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. 1896 Excerpt: ...P is an arc. 374. Proceeding as in exs. 372, 373, P probably lies on an arc; this is so because Z P = 180--% Z A--i Z B = 90;.-.the arc is a semicircumference. The locus would also be seen from ex. 118, p. 47. 375. Take four concentric (r) with radii say 1, 3, 6, 10, and draw the tangents from a point say 12 from the ...
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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. 1896 Excerpt: ...P is an arc. 374. Proceeding as in exs. 372, 373, P probably lies on an arc; this is so because Z P = 180--% Z A--i Z B = 90;.-.the arc is a semicircumference. The locus would also be seen from ex. 118, p. 47. 375. Take four concentric (r) with radii say 1, 3, 6, 10, and draw the tangents from a point say 12 from the center. The points of contact r appear to lie on an arc. If r = 0, the arc passes through the given center; if r = 12, it passes through the given point. Hence if O is the center, P the point, and T a point of tangency, / OPT must be constant; but it is constant v it = 90..-. the locus is a semicircumference on each side of OP, i.e. a circumference. 376. See ex. 305. 377. Suppose A and B the points on circumference c, and k the given arc; after fixing a few points P, P seems to lie on an arc; but then /_ P would be constant; now P is constant, -./_S--i cent. /_ on (BA 1(). The sign depends on whether P is within or without the circle. Page 137.--383. Proof substantially like 382. 384. Let AB be the line, O the center of the 0. If A B is translated to its original position, to the required position A'B', a _L, OM', frontO will bisect A'B' at M', and if produced will be _L AB at X; if X coincides with M, the mid-point of AB, the construction is readily seen. If X does not coincide with M, lay off XB" = MB, and XA"= MA, on XB and XA; then from A," B" draw A"A' XO, B"B' XO. 385. Let Si, S2 be the two positions of the ship. Then SiS2 is given in length and direction;.-. SiS2 is parallel and equal to a known line;.-. the problem reduces to ex. 379, and both positions of the ship are known. 386. Construct Q AA'CC as in ex. 381; then AA' is one diagonal, CA the other, /_ A an included angle. On opposite side.
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Add this copy of Key to Beman and Smith's Plane and Solid Geometry: to cart. $44.02, good condition, Sold by Bonita rated 4.0 out of 5 stars, ships from Newport Coast, CA, UNITED STATES, published 2012 by Nabu Press.
