ISBN: 1130889955

Paperback, [EAN: 9781130889956], RareBooksClub.com, RareBooksClub.com, Book, [PU: RareBooksClub.com], RareBooksClub.com, This historic book may have numerous typos and missing text. Purch… More...

Paperback, [EAN: 9781130889956], RareBooksClub.com, RareBooksClub.com, Book, [PU: RareBooksClub.com], RareBooksClub.com, 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. 1806 Excerpt: ...on which the elliptical compasses arc constructed, fee Robertson's Conies, Prop. 24, B II.) that IR is equal to the sum or difference of the semi-axis CB, CD, that is, equal to CH. Join GH, FH, and FC; then (lie angles CFH, ICR, being right angles, and the angles HFB, HCQ equal (Eu. 2t and 22. III.), and CH = IR and CB = IQ, the triangles BCF, QCI will be equal inevery respect as may be very easily proved; and, consequently, BF is equal to CQ. It may be proved in a like manner that BG is equal to CP, and therefore FG is equal to the sum or difference of CP and GQ. Again, let a circle be described through the points C, L, K, cutting CB in O, and CD in T, and join TO, OM, OK, and LK. Then because FG and LK are equal, and subtend the same angle at C, the circles CGFH, CKL will be equal; and the angles FCH, OTK, on the fame arc OK, are equal. And the angles OTL, HCG, on the arc OL are equal, therefore; the angles OKM, HFB, are equal (Eu. 21 and 22, 111.); and the triangles OKM, HFB, having the side OK--HF, the fide KM = FB or CO, and the angle OKM = HFB, will have OM = HB = CD, and the angle KOM--FHB. Also the angle TOK is equal to the angle CHF, each being the complement of the angle OCF, and therefore the angles KOM, TOK, together are equal to two right angles, and the points T, O, M, are in a straight line. But since TCO is a right angle, TO is the diameter of the circle CKL, and equal to CH; and, consequently, TM and OM are respectively equal to the axis CB, CD, and therefore the point M is in the ellipse by the proposition referred to above. It is evident from what has been already done, that the first part of the proposition is true when AB, DE, are any two diameters whatever, if instead of a circle being described on CH, as a diameter, a segment CGFH be ..., 5035, World, 15812171, Civilization & Culture, 3825131, Expeditions & Discoveries, 4992, Jewish, 6343224011, Religious, 16244141, Slavery & Emancipation, 15812211, Women in History, 9, History, 1000, Subjects, 283155, Books<

ISBN: 1130889955

Paperback, [EAN: 9781130889956], RareBooksClub.com, RareBooksClub.com, Book, [PU: RareBooksClub.com], RareBooksClub.com, 65, History, 271237, Ancient History & Civilisation, 276416, Archa… More...

Paperback, [EAN: 9781130889956], RareBooksClub.com, RareBooksClub.com, Book, [PU: RareBooksClub.com], RareBooksClub.com, 65, History, 271237, Ancient History & Civilisation, 276416, Archaeology, 271343, Britain & Ireland, 271480, Europe, 271565, North America, 271411, Military History, 772016, Maritime History, 771710, Political History, 772148, Social & Economic History, 772162, Cultural History, 771714, Religious History, 271389, Other Historical Subjects, 771724, Essays, Journals, Letters & True Accounts, 271605, Reference, 771718, World History, 771708, Countries & Regions, 4119251, Academic History, 1025612, Subjects, 266239, Books<

ISBN: 9781130889956

Thomas Leybourn, Paperback, English-language edition, Pub by General Books LLC Books History~~General New-series-of-The-mathematical-repository~~Thomas-Leybourn General Books LLC