Elements of Geometry |
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Results 6-10 of 38
Page 39
... 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 , DH = C , 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 , DH = C , and draw GH ; DGH will be the triangle required ( 36 ) . PROBLEM . 142. One side and two angles of a ...
Page 40
... Solution . The problem admits of two cases . Fig . 81 . Fig . 82 . Fig . 83 . It is necessary , in this case , that the side B should be greater than A , for the angle C being a right or an obtuse angle , it is the greatest of the ...
... Solution . The problem admits of two cases . Fig . 81 . Fig . 82 . Fig . 83 . It is necessary , in this case , that the side B should be greater than A , for the angle C being a right or an obtuse angle , it is the greatest of the ...
Page 41
... 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 42
... 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 43
... 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 BAC 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 angles equiangular equilateral equivalent faces figure formed four right angles frustum GEOM given point gles greater hence homologous sides hypothenuse inclination intersection isosceles triangle JOHN CRERAR LIBRARY join less Let ABC let fall line AC mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism produced proposition radii radius ratio rectangle regular polygon right angles Scholium sector segment semicircle semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three angles triangle ABC triangular prism triangular pyramids vertex vertices whence