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 11
... opposite the right angle is called the hypothenuse . Thus , in the triangle ABC , right - angled at A , the side BC is the hypothenuse . B 17. Among the quadrilaterals , we distinguish : · The square , which has its sides equal , and ...
... opposite the right angle is called the hypothenuse . Thus , in the triangle ABC , right - angled at A , the side BC is the hypothenuse . B 17. Among the quadrilaterals , we distinguish : · The square , which has its sides equal , and ...
Side 15
... and the same straight line . PROPOSITION IV . THEOREM . When two straight lines intersect each other , the opposite or ver- tical angles , which they form , are equal . Let AB and DE be two straight A lines , BOOK I. 15.
... and the same straight line . PROPOSITION IV . THEOREM . When two straight lines intersect each other , the opposite or ver- tical angles , which they form , are equal . Let AB and DE be two straight A lines , BOOK I. 15.
Side 16
... opposite or vertical angle ECB ( Ax . 3. ) . Scholium . The four angles formed about a point by two straight lines , which intersect each other , are together equal to four right angles : for the sum of the two angles ACE , ECB , is ...
... opposite or vertical angle ECB ( Ax . 3. ) . Scholium . The four angles formed about a point by two straight lines , which intersect each other , are together equal to four right angles : for the sum of the two angles ACE , ECB , is ...
Side 20
... opposite the equal sides are equal . Let the side BA be equal to the side AC ; then will the angle C be equal to the angle B. A For , join the vertex A , and D the middle point of the base BC . Then , the triangles BAD , DAC , will have ...
... opposite the equal sides are equal . Let the side BA be equal to the side AC ; then will the angle C be equal to the angle B. A For , join the vertex A , and D the middle point of the base BC . Then , the triangles BAD , DAC , will have ...
Side 21
... opposite to the greater an- gle ; and conversely , the greater angle is opposite to the greater side . First , Let the angle C be greater than the angle B ; then will the side AB , opposite C , be greater than AC , opposite B. For ...
... opposite to the greater an- gle ; and conversely , the greater angle is opposite to the greater side . First , Let the angle C be greater than the angle B ; then will the side AB , opposite C , be greater than AC , opposite B. For ...
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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...