Geometrical Exercises for BeginnersMacmillan, 1882 - 203 sider |
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Side 67
... to find the locus of the middle points of all chords , BC , subtending a right angle at P. Join O to D , the middle point of one such chord BC . Join OP . Since the angle BPC is right , and D is 5-2 FOR BEGINNERS . 67.
... to find the locus of the middle points of all chords , BC , subtending a right angle at P. Join O to D , the middle point of one such chord BC . Join OP . Since the angle BPC is right , and D is 5-2 FOR BEGINNERS . 67.
Side 68
... chord of a given circle passes through a fixed point : to find the locus of the intersection of the tangents at its extremities . R B A D O Let AB be any chord of the circle ( 0 68 GEOMETRICAL EXERCISES.
... chord of a given circle passes through a fixed point : to find the locus of the intersection of the tangents at its extremities . R B A D O Let AB be any chord of the circle ( 0 68 GEOMETRICAL EXERCISES.
Side 69
Samuel Constable. Let AB be any chord of the circle ( 0 ) , passing through the fixed point Q ; let the tangents at A , B intersect in P : it is required to find the locus of the point P. Join OP , cutting AB in D. Join OA , OB ; and ...
Samuel Constable. Let AB be any chord of the circle ( 0 ) , passing through the fixed point Q ; let the tangents at A , B intersect in P : it is required to find the locus of the point P. Join OP , cutting AB in D. Join OA , OB ; and ...
Side 71
... chord of the circles , produced , is the required locus . For , taking any point , P , on the production of ST , and drawing tangents PL , PM , to the circles ( A ) , ( B ) , we have ( III . 36 ) PS . PTPL = PM2 , that is , the tangents ...
... chord of the circles , produced , is the required locus . For , taking any point , P , on the production of ST , and drawing tangents PL , PM , to the circles ( A ) , ( B ) , we have ( III . 36 ) PS . PTPL = PM2 , that is , the tangents ...
Side 81
... chords subtending a right angle at a fixed point being determined by Prop . 57 , the middle point of a chord , satisfying both conditions , is therefore determined , and hence the chord itself . ] 3. Given the base , vertical angle ...
... chords subtending a right angle at a fixed point being determined by Prop . 57 , the middle point of a chord , satisfying both conditions , is therefore determined , and hence the chord itself . ] 3. Given the base , vertical angle ...
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Geometrical Exercises: For Beginners (1882) Samuel Constable Ingen forhåndsvisning tilgjengelig - 2008 |
Geometrical Exercises: For Beginners (1882) Samuel Constable Ingen forhåndsvisning tilgjengelig - 2008 |
Vanlige uttrykk og setninger
AD² angle BAC angle BCD angle equal Assistant-Master base angles base BC BD² Bisect Cambridge centre chord circumference circumscribing circle Clifton College Conic Sections Consider Prop construct Crown 8vo draw ELEMENTARY TREATISE English equal to half escribed circle Eton College Extra fcap Fellow of St find the locus fixed point GEOMETRY given base given circle given difference given ratio given square given straight line Given the base given vertical angle GRAMMAR GREEK half a right half the given Hence HISTORY hypotenuse inscribed internal bisector intersection isosceles J. P. MAHAFFY John's College late Fellow LATIN Let ABC Mathematical meet middle point nine-point circle numerous Illustrations Owens College Oxford parallel perpendicular perpendicular to BC Professor prove quadrilateral R. C. JEBB radius rectangle rectangle contained right angle right-angled triangle School segment semicircle tangent triangle ABC Trinity College vertex W. K. CLIFFORD
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