A Popular Course of Pure and Mixed Mathematics ...: With Tables of Logarithms, and Numerous Questions for Exercise |
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Αποτελέσματα 1 - 5 από τα 100.
Σελίδα 164
to AF , the less , and join FC , GB . Because AF is equal to AG , and AB to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common to the two triangles AFC , AGB ; therefore the ...
to AF , the less , and join FC , GB . Because AF is equal to AG , and AB to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common to the two triangles AFC , AGB ; therefore the ...
Σελίδα 165
off DB equal AC , the less , and join DC ; therefore , because in the triangles DBC , ACB , DB is equal to AC , A and BC common to both the two sides , DB , BC , are equal to the two AC , CB each to each ; and the angle DBC is equal to ...
off DB equal AC , the less , and join DC ; therefore , because in the triangles DBC , ACB , DB is equal to AC , A and BC common to both the two sides , DB , BC , are equal to the two AC , CB each to each ; and the angle DBC is equal to ...
Σελίδα 167
off AE equal to AD ; join DE , and apon it describe ( 1. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the angle BAC . Because AD is equal to AE , and AF is com . mon to the two triangles DAF , EAF ; the ...
off AE equal to AD ; join DE , and apon it describe ( 1. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the angle BAC . Because AD is equal to AE , and AF is com . mon to the two triangles DAF , EAF ; the ...
Σελίδα 168
FG in H , and join CF , CH , CG ; the straight line CH , drawn from the given point C , is perpendicular to the given straight line AB . Because FH is equal to HG , and HC common to the two triangles FHC , GHC , the two sides FH ...
FG in H , and join CF , CH , CG ; the straight line CH , drawn from the given point C , is perpendicular to the given straight line AB . Because FH is equal to HG , and HC common to the two triangles FHC , GHC , the two sides FH ...
Σελίδα 170
AC in E , join BE and produce it to F , and make EF equal to BE ; join also FC , and produce AC to G. Because AE is equal to EC , and BE to EF ; AE , EB are equal to CE , EF , each to each ; the angle AEB is equal ( 15.1 . ) ...
AC in E , join BE and produce it to F , and make EF equal to BE ; join also FC , and produce AC to G. Because AE is equal to EC , and BE to EF ; AE , EB are equal to CE , EF , each to each ; the angle AEB is equal ( 15.1 . ) ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD altitude axis base become called centre circle circumference co-efficient common cone consequently contained convergency curve denominator describe diameter difference distance divided draw drawn equal equation Examples expression factors feet figure fluxion four fraction function given gives greater half Hence infinite integral join less magnitudes manner measure meet meridian method Multiply opposite original parallel parallelogram pass perpendicular plane pole Prob probability Problem projection Prop proportional Proposition pyramid quantity radius ratio rectangle Reduce remaining represented result right angles root rule sides similar sine solid sphere square straight line subtract suppose taken tang tangent Theorem things third triangle vanishing whence Wherefore whole
Δημοφιλή αποσπάσματα
Σελίδα 172 - If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a greater angle. Let...
Σελίδα 191 - 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 on the line between the points of section, is equal to the square on half the line.
Σελίδα 190 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Σελίδα 196 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Σελίδα 192 - If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together •with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.
Σελίδα 177 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
Σελίδα 209 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...
Σελίδα 284 - The bases of a cylinder are the circles described by the two revolving opposite sides of the parallelogram.
Σελίδα 286 - 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. either the sides adjacent to the equal...
Σελίδα 179 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Πληροφορίες βιβλιογραφίας
| Τίτλος | A Popular Course of Pure and Mixed Mathematics ...: With Tables of Logarithms, and Numerous Questions for Exercise |
| Συγγραφέας | Peter Nicholson |
| Εκδότης | G. B. Whittaker, 1825 |
| Πρωτότυπο από | το Πανεπιστήμιο του Μίτσιγκαν |
| Ψηφιοποιήθηκε στις | 11 Ιουν. 2007 |
| Μέγεθος | 372 σελίδες |
| Εξαγωγή αναφοράς | BiBTeX EndNote RefMan |