Key to exercises in Euclid [book 1-6 and parts of book 11,12]. - Brossura

Todhunter, Isaac

Brossura

ISBN 10: 1235969177 ISBN 13: 9781235969171

Casa editrice: RareBooksClub.com, 2012

Riassunto

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1880 Excerpt: ... OB, OC be the three equal straight lines, meeting at 0. From 0 draw OP perpendicular to the plane ABC; join PA, PB, PC. The angles OPA, OPB, OPC are right angles. Hence by I. 47 it can be shewn that PA, PB, PC are all equal; so that P is the centre of the circle described round the triangle ABC. 424. Let the three straight lines meet at 0. Take on these straight lines equal lengths OA, OB, OC. From 0 draw OP perpendicular to the plane ABC. Then OP is the required straight line. For PA, PB, PC are all equal by Exercise 423; therefore the angles POA, POB, POC are all equal, by I. 8. 425. Since EC, DF are perpendicular to the same plane they are parallel, by XI. 6: therefore the points E, C, D, F are in one plane. Let CF, produced if necessary, meet AB at G. Draw the straight line GH parallel to EC or DF. Then GH is at right angles to the plane CAB, by XI. 8; and therefore the angle AGH is a right angle. Similarly a straight line GK drawn-parallel to ED will lie in the plane ECD, and will be at right angles to the plane DAB; therefore the angle AGK is a right angle. Thus the straight line AB will be at right angles to the plane in which GH and GK lie, by XI. 4; and will therefore be at right angles to the straight line CFG which lies in that plane. 426. From a point O draw a straight line OP perpendicular to a given plane, and a straight line OQ perpendicular to the straight line AB lying in that plane. Then PQ will be perpendicular to AB. From Q draw QL parallel to OP; then QL is at-right angles to the plane, by XL 8; so that the angle LQA is a right angle. Also AQO is a right angle, by supposition. Therefore AQ is at right angles to the plane containing OQ and LQ, by XI. 4: so that the angle AQP is a right angle. XI. 13 to 21. 427. AB is perpendicular to th...

Dettagli dell'edizione

• Editore: RareBooksClub.com
• Data di pubblicazione: 2012
• Lingua: Inglese
• ISBN 10: 1235969177
• ISBN 13: 9781235969171
• Rilegatura: Copertina flessibile
• Numero di pagine: 76
• Contatto del produttore: non disponibile
• Persona responsabile: non disponibile