Elements of Geometry and Trigonometry |
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Page 14
B At the point C , erect CE perpendicular to AB . The angle ACD is the sum of the angles ACE , ECD : therefore ACD + DCB is the sum of the three angles ACE , ECD , DCB : but the first of these three angles is a right angle , and the ...
B At the point C , erect CE perpendicular to AB . The angle ACD is the sum of the angles ACE , ECD : therefore ACD + DCB is the sum of the three angles ACE , ECD , DCB : but the first of these three angles is a right angle , and the ...
Page 16
... if four right angles were formed about the point C , by two lines perpendicular to each other , the same space would be occupied by the four right angles , as by the successive angles ACB , BCD , DCE , ECF , FCA .
... if four right angles were formed about the point C , by two lines perpendicular to each other , the same space would be occupied by the four right angles , as by the successive angles ACB , BCD , DCE , ECF , FCA .
Page 20
... hence the latter two are right angles ; therefore , the line drawn from the vertex of an isosceles triangle to the middle point of its base , is perpendicular to the base , and divides the angle at the vertex into two equal parts .
... hence the latter two are right angles ; therefore , the line drawn from the vertex of an isosceles triangle to the middle point of its base , is perpendicular to the base , and divides the angle at the vertex into two equal parts .
Page 22
From a given point , without a straight line , only one perpendicular can be drawn to that line . Let A be the point , and DE the given line . A Let us suppose that we can draw two perpendiculars , AB , AC .
From a given point , without a straight line , only one perpendicular can be drawn to that line . Let A be the point , and DE the given line . A Let us suppose that we can draw two perpendiculars , AB , AC .
Page 23
Let A be the given point , DE the given line , AB the perpendicular , and AD , AC , AE , the oblique lines . A Produce the perpendicular AB till BF is equal to AB , and draw FC , FD . D F First . The triangle BCF , is equal to the ...
Let A be the given point , DE the given line , AB the perpendicular , and AD , AC , AE , the oblique lines . A Produce the perpendicular AB till BF is equal to AB , and draw FC , FD . D F First . The triangle BCF , is equal to the ...
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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 |