Weekly problem papers, with notes. [With] Solutions1885 |
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Page 4
... centre of the circle , OG the radius perpendicular to BC . Make the angle GON equal to the angle BẶC . Draw NM parallel to BC , MP parallel to AC . Then MNP is the triangle required . For arc MG = arc GN , .. angle MOG .. angle MPN ...
... centre of the circle , OG the radius perpendicular to BC . Make the angle GON equal to the angle BẶC . Draw NM parallel to BC , MP parallel to AC . Then MNP is the triangle required . For arc MG = arc GN , .. angle MOG .. angle MPN ...
Page 5
... centre of the square , T the tension of the string , W the weight of the square , R the reaction at A. CL and CM are perpendiculars on the lines of action of T and R. If the vertical through G cuts CN ' in N and AM in F , the line of ...
... centre of the square , T the tension of the string , W the weight of the square , R the reaction at A. CL and CM are perpendiculars on the lines of action of T and R. If the vertical through G cuts CN ' in N and AM in F , the line of ...
Page 7
... centre , R the radius of the circle round ABCD , and let G be the other point of intersection of the circles round FDC , BCE . Then the angle FGC = ADC CRE . = = .. EGC + FGC .. EG and GF are in one straight line . EGC + EBC = two right ...
... centre , R the radius of the circle round ABCD , and let G be the other point of intersection of the circles round FDC , BCE . Then the angle FGC = ADC CRE . = = .. EGC + FGC .. EG and GF are in one straight line . EGC + EBC = two right ...
Page 8
... centre L and radius LE or LF passes through M. And the radii LM , MO are at right angles , .. the circles cut orthogonally . μ μ -9 6. Let the forces be denoted by AB BC Then resolving along the normal at B we have sin C sin A μ μ ...
... centre L and radius LE or LF passes through M. And the radii LM , MO are at right angles , .. the circles cut orthogonally . μ μ -9 6. Let the forces be denoted by AB BC Then resolving along the normal at B we have sin C sin A μ μ ...
Page 10
... centre of the circum - circle .. PQR : BC : CA : AB :: sin A : sin B : sin C = D , which is Their directions pass ... centres of the 10 SOLUTIONS OF.
... centre of the circum - circle .. PQR : BC : CA : AB :: sin A : sin B : sin C = D , which is Their directions pass ... centres of the 10 SOLUTIONS OF.
Common terms and phrases
a+b+c ABCD axis bisects the angle Cambridge centre chord circle round circle will go circum-circle Clifton College coef College conic const coordinates cos² cosec Crown 8vo denote diameter directrix draw Edition ellipse envelope equal equation Fcap fixed point geometrical given expression hyperbola inscribed circle intersect Join Let ABC Let the tangent locus Mathematics meet middle point Monday afternoon morning nine-point circle nth term orthocentre PAPER parabola parallel perpendicular plane Professor quadrilateral radical axis radius right angles shew shewn sides Similarly sin² sin³ straight line tan² tangent Todh triangle Trinity College Tripos velocity vertex vertical
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