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. 1869 edition. Excerpt: ...as stated, (Prop. 6), is 3.141592653. The value of the half circumference being now determined, if that of any arc whatever be required, we have merely to divide 3.141592, etc., by 10800, the number of minutes in a semi-circumference, and multiply the quotient by the number of minutes in the arc whose length ...
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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. 1869 edition. Excerpt: ...as stated, (Prop. 6), is 3.141592653. The value of the half circumference being now determined, if that of any arc whatever be required, we have merely to divide 3.141592, etc., by 10800, the number of minutes in a semi-circumference, and multiply the quotient by the number of minutes in the arc whose length is required. But this investigation has been carried far enough for our present purposes. It will be resumed under the subject of Trigonometry. We insert the following beautiful theorem for the trisection of an arc, although not necessary for practical application. Those not acquainted with cubic equations may omit it. PROPOSITION VIII.--THEOREM. Given, the chord of any arc, to determine the chord of one third of such arc. Let AE be the given chord, and conceive its arc divided into three equal parts, as represented by AB, BD, and DE. Through the center draw BCG-, and draw AB. The two A's, CAB and ABF, are equiangular; for, the angle FAB, being at the circumference, is measured by one half the arc BE, which is equal to AB, and the angle BCA, being at the center, is measured by the arc AB; therefore, the angle FAB = the angle BCA; but the angle CBA or FBA, is common to both triangles; therefore, the third angle, CAB, of the one triangle, is equal to the third angle, AFB, of the other, (Th. 12, B. I, Cor. 2), and the two triangles aie equiangular and similar. But the A ACB is isosceles; therefore, the A AFB is also isosceles, and AB = AF, and we have the following proportions: CA: AB:: AB: BF. Now, let AE = e, AB = x, A C = 1. Then AF = x, and EF= c--x, and the proportion becomes, 1: x:: x: BF. Hence, BF= x Also, Fa = 2--x As AE and BG are two chords intersecting each other at the point F, we have, GFx FB = AFx FE, (Th. 17, B. in). That is,
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Add this copy of Elements of Geometry, Plane and Spherical; With to cart. $10.00, good condition, Sold by 3rd St. Books rated 4.0 out of 5 stars, ships from Lees Summit, MO, UNITED STATES, published 1868 by American Book Co..
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Good. Text neat and clean. Binding tight. Worn corners. Leather spine is good but chipped at top and bottom. Some marks on the inner fly leaf. Overall a good copy. Professional book dealer with storefront since 1999. All orders are processed promptly and carefully packaged.
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