Elements of Geometry and Trigonometry |
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Page 19
... divided into two equal parts . 5. An angle may be bisected . 6. A perpendicular may be drawn to a given straight line , either from a point without , or from a point on the line . 7. A straight line may be drawn , making with a given ...
... divided into two equal parts . 5. An angle may be bisected . 6. A perpendicular may be drawn to a given straight line , either from a point without , or from a point on the line . 7. A straight line may be drawn , making with a given ...
Page 45
... divided into as many triangles , less two , as it has sides , having the point A for a common vertex , and for bases , the sides of the polygon , except the two which form the angle A. It is evident , also , that the sum of the angles ...
... divided into as many triangles , less two , as it has sides , having the point A for a common vertex , and for bases , the sides of the polygon , except the two which form the angle A. It is evident , also , that the sum of the angles ...
Page 46
... divided by the number of angles . PROPOSITION XXVII . THEOREM . The sum of the exterior angles of a polygon is equal to four right angles . Let the sides of the polygon ABCDE be prolonged , in the same order , forming the exterior ...
... divided by the number of angles . PROPOSITION XXVII . THEOREM . The sum of the exterior angles of a polygon is equal to four right angles . Let the sides of the polygon ABCDE be prolonged , in the same order , forming the exterior ...
Page 74
... the angle DOE ACB be divided into 7 angles , by the radii & c .; and DOE into 4 angles , by the radii Oz , each equal to the unit M. Then , suppose Cm , Cn , Cp , Ox , Oy , and 12 )に From the last proposition , the arcs 74 GEOMETRY .
... the angle DOE ACB be divided into 7 angles , by the radii & c .; and DOE into 4 angles , by the radii Oz , each equal to the unit M. Then , suppose Cm , Cn , Cp , Ox , Oy , and 12 )に From the last proposition , the arcs 74 GEOMETRY .
Page 76
... divided into equal parts , each less than DO : there will be at division between D and 0 ; let I draw CI Then the arcs AB , AI , ble , and we shall have ( P. XVI . ) , I • be that point ; and will be commensura- angle ACB : : angle ACI ...
... divided into equal parts , each less than DO : there will be at division between D and 0 ; let I draw CI Then the arcs AB , AI , ble , and we shall have ( P. XVI . ) , I • be that point ; and will be commensura- angle ACB : : angle ACI ...
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Common terms and phrases
ABCD AC² 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