## 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 pages |

### From inside the book

Results 6-10 of 30

Page 12

...

...

**given point**without it . ( XI . 11. ) ( B ) . Prove that equal right lines drawn from a**given point**to a given plane are equally inclined to the plane . Let P be the**given point**; PA , PA ' , two equal straight lines drawn from P to ... Page 13

...

...

**point**P of**a**parabola to make equal angles with the focal distance SP and the diameter at that**point**, prove that SY ...**given**line , shew that the axis - major is equal to the same line . ( B ) . Shew that the axis - major is ... Page 45

...

...

**a point**; and in both cases , shew that any two of the forces are inversely proportional to the perpendiculars drawn on their respective lines of action from any point in the line of action of the third . ( B ) . An uniform heavy rod of ... Page 46

...

...

**a point**( because they are not parallel ) in order that there may be equilibrium . We must therefore suppose the string BP placed in such a position that it shall pass through O the intersection of AR and WG pro- duced . When this is ... Page 47

Francis James Jameson. Let F be the

Francis James Jameson. Let F be the

**given**force acting along CA ( fig . 34 ) , the pressure ( R ) at the hinge B or ...**point**of the plane the axis passes . The result obtained in (**A**) is that the sum of the moments about the fixed ...### Other editions - View all

The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson No preview available - 2018 |

The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson No preview available - 2015 |

### Common terms and phrases

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

### Popular passages

Page 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.

Page 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.

Page 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.

Page 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...

Page 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.