Elements of Geometry and Trigonometry: From the Works of A. M. Legendre |
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Page 90
... inscribe a circle in a given triangle . Let ABC be the given triangle . B E D 0 Bisect the angles A. and B , by the lines 40 and BO , meeting in the point ( Prob . V. ) ; from the point O A let fall the perpendiculars OD , OE , OF , .90 ...
... inscribe a circle in a given triangle . Let ABC be the given triangle . B E D 0 Bisect the angles A. and B , by the lines 40 and BO , meeting in the point ( Prob . V. ) ; from the point O A let fall the perpendiculars OD , OE , OF , .90 ...
Page 92
... inscribed angle AMB is measured by half of the same arc : hence , the angle AMB is equal to the angle EBD , and conse- quently , to the given angle . BOOK IV . MEASUREMENT AND RELATION OF POLYGONS . DEFINITIONS 92 GEOMETRY .
... inscribed angle AMB is measured by half of the same arc : hence , the angle AMB is equal to the angle EBD , and conse- quently , to the given angle . BOOK IV . MEASUREMENT AND RELATION OF POLYGONS . DEFINITIONS 92 GEOMETRY .
Page 100
... inscribed circle . For , let DEF be a circle inscribed in the triangle ABC . Draw OD , OE , and OF , to the points of contact , and OA , OB , and OC , to the verti- ces . D B The area of OBC will be equal to OE × BC ; the A area of OAC ...
... inscribed circle . For , let DEF be a circle inscribed in the triangle ABC . Draw OD , OE , and OF , to the points of contact , and OA , OB , and OC , to the verti- ces . D B The area of OBC will be equal to OE × BC ; the A area of OAC ...
Page 137
... inscribed in it . 1o . Let ABCF be a regular polygon : then can the circumference of a circle be circumscribed about it . For , through three consecutive ver- tices A , B , C , describe the circum- ference of a circle ( B. III ...
... inscribed in it . 1o . Let ABCF be a regular polygon : then can the circumference of a circle be circumscribed about it . For , through three consecutive ver- tices A , B , C , describe the circum- ference of a circle ( B. III ...
Page 138
... inscribed polygon . For , the sides are equal , because they are chords of equal arcs , and the angles are equal , because they are measured by halves of equal arcs . If the vertices A , B , C , & c . , of a regular inscribed polygon be ...
... inscribed polygon . For , the sides are equal , because they are chords of equal arcs , and the angles are equal , because they are measured by halves of equal arcs . If the vertices A , B , C , & c . , of a regular inscribed polygon be ...
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Common terms and phrases
AB² ABCD altitude apothem Applying logarithms centre chord circle circumference cone consequently convex surface cosec Cosine Cotang cylinder demonstrated in Book denote diameter distance divided draw edges Equation feet find the area Find the logarithmic following RULE frustum given angle greater hence homologous hypothenuse included angle inscribed intersection less Let ABC linear units log cot log sin lower base lune mantissa multiplied number of sides opposite parallel parallelogram parallelopipedon perpendicular plane MN polar triangle polyedral angle polyedron principle demonstrated prism proportional PROPOSITION proved pyramid quadrant radii radius rectangle regular polygon right angles right-angled triangle Scholium segment similar six right slant height solution sphere spherical angle spherical excess spherical polygon spherical triangle square straight line subtracting Tang tangent THEOREM triangle ABC triangular prism upper base vertex volume whence write the following