1846.] 



THE CIVIL ENGINEEERAND ARCHITECT'S JOURNAL 



273 



given by me in your Journal, were soWed from the following principle ; 

 " The angles of the figure can be ascertained in terms of those observed 

 witliout ioTolving any of the containing or subtending parts, and hence 

 the distance between two remote objects become known in terms of the 

 observed angles and the measured base. If to all the angles of any plane 

 polygon figure, right lines be drawn from a point within it the product of 

 the sines of the alternate angles will be equal to each other." From what 

 Mr. TumbuU calls an obscure and neglected proposition in Emerson's 

 Trigonometry, the abnre principle was tery readily drawn. It will be found 

 that the principles on which Emerson, Gregory, and the writer in Col- 

 burn's Journal, proceeded, are but particalar cases of the very extensive 

 one, upon which I have based my solutions ; this will become very evident 

 when the whole of my paper ii published. You will find that I Worded 

 my general principle tiius: — "If any number of lines A, B, C, D, kc, be 

 drawn, the ratio propounded of the ratios of A ; B, B ; C, C ; D, &c., con- 

 tinued in order to A, is a ratio of equality : or which is the same thing, 

 when each becomes an antecedent and a consequent, taken in the above 

 mentioned order, the continied product of the antecedents is equal to the 

 continued product of the consequents." My application of this proposi- 

 tion I consider entirely original ; I have used it in the solution of numerous 

 problems, a few of which I have submitted to you for publication. 

 TVhen the whole of them appear, I will revert to the subject again. 

 I am, Sir, your's sincerely, 



Oliver Byrne. 



Continued from Part XCVII., Oct., 1845. 



(5.) In order to determine the horizontal distance between two remote 

 objects, O, B, a base line A C, of 500 chains was measured, then at each 

 extremity of this base the following angular distances were taken : — At 

 A,OAB = 75=50' = a, BAC = 45°03' = i ; at C, OCB = 75°30' = c, 

 and OCA =: 40" 20' = d. Required the distance between the two objects ? 

 From having the angular distance a, b, c, d, all the other angles of the 

 Fig. 1. figure may be found, without knowing 



the lengths of any of the lines; there- 

 fore, from having the length of any of 

 the lines, under such circumstances the 

 remaining linear distances can easily 

 be determined. Let angle OBA = x, 

 AhC=lSO° — b — c — d = e. Then 

 OBC = e-fj-. .'.OA; OB =sin.j: 

 : sin. a ; OB : OC = sin. c ; sin. 

 (e + x); and OC : OA = sin. (a + b) 

 : sin. d. Hence, sin. (a -f b) sin. c 

 sin. d sin.(e-f i); sin. (a -|- i) sin. c — sin. a sin. d 



sin. (c -f- x) sin. e cos, x + cos, e sin. x 



sm. .r = sm. a 



sin. (e + x) 



But, 



sin. j: 



cot. X -\- cos. e. . • . sin. (o + b) sin. c = sin. a sin. d (sin. e cot. x -f cos. e) ; 

 sin. (a + b) sin. c ^ sin. a sin. d cos.e = sin. a sin. d sin. e cot. x. . • .cot. 



sin. (n 4- ') sin. e 



X — :; — cot. e; or, cot.x^cosec.a sin. (a-fo) sin. ccosec. 



sin. a sin. o sin. e ' ' \ i / 



rfc osec. e — cot. c. Rcle. — Add together, the log. cosec. »fa, the log. sin. of 

 the »um of a und b, the tog. sin. of]c, the log. cosec. ofd, and the log. cusec. of 

 e; the natural number corresponding to this sum, rejecting 50 in the index, 

 mide less by the natural co-tangent of e, will give the natural co-tangent 

 of X. 



log, 



(« + *) 



= 75° 50' 

 = 120 

 — 75 

 = 40 

 = 19 



65 , log. 



30 , log. 



20 , log. 



05 , log. 



cosec. 



sin. 



sin. 



cosec. 



cosec. 



The natural number corresponding =: 

 The natural cot. of 19° 05' (e) =: 



= 100134127 

 = 9 0334445 

 = 99S59416 

 = 10-1889391 

 = 10-4855279 



0-6072758 

 4-0483290 

 2-8905407 



Natural cot. of x = 1-1577823 

 . • . ir = 40' 49' 04" = ABO ; and c -f x = 59" 54' 04" = OBC. Hence, 

 OC = 1JS4-49, and OB = 1493-34 chains respectively, which may be 

 found by Plane Trigonometry. 



