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 1848-1851 |
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Page 2
... angles , each to each . Therefore ABD = 4CDB and ADB = △ CBD ) Hence ABCD is a parallelogram . AB || DC , whence AD || BC . 2. Let ABC = < ADC , and 2BAD = 2BCD . The three angles of triangle ABD = two right - angles = three angles of ...
... angles , each to each . Therefore ABD = 4CDB and ADB = △ CBD ) Hence ABCD is a parallelogram . AB || DC , whence AD || BC . 2. Let ABC = < ADC , and 2BAD = 2BCD . The three angles of triangle ABD = two right - angles = three angles of ...
Page 5
... right angles to AB , equal to a side of the given square . Join AH ; make AHE = < EAH ; and with centre E and radius EA or EH describe the circle AHD . Then , i . e . · AD is bisected in E , rect . of AB , BD + sq . on EB sq . on ED sq ...
... right angles to AB , equal to a side of the given square . Join AH ; make AHE = < EAH ; and with centre E and radius EA or EH describe the circle AHD . Then , i . e . · AD is bisected in E , rect . of AB , BD + sq . on EB sq . on ED sq ...
Page 6
... right angles to it , each will be a tangent to the second circle . 1848 . ( A ) . 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 ...
... right angles to it , each will be a tangent to the second circle . 1848 . ( A ) . 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 7
... right angles , 2 right angles , <  ̧ ̧ ‚ + ¿  ̧A , = 2 right angles , 1 7 ¿  ̧à „ + ¿ ÂÂ1A , = 2 right angles . 9 10 1 sum of alternate angles A , A , A , & c . = twice as many right angles as there are quadrilaterals 2 ( r − 1 ) ...
... right angles , 2 right angles , <  ̧ ̧ ‚ + ¿  ̧A , = 2 right angles , 1 7 ¿  ̧à „ + ¿ ÂÂ1A , = 2 right angles . 9 10 1 sum of alternate angles A , A , A , & c . = twice as many right angles as there are quadrilaterals 2 ( r − 1 ) ...
Page 8
... angle BAC by the line AD . 1 < DAB = CAB . Then But CAB of 2 right angles ; ... DAB of 2 right angles ; ✩ .. DAG 1⁄2 of 2 right angles : whence But ZDAG 5 4 DAB . DAG in segment AFD , = 2 and DAB = △ in segment AED ; ..in segment AFD ...
... angle BAC by the line AD . 1 < DAB = CAB . Then But CAB of 2 right angles ; ... DAB of 2 right angles ; ✩ .. DAG 1⁄2 of 2 right angles : whence But ZDAG 5 4 DAB . DAG in segment AFD , = 2 and DAB = △ in segment AED ; ..in segment AFD ...
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.