Elements of Geometry and Trigonometry |
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Page 24
Then , in the two triangles BAG , DEF , the angles R and E are equal , being right angles , the side BA - ED by hypothesis , and the side BG - EF by construction : consequently , AG - DF ( Prop . V. Cor . ) . But , by hypothesis AC = DF ...
Then , in the two triangles BAG , DEF , the angles R and E are equal , being right angles , the side BA - ED by hypothesis , and the side BG - EF by construction : consequently , AG - DF ( Prop . V. Cor . ) . But , by hypothesis AC = DF ...
Page 32
angles , is equal to twice as many right angles as the polygon has sides , and consequently , equal to the sum of the interior angles plus the exterior angles . Taking from each the sum of the interior angles , and there remains the ...
angles , is equal to twice as many right angles as the polygon has sides , and consequently , equal to the sum of the interior angles plus the exterior angles . Taking from each the sum of the interior angles , and there remains the ...
Page 40
Let M and N be any two magnitudes , and- parts of each : then will MN : M ± M + M N m M N and m N m be like m since For , it is obvious that M × ( N ± N ) = N × ( M ± M ) each is equal to M.N ± N.M. Consequently , the four quan- m ...
Let M and N be any two magnitudes , and- parts of each : then will MN : M ± M + M N m M N and m N m be like m since For , it is obvious that M × ( N ± N ) = N × ( M ± M ) each is equal to M.N ± N.M. Consequently , the four quan- m ...
Page 41
In all cases , the same chord FG belongs to two arcs , FGH , FEG , and consequently also to two segments : but the smaller one is always meant unless the contrary is expressed . An inscribed triangle is one which , like BAC , D * 6 BOOK ...
In all cases , the same chord FG belongs to two arcs , FGH , FEG , and consequently also to two segments : but the smaller one is always meant unless the contrary is expressed . An inscribed triangle is one which , like BAC , D * 6 BOOK ...
Page 44
... the triangles ACD , EOG , will have all their sides equal , each to each , namely , AC = EO , CD = OG , and AD = EG ; hence the triangles are themselves equal ; and , consequently , the angle ACD is equal EOG ( Book I. Prop . X. ) .
... the triangles ACD , EOG , will have all their sides equal , each to each , namely , AC = EO , CD = OG , and AD = EG ; hence the triangles are themselves equal ; and , consequently , the angle ACD is equal EOG ( Book I. Prop . X. ) .
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ABCD adjacent altitude base become Book called centre chord circle circumference circumscribed common cone consequently construction contained corresponding cosine Cotang cylinder described diameter difference distance divided draw drawn equal equation equivalent evident expressed extremities fall figure follows formed formulas four frustum give given gles greater half hence homologous included inscribed intersection less likewise logarithm manner means measured meet middle multiplied opposite parallel parallelogram parallelopipedon pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment shown sides similar sine solid solid angle sphere spherical triangle square straight line Suppose surface taken tang tangent THEOREM third triangle triangle ABC vertex whole
Bibliographic information
| Title | Elements of Geometry and Trigonometry |
| Author | Adrien Marie Legendre |
| Editor | Charles Davies |
| Translated by | David Brewster |
| Edition | 4 |
| Publisher | A.S. Barnes and Company, 1834 |
| Original from | the University of California |
| Digitized | 14 Oct 2010 |
| Length | 359 pages |
| Export Citation | BiBTeX EndNote RefMan |