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 σελίδες |
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Σελίδα 1
... triangle upon a given finite straight line . ( Euc . I. 1. ) ( B ) . By a method similar to that used in this problem , describe on a given finite straight line an isosceles tri- angle , the sides of which shall be each equal to twice ...
... triangle upon a given finite straight line . ( Euc . I. 1. ) ( B ) . By a method similar to that used in this problem , describe on a given finite straight line an isosceles tri- angle , the sides of which shall be each equal to twice ...
Σελίδα 2
... triangles ABD , CDB , two sides in one are equal to two sides in the other , each to each , and the third side BD is ... triangle CDB . But Also BAD BCD ; .. LABD + ZADB = BDC + 4DBC . ZABD + 2DBC = 4BDC + ZADB . Adding these equals , 24 ...
... triangles ABD , CDB , two sides in one are equal to two sides in the other , each to each , and the third side BD is ... triangle CDB . But Also BAD BCD ; .. LABD + ZADB = BDC + 4DBC . ZABD + 2DBC = 4BDC + ZADB . Adding these equals , 24 ...
Σελίδα 3
... 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 triangles BCE , ECD , being upon equal ...
... 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 triangles BCE , ECD , being upon equal ...
Σελίδα 4
... triangle whose angle A is a right angle , and BE , CF be drawn bisecting the opposite sides respectively ; shew that ... triangles BHL , BAE , are 4 RIDERS ' . SOLUTIONS OF SENATE - HOUSE 6.
... triangle whose angle A is a right angle , and BE , CF be drawn bisecting the opposite sides respectively ; shew that ... triangles BHL , BAE , are 4 RIDERS ' . SOLUTIONS OF SENATE - HOUSE 6.
Σελίδα 5
Francis James Jameson. Moreover , since the triangles BHL , BAE , are similar , BE , BA are divided , similarly in the points L , K. * Hence the lines CF , KG , CD , EB are divided in the same manner as the line AB . 1850. ( A ) ...
Francis James Jameson. Moreover , since the triangles BHL , BAE , are similar , BE , BA are divided , similarly in the points L , K. * Hence the lines CF , KG , CD , EB are divided in the same manner as the line AB . 1850. ( A ) ...
Άλλες εκδόσεις - Προβολή όλων
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2015 |
Συχνά εμφανιζόμενοι όροι και φράσεις
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
Δημοφιλή αποσπάσματα
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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...
Σελίδα 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.