Elements of Geometry and Trigonometry Translated from the French of A.M. Legendre by David Brewster: Revised and Adapted to the Course of Mathematical Instruction in the United StatesA.S. Barnes & Company, 1849 - 359 sider |
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Side 5
... Problems relating to the First and Third Books , BOOK IV . The Proportions of Figures and the Measurement of Areas , - Problems relating to the Fourth Book , 41 57 68 8888 98 BOOK V. Regular Polygons and the Measurement of the Circle ...
... Problems relating to the First and Third Books , BOOK IV . The Proportions of Figures and the Measurement of Areas , - Problems relating to the Fourth Book , 41 57 68 8888 98 BOOK V. Regular Polygons and the Measurement of the Circle ...
Side 12
... problem . The common name , proposition , is applied indifferently , to theorems , problems , and lemmas . A corollary is an obvious consequence , deduced from one or several propositions . A scholium is a remark on one or several ...
... problem . The common name , proposition , is applied indifferently , to theorems , problems , and lemmas . A corollary is an obvious consequence , deduced from one or several propositions . A scholium is a remark on one or several ...
Side 57
... PROBLEMS RELATING TO THE FIRST AND THIRD BOOKS . PROBLEM I. To divide a given straight line into two equal parts . Let AB be the given straight line . From the points A and B as centres , with a radius greater than the half of AB ...
... PROBLEMS RELATING TO THE FIRST AND THIRD BOOKS . PROBLEM I. To divide a given straight line into two equal parts . Let AB be the given straight line . From the points A and B as centres , with a radius greater than the half of AB ...
Side 58
... PROBLEM II . At a given point , in a given straight line , to erect a perpendicu- lar to this line . Let A be the given point , and BC the given line . A ct Take the points B and C at equal dis- tances from A ; then from the points B ...
... PROBLEM II . At a given point , in a given straight line , to erect a perpendicu- lar to this line . Let A be the given point , and BC the given line . A ct Take the points B and C at equal dis- tances from A ; then from the points B ...
Side 59
... PROBLEM V. To divide a given arc , or a given angle , into two equal parts . First . Let it be required to divide the arc AEB into two equal parts . From the points A and B , as centres , with the same radius , describe two arcs cutting ...
... PROBLEM V. To divide a given arc , or a given angle , into two equal parts . First . Let it be required to divide the arc AEB into two equal parts . From the points A and B , as centres , with the same radius , describe two arcs cutting ...
Andre utgaver - Vis alle
Elements of Geometry and Trigonometry Translated from the French of A.M ... Charles Davies,Adrien Marie Legendre Uten tilgangsbegrensning - 1842 |
Elements of Geometry and Trigonometry: Translated from the French of A.M ... Adrien Marie Legendre,Charles Davies Uten tilgangsbegrensning - 1848 |
Elements of Geometry and Trigonometry Translated from the French of A.M ... Charles Davies Uten tilgangsbegrensning - 1835 |
Vanlige uttrykk og setninger
adjacent altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone consequently convex surface cosine Cotang 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 Scholium secant segment side BC 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
Populære avsnitt
Side 241 - In every plane triangle, the sum of two sides is to their difference as the tangent of half the sum of the angles opposite those sides is to the tangent of half their difference.
Side 251 - Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51° ; then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45' ; required the height of the tower.
Side 109 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their surfaces are to each other as the squares of those sides (Book IV.
Side 91 - Two similar triangles are to each other as the squares described on their homologous sides. Let ABC, DEF, be two similar triangles, having the angle A equal to D, and The angle B=E.
Side 169 - THEOREM. 7?/6 convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude.
Side 41 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Side 155 - AK. The two solids AG, AQ, having the same base AEHD are to each other as their altitudes AB, AO ; in like manner, the two solids AQ, AK, having the same base AOLE, are to each other as their altitudes AD, AM. Hence we have the two proportions, sol.
Side 86 - 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'.
Side 282 - ... 1. To find the length of an arc of 30 degrees, the diameter being 18 feet. ' Ans. 4.712364. 2. To find the length of an arc of 12° 10', or 12£°, the diameter being 20 feet.
Side 93 - ABC : FGH : : ACD : FHI. By the same mode of reasoning, we should find ACD : FHI : : ADE : FIK; and so on, if there were more triangles. And from this series of equal ratios, we conclude that the sum of the antecedents...