Plane and Spherical Trigonometry in Three Parts

This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1907 edition. Excerpt: ...as O. Therefore BD must be a diameter of the sphere, and of each of the two circles BAD, BFD. Therefore the two circles BAD, BFD bisect each other. 13. If the arcs of great circles joining a point on the surface of a sphere with two other points on the surface of the sphere, u'hich are not at opposite extremities of the same diameter, be each of them quadrants, then the first point is a pole of the great circle through the last two points. Let O be the centre of the sphere, P a point on the surface, and PA, PB be two arcs, each equal to a quadrant. Through the points A, B let a great circle be described, and join OA, OB, OP. Then, because PA, PB are quadrants, therefore POA, POB are each right angles. Therefore PO stands at right angles to each of the straight lines OA, OB at the point of their intersection. Therefore PO is at right angles to the plane AOB in which they are (Euc. XI. 4). Therefore OP is a portion of the axis of the circle AB (art. 9), and P, being an extremity of the axis, is a pole of the circle AB. 14. If from a point on the surface of a sphere there can be drawn two arcs of great circles, not parts of the same qreat circle, the planes of which are at right angles to the plane ofa qiven circle, that point is a pole of the given circle. Let O be the centre of the sphere, AB a great circle of the sphere, and PA, PB two great circles whose planes are at right angles to the plane of AB. Then, since the planes of the circles PA, PB are at right angles to the circle AB, therefore PO, their common section, is at right angles to AB (Euc. XL 19). Therefore P is a pole of the great circle AB. Similarly P will be the pole of any small circle which has its plane parallel to that of AB, so that the planes of PA, PB are at right...