Elements of Geometry and Trigonometry |
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Page 11
... gles four fute , and in Sanct Gabryell's ile vj fute new gles , price of the fute xxa ; summa , Item , thre fute auld gles , . xija ij " vijs vja xxiiij vjd xijs vijs iiij " xix * XXX * v " viija xviijs viija xviijd Item , for ane stane ...
... gles four fute , and in Sanct Gabryell's ile vj fute new gles , price of the fute xxa ; summa , Item , thre fute auld gles , . xija ij " vijs vja xxiiij vjd xijs vijs iiij " xix * XXX * v " viija xviijs viija xviijd Item , for ane stane ...
Page 1425
... GLES occurrence . Fig . 4. Time history of GLES ( black rhombs. various classes is considered in more detail for 1975-2001 in ( Bazilevskaya et al . , 2004 ) . 3. Features of X - ray bursts followed by ground level enhancements It is ...
... GLES occurrence . Fig . 4. Time history of GLES ( black rhombs. various classes is considered in more detail for 1975-2001 in ( Bazilevskaya et al . , 2004 ) . 3. Features of X - ray bursts followed by ground level enhancements It is ...
Page
... gles . See Figure 4.13 for corresponding probe sampling measure- ments . FDDI reconstructions are calculated and plotted at probe sampling points . 4.20 Adaptive FDDI reconstructions . 324 lines - of - sight over 12 view an- gles . See ...
... gles . See Figure 4.13 for corresponding probe sampling measure- ments . FDDI reconstructions are calculated and plotted at probe sampling points . 4.20 Adaptive FDDI reconstructions . 324 lines - of - sight over 12 view an- gles . See ...
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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.