A Companion to Wrigley's Collection of Examples: Being Illustrations of Mathematical Processes and Methods of Solution |
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Page 99
... ( Eucl . II . 12 ) ABAD + DB2 + 2BD × DE , B and ( Eucl . II . 13 ) E AC2 = AD2 + DC3 — 2 CD × DE = AD2 + DB3- 2BD × DE , whence AB + AC2 = 2AD3 + 2BD3 . D 32. Let ABC be any triangle ; and let BP , CQ be the pendiculars on AC , AB . Then ( ...
... ( Eucl . II . 12 ) ABAD + DB2 + 2BD × DE , B and ( Eucl . II . 13 ) E AC2 = AD2 + DC3 — 2 CD × DE = AD2 + DB3- 2BD × DE , whence AB + AC2 = 2AD3 + 2BD3 . D 32. Let ABC be any triangle ; and let BP , CQ be the pendiculars on AC , AB . Then ( ...
Page 101
... ( Eucl . 1. 32 ) the angle DOF is a right angle . 48. Let A be the centre of the central circle , and B , C the centres of two others . Then , the triangle ABC being equilateral , the angle BAC is one - third of two right an- gles ; and ( ...
... ( Eucl . 1. 32 ) the angle DOF is a right angle . 48. Let A be the centre of the central circle , and B , C the centres of two others . Then , the triangle ABC being equilateral , the angle BAC is one - third of two right an- gles ; and ( ...
Page 102
... ( Eucl . III . 7 ) , therefore OH is less than OL ; and therefore BC is greater than EF . E G L D H F Similarly , if any other line were taken , it might be shown that BC is greater than that line ; therefore , & c . 53. Let the circles ...
... ( Eucl . III . 7 ) , therefore OH is less than OL ; and therefore BC is greater than EF . E G L D H F Similarly , if any other line were taken , it might be shown that BC is greater than that line ; therefore , & c . 53. Let the circles ...
Page 103
... ( Eucl . III . 31 ) the angle AEB is a right angle ; therefore AE is parallel to CD ; and AC is parallel to ED ... ( Eucl . II . 12 ) , MP MO + OP2 + 20F × OP , A and ( Eucl . II . 13 ) NP2 = NO2 + OP2 – 2OG × OP , B whence the squares of MP ...
... ( Eucl . III . 31 ) the angle AEB is a right angle ; therefore AE is parallel to CD ; and AC is parallel to ED ... ( Eucl . II . 12 ) , MP MO + OP2 + 20F × OP , A and ( Eucl . II . 13 ) NP2 = NO2 + OP2 – 2OG × OP , B whence the squares of MP ...
Page 104
... ( Eucl . III . 3. ) A D C B E 65. Let BAC be a circle ; AB any chord in it ; and Cany point in its circumference . Let ... ( Eucl . III . 31 ) the angle CDB is a right angle ; and ( Eucl . II . 13 ) BC2 is less than AB + AC2 by 2AB.AD ; i ...
... ( Eucl . III . 3. ) A D C B E 65. Let BAC be a circle ; AB any chord in it ; and Cany point in its circumference . Let ... ( Eucl . III . 31 ) the angle CDB is a right angle ; and ( Eucl . II . 13 ) BC2 is less than AB + AC2 by 2AB.AD ; i ...
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A Companion to Wrigley's Collection of Examples: Being Illustrations of ... John Thompson Platts,Alfred Wrigley No preview available - 2016 |
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a+b+c a²+b² a²b² AB² AC² axis BC² beam bisects centre of gravity chord circle co-ordinates coefficient cos² cos³ cosec curve diameter draw ellipse equal Eucl Hence hyperbola Join latus rectum Let ABC middle point MULTINOMIAL THEOREMS parabola parallel perpendicular plane point of contact r₁ radius required locus right angle roots S₁ S₂ sec² segment sides Similarly sin² sin³ square Taking the moments tan¹ tan² tangent term transformed equation triangle ABC values velocity vertical weight whence X₁ ΙΟ