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 41
... inscribed in a circle , when its extremities are in the cir- cumference , as AB . An inscribed angle is one which , like BAC , has its vertex in the circumference , and is formed by two chords . C B * Note . In common language , the ...
... inscribed in a circle , when its extremities are in the cir- cumference , as AB . An inscribed angle is one which , like BAC , has its vertex in the circumference , and is formed by two chords . C B * Note . In common language , the ...
Side 42
... inscribed triangle is one which , like BAC , has its three angular points in the circumference . And , generally , an inscribed figure is one , of which all the angles have their vertices in the circumference . The circle is then said ...
... inscribed triangle is one which , like BAC , has its three angular points in the circumference . And , generally , an inscribed figure is one , of which all the angles have their vertices in the circumference . The circle is then said ...
Side 43
... inscribed in a circle is its diameter . PROPOSITION III . THEOREM . A straight line cannot meet the circumference of a circle in more than two points . For , if it could meet it in three , those three points would be equally distant ...
... inscribed in a circle is its diameter . PROPOSITION III . THEOREM . A straight line cannot meet the circumference of a circle in more than two points . For , if it could meet it in three , those three points would be equally distant ...
Side 54
... inscribed angle is measured by half the arc included between its sides . Let BAD be an inscribed angle , and let us first suppose that the centre of the cir- cle lies within the angle BAD . Draw the diameter AE , and the radii CB , CD ...
... inscribed angle is measured by half the arc included between its sides . Let BAD be an inscribed angle , and let us first suppose that the centre of the cir- cle lies within the angle BAD . Draw the diameter AE , and the radii CB , CD ...
Side 55
... inscribed in the same segment are equal ; because they are all measured by the half of the same arc BOC . B Cor . 2. Every angle BAD , inscribed in a semicircle is a right angle ; because it is mea- sured by half the semicircumference ...
... inscribed in the same segment are equal ; because they are all measured by the half of the same arc BOC . B Cor . 2. Every angle BAD , inscribed in a semicircle is a right angle ; because it is mea- sured by half the semicircumference ...
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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...