Synopsis
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. 1870 Excerpt: ...Any angle in the segment A D B must be 1800--51, or 1490; hence we may make the following construction (fig. 183). Through the point A draw any straight line, and from the point B draw another, making with the first an angle of 149; the point M where they cut will be in the arc, and in this way we shall be able to find as many points as we please. Practically, the operation will be rendered easy by cutting out from a piece of pasteboard an angle of 149 then placing the pasteboard in such a manner that its two sides pass through the points A and B, we shall obtain the position of the apex M, and if we move the pasteboard, the point M will describe the arc required. We may fix two pins at A and B and move the pasteboard so that its sides always touch the pins. The same process might be applied if we wished to describe an arc three points in which are given. We should take the angle A M B, and proceed as above described. At one of the extremities of a straight line of limited length to raise a perpendicular. 210. We have already pointed out (§ 79) one solution Fig. 183. Fig. 184. of this problem. Let it be required to raise a perpendicular at A to the straight line A B (fig. 184). Place one point of the compasses at a point O, so that in describing the circle, the other may cut the straight line AB in the point A and some other point B; join B O and produce it till it meets the circle in C, the line C A will be the perpendicular required; for since B C is a diameter, the angle B A C inscribed in a semicircle is a right-angle. When two secants intersecting one another are drawn in the plane of a circle, there are important relations between the segments of the secants determined by the circle. We will proceed to consider them. When two secants start from t...
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