Elements of Geometry and Plane Trigonometry: With an Appendix, and Copious Notes and IllustrationsA. Constable & Company, 1817 - 432 Seiten |
Im Buch
Ergebnisse 6-10 von 100
Seite 20
... join AD . Comparing now the triangles BAC and DCA , the side AB is by supposition equal to CD , AC is common to both , and the contained angle BAC is equal to DCA ; the two triangles ( I. 3. ) are , therefore , equal . But this con ...
... join AD . Comparing now the triangles BAC and DCA , the side AB is by supposition equal to CD , AC is common to both , and the contained angle BAC is equal to DCA ; the two triangles ( I. 3. ) are , therefore , equal . But this con ...
Seite 21
... join AD . The angle CAB is evidently greater than DAB ; but since BA is equal to BD , this angle DAB ( I. 10. ) is equal to ADB , and consequently CAB is greater than ADB . Again , the angle ADB , being an exte- rior angle of the ...
... join AD . The angle CAB is evidently greater than DAB ; but since BA is equal to BD , this angle DAB ( I. 10. ) is equal to ADB , and consequently CAB is greater than ADB . Again , the angle ADB , being an exte- rior angle of the ...
Seite 22
... join CD . B Because BC is equal to BD , the angle BCD is equal to BDC ( I. 10. ) ; but the angle ACD is greater than BCD , and therefore greater than BDC , or ADC ; con- sequently the opposite side AD is greater than AC ( I. 13. ) ; and ...
... join CD . B Because BC is equal to BD , the angle BCD is equal to BDC ( I. 10. ) ; but the angle ACD is greater than BCD , and therefore greater than BDC , or ADC ; con- sequently the opposite side AD is greater than AC ( I. 13. ) ; and ...
Seite 26
... join AG and GC . Because AB and BG are equal to DE and EF , and the contained angle ABG is equal to DEF ; the triangles ABG and DEF ( I. 3. ) are equal , and have equal bases AG and DF .. First , let the triangles ABC and DEF be ...
... join AG and GC . Because AB and BG are equal to DE and EF , and the contained angle ABG is equal to DEF ; the triangles ABG and DEF ( I. 3. ) are equal , and have equal bases AG and DF .. First , let the triangles ABC and DEF be ...
Seite 31
... , a straight line parallel to AB . In AB take any point D , join CD , and at the point C make ( I. 4. ) an an- gle DCE equal to CDA ; CE is paral- lel to AB . E D B For the angles CDA and DCE , thus formed equal ELEMENTS OF GEOMETRY . 31.
... , a straight line parallel to AB . In AB take any point D , join CD , and at the point C make ( I. 4. ) an an- gle DCE equal to CDA ; CE is paral- lel to AB . E D B For the angles CDA and DCE , thus formed equal ELEMENTS OF GEOMETRY . 31.
Inhalt
219 | |
225 | |
226 | |
231 | |
234 | |
237 | |
251 | |
256 | |
101 | |
110 | |
128 | |
142 | |
155 | |
181 | |
199 | |
200 | |
204 | |
257 | |
260 | |
262 | |
264 | |
283 | |
318 | |
338 | |
360 | |
365 | |
Andere Ausgaben - Alle anzeigen
Elements of Geometry, and Plane Trigonometry: With an Appendix, and Very ... University Professor Emeritus John Leslie, Sir Keine Leseprobe verfügbar - 2016 |
Häufige Begriffe und Wortgruppen
ABCD adjacent angle altitude angle ABC angle ADB angle BAC base AC bisect centre chord circle circumference consequently construction contained angle cosine decagon denote describe diameter difference distance diverging lines divided draw equal to BC equilateral triangle equivalent to twice evidently exterior angle given greater half Hence hypotenuse inscribed isosceles triangle join let fall likewise measure parallel perpendicular point G polygon PROB PROP Proposition quadrilateral figure quantities radius ratio rectangle rectangle contained rectilineal figure rhomboid right angles right-angled triangle Scholium segments semicircle semiperimeter sequently side AC sinB sine square of AB square of AC straight line tangent THEOR tion triangle ABC twice the rectangle twice the square vertex vertical angle whence Wherefore