Solutions of the Cambridge Senate-house Problems and Riders for the Year 1875Sir George Greenhill Macmillan and Company, 1876 - 236 páginas |
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... OF HORACE APPLETON HAVEN , OF PORTSMOUTH , N. H. ( Class of 1842. ) 14 April , 1886 . SCIENCE CENTER LIBRARY CAMBRIDGE SENATE HOUSE PROBLEMS AND RIDERS FOR THE YEAR 1875. 0.99 SIGILL COL OV ANGLE ΛΟΝ CHRISTO LEVINVO HARVARD. Math 395.1.
... OF HORACE APPLETON HAVEN , OF PORTSMOUTH , N. H. ( Class of 1842. ) 14 April , 1886 . SCIENCE CENTER LIBRARY CAMBRIDGE SENATE HOUSE PROBLEMS AND RIDERS FOR THE YEAR 1875. 0.99 SIGILL COL OV ANGLE ΛΟΝ CHRISTO LEVINVO HARVARD. Math 395.1.
Página 1
... away the common part ◊ Bba ; And △ Bbc therefore = abcaBC = £ ^ ABC . Δ Also Aab + Aac = AaB + △ AaC = ABC = abc ; therefore bAc is a straight line . B 2. The angles in the same segment of a circle SOLUTIONS OF SENATE-HOUSE ...
... away the common part ◊ Bba ; And △ Bbc therefore = abcaBC = £ ^ ABC . Δ Also Aab + Aac = AaB + △ AaC = ABC = abc ; therefore bAc is a straight line . B 2. The angles in the same segment of a circle SOLUTIONS OF SENATE-HOUSE ...
Página 2
... angles in a fixed point P , prove that the feet of the perpendiculars drawn from O and P to the sides of the ... angle . Through a fixed point O any straight line OPQ is drawn cutting a fixed circle in P and Q , and upon OP and ...
... angles in a fixed point P , prove that the feet of the perpendiculars drawn from O and P to the sides of the ... angle . Through a fixed point O any straight line OPQ is drawn cutting a fixed circle in P and Q , and upon OP and ...
Página 3
... angle ORC is a right angle , and there- fore the segments intersect on a circle described on OC as diameter . iv . Describe an isosceles triangle having each of the angles at the base double of the third angle . Prove that the circle ...
... angle ORC is a right angle , and there- fore the segments intersect on a circle described on OC as diameter . iv . Describe an isosceles triangle having each of the angles at the base double of the third angle . Prove that the circle ...
Página 4
... angle between the planes PAC , PBC and the angle between the planes PAB , PBC are together equal to the angle between the planes РАС , РАВ . For ( fig . 6 ) PA2 = PB2 = PD2 + DB2 = PD2 + DA ” ; therefore PDA is a right angle , and since ...
... angle between the planes PAC , PBC and the angle between the planes PAB , PBC are together equal to the angle between the planes РАС , РАВ . For ( fig . 6 ) PA2 = PB2 = PD2 + DB2 = PD2 + DA ” ; therefore PDA is a right angle , and since ...
Otras ediciones - Ver todas
Solutions of the Cambridge Senate-House Problems and Riders for the Year 1875 George Greenhill Sin vista previa disponible - 2016 |
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apse line Arabic numbers axes axis Cambridge centre of inertia circle coefficient conic section coordinates cos² cose cosec Crown 8vo curvature curve density diameter differential equation distance dx dy dz dx² Edition electricity ELEMENTARY TREATISE ellipse ellipsoid equal equations of motion equilibrium Extra fcap Fellow of St fixed fluid force given GREEK harmonic Hence hyperbola hypocycloid integral intersect J. P. MAHAFFY John's College late Fellow LATIN length lines of curvature liquid lunar precession magnets middle points normal numerous Illustrations orbit Owens College parabola parallel particle perpendicular plane polar pressure prism Professor prove radius ratio refracting right angles Roman numbers School shew sin sin sin sin² sphere straight line surface tangent triangle ABC Trinity College velocity vertical
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