The Principles of the Solution of the Senate-house 'riders,' Exemplified by the Solution of Those Proposed in the Earlier Parts of the Examinations of the Years 1848-1851Macmillan & Company, 1851 - 116 sider |
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Resultat 1-5 av 13
Side 41
... horizontal plane from observations made at two given stations in the plane . ( B ) . The angular elevation of a tower at a place A due south of it is 30 ° , and at a place B , due west of A , and at a distance a from it the elevation is ...
... horizontal plane from observations made at two given stations in the plane . ( B ) . The angular elevation of a tower at a place A due south of it is 30 ° , and at a place B , due west of A , and at a distance a from it the elevation is ...
Side 43
... horizontal plane , the distance between the stations being known . ( B ) . If the stations are in the same vertical plane passing through the summit , and the summit ( S ) is ob- served from the further station , but a lower point ( S ) ...
... horizontal plane , the distance between the stations being known . ( B ) . If the stations are in the same vertical plane passing through the summit , and the summit ( S ) is ob- served from the further station , but a lower point ( S ) ...
Side 48
... horizontal plane sufficiently rough to prevent all sliding . A force acts upon it in its own plane and in a given line drawn through the vertex and without the triangle : find by a geometrical construc- tion , or otherwise , the limits ...
... horizontal plane sufficiently rough to prevent all sliding . A force acts upon it in its own plane and in a given line drawn through the vertex and without the triangle : find by a geometrical construc- tion , or otherwise , the limits ...
Side 53
... horizontal table , find the horizontal force which must act on the wedge to keep it at rest . From the equilibrium of the particle O ( fig . 41 ) we have , by the same method as is pursued in ( 4 ) , R = W.cosa ( 1 ) . From the ...
... horizontal table , find the horizontal force which must act on the wedge to keep it at rest . From the equilibrium of the particle O ( fig . 41 ) we have , by the same method as is pursued in ( 4 ) , R = W.cosa ( 1 ) . From the ...
Side 60
... horizontal line through B in C. AC will be the position required . For , by ( 4 ) , 1848 . velocity at C = velocity ... horizontally through a given point . Since the proposition teaches us that the path of a projectile is a parabola ...
... horizontal line through B in C. AC will be the position required . For , by ( 4 ) , 1848 . velocity at C = velocity ... horizontally through a given point . Since the proposition teaches us that the path of a projectile is a parabola ...
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The Principles of the Solution of Senate-house 'riders': Exemplified by the ... Francis James Jameson Uten tilgangsbegrensning - 1851 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning tilgjengelig - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning tilgjengelig - 2015 |
Vanlige uttrykk og setninger
AC² AN.NM Arithmetic arithmetical progression axis bisects body C₁ Cambridge centre of gravity chord CHURCHILL BABINGTON circle cloth cone Conic Sections conjugate hyperbola constant curvature curve cycloid describe diameter direction directrix distance drawn Edition ellipse equations equilibrium Fellow of St fluid focus geometrical given point Hence horizontal hyperbola inches inclined inscribed John's College joining latus-rectum least common multiple Lemma length locus meet mirror move number of seconds oscillation parabola parallel parallelogram particle perpendicular plane polygon pressure prop proportional proposition prove pullies quadrilateral quantity radius ratio rays rectangle refraction right angles sewed shew sides specific gravity spherical square straight line string surface tan² tangent triangle ABC Trinity College tube V₁ vary vertex vertical W₁ weight
Populære avsnitt
Side 4 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.
Side 6 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Side 11 - AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary ; if from any point C of AB, a perpendicular be drawn to AB meeting AP and .BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional of CE and CF.
Side 9 - IF the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Side 4 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.