Elements of Geometry and Trigonometry |
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Page 46
... circumference described from the centre O , with the radius OB , will pass through the three given points A , B , C. We have now shown that one circumference can always be made to pass through three given points , not in the same ...
... circumference described from the centre O , with the radius OB , will pass through the three given points A , B , C. We have now shown that one circumference can always be made to pass through three given points , not in the same ...
Page 47
... circumference . Let BD be perpendicular to the B radius CA , at its extremity A ; then will it be tangent to the circumfe- rence . For , every oblique line CE , is longer than the perpendicular CA ( Book I. Prop . XV . ) ; hence the ...
... circumference . Let BD be perpendicular to the B radius CA , at its extremity A ; then will it be tangent to the circumfe- rence . For , every oblique line CE , is longer than the perpendicular CA ( Book I. Prop . XV . ) ; hence the ...
Page 48
... circumference ; for , if another could be drawr , it would not be perpendicular to the radius CA ( Book I. Prop . XIV . Sch . ) ; hence in reference to this new tangent , the radius AC would be an oblique line , and the perpendicular ...
... circumference ; for , if another could be drawr , it would not be perpendicular to the radius CA ( Book I. Prop . XIV . Sch . ) ; hence in reference to this new tangent , the radius AC would be an oblique line , and the perpendicular ...
Page 49
... circumferences will cut each other . For , to make an intersection possible , the triangle CAD must be possible . Hence , not only must we have CD < AC + AD , but also the greater radius AD < AC + CD ( Book I. Prop . VII . ) . And ...
... circumferences will cut each other . For , to make an intersection possible , the triangle CAD must be possible . Hence , not only must we have CD < AC + AD , but also the greater radius AD < AC + CD ( Book I. Prop . VII . ) . And ...
Page 51
... circumference : and conversely , if the arcs intercepted are equal , the angles contained by the radii will also be equal . Let C and C be the centres of equal circles , and the angle ACB = DCE . First . Since the angles ACB , DCE , are ...
... circumference : and conversely , if the arcs intercepted are equal , the angles contained by the radii will also be equal . Let C and C be the centres of equal circles , and the angle ACB = DCE . First . Since the angles ACB , DCE , are ...
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Common terms and phrases
adjacent altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone convex surface cosine cotangent cylinder diagonal diameter dicular distance divided draw drawn equal angles equally distant equations equivalent feet figure find the area formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC logarithm measured by half number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE PROBLEM proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABCDE Scholium secant segment similar sine slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex