Euclid's Elements of Geometry: From the Latin Translation of Commandine. To which is Added, a Treatise of the Nature and Arithmetic of Logarithms ; Likewise Another of the Elements of Plain and Spherical Trigonometry : with a Preface ... |
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Page 269
... Quadrant of the Circle BE , be the Right Lines BK , KL , LM , ME ; and produce the Line joining the Points K , A ... Quadrants BÊ , BX , KX , shall be alfo equal 2. 6 . * 6. 11 . equal . And Book XII . Euclid's ELEMENTS . 269.
... Quadrant of the Circle BE , be the Right Lines BK , KL , LM , ME ; and produce the Line joining the Points K , A ... Quadrants BÊ , BX , KX , shall be alfo equal 2. 6 . * 6. 11 . equal . And Book XII . Euclid's ELEMENTS . 269.
Page 270
... Quadrant BE has , fo many , Sides may there be in the Quadrants BX , KX , equal to the Sides BK , KL , LM , ME . Let thofe Sides be BO , OP , PR , RX , KS , ST , TY , YX : And join SO , TP , YR ; and let Perpendiculars be drawn from O ...
... Quadrant BE has , fo many , Sides may there be in the Quadrants BX , KX , equal to the Sides BK , KL , LM , ME . Let thofe Sides be BO , OP , PR , RX , KS , ST , TY , YX : And join SO , TP , YR ; and let Perpendiculars be drawn from O ...
Page 278
... Quadrant will be 25 of these Parts . The Complement of an Arc is the Difference there- of from a Quadrant . A Chord , or Subtenfe , is a Right Line drawn from one End of the Arc to the other . The Right Sine of any Arc , which alfo is ...
... Quadrant will be 25 of these Parts . The Complement of an Arc is the Difference there- of from a Quadrant . A Chord , or Subtenfe , is a Right Line drawn from one End of the Arc to the other . The Right Sine of any Arc , which alfo is ...
Page 282
... Quadrant to any Part of the Quadrant , diftant from each other , by a given Interval be given , thence we may find the Sines of all Arcs to the Double of that Part . For Example , Let all the Sines to 15 Degrees be given ; then , by the ...
... Quadrant to any Part of the Quadrant , diftant from each other , by a given Interval be given , thence we may find the Sines of all Arcs to the Double of that Part . For Example , Let all the Sines to 15 Degrees be given ; then , by the ...
Page 283
... Quadrant be found , by having the Sines and Cofines of one and two Minutes given . For as the Radius is to double the Cofine of 2 ' :: Sine i ' : Difference of the Sines of 1 ' and 3 ' :: Sine 2 ' : Dif- ference of the Sines of o ' and ...
... Quadrant be found , by having the Sines and Cofines of one and two Minutes given . For as the Radius is to double the Cofine of 2 ' :: Sine i ' : Difference of the Sines of 1 ' and 3 ' :: Sine 2 ' : Dif- ference of the Sines of o ' and ...
Common terms and phrases
alfo equal alſo Angle ABC Angle BAC Baſe becauſe bifected Center Circle ABCD Circle EFGH Circumference Cofine Cone confequently contain'd Coroll Cylinder defcrib'd defcribed demonftrated Diameter Diſtance drawn thro equal Angles equiangular equilateral Equimultiples faid fame Altitude fame Multiple fame Plane fame Proportion fame Reafon fecond fhall be equal fimilar fince firft firſt folid Parallelepipedon fome fore ftand fubtending given Right Line Gnomon greater join leffer lefs likewife Logarithm Magnitudes Meaſure Number paffing thro Parallelogram perpendicular Polygon Prifm Priſms Prop PROPOSITION Pyramid Quadrant Ratio Rectangle remaining Angle Right Angles Right Line A B Right Line AB Right-lin'd Figure Right-lin❜d Segment ſhall Sine Solid Sphere Subtangent thefe THEOREM theſe thofe Triangle ABC triplicate Proportion Unity Vertex the Point Wherefore whofe Bafe whole
Popular passages
Page 190 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Page 160 - IF two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals : the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.
Page 63 - DBA ; and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater (16.
Page 152 - ... therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E : but the angle DFG is equal to the angle ACB...
Page 100 - About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle; it is required to describe a triangle about the circle ABC equiangular to the triangle DEF.
Page 17 - CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which...
Page 210 - CD; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, AE, is a parallelogram...
Page 229 - If two right-angled triangles have their hypotenuses equal, and one side of the one equal to one side of the other, the triangles are congruent.
Page 164 - ABG ; (vi. 1.) therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF: but the triangle ABG is equal to the triangle DEF; therefore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. Therefore similar triangles, &c.
Page 93 - If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.