Suppose the angles a, b, c, d, remain as before, but it is found impossible 

 to measure AC with any degree of accuracy, OB being on a plane it is 

 measured and found to be 18G09 links. Required the angle x, and the 

 distance AC in feet? In this instance, x = 40" 49' 04" ; and 4C = 

 4112-227 feet. This could not be solved by the rules of pla'ne trigonome- 

 try. 



(6.) There are four stations on the same plane, the linear distance be- 

 tween every two of any three of them being given, as well as the angular 

 distance of the fourth from each of the other three ; to find the remaining 

 parts. 



Let A, B, and C, be the three stations whose distances are known ; 

 ^en<:e the three angles a, b, and c, of 

 the triangle ABC may be found. [See 

 Problem III, in the Journal for Octo- 

 ber, 1845.] Having these three angles, 

 and also m, and n, taken at the fourth 

 station, we can find all the angles of 

 the figure without any reference to the 

 lengths of the lines. 



Let X be the angle made by the dian- 

 onal .4D and the side AB ; then will 

 the angle BCD = 6 — n -f x. Hence, AB ; BC = sin. c ; sin. a ■ CC : 

 BD = sin. (m -f n) ; sin. (6 — n) + x ; and BD : AB = sin. x ; sin. m. 



. - . sin. c sin. (m + n) sin. x =z sin. a sin. : (4 — n) -|- x i sin. m. 



sin. c sin. (m-f n) sin. {(«.— n) -f x} \ » , 



-^ -' -' = — = sin. (0 — «) cot. X -L COS. 



sm, a sin. m 



{I' 



sin. csin,(m + n\ 

 - ")• • ■ • sin. a sin. m " "'"■ ^* ""^ = ''°- (^ " ") ™'- •^- 



sin. c sin. (m + n) 



^ : — r, : — cot. (b — n) = cot. x. .• . cot x=cosec. a sic. 



sin. a sin. m sin. i,i — n) ^ ' 



c cosec. tn sin. (m -{- n) cosec. (6 — n)— cot. (i — n). If i=n, the problem 

 will be indeterminate, for cosec.(i — k)i or cosec. o,is infinite. It also shows 

 that a circle can be described through the four points .4, B, C, D ; so that 

 the point D, when such is the case, will have an infinite number of posi- 

 tions all satisfying the question. When A, B, C, are in the same right 

 line, this problem is readily solved by problem 2. Or when any two of tlie 

 stations A, B, C, and the station D, are in a straight line, the problem 

 falls under the resolution of plane triangles. It may also be remarked, 

 that the angles b and c, are negative when the point A is below the line 

 BC. Rule. — Add together the log. cosec. of a, the log. sine of c, the ksr. 

 cosec. of m, the log. sine nf{m-^n), and the log. cosec. of(b—n) ; the natu- 

 ral number corresponding to this sum after a proper allowance is made in 

 the index, made less by the natural cotangent if (i — n) u'ill give the naln- 

 ral cotangent of x. 



This problem, of which the three following examples are particular 

 cases, was first proposed by Richard Townley, in the Philosophical Tram, 

 actions, where also is inserted solutions to the different cases by John Col- 

 lins, No. 69, 1671. The first of these examples is given in the Lady's 

 Diary for 1723, by John Richards, and answered the following year ty 

 John Topham. Dr. Hutton in his edition of the Diaries gives an addi- 

 tional solution, with a geometrical construction. Professor Leybourn, in 

 his edition of the Diaries, gives a second additional solution, also with a 

 geometrical construction ; and in his appendix to the same work, he gives 

 a general one from " Cagnoh's Trigonometry." However, none of lhe.se 

 solutions are as simple or as practical as the one htre given. Cao-noli 

 < sin. c sin. (in -f n) 



makes cot. x = cot. (B_«) < ^Z'HACTin. m sin. (B - „) ~ ^ i 



It 



sin.c sin, (m -|- «) 



.-1? 



should be cot. x = cot. (B — n) , .. , ,, — n 



^ ' ( 5111. bAC sin. m cos. (B — n) 



This must be a misprint, for it cannot be supposed that either Cagnoli or 

 Leybourn could make such a mistake. 



I. It is required to find the distance from the Edysfone Lighthouse to 

 Plymouth, Start Point, and the Lizard, respectively, from the following 

 data: — The distance of Plymouth from the Lizard CO miles, from Lizard 

 to Start Point 70 miles, and from Start Point to Plymouth 20 miles; also 

 Plymouth bears due north from the Edystone Rock, the Lizard W.S.M'. 

 and Start Point N. by E. Let E represent the position of the Edystone, 

 L the Lizard, P Plymouth, and S Start Point. When a right line is sui:- 



