Elements of Geometry and Trigonometry |
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Page 160
... plane hence , two parallels determine the position of a plane . A B C- D F PROPOSITION III . THEOREM . The ... MN be the plane of the two lines BB , CC , and let AP be perpendicular to these lines at P : then will AP be perpendicular to ...
... plane hence , two parallels determine the position of a plane . A B C- D F PROPOSITION III . THEOREM . The ... MN be the plane of the two lines BB , CC , and let AP be perpendicular to these lines at P : then will AP be perpendicular to ...
Page 161
Adrien Marie Legendre. AP be perpendicular to every straight line of the plane which passes through P , and consequently , to the plane itself . For , through P , draw in the plane MN , any line PQ ; through any point of this line , as Q ...
Adrien Marie Legendre. AP be perpendicular to every straight line of the plane which passes through P , and consequently , to the plane itself . For , through P , draw in the plane MN , any line PQ ; through any point of this line , as Q ...
Page 162
... plane from a point of that plane . For , suppose that two perpen- diculars could be drawn to the plane MN , from the point P. Pass a plane through the perpendiculars , and let PQ be its intersection with MN ; then we should have two per ...
... plane from a point of that plane . For , suppose that two perpen- diculars could be drawn to the plane MN , from the point P. Pass a plane through the perpendiculars , and let PQ be its intersection with MN ; then we should have two per ...
Page 163
... plane MN in the circumference of a circle , whose centre is P , and whose radius , is PB : hence , to draw a perpendi cular to a given plane MN , from a point A , without that plane , find three points B , C , D , of the plane equally ...
... plane MN in the circumference of a circle , whose centre is P , and whose radius , is PB : hence , to draw a perpendi cular to a given plane MN , from a point A , without that plane , find three points B , C , D , of the plane equally ...
Page 164
... plane MN , Pits foot , BC the given line , and A any point of the perpendicular ; draw PD at right angles to BC , and join the point D with A then will AD be perpendicular to BC . For , lay off DB equal to DC , and draw PB , PC , AB ...
... plane MN , Pits foot , BC the given line , and A any point of the perpendicular ; draw PD at right angles to BC , and join the point D with A then will AD be perpendicular to BC . For , lay off DB equal to DC , and draw PB , PC , AB ...
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Common terms and phrases
ABCD ACĀ² adjacent angles altitude angle ACB apothem Applying logarithms base and altitude bisect centre chord circle circumference circumscribed coincide cone consequently convex surface corresponding cosec cosine Cotang cylinder denote diagonals diameter distance divided draw drawn edges equally distant feet find the area Formula frustum given angle given line given point greater hence homologous hypothenuse included angle interior angles intersection less Let ABC log sin logarithm lower base mantissa mean proportional measured by half middle point number of sides opposite parallel parallelogram parallelopipedon perimeter perpendicular plane MN polyedral angle polyedron prism PROBLEM PROPOSITION proved pyramid quadrant radii radius rectangle regular polygons right angles right-angled triangle Scholium secant segment side BC similar sine slant height sphere spherical polygon spherical triangle square Tang tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence