Some Propositions in Geometry: In Five PartsWertheimer, Lea and Company, 1884 - 144 páginas |
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Página 22
... sine is Prop . 1. Theorem : - an aliquot part of the are Repetition of the Demonstration - Varied 136 • · 137 2. Theorem : - The part by which an octantal arc is greater than its sine is the tenth part of the arc length 139 Three Modes ...
... sine is Prop . 1. Theorem : - an aliquot part of the are Repetition of the Demonstration - Varied 136 • · 137 2. Theorem : - The part by which an octantal arc is greater than its sine is the tenth part of the arc length 139 Three Modes ...
Página 112
... sine of an arc and the arc itself can never become coincident , nor can the ratio of any fraction of the lesser sine EK to a similar fraction of the greater sine F1 , ever become equal to the ratio of the former to a similar fraction of ...
... sine of an arc and the arc itself can never become coincident , nor can the ratio of any fraction of the lesser sine EK to a similar fraction of the greater sine F1 , ever become equal to the ratio of the former to a similar fraction of ...
Página 135
... SINE . Rolling a Circle or Arc on a Straight Line . - Def .: Let B and C be any two points at a distance less than or not exceeding a quadrantal arc , in the circumference of a circle of which the radius is A B ; and let B be the point ...
... SINE . Rolling a Circle or Arc on a Straight Line . - Def .: Let B and C be any two points at a distance less than or not exceeding a quadrantal arc , in the circumference of a circle of which the radius is A B ; and let B be the point ...
Página 136
... sine N M , of the first arc BM , and the sine np , of the second arc Bn . Then , because the first octantal arc ( BM ) is to the second ( Bn ) as n to n - 1 , the advance or diff DO of the first is to Cd that of the second as n to n - 1 ...
... sine N M , of the first arc BM , and the sine np , of the second arc Bn . Then , because the first octantal arc ( BM ) is to the second ( Bn ) as n to n - 1 , the advance or diff DO of the first is to Cd that of the second as n to n - 1 ...
Página 137
... sine NM ( or BD ) of the first octant into n - 1 equal parts , and divide the sine p n ( equal to BC ) of the second octant into n - 1 equal parts ; and suppose the diff . ( Cd ) belonging to the second octant also divided into n - 1 ...
... sine NM ( or BD ) of the first octant into n - 1 equal parts , and divide the sine p n ( equal to BC ) of the second octant into n - 1 equal parts ; and suppose the diff . ( Cd ) belonging to the second octant also divided into n - 1 ...
Otras ediciones - Ver todas
Some Propositions in Geometry: In Five Parts (Classic Reprint) John Harris Sin vista previa disponible - 2017 |
Some Propositions in Geometry: In Five Parts Associate Professor University of Alberta Canada John Harris Sin vista previa disponible - 2016 |
Some Propositions in Geometry: In Five Parts (Classic Reprint) John Harris Sin vista previa disponible - 2017 |
Términos y frases comunes
arc FM arc q arc-length areally equal centre of description circle's circumference circumscribed connecting arc curvilineal cut off one-third demonstration describe the arc diagonal diameter diff Dinostratus distance divided divisional arc divisional points Draw the chord duplicate ratio entire arc equal angles equals arc equals twice equilateral triangle Euclid's Euclid's Elements geometrical given angle given circle given cube given straight line greater octant half half-arc isosceles triangle Join lesser octant lineal magnitude lineal side manifestly mean proportional number of equal octagon octantal segments one-nth parallelogram point D point g point of bisection polygon Polysection primary arc primary octant PROBLEM Produce Prop quadrantal arc quadratrix radial line rectangle contained regular polygon required number rhombus right angles Scholium semicircle similar triangles sine tangent line terminal point Theorem transverse arc trisect unital increment vertex vertical angle Wherefore
Pasajes populares
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Página 73 - To describe an isosceles triangle, having each of the angles at the base double of the third angle.
Página 41 - To find a fourth proportional to three given straight lines. Let A, B, C be the three given straight lines ; it is required to find a fourth proportional to A, B, C. Take two straight lines DE, DF, containing any angle Book VI. EDF ; and upon these make DG equal to A, GE equal to B, and DH equal to C : and having joined GH, draw EF parallel...
Página 40 - To find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines ; it is required to find a mean proportional between them. Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B draw (9.
Página 50 - Three numbers may be in proportion when the first is to the second as the second is to the third.
Página 106 - PKOPOSITION 46. PROBLEM. To describe a square on a given straight line. Let AB be the given straight line : it is required to describe a square on AB.
Página 29 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Página 74 - ... To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square on the other part.
Página 45 - To inscribe a circle in a given square. Let ABCD be the given square ; it is required to inscribe a circle in ABCD.
Página 128 - CD the triplicate ratio of that • which AE has to CF. Produce AE, GE, HE, and in these produced take EK equal to CF, EL equal to FN, and EM equal to FR; and complete the parallelogram...