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 15
... suppose intersects the circle in Y , Z. Join SY ' , HZ . By ( A ) , SY ' , HZ ' are perpendicular to QZ ' ; and therefore QY ' = SY , QZ ' = HZ . Hence , drawing QKK ' through the centre C , QK.QK ' = QY'.QZ ' SY.HZ BC2 . Therefore the ...
... suppose intersects the circle in Y , Z. Join SY ' , HZ . By ( A ) , SY ' , HZ ' are perpendicular to QZ ' ; and therefore QY ' = SY , QZ ' = HZ . Hence , drawing QKK ' through the centre C , QK.QK ' = QY'.QZ ' SY.HZ BC2 . Therefore the ...
Page 25
... suppose . 1 ; then A becomes the area of a circle whose .. 3 · 14 ... p ; .. A = ( 3 · 14 ... ) ab . Now Therefore 1851 . 3 * α 3297 21 b А 15 ( 3.14 ... ) 5 , 11.77 ... ( A ) . Find the sum of a series of quantities in arith- metical ...
... suppose . 1 ; then A becomes the area of a circle whose .. 3 · 14 ... p ; .. A = ( 3 · 14 ... ) ab . Now Therefore 1851 . 3 * α 3297 21 b А 15 ( 3.14 ... ) 5 , 11.77 ... ( A ) . Find the sum of a series of quantities in arith- metical ...
Page 37
... suppose , ¿ C'OA ' = 2B ' , LA'OB ' = 2C ' . Now LB ' OC ' + ¿ C ' OA ' + LA'OB ′ = 360 ° ; therefore A ' + B'C ' = 180 ° . Let ОA ' = r = OB ' OB ' = OC ' . We have also AB ' AC ' BC ' BA ' B1 CA ' CB ' , and the geometry of the figure ...
... suppose , ¿ C'OA ' = 2B ' , LA'OB ' = 2C ' . Now LB ' OC ' + ¿ C ' OA ' + LA'OB ′ = 360 ° ; therefore A ' + B'C ' = 180 ° . Let ОA ' = r = OB ' OB ' = OC ' . We have also AB ' AC ' BC ' BA ' B1 CA ' CB ' , and the geometry of the figure ...
Page 40
... suppose . Now the rule of proportional parts gives ( if consecutive angles in the tables differ by 1 ' ) therefore Hence S L tan ( 15 ° .53 ' 8 " ) Z tan 15 ° .53 ' 610 60 L tan 15 ° .54 ' L tan 15 ° .53 ' 1588 ; 563 97 8 = 20.158819 ...
... suppose . Now the rule of proportional parts gives ( if consecutive angles in the tables differ by 1 ' ) therefore Hence S L tan ( 15 ° .53 ' 8 " ) Z tan 15 ° .53 ' 610 60 L tan 15 ° .54 ' L tan 15 ° .53 ' 1588 ; 563 97 8 = 20.158819 ...
Page 46
... 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 done , all the statical conditions have been brought in , and the problem of finding the position of ...
... 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 done , all the statical conditions have been brought in , and the problem of finding the position of ...
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.