Elements of Geometry and Trigonometry |
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Results 6-10 of 41
Page 40
... corresponding terms will be proportional . Let and then will For since and or therefore , MN : P : Q R : S : : T : V MXR NxS : P × T : Q × V. MxQ = NxP RXV = SXT , we shall have MXQXRXV = NxPxSxT MXRxQxV = N × S × P × T MXR NXS : P × T ...
... corresponding terms will be proportional . Let and then will For since and or therefore , MN : P : Q R : S : : T : V MXR NxS : P × T : Q × V. MxQ = NxP RXV = SXT , we shall have MXQXRXV = NxPxSxT MXRxQxV = N × S × P × T MXR NXS : P × T ...
Page 51
... corresponding arc to the corresponding arc . Suppose , for example , that the angles ACB , BOOK III . 51.
... corresponding arc to the corresponding arc . Suppose , for example , that the angles ACB , BOOK III . 51.
Page 68
... correspond to equal angles at the centre . Thus , if the angles A and O are equal , the arc BC will be similar to DE , the sector BAC to the sector DOE , and the segment whose chord is BC , to the seg- ment whose chord is DE . B E C 4 ...
... correspond to equal angles at the centre . Thus , if the angles A and O are equal , the arc BC will be similar to DE , the sector BAC to the sector DOE , and the segment whose chord is BC , to the seg- ment whose chord is DE . B E C 4 ...
Page 72
... corresponding terms of these proportions together , and observing that the term AEHD may be omit- ted , since it is a multiplier of both the antecedent and the con- sequent , we shall have ABCD : AEGF :: AB × AD : AE × AF . Scholium ...
... corresponding terms of these proportions together , and observing that the term AEHD may be omit- ted , since it is a multiplier of both the antecedent and the con- sequent , we shall have ABCD : AEGF :: AB × AD : AE × AF . Scholium ...
Page 81
... corresponding members together , and observing that BE and DE are equal , we shall have AB2 + AD2 + DC2 + BC2-4AE2 + 4DE . But 4AE2 is the square of 2AE , or of AC ; 4DE2 is the square of BD ( Prop . VIII . Cor . ) : hence the squares ...
... corresponding members together , and observing that BE and DE are equal , we shall have AB2 + AD2 + DC2 + BC2-4AE2 + 4DE . But 4AE2 is the square of 2AE , or of AC ; 4DE2 is the square of BD ( Prop . VIII . Cor . ) : hence the squares ...
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Common terms and phrases
adjacent altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone convex surface cosine Cotang cylinder diagonal diameter dicular distance divided draw drawn equally distant equations equivalent feet figure find the area formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC logarithm number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular perpendicular let fall plane MN polyedron polygon ABCDE PROBLEM proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABCDE Scholium secant segment similar sine slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex
Popular passages
Page 19 - 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.
Page 232 - ... the logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator.
Page 11 - A right-angled triangle is one which has a right angle. The side opposite the right angle is called the hypothenuse.
Page 168 - The radius of a sphere is a straight line drawn from the centre to any point of the surface ; the diameter or axis is a line passing through this centre, and terminated on both sides by the surface.
Page 31 - Hence, the interior angles plus four right angles, is equal to twice as many right angles as the polygon...
Page 18 - America, but know that we are alive, that two and two make four, and that the sum of any two sides of a triangle is greater than the third side.
Page 20 - In an isosceles triangle the angles opposite the equal sides are equal.
Page 86 - 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'.
Page 159 - S-ahc be the smaller : and suppose Aa to be the altitude of a prism, which having ABC for its base, is equal to their difference. Divide the altitude AT into equal parts Ax, xy, yz, &c. each less than Aa, and let k be one of those parts ; through the points of division...
Page 64 - To inscribe a circle in a given triangle. Let ABC be the given triangle. Bisect the angles A and B by the lines AO and BO, meeting at the point 0.