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, Edlemnts of Plane and Spherical TrigonometryW.E. Dean, 1836 - 311 sider |
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Resultat 1-3 av 55
Side 61
... shewn that FD is greater than CD . Therefore DA is the greatest ; and DE greater than DF , and DF than DC . H LK GB N And because MK , KD are greater ( 13. 1. ) than MD , and MK is equal to MG , the remainder KD is greater ( 5 . Ax ...
... shewn that FD is greater than CD . Therefore DA is the greatest ; and DE greater than DF , and DF than DC . H LK GB N And because MK , KD are greater ( 13. 1. ) than MD , and MK is equal to MG , the remainder KD is greater ( 5 . Ax ...
Side 172
... shewn to be less than ten of those of which the diameter contains 70 . COR . 1. Hence the diameter of a circle being given , the circumference may be found nearly , by making as 7 to 22 , so the given diameter to a fourth proportional ...
... shewn to be less than ten of those of which the diameter contains 70 . COR . 1. Hence the diameter of a circle being given , the circumference may be found nearly , by making as 7 to 22 , so the given diameter to a fourth proportional ...
Side 231
... shewn that the radius is a mean proportional between the cosine and the secant of any arc , so that if A be any arc , sec . A = - 1 1 1 cos . A The versed sines are found by subtracting the cosines from the radius . 5. The preceding ...
... shewn that the radius is a mean proportional between the cosine and the secant of any arc , so that if A be any arc , sec . A = - 1 1 1 cos . A The versed sines are found by subtracting the cosines from the radius . 5. The preceding ...
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Vanlige uttrykk og setninger
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