First lessons in Plane Geometry. Together with an application of them to the solution of problems, etcCarter and Hendee, 1830 - 12 sider |
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Resultat 1-5 av 44
Side 6
... divided into 360 equal parts , called degrees ; each degree , again , into 60 equal parts , called min- utes ; a minute , again , subdivided into 60 equal parts , called seconds , & c .; and that the magnitude of an angle can thus be ...
... divided into 360 equal parts , called degrees ; each degree , again , into 60 equal parts , called min- utes ; a minute , again , subdivided into 60 equal parts , called seconds , & c .; and that the magnitude of an angle can thus be ...
Side 58
... divided into two equal parts . Q. How can you prove this ? A. The two triangles , ABC and CDB , have the side CB common ; and the angle y is equal to the angle w ; because y and w , are alternate angles , formed by the intersection of ...
... divided into two equal parts . Q. How can you prove this ? A. The two triangles , ABC and CDB , have the side CB common ; and the angle y is equal to the angle w ; because y and w , are alternate angles , formed by the intersection of ...
Side 61
... divided ? E B A. Into as many , as the figure has sides less two . For it is evident , that if from the vertex A , for instance , you draw the diagonal AF , AE , AD , AC , to the vertices F , E , D , C , each of the two triangles AGF ...
... divided ? E B A. Into as many , as the figure has sides less two . For it is evident , that if from the vertex A , for instance , you draw the diagonal AF , AE , AD , AC , to the vertices F , E , D , C , each of the two triangles AGF ...
Side 79
... divided the line AB into equal parts ) ; and the angle x is equal to the angle z ; because these angles are form- ed by the two parallels , DL and BC , being in- tersected by the straight line AB ( Query 11 , Sect . I. ) ; and the angle ...
... divided the line AB into equal parts ) ; and the angle x is equal to the angle z ; because these angles are form- ed by the two parallels , DL and BC , being in- tersected by the straight line AB ( Query 11 , Sect . I. ) ; and the angle ...
Side 80
Francis Joseph Grund. divided into the same number of equal parts as the line AB . Q. Will you now prove the same principle in the case where the line AB is divided into five , six , or more equal parts ? * * QUERY XV . If in a triangle ...
Francis Joseph Grund. divided into the same number of equal parts as the line AB . Q. Will you now prove the same principle in the case where the line AB is divided into five , six , or more equal parts ? * * QUERY XV . If in a triangle ...
Andre utgaver - Vis alle
First lessons in Plane Geometry. Together with an application of them to the ... Francis Joseph Grund Uten tilgangsbegrensning - 1830 |
First Lessons in Plane Geometry: Together with an Application of Them to the ... Francis Joseph Grund Ingen forhåndsvisning tilgjengelig - 2009 |
First Lessons in Plane Geometry: Together with an Application of Them to the ... Francis Joseph Grund Ingen forhåndsvisning tilgjengelig - 2009 |
Vanlige uttrykk og setninger
adjacent angles angle ABC angle ACB angle opposite angle x basis and height bisected called centre chord circum circumference circumscribed circle consequently construct the triangle DEMON diagonal diameter draw the lines equal angles exterior angle figure ABCDEF found by multiplying fourth term geometrical figures geometrical proportion given angle given circle given straight line given triangle hypothenuse isosceles triangle line AB line AC line CD line MN LUDOLPH VAN CEULEN mean proportional number of sides parallel lines parallelogram perpendicular points of division PROBLEM prove quadrilateral Query 11 radii radius ratio rectilinear figure regular polygon ABCDEF remaining sides Remark right angles right-angular triangle Sect semicircle side AB side AC similar triangles smaller SOLUTION square feet square inches square seconds tangent third line three angles three sides trapezoid trian triangle ABC triangle are equal vertex zoid
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Side 211 - HDG, the corresponding sides are proportional (page 70, 4thly) ; therefore we have the proportion ED : DG =AD : HD (II.) This last proportion has the first ratio common with the first proportion ; consequently the two remaining ratios are in a geometrical proportion (Theory of Prop., Prin. 3d) ; that is, we have AD : HD = DG: BD; and as, in...
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