Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids ; to which are Added, Elements of Plane and Spherical TrigonometryW.E. Dean, 1836 - 311 páginas |
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Página 88
... angle GAB equal to the angle DFE , and join BC . Therefore , because HAG touches the circle ABC , and AC is drawn from the point of con- tact , the angle HAC is equal ( 30. 3. ) to the angle ABC in the alternate segment of the circle ...
... angle GAB equal to the angle DFE , and join BC . Therefore , because HAG touches the circle ABC , and AC is drawn from the point of con- tact , the angle HAC is equal ( 30. 3. ) to the angle ABC in the alternate segment of the circle ...
Página 122
... angles equal about which the sides are propor- tionals . Let the two triangles ABC , DEF have one angle in the one equal to one angle in the other , viz . the angle BAC to the angle EDF , and the sides about two other angles ABC , DEF ...
... angles equal about which the sides are propor- tionals . Let the two triangles ABC , DEF have one angle in the one equal to one angle in the other , viz . the angle BAC to the angle EDF , and the sides about two other angles ABC , DEF ...
Página 214
... angle and a right angle , or between any arc and a quadrant , is called the Complement of that angle , or of that arc . Thus , if BH be perpendicular to AB , the angle CBH is the com- plement of the angle ABC , and the arc HC the ...
... angle and a right angle , or between any arc and a quadrant , is called the Complement of that angle , or of that arc . Thus , if BH be perpendicular to AB , the angle CBH is the com- plement of the angle ABC , and the arc HC the ...
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ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal arc AC base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated diameter divided draw Prob equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROP proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC wherefore