1010 



NAVIGATION. 





Snmplt L In tlie right-angled triangle ABC (Via. 

 S), ant Kiren A B - 480 feet, and the angle A - :. 

 ' required the perpendicular BC, and the base AC. 



J To find the perpendicular EC. 



Without logarithms. 



BC AB sin. A. 



Sin. A, 63 6', - 7995 



AB- 480 



MMM 



31080 



.'. BC -3837600 



With logarithms. 

 Lo~. B C log. A B + log. sin. A - 10. 



-10 



Log. A B - log. 480 = 2 

 Log. tin. A, 53 6', - 9 



Log. BC log. - 383-8 - 2-6840 

 Hence B C is o83'8 feet 



2. ro find the bate AC. 

 \\ ithout logarithms. 



A C - A B cos. A. 

 Cos. A, 53 6', - -6006 



AB 



480 



480480 

 MOM 



/.AC- 288-2880 



With logarithms. 



Log. AC = log. A B 4- log. cos. A - 10. 



-10 



Log. A B = log. 480 = 2-6812 

 Log. cos. A, 63 6', = 97786 



Log. AC = log. 288 3= 2-4008 

 Hence A C is 28d-3 feet 



In solving the preceding example we havo taken only 

 four places of decimals from the tables ; this is a number 

 quite sufficient for all the ordinary purposes of naviga- 

 tion, and it will be remembered that the present intro- 

 ductory article is only preparatory to the discussion of 

 that subject. 



If, instead of the angle A at the base, the angle B at 

 the vertex had been given, the operation would have 

 been much the same as that above ; for, since A = 90 

 B, sin. A cos. B ; and cos. A = sin. B, because the 

 sine and cosine of an angle are respectively the same as 

 the cosine and sine of the complement of that angle ; so 

 that where we have used above sin. A, we should have 

 used it equal cos. B ; and where we have used cos. A 

 we should have substituted its equal sin. B. This must 

 be observed in solving the last three of the following 

 examples. 



2. In the triangle ABC (last figure) are given 



A B= 291, angle A-47" 55', to find A C, B C. 

 Ans. AC = 195, BC-216. 



3. Given A B- 480, angle A=63 8', to find AC, B C. 



Ans. AC-288, BC-384. 



4. Given A B- 521, angle B-3C C', to find A C, B C. 



Ans. AC-307, BC-421. 

 B. Given A B- 645, angle B- 60 50' to find AC, BC. 



Ans. AC-600, BC-407'4. 

 6. Given AB-08, angle B-33" 12 7 , to find AC, BC. 



Ans. AC-63-66, BC-82.01. 



II. TF*m th ban or perpendicular, and one of tin acute 

 angle* an given. 



Lt the base A 0, and the angle A be given ; then, refer- 

 ring to the diagram at p. 1037, Fig. 1, we are to compare the 

 given side A C with the radius, or unit-line, O A; so that 

 Of A is the tabular triangle now to be referred to. As 

 A C is 6 times O A, therefore, because the proposed and 



tabular triangles are equiangular, B C must be 6 times 

 i A IJ /. time* (( ; that is, 



a 6 tan. A, and c I sec. A. 

 II. -nro tlic following rule : 



I.E. Multiply the given base by the tangent of 

 the angle at the base : the product will bo tl> 

 dicular. Multiply the given base by the secant of the 

 angle at the base : the product will be tl iae. 



In a right-angled triangle either of the two i>erpen<ii- 

 cular sides may be regarded as tin- the other 



as the perpendicular ; so that two distinct rules would be 

 unnecessary. And as in the former case, when the ver- 

 tical angle B is given instead of the base angle A, the 

 former may take the place of the hitt--r in the operation, 

 provided only that we write cotan. for tan., and cosec. 

 for sec. 



But, as already observed at page 1038, the tables of 

 natural sines and co*iiios, which accompany books on 

 navigation, do not, in general, furnish the natural 

 tangents and secants ; so that, if this tal.Ie IHJ employed 

 in tlir work, we inu-t make use of the relations established 

 at page ItKM, namely, 



sin. A 1 _ cos. B 



*" A = cos. A' 8CC ' A * c-oTA : 00tan - B " *>. I f 



C08ea B - slnTB 



The working formula) for the present case are as 

 follows : 

 perp. = base X tan. ang. at base ; hyp. base X sec. 



ang. at base. 

 Or, by logarithms, 



log. perp. = log. base + log. tan. ang. at base 10 ; 

 Io_r. hyp. = log. base + lg- sec - an 8- a * b* 86 10. 

 The first of these formate, if the table referred to be 

 limited to natural sines and cosines only, must be 

 replaced by 



base X sin, ang. at base . base 



P er P- = cos. ang. at base ' ^' = cos. ang. at base" 

 agreeably to the relations given above. 



1. In the right-angled triangle ABC, Fig. 3, arc givon 

 the base AC = 327, and the angle A = 51 17': required 

 the perpendicular B C, and the hypotenuse A B. 



1. To find the perpendicular B C. 



Without logarithms. 

 AC sin. A 



BC 



sin. A 54 C 



cos. A 



17', - '8119 

 AC = 327 



60833 

 18238 

 24357 



(See the remarks 

 at page 10-1-'.) 



cos. 54 17' = -5838)205 I'.i:iii54 -8 = B C. 

 With logarithms. 



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



10 

 AC -327 ...... L'.MT, 



tan. A, 5 17' ..... 10 I 



loj. 150, i:.l-8 ..... 2-6573 



Hence B C * 45 1 -S 

 2. To find the hypotenuse A B. 



583.8) 327 (660- AB 

 9B19 



351 

 360 



