Elements of Geometry and Trigonometry: With Practical ApplicationsR.S. Davis & Company, 1862 - 490 sider |
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Resultat 1-5 av 55
Side 41
... consequently A D is parallel to BC ( Prop . XX . ) ; therefore the figure ABCD is a parallelogram . PROPOSITION XXXIV . - THEOREM . 113. The diagonals of every parallelogram bisect each other . Let ABCD be a parallelogram , and A C , DB ...
... consequently A D is parallel to BC ( Prop . XX . ) ; therefore the figure ABCD is a parallelogram . PROPOSITION XXXIV . - THEOREM . 113. The diagonals of every parallelogram bisect each other . Let ABCD be a parallelogram , and A C , DB ...
Side 78
... consequently equal ( Prop . XVIII . Bk . I. ) . If from the quadrilateral A B E D , we take away the tri- angle A D F , there will remain the parallelogram ABEF ; and if from the same quadrilateral A B E D , we take away the triangle ...
... consequently equal ( Prop . XVIII . Bk . I. ) . If from the quadrilateral A B E D , we take away the tri- angle A D F , there will remain the parallelogram ABEF ; and if from the same quadrilateral A B E D , we take away the triangle ...
Side 88
... consequently the tri- angle and the square have the same altitude A B ( Prop . L A K I B D C F E G XXV . Bk . I. ) ; and they also have the same base BH ; hence the triangle is equivalent to half the square ( Prop . II . ) . The ...
... consequently the tri- angle and the square have the same altitude A B ( Prop . L A K I B D C F E G XXV . Bk . I. ) ; and they also have the same base BH ; hence the triangle is equivalent to half the square ( Prop . II . ) . The ...
Side 90
... consequently , since the triangles HBN , A B C are both right - angled , and have also the sides BH , B A equal , their hypothenuses BN , B C are equal ( Prop . VI . Cor . , Bk . I. ) . But BC is equal to BF ; therefore BN is equal to ...
... consequently , since the triangles HBN , A B C are both right - angled , and have also the sides BH , B A equal , their hypothenuses BN , B C are equal ( Prop . VI . Cor . , Bk . I. ) . But BC is equal to BF ; therefore BN is equal to ...
Side 91
... consequently , BD2 BC2 + CD - 2 BCX CD ( Prop . IX . ) . By adding A D2 to each of these equals , we have 2 = 2 BD2 + AD2 = BC2 + CD2 + A D2 — 2 BC × CD . But the two right - angled triangles A D B , A D C give 2 A B2 = BD2 + AD , and ...
... consequently , BD2 BC2 + CD - 2 BCX CD ( Prop . IX . ) . By adding A D2 to each of these equals , we have 2 = 2 BD2 + AD2 = BC2 + CD2 + A D2 — 2 BC × CD . But the two right - angled triangles A D B , A D C give 2 A B2 = BD2 + AD , and ...
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Elements of Geometry and Trigonometry;: With Practical Applications Benjamin Greenleaf Uten tilgangsbegrensning - 1863 |
Elements of Geometry and Trigonometry: With Practical Applications Benjamin Greenleaf Uten tilgangsbegrensning - 1869 |
Elements of Geometry and Trigonometry: With Practical Applications Benjamin Greenleaf Uten tilgangsbegrensning - 1867 |
Vanlige uttrykk og setninger
A B C ABCD adjacent angles altitude angle equal base bisect centre chord circle circumference circumscribed cone convex surface cosec cosine Cotang cylinder diagonal diameter distance divided drawn equal Prop equilateral triangle equivalent exterior angle feet formed frustum gles greater half the sum hence homologous hypothenuse inches included angle inscribed less Let ABC line A B logarithm logarithmic sine mean proportional measured by half multiplied number of sides parallel parallelogram parallelopipedon pendicular perimeter perpendicular polyedron prism PROBLEM PROPOSITION pyramid quadrantal radii radius ratio rectangle regular polygon right angles right-angled triangle rods Scholium secant segment side A B similar sine slant height solidity solve the triangle sphere spherical polygon spherical triangle Tang tangent THEOREM triangle ABC triangle equal trigonometric functions vertex
Populære avsnitt
Side 35 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 57 - 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.
Side 117 - Through a given point to draw a straight line parallel to a given straight line, Let A be the given point, and BC the given straight line : it is required to draw through the point A a straight line parallel to BC.
Side 50 - If any number of magnitudes are proportional, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let A : B : : C : D : : E : F; then will A : B : : A + C + E : B + D + F.
Side 77 - Two rectangles having equal altitudes are to each other as their bases.
Side 158 - If a straight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of those lines.
Side 313 - FRACTION is a negative number, and is one more tftan the number of ciphers between the decimal point and the first significant figure.
Side 314 - The logarithm of any POWER of a number is equal to the product of the logarithm of the number by the exponent of the power. For let m be any number, and take the equation (Art.
Side 100 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Side 244 - RULE. — Multiply the base by the altitude, and the product will be the area.