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 Seiten |
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Seite 41
... radii are equal ; that all the diameters are equal also , and each double of the radius . 3. A portion of the circumference , such as FHG , is called an arc . The chord , or subtense of an arc , is the straight line FG , which joins its ...
... radii are equal ; that all the diameters are equal also , and each double of the radius . 3. A portion of the circumference , such as FHG , is called an arc . The chord , or subtense of an arc , is the straight line FG , which joins its ...
Seite 43
... radii We shall then have AD < AC + CD ( Book I. Prop . VII . * ) ; A or AD < AB . Cor . Hence the greatest line which can be inscribed in a circle is its diameter . PROPOSITION III . THEOREM . A straight line cannot meet the ...
... radii We shall then have AD < AC + CD ( Book I. Prop . VII . * ) ; A or AD < AB . Cor . Hence the greatest line which can be inscribed in a circle is its diameter . PROPOSITION III . THEOREM . A straight line cannot meet the ...
Seite 44
... radii AC , EO , to be equal , if the chord AD is equal to the chord EG , the arcs AMD , ENG will also be equal . For , if the radii CD , OG , be drawn , the triangles ACD , EOG , will have all their sides equal , each to each , namely ...
... radii AC , EO , to be equal , if the chord AD is equal to the chord EG , the arcs AMD , ENG will also be equal . For , if the radii CD , OG , be drawn , the triangles ACD , EOG , will have all their sides equal , each to each , namely ...
Seite 45
... radii CA , CB . Then the two right angled triangles ADC , CDB , will have AC = CB , and CD com- mon ; hence , AD is equal to DB ( Book I. Prop . XVII . ) . Again , since AD , DB , are equal , CG is a perpendicular erected from the mid ...
... radii CA , CB . Then the two right angled triangles ADC , CDB , will have AC = CB , and CD com- mon ; hence , AD is equal to DB ( Book I. Prop . XVII . ) . Again , since AD , DB , are equal , CG is a perpendicular erected from the mid ...
Seite 47
... radii CA , CD . D M A In the right angled triangles CAF , DCG , the hypothenuses CA , CD , are equal ; and the side AF , the half of AB , is equal to the side DG , the half of DE : hence the triangles are equal , and CF is equal to CG ...
... radii CA , CD . D M A In the right angled triangles CAF , DCG , the hypothenuses CA , CD , are equal ; and the side AF , the half of AB , is equal to the side DG , the half of DE : hence the triangles are equal , and CF is equal to CG ...
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Häufige Begriffe und Wortgruppen
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
Beliebte Passagen
Seite 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.
Seite 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.
Seite 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.
Seite 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.
Seite 169 - THEOREM. 7?/6 convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude.
Seite 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.
Seite 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.
Seite 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'.
Seite 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.
Seite 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...