Plane and Solid Geometry: To which is Added Plane and Spherical Trigonometry and Mensuration. Accompanied with All the Necessary Logarithmic and Trigonometric TablesD. Appleton & Company, 1856 - 235 σελίδες |
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Αποτελέσματα 1 - 5 από τα 100.
Σελίδα 14
... consequently , BFC + CFA = CFA + AFD ( A. I. ) : from each taking CFA , we have BFC equal to its opposite angle AFD . In a similar manner we have CFA - DFB . Cor . I. If either of the four angles , formed by the intersection of two ...
... consequently , BFC + CFA = CFA + AFD ( A. I. ) : from each taking CFA , we have BFC equal to its opposite angle AFD . In a similar manner we have CFA - DFB . Cor . I. If either of the four angles , formed by the intersection of two ...
Σελίδα 15
... consequently supplementary with its equal angle BAC . E THEOREM III . Two angles having their corresponding sides ... consequently the angle BAC is equal to KGC . And since AC and DF are parallel they have the same direc- tion in ...
... consequently supplementary with its equal angle BAC . E THEOREM III . Two angles having their corresponding sides ... consequently the angle BAC is equal to KGC . And since AC and DF are parallel they have the same direc- tion in ...
Σελίδα 16
... consequently , the angle KGC is equal to EDF ; therefore ( A. I. ) , the angle BAC is equal to EDF . Secondly . When the sides AB and AC are respectively in opposite directions to DE and DF . As before , we have the angle BAC equal to ...
... consequently , the angle KGC is equal to EDF ; therefore ( A. I. ) , the angle BAC is equal to EDF . Secondly . When the sides AB and AC are respectively in opposite directions to DE and DF . As before , we have the angle BAC equal to ...
Σελίδα 19
... consequently ( A. IV . ) , AD + DB < AD + DE + EB , that is , AD + DB < AE + EB . Again , we have AE < AC + CE , consequently , AE + EB < AC + CE + EB , that is , AE + EB < AC + CB . Comparing these conditions , we have AD + DB < AC + ...
... consequently ( A. IV . ) , AD + DB < AD + DE + EB , that is , AD + DB < AE + EB . Again , we have AE < AC + CE , consequently , AE + EB < AC + CE + EB , that is , AE + EB < AC + CB . Comparing these conditions , we have AD + DB < AC + ...
Σελίδα 23
... Consequently the perpendiculars DE and DF will coincide ( T. XI . ) , and are therefore equal . Secondly . Suppose G to be a point without the bisecting line . If we draw the perpendiculars GF and GH , one of these must cut the ...
... Consequently the perpendiculars DE and DF will coincide ( T. XI . ) , and are therefore equal . Secondly . Suppose G to be a point without the bisecting line . If we draw the perpendiculars GF and GH , one of these must cut the ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
a+b+c altitude apothem bisect centre chord circumference circumscribed cone consequently corresponding cosec Cosine Cotang cube cubic cylinder decimal denote diameter dicular divided draw drawn equation equivalent exterior angles feet figure frustum Geom give greater half hence hypotenuse inches intersection logarithm measure multiplied number of sides opposite parallel parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedral angle polyedron prism PROBLEM proportion pyramid quadrant radii radius ratio rectangle regular inscribed regular polygon respectively equal right angles right-angled triangle Scholium secant sector similar similar triangles Sine slant height solid solve the triangle sphere spherical triangle square straight line subtract suppose surface Tang tangent THEOREM three sides triangle ABC triangular prism volume ΙΟ
Δημοφιλή αποσπάσματα
Σελίδα 35 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Σελίδα 80 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Σελίδα 139 - If a straight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of those lines.
Σελίδα 17 - The sum of all the angles of a polygon is equal to twice as many right angles as the polygon has sides, less two.
Σελίδα 176 - The radius of a sphere is a straight line, drawn from the centre to any point of the...
Σελίδα 182 - Every section of a sphere, made by a plane, is a circle.
Σελίδα 28 - If two triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given A ABC and A'B'C ' with Proof STATEMENTS Apply A A'B'C ' to A ABC so that A'B
Σελίδα 165 - ... bases simply : hence two prisms of the same altitude are to each other as their bases. For a like reason, two prisms of the same base are to each other as their altitudes.
Σελίδα 29 - ... to two sides of the other, but the third side of the first greater than the third side of the second, the angle opposite the third side of the first is.
Σελίδα 13 - If equals be added to unequals, the wholes are unequal. V. If equals be taken from unequals, the remainders are unequal. VI. Things which are double of the same, are equal to one another.