Elements of Geometry |
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Results 6-10 of 38
Page 37
... Solution . Draw the indefinite line DE , and make at the point D the angle EDF equal to the given angle A ; then take DG = B , DHC , and draw GH ; DGH will be the triangle required ( 36 ) . PROBLEM . 142. One side and two angles of a ...
... Solution . Draw the indefinite line DE , and make at the point D the angle EDF equal to the given angle A ; then take DG = B , DHC , and draw GH ; DGH will be the triangle required ( 36 ) . PROBLEM . 142. One side and two angles of a ...
Page 38
... Solution . The problem admits of two cases . 1. If the angle Fig . 80. C ( fig . 80 ) is a right angle , or an obtuse angle , make the angle EDF equal to the angle C ; take DE = A , from the point E , as a centre , and with a radius ...
... Solution . The problem admits of two cases . 1. If the angle Fig . 80. C ( fig . 80 ) is a right angle , or an obtuse angle , make the angle EDF equal to the angle C ; take DE = A , from the point E , as a centre , and with a radius ...
Page 39
... Solution . Take at pleasure three points A , B , C ( fig . 84 ) , in Fig . 84 the circumference of the circle or in the given arc ; join AB and BC , and bisect them by the perpendiculars DE , FG ; the point O , in which these ...
... Solution . Take at pleasure three points A , B , C ( fig . 84 ) , in Fig . 84 the circumference of the circle or in the given arc ; join AB and BC , and bisect them by the perpendiculars DE , FG ; the point O , in which these ...
Page 40
... Solution . Produce AB toward D , make at the point B the angle DBE = C , draw BO perpendicular to BE , and GO per- pendicular to AB , G being the middle of AB ; from the point of meeting O , as a centre , and with the radius OB ...
... Solution . Produce AB toward D , make at the point B the angle DBE = C , draw BO perpendicular to BE , and GO per- pendicular to AB , G being the middle of AB ; from the point of meeting O , as a centre , and with the radius OB ...
Page 41
... Solution . Apply the smaller CD to the greater AB , as many times as it will admit of , for example , twice with a remainder BE . Apply the remainder BE to the line CD , as many times as it will admit of , for example , once with a ...
... Solution . Apply the smaller CD to the greater AB , as many times as it will admit of , for example , twice with a remainder BE . Apply the remainder BE to the line CD , as many times as it will admit of , for example , once with a ...
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Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAD base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal and parallel equiangular equilateral equivalent faces figure four right angles frustum Geom gles greater hence homologous sides hypothenuse inclination inscribed circle isosceles join less let fall line AC manner mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism proposition pyramid S-ABC quadrilateral radii radius ratio rectangle regular polygon right angles right-angled triangle Scholium segment semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three plane angles triangle ABC triangular prism triangular pyramids vertex vertices whence