Elements of Geometry, and Plane and Spherical Trigonometry: With Numerous Practical ProblemsIvison, Phinney, Blakeman & Company, 1865 - 444 σελίδες |
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Σελίδα iii
... construction . ut in adapting the work to the present advanced state of Mathematical du- cation in our best Institutions , it was found necessary to so alter the plan , and the arrangement of subjects , as to make this essentially a new ...
... construction . ut in adapting the work to the present advanced state of Mathematical du- cation in our best Institutions , it was found necessary to so alter the plan , and the arrangement of subjects , as to make this essentially a new ...
Σελίδα v
... 19 Proportion , and its Application to Geometrical Investigations .... 59 BOOK III . Of the Circle , and the Investigation of Theorems dependent on its Properties ....... 88 BOOK IV . Problems in the Construction of Figures in 1 * ( v )
... 19 Proportion , and its Application to Geometrical Investigations .... 59 BOOK III . Of the Circle , and the Investigation of Theorems dependent on its Properties ....... 88 BOOK IV . Problems in the Construction of Figures in 1 * ( v )
Σελίδα vi
With Numerous Practical Problems Horatio Nelson Robinson. BOOK IV . Problems in the Construction of Figures in Plane Geometry .... 111 BOOK V. On the Proportionalities and Measurement of Polygons and Circles . 130 Practical Problems ...
With Numerous Practical Problems Horatio Nelson Robinson. BOOK IV . Problems in the Construction of Figures in Plane Geometry .... 111 BOOK V. On the Proportionalities and Measurement of Polygons and Circles . 130 Practical Problems ...
Σελίδα 47
... construction . = = As AB = BD , and CB = BH , by subtraction , AB — CB = BD - BH ; or AC HD . But NK = AC , being opposite sides of a parallelogram ; and for the same rea- son , KM HD . Therefore , ( Ax . 1 ) , NK = KM , and the figure ...
... construction . = = As AB = BD , and CB = BH , by subtraction , AB — CB = BD - BH ; or AC HD . But NK = AC , being opposite sides of a parallelogram ; and for the same rea- son , KM HD . Therefore , ( Ax . 1 ) , NK = KM , and the figure ...
Σελίδα 57
... construction of the rectangle ADEK Then we are to prove that the rectangle , AE , and the square , GL , are together equivalent to the square , CDFG . The two complementary rectangles , CL and LF , are equal , ( Th . 31 ) . But CL = AH ...
... construction of the rectangle ADEK Then we are to prove that the rectangle , AE , and the square , GL , are together equivalent to the square , CDFG . The two complementary rectangles , CL and LF , are equal , ( Th . 31 ) . But CL = AH ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD altitude angle opposite axis bisected chord circle circumference circumscribed common cone convex surface cos.a cos.c Cosine Cotang diagonal diameter dicular difference distance divided draw equal angles equation equiangular equivalent find the angles four magnitudes frustum given line greater half Hence the theorem homologous hypotenuse included angle inscribed intersect isosceles less Let ABC logarithm measured multiplied N.sine number of sides opposite angles parallelogram parallelopipedon pendicular perpen perpendicular plane ST polyedron PROB PROBLEM produced Prop proportion PROPOSITION prove pyramid quadrantal radii radius rectangle regular polygon right angles right-angled spherical triangle right-angled triangle SCHOLIUM secant segment similar sin.a sin.b sin.c sine solid angles sphere SPHERICAL TRIGONOMETRY square straight line Tang tangent three angles three sides triangle ABC triangular prisms Trigonometry vertex vertical angle volume
Δημοφιλή αποσπάσματα
Σελίδα 322 - 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.
Σελίδα 30 - Therefore all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Σελίδα 123 - In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle ; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.
Σελίδα 58 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Σελίδα 29 - If one side of a triangle is produced, the exterior angle is equal to the sum of the two interior and opposite, angles; and the three interior angles of every triangle are equal to two right angles.
Σελίδα 41 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Σελίδα 96 - The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.
Σελίδα 65 - In a series of equal ratios, any antecedent is to its consequent, as the sum of all the antecedents is to the sum of all the consequents. Let a: 6 = c: d = e :/. Then, by Art.
Σελίδα 77 - FGL ; (vi. 6.) and therefore similar to it ; (vi. 4.) wherefore the angle ABE is equal to the angle FGL: and, because the polygons are similar, the whole angle ABC is equal to the whole angle FGH ; (vi.
Σελίδα 113 - From a given point, to draw a line parallel to a given line. Let A be the given point, and BC the given line.