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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. 1819. Excerpt: ... s = d 1-&c. 6 T c =1--iL- &c. t = d ---V &c. 3 Hence a small arc exceeds its sine by--nearly the radius exceeds the cosine by-----nearly and the tangent exceeds its arc by--nearly From whence by the trigonometrical tables it will easily be found that the difference between the arch of J.4-6'' and its sine is only 1 and therefore in many cases one may be safely substituted for the other. The difference between the tangent of 53'' and the arc is only 1. It is often of use in these problems to reduce a small arc, sine, &c. expressed in decimals of the radius (unity) to seconds, and the contrary. Which may be done as follows, let d = arc in decimals of the radius, a = the seconds in the arc, then as the sine of l and the arc of lit are nearly equal, we have sin 1: I-:: d:-----= a. Hence also d = a sin 1. sin 1 PROBLEM XL To find how much the time of rising of the sun or a star is advanced by refraction. Solution. Let RS represent part of the sun''s parallel of declination, R the true place of the sun, when it Pig. 56. appears rising at D. Then, the time of rising is advanced by the angle SPR, the measure of which is the arc of the equator HL; and also the arc DR = the horizontal refraction. These expressions are easily proved by fluxions, and some writer! have proved them by principles purely trigonometrical. The small triangle SRD may be considered as a plane triangle and RD: SR:: sin RSD = cos PSQ: rad SR; HL:: rad parallel: rad equat:: sin SP: rad. therefore RD: HL::''cos PSQ X sin SP; rad RD 13 Vcos (lat--decl) X cos (lat decl) See remarks on this problem in Lalande''s Astr. 3d edition, vol. 3. art. 4028. Cagnoli Trig. p. 368. Cor. 1. If RD be taken equal to the diameter of the sun, the time the sun takes in rising will be had. Cor. 2. If RD be ta...
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