Mathematical Problems on the First and Second Divisions of the Schedule of Subjects for the Cambridge Mathematical Tripos ExaminationMacmillan and Company, 1878 - 480 páginas |
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Página ix
... RIGID BODY . Moments of Inertia , Principal Axes Motion about a fixed Axis Motion in two Dimensions Miscellaneous 2693-2764 . HYDROSTATICS . 2765-2784 . GEOMETRICAL OPTICS 2785-2815 . SPHERICAL TRIGONOMETRY AND ASTRONOMY CONTENTS . ix.
... RIGID BODY . Moments of Inertia , Principal Axes Motion about a fixed Axis Motion in two Dimensions Miscellaneous 2693-2764 . HYDROSTATICS . 2765-2784 . GEOMETRICAL OPTICS 2785-2815 . SPHERICAL TRIGONOMETRY AND ASTRONOMY CONTENTS . ix.
Página 8
... the middle points of these six edges lie on one sphere which also passes through the feet of the shortest listances between the opposite edges . 87. In a certain tetrahedron each edge is perpendicular to 8 GEOMETRY .
... the middle points of these six edges lie on one sphere which also passes through the feet of the shortest listances between the opposite edges . 87. In a certain tetrahedron each edge is perpendicular to 8 GEOMETRY .
Página 9
... sphere to the middle point of any edge will be equal and parallel to the straight line joining the centre of ... sphere is described touching three given spheres : prove that the plane passing through the points of contact contains one ...
... sphere to the middle point of any edge will be equal and parallel to the straight line joining the centre of ... sphere is described touching three given spheres : prove that the plane passing through the points of contact contains one ...
Página 96
... sphere . ] A triangle is formed by joining the feet of the perpendiculars of the triangle ABC , and the circle inscribed in this triangle touches the sides in A ' , B ' , C ' : prove that B'C ' C'A ' A'B ' = = BC CA AB = 2 cos A cos B ...
... sphere . ] A triangle is formed by joining the feet of the perpendiculars of the triangle ABC , and the circle inscribed in this triangle touches the sides in A ' , B ' , C ' : prove that B'C ' C'A ' A'B ' = = BC CA AB = 2 cos A cos B ...
Página 320
... sphere is described touching a given plane at a given point , and a segment of given curve surface is cut off by a plane parallel to the former prove that the locus of the circular boundary of this segment is a sphere . 1864. Two ...
... sphere is described touching a given plane at a given point , and a segment of given curve surface is cut off by a plane parallel to the former prove that the locus of the circular boundary of this segment is a sphere . 1864. Two ...
Otras ediciones - Ver todas
Mathematical Problems on the First and Second Divisions of the Schedule of ... Joseph Wolstenholme Sin vista previa disponible - 2012 |
Mathematical Problems: On the First and Second Divisions of the Schedule of ... Joseph Wolstenholme Sin vista previa disponible - 2017 |
Términos y frases comunes
angular points angular velocity asymptotes ax² axes bisected cardioid centre of perpendiculars chord circumscribed circle co-ordinates common point common tangents confocal conicoid conjugate diameters constant continued fraction cos² curve described diagonals directrix envelope equal excentric angles fixed circle fixed point fixed straight line foci focus four points given circle given conic given ellipse given point given the equations inscribed latus rectum length locus major axis meet minor axis normal parabola parallel particle passes plane point of intersection points of contact polar pole prove radical axis radius of curvature ratio rectangle rectangular hyperbola respectively right angles roots self-conjugate sin² sin³ straight line joining subtends a right tangents drawn tetrahedron triangle ABC velocity vertex vertical
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