of a great circle which joins any two points A and B on the surface of a sphere; and let AC PQB be any other line on the surface between A and B. We are to prove arc A B A C P Q B. Let P be any point in ACP Q B. Pass arcs of great circles through A and P, and P and B. § 704 Join A, P and B with the centre of the sphere 0. hedral whose vertex is at 0. The arcs A B, A P and P B are measures of these Now AO B < ZA OP+▲ PO B, .. arc AB < arc A P + arc P B. . § 202 § 487 In like manner, joining any point in A CP with A and P by arcs of great, their sum would be greater than arc A P; and, joining any point in PQB with P and B by arcs of great, the sum of these arcs would be greater than arc P B. If this process be indefinitely repeated the distance from A to B on the arcs of the great will continually increase and approach to the line A CPQ B. .. arc A B A C P Q B. Q. E. D. PROPOSITION IX. THEOREM. 708. Every point in an arc of a great circle which bisects a given are at right angles is equally distant from the extremities of the given arc. Let arc CD bisect arc AB at right angles. We are to prove any point 0 in CD is equally distant from A and B. Since great circle CDE bisects arc AB at right angles, it also bisects chord A B at right angles. Hence, chord AB is to the plane C D E at K. B E § 430 $58 .. OK is to chord A B at its middle point. § 182 Q. E. D. PROPOSITION X. PROBLEM. 709. To pass the circumference of a small circle through any three points on the surface of a sphere. B Let A, B and C be any three points on the surface of a sphere. It is required to pass the circumference of a small circle through the points A, B and C. Pass arcs of great circles through A and B, A and C, B and C. Arcs of great circles ao 1 to AC and BC at their middle points intersect at o. Then o is equally distant from A, B and C. $704 and bo $ 708 .. the circumference of a small circle drawn from o as a pole, with an arc o A will pass through A, B and C, and be the circumference required. Q. E. D. ON SPHERICAL ANGLES. 710. DEF. The angle of two curves which have a common point is the angle included by the two tangents to the two curves at that point. 711. DEF. A spherical angle is the angle included between two arcs of great circles. 712. The angle of two curves which intersect on the surface of a sphere is equal to the dihedral angle between the planes passed through the centre of the sphere, and the tangents of the two curves at their point of intersection. Let the curves A B and AC intersect at A on the surface of a sphere whose centre is 0; and let AT and AS be the tangents to the two curves respectively. We are to prove TAS equal to the dihedral angle formed by the planes 0 A T and O A S. Since A T and AS do not cut the curves at A, they do not cut the surface of the sphere, and are therefore tangents to the sphere. .. A T and AS are to the radius O A, drawn to the point of contact. ..TAS measures the dihedral and O AS, passed through the radius OA and AS. § 186 of the planes O A T and the tangents A T $ 470 But TAS is the of the two curves A B and A C. §710 .. the of the two curves A B and A C of the planes O A T and O A S. == the dihedral Q. E. D. PROPOSITION XII. THEOREM. 713. A spherical angle is equal to the measure of the dihedral angle included by the great circles whose arcs form the sides of the angle. T P T E Let BPC be any spherical angle, and BPDP' and CPEP the great circles whose arcs BP and C P include the angle. We are to prove BPC equal to the measure of the dihedral C-P P'-B. Since two great intersect in a diameter, PP' is a diameter. Draw PT tangent to the O B P D P'. $685 Then PT lies in the same plane as the OBPD P', and is I to P P' at P. In like manner draw P T' tangent to the OG PEP'. Then PT lies in the same plane as the OC PE P', and is I to P Pat P. ../ TPT is the measure of the dihedral ≤ C-P P'-B. § 470 But spherical BPC is the same as plane TPT'; §710 •*. spherical ▲ BPC is equal to the measure of dihedral C-P P-B. Q. E. D. 714. COROLLARY. A spherical angle is measured by the arc of a great circle described about its vertex as a pole and intercepted by its sides (produced if necessary). For, if BC be the arc of a great circle described about the vertex P as a pole, PB and PC are quadrants. Hence, BO and CO are perpendicular to PP. Therefore BOC measures the dihedral angle B-PO-C, and, hence, the spherical angle BPC. Therefore, arc BC, which measures the angle BOC, measures the spherical angle BP C. |