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 33
Side 1
... centre A and radius equal to 2AB describe a circle CDF ; and with centre B and radius equal to 2AB describe a circle CEF . Let the circles intersect in C. Join AC , BC . Then AC , BC being radii of the two circles are each equal to 2AB ...
... centre A and radius equal to 2AB describe a circle CDF ; and with centre B and radius equal to 2AB describe a circle CEF . Let the circles intersect in C. Join AC , BC . Then AC , BC being radii of the two circles are each equal to 2AB ...
Side 5
... centre E and radius EA or EH describe the circle AHD . Then , i . e . · AD is bisected in E , rect . of AB , BD + sq ... centre ; and , conversely , those which are equally distant from the centre are equal to one another . ( III.14 ...
... centre E and radius EA or EH describe the circle AHD . Then , i . e . · AD is bisected in E , rect . of AB , BD + sq ... centre ; and , conversely , those which are equally distant from the centre are equal to one another . ( III.14 ...
Side 6
... centre at the centre of the given circle : and since each of the equal straight lines is thus drawn through the extremity of a diameter at right angles to it , each will be a tangent to the second circle . 1848 . ( A ) . The angle at ...
... centre at the centre of the given circle : and since each of the equal straight lines is thus drawn through the extremity of a diameter at right angles to it , each will be a tangent to the second circle . 1848 . ( A ) . The angle at ...
Side 7
... is a right angle , ABC is a semicircle . Draw BE perpendicular to AC , and take F the centre . * If the polygon has six sides , the lines A4 , A4 , will coincide . Then rectangle ABCD = 2 triangle ABC = rectangle on EUCLID . 7.
... is a right angle , ABC is a semicircle . Draw BE perpendicular to AC , and take F the centre . * If the polygon has six sides , the lines A4 , A4 , will coincide . Then rectangle ABCD = 2 triangle ABC = rectangle on EUCLID . 7.
Side 9
... centre of the small circle , ¿ COD = 2 ≤CAD . But by ( 4 ) , the angles ABD , ADB are each double of BAD ; therefore the sum of these 3 angles , or 2 right angles , = 5 ≤ BAD ; .. BAD ( 2 right angles ) , and / < COD = 2 BAD ( 4 right ...
... centre of the small circle , ¿ COD = 2 ≤CAD . But by ( 4 ) , the angles ABD , ADB are each double of BAD ; therefore the sum of these 3 angles , or 2 right angles , = 5 ≤ BAD ; .. BAD ( 2 right angles ) , and / < COD = 2 BAD ( 4 right ...
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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.