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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Side 1
... parallelograms are equal to one another ; and the diameter bisects them . ( Euc . I. 34. ) ( B ) . If the opposite sides ... parallelogram . * 1848. ( C ) . If the two diameters be drawn , shew that a paral- lelogram will be divided into ...
... parallelograms are equal to one another ; and the diameter bisects them . ( Euc . I. 34. ) ( B ) . If the opposite sides ... parallelogram . * 1848. ( C ) . If the two diameters be drawn , shew that a paral- lelogram will be divided into ...
Side 2
... 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 triangle CDB . But Also BAD BCD ; .. LABD + ZADB = BDC + 4DBC . ZABD + 2DBC ...
... 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 triangle CDB . But Also BAD BCD ; .. LABD + ZADB = BDC + 4DBC . ZABD + 2DBC ...
Side 3
... parallelogram , ADAB . But the triangle AED is common ; therefore ADEC △ BEA . By exactly similar reasoning , AAED ACEB , Again , in the triangles AEB , CED , two angles are equal and AB = DC , therefore EB = ED . Therefore the ...
... parallelogram , ADAB . But the triangle AED is common ; therefore ADEC △ BEA . By exactly similar reasoning , AAED ACEB , Again , in the triangles AEB , CED , two angles are equal and AB = DC , therefore EB = ED . Therefore the ...
Side 16
... parallelograms have the least area of all which circumscribe the ellipse . Let TTTT ( fig . 21 ) be a parallelogram circumscribing an ellipse at the extremities of conjugate diameters CP , CD ; t ̧tëtât another parallelogram ...
... parallelograms have the least area of all which circumscribe the ellipse . Let TTTT ( fig . 21 ) be a parallelogram circumscribing an ellipse at the extremities of conjugate diameters CP , CD ; t ̧tëtât another parallelogram ...
Side 17
... parallelograms circumscribing the ellipse , this constant area is the least . 1850. ( A ) . In the hyperbola the rectangle under the lines intercepted between the centre and the intersections of the axis with the ordinate and tangent ...
... parallelograms circumscribing the ellipse , this constant area is the least . 1850. ( A ) . In the hyperbola the rectangle under the lines intercepted between the centre and the intersections of the axis with the ordinate and tangent ...
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The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning - 2015 |
Almindelige termer og sætninger
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 passager
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.