Elements of Geometry, and Plane and Spherical Trigonometry: With Numerous Practical ProblemsIvison, Blakeman, Taylor, 1860 - 453 sider |
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Resultat 1-5 av 34
Side 13
... Parallelogram is a quadrilateral which has its opposite sides parallel . Parallelograms are denominated from the rela- tions both of their sides and angles . 34. A Rectangle is a parallelogram hav- ing its angles right angles . 35. A ...
... Parallelogram is a quadrilateral which has its opposite sides parallel . Parallelograms are denominated from the rela- tions both of their sides and angles . 34. A Rectangle is a parallelogram hav- ing its angles right angles . 35. A ...
Side 15
... parallelogram ABDC is mea- sured by the number of linear units in CD , mul- tiplied by the number of linear units in AC or BD ; the product is the square units in ABDC . For , conceive CD to be composed of any number A B of equal parts ...
... parallelogram ABDC is mea- sured by the number of linear units in CD , mul- tiplied by the number of linear units in AC or BD ; the product is the square units in ABDC . For , conceive CD to be composed of any number A B of equal parts ...
Side 16
With Numerous Practical Problems Horatio Nelson Robinson. as an axiom that the parallelogram will contain 5 × 3 = 15 square units . Hence , to find the areas of right - angled parallelograms , mul tiply the base by the altitude ...
With Numerous Practical Problems Horatio Nelson Robinson. as an axiom that the parallelogram will contain 5 × 3 = 15 square units . Hence , to find the areas of right - angled parallelograms , mul tiply the base by the altitude ...
Side 25
... parallelogram . Draw the diagonal GE ; thus di- viding the parallelogram into two triangles , and the opposite angles G and E each into two angles . A G E Because GB and AE are parallel , the alternate interior angles BGE and GEA are ...
... parallelogram . Draw the diagonal GE ; thus di- viding the parallelogram into two triangles , and the opposite angles G and E each into two angles . A G E Because GB and AE are parallel , the alternate interior angles BGE and GEA are ...
Side 26
... parallelogram are equal . Let AEBG be a parallel- ogram . Then we are to prove that the angle GBE is equal to its ... parallelogram AEBG , are equal . In like manner , we can prove the angle E equal to the angle G. Hence the theorem ...
... parallelogram are equal . Let AEBG be a parallel- ogram . Then we are to prove that the angle GBE is equal to its ... parallelogram AEBG , are equal . In like manner , we can prove the angle E equal to the angle G. Hence the theorem ...
Andre utgaver - Vis alle
Elements of Geometry and Plane and Spherical Trigonometry: With Numerous ... Horatio Nelson Robinson Uten tilgangsbegrensning - 1867 |
Elements of Geometry, and Plane and Spherical Trigonometry: With Numerous ... Horatio Nelson Robinson Uten tilgangsbegrensning - 1865 |
Elements of Geometry, and Plane and Spherical Trigonometry: With Numerous ... Horatio Nelson Robinson Uten tilgangsbegrensning - 1860 |
Vanlige uttrykk og setninger
ABCD altitude angle opposite axis bisected chord circle circumference circumscribed common cone convex surface cos.a cos.b cos.c Cosine Cotang diagonal diameter dicular difference distance divided draw equal angles equation equiangular equivalent find the angles formulæ 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 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 straight line subtracting Tang tangent three angles three sides triangle ABC triangular prisms TRIGONOMETRY vertex vertical angle volume
Populære avsnitt
Side 320 - 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.
Side 65 - If four magnitudes are in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference.
Side 121 - 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.
Side 56 - 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.
Side 34 - Conversely: if two angles of a triangle are equal, the sides opposite to them are equal, and the triangle it itosceles.
Side 126 - To inscribe a regular polygon of a certain number of sides in a given circle, we have only to divide the circumference into as many equal parts as the polygon has sides : for the arcs being equal, the chords AB, BC, CD, &c.
Side 22 - If two parallel lines are cut by a third straight line, the sum of the two interior angles on the same side of the secant line is equal to two right angles.
Side 277 - The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power.
Side 94 - 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.
Side 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.