Elements of Geometry and Trigonometry |
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Page 2
... EQUATIONS OF THE POINT AND STRAIGHT LINE -- of the CONIC SECTIONS of the LINE AND PLANE IN SPACE - also , the discussion of the GENERAL EQUATION of the second degree , and of SURFACES OF THE SECOND ORDER .
... EQUATIONS OF THE POINT AND STRAIGHT LINE -- of the CONIC SECTIONS of the LINE AND PLANE IN SPACE - also , the discussion of the GENERAL EQUATION of the second degree , and of SURFACES OF THE SECOND ORDER .
Page 38
... from the first proportion , we have MxQ = NxP , or Nx P = M × Q ; Add each of the members of the last equation to , or subtract it from M.P , and we shall have , M.P ± N.P = M.P ± M.Q ; or ( MN ) x P ( P ± Q ) × M. - But M ± N and P ...
... from the first proportion , we have MxQ = NxP , or Nx P = M × Q ; Add each of the members of the last equation to , or subtract it from M.P , and we shall have , M.P ± N.P = M.P ± M.Q ; or ( MN ) x P ( P ± Q ) × M. - But M ± N and P ...
Page 128
B A D AC2 PC - AP2 , and AB2 - PB2 = AP2 ; we shall have Taking the first equation from the second , and observing that the triangles APC , APB , which are both right angled at P , give AP2 + AP2 = 2AQ2 — 2PQ2 .
B A D AC2 PC - AP2 , and AB2 - PB2 = AP2 ; we shall have Taking the first equation from the second , and observing that the triangles APC , APB , which are both right angled at P , give AP2 + AP2 = 2AQ2 — 2PQ2 .
Page 207
The solution of these equations , when so formed , gives the solution of the problem . No general rule can be given for forming the equations . The equations must be independent of each other , and their number equal to that of the ...
The solution of these equations , when so formed , gives the solution of the problem . No general rule can be given for forming the equations . The equations must be independent of each other , and their number equal to that of the ...
Page 208
hence , Therefore From 1st equ : x = s — y x2 = s2 — Qsy + y2 and By subtracting , 0 = s - 2sy - c2 2sy = s2 ― c2 or PROBLEM I. or Hence , or and From first equation or c2 y = 2s = 4 = AC + 4-9 or x = 5 = BC .
hence , Therefore From 1st equ : x = s — y x2 = s2 — Qsy + y2 and By subtracting , 0 = s - 2sy - c2 2sy = s2 ― c2 or PROBLEM I. or Hence , or and From first equation or c2 y = 2s = 4 = AC + 4-9 or x = 5 = BC .
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Common terms and phrases
ABCD adjacent altitude base become Book called centre chord circle circumference circumscribed common cone consequently contained Cosine Cotang cylinder described determine diameter difference distance divided draw drawn equal equations equivalent expressed extremities faces feet figure follows formed four frustum give given gles greater half hence homologous hypothenuse included inscribed intersection less let fall logarithm manner means measured meet middle multiplied number of sides opposite parallel parallelogram pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quadrant quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment sides similar Sine solid solid angle sphere spherical triangle square straight line suppose taken Tang tangent THEOREM third triangle triangle ABC unit vertex whole
Bibliographic information
| Title | Elements of Geometry and Trigonometry DAVIES' COURSE OF MATHEMATICS |
| Author | Adrien Marie Legendre |
| Editor | Charles Davies |
| Translated by | David Brewster |
| Publisher | A.S. Barnes and Company, 1839 |
| Original from | Harvard University |
| Digitized | 14 Jan 2008 |
| Length | 359 pages |
| Export Citation | BiBTeX EndNote RefMan |