1 :.' 



NAVIGATION. 



[SOLCTIOX OF TRIASdLRS. 



Or, by logarithms, 



log. tan. ang. at bane-log, porp. log. base + 10 ; 

 log. tan. ang. at Yertex log. bao log. perp. + 10. 



EXAMPLES. 



1. In the right-angled triangle ABC (Fig. 3), page 1039, 

 are given the bate AC- 288. and the perpendicular B C 

 364, to find the angles and the hypotenuse. 



1 To find the angles A, D. 

 Without logarithms. 

 BO 

 AC 



888)384(1 -3333 . . . - tan. A, 53 8'. 

 988 



tan. A- 



96 



864 



90 



With logarithms. 



log. tan. A = log. B C log. A C + 10. 



BC, 384 2-5843 



AC, 288 - 2-459-1 



tan. A. 63 8' 10-1249 



,'. A - 53 8', and B = 90 - A = 30 62-. 



2. To find the hypotenuse A B. 

 AB= ^(AC' 

 288'= 82944 

 384'= 14745G 



230400(430 = AB 

 10 



88) 701 

 704 



00 



AB= AC sec. A. 

 .*. log. AB - log. A C + log. sec. A -10. 



AC, 288 2-4594 



sec. A, 63 8' 10 2219 



AB, 480 2-C813 



In the foregoing example, a table of natural tangents 

 and secants in necessary to enable us to find the required 

 parts without logarithms, and in the shortest manner ; 

 tan. A being determined as above, and then A B from the 

 formula A C eo. A. 



In all works on Navigation, as also in most books on 

 Trigonometry, the calculation of the several cases of right- 

 angled triangles is performed exclusively by logarithms. 

 mer will perceive, from the illustrations furnished 

 that, in general, the work is more expeditious 

 without logarithms than with them ; and that it would 

 be of advantage to the practical navigator if tables more 

 comprrhuiuive than those generally given that is, tables 

 including the natural tangents and secants were bound 

 up with books on navigation. As the preceding examples 

 snow, a single reference to such a table suffices for the 

 discovery of the unknown part ; whereas three references 

 to the logarithmio tables are necessary. In the latter 

 mode of working, fewer figures may appear on the paper, 

 b it tlit time !< npirrl in two soarchings in the table, out 

 , is saved ; and, in nitrations of this kind, re- 

 ferring to tables, and transcriliing the figures, constitute 

 the principal part of the tr< 



The greatest amount of arithmetical work involved in 



any of the preceding solutions occurs in the example at 

 p. 1040 ; but the mul- cin. 64' 17' - *l !' 



tiplication and divi>i< >n 

 operations there indi- 

 cated, according to the 

 common beaten track, 

 may bo pruned of many 

 superfluous figures by 

 using the contracted 

 methods : thus to mul- 

 tiply -8119 by 327, in 

 the shortest way, with- 

 out sacrificing accuracy 

 in the result, write the 

 figuresofthomultiplier 

 in reverse order thus, 

 723 ; and multiplyas in 

 the margin, rejt'i-tini; 

 from the multiplicand 



cos. 64 17'- -6838)205 -49(454 8 



3197 

 2919 



278 

 233 



45 

 46 



723 



again herein the margin. It will be observed, however, t ha t 

 the whole of the division part of this work would have 

 been saved if tan. 64 17' had been supplied by the table s : 

 the entire work would then have stood as hero amu-xi'd. 

 A person but moderately expert in the simple opcra- 

 tau. 64 17' = 1 '3908 tious of arithmetic could execute 

 the work in the margin in less 

 time than it would take to rrfrr 

 to the logarithmic tables, and to 

 transcribe therefrom a single 

 number. And then the nunn ri- 

 cal process, being placed per- 

 B C = 454-80 maneutly before the eye, is very 

 easily revised if error bo sus- 

 pected. These obvious advantages are of sufficient im- 

 portance to justify the entire abandonment of logarithms 

 in those calculations of Navigation and they include 

 nearly all in which right-angled triangles only are con- 

 cerned. When oMique-angled triangles are the sul>. ; 

 of computation, the case is different ; as the logarithmic 

 operation has then x with few exceptions, the ad van 

 over that with the natural numbers, as will bo sufficiently 

 seen in the next article. 



2. In the right-angled triangle ABC (Fig. 3), are given 

 the base A C = 101-9, and the perpendicular B C 195-4, 

 to find the angles ana the hypnU'iiuso. 



Ans. A = 62 27', B = 27 33', A B = 220-3. 



3. Given the base A C = 659-8, and the perpendicular 

 B C = 520-5, to find the other parts. 



Ans. A = 38 1C', B = 51 44', A B = 840'4. 



4. Given the base AC = 35 '5, and the perpendicular 

 B C = 41 -6, to find the other parts. 



Ans. A = 49 31', B =40 29', A B = 54-68. 



5. Given the base A C = 32-76, and the perpendicular 

 B C = 46 '58, to find the vertical angle B. 



Ans. B = 357'. 



6. Given the base AC = 53-66, and the perpendicular 

 B C = 82-01, to find the angles and the hypotenuse. 



Ans. A = 56 48', B - 33 1^, A C = 98. 

 Miscellaneous Examples in the Calculation of Riyht- 



augled Triangles. 



1. The angle of elevation A C D (Fig. 4), of tin 

 of a tower, was found to be 55 64' ; and bom the station 

 B, 100 feet from C in the same straight lino D B, the 

 of elevation ABD was found to bo 33 i!t>' : what 

 was the height of the tower I 



Solution without Logarithms. 

 By right-angled tri:i 

 DC = AD tan. CAD, DB- AD 



tan. BAD, 

 .-. DB DC = BCAD (tan. 



BAD tan. CAD). 

 l!nt BAD- 20' = 56 



40', and C A D- 90 -56 54' =34" 6', 

 . 1! C= 100= AD (tan. 66 40' 

 tan. 34 6'). 



Fig. 4. 



