A Companion to Wrigley's Collection of Examples: Being Illustrations of Mathematical Processes and Methods of Solution |
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Page 102
... segment containing an angle equal to the supplement of the given angle , in which are take any point C ; join CA , CB , and pro- duce them to meet the circumfer- ences in D and E ; at which points let the tangents DE , EF be drawn ...
... segment containing an angle equal to the supplement of the given angle , in which are take any point C ; join CA , CB , and pro- duce them to meet the circumfer- ences in D and E ; at which points let the tangents DE , EF be drawn ...
Page 112
... segment ; also the angle ACD being a right angle is equal to the angle CAB ; therefore the triangles ACD , ABC are equiangular , and BA : AC AC : CD . = D 113. At C and D the extremities of the 112 MISCELLANEOUS THEOREMS AND PROBLEMS .
... segment ; also the angle ACD being a right angle is equal to the angle CAB ; therefore the triangles ACD , ABC are equiangular , and BA : AC AC : CD . = D 113. At C and D the extremities of the 112 MISCELLANEOUS THEOREMS AND PROBLEMS .
Page 115
... segment containing an angle equal to the given angle ; bisect the arc ABC in B , and join AB , BC . The triangle ABC is a max- imum . B E D Through B draw BE parallel to AC , and therefore a tangent to the circle at B. Take any point D ...
... segment containing an angle equal to the given angle ; bisect the arc ABC in B , and join AB , BC . The triangle ABC is a max- imum . B E D Through B draw BE parallel to AC , and therefore a tangent to the circle at B. Take any point D ...
Page 117
... segment of a circle , containing an angle equal to half a right angle , in which place the straight line AD equal to the sum of the sides . Join DC , and make the angle DCB equal to the angle ADC . Then ABC is the triangle re- quired ...
... segment of a circle , containing an angle equal to half a right angle , in which place the straight line AD equal to the sum of the sides . Join DC , and make the angle DCB equal to the angle ADC . Then ABC is the triangle re- quired ...
Page 118
... segment ACB containing an angle equal to the given vertical angle , and com- plete the circle . Draw the diameter FE perpendicular to AB , and there- fore bisecting the arc AEB in the point E. Divide AB in the point D in the ratio of ...
... segment ACB containing an angle equal to the given vertical angle , and com- plete the circle . Draw the diameter FE perpendicular to AB , and there- fore bisecting the arc AEB in the point E. Divide AB in the point D in the ratio of ...
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A Companion to Wrigley's Collection of Examples: Being Illustrations of ... John Thompson Platts,Alfred Wrigley No preview available - 2016 |
Common terms and phrases
a+b+c a²+b² a²b³ AB² ABCD AC² angle ABC axis BC² beam centre of gravity chord circle co-ordinates coefficient cos² cos³ cosec² curve diameter draw ellipse equal equilateral Eucl Hence hyperbola Join latus rectum Let ABC middle point MULTINOMIAL THEOREMS parabola parallel perpendicular plane point of contact point of intersection r₁ radius required locus right angle roots S₁ sec² segment sides Similarly sin² sin³ square straight line tan² tangent term transformed equation triangle ABC values vertical weight whence X₁