SOLUTION OF TRIANGLES.] 



NAVIGATION. 



104? 



By formula. (II). 



cos. J A = 



o= 95-12 

 6 = 162 34 

 e = 98 



Arith. Comp. 7 '7896 

 Arith 1 . Comp. 8-0088 



2)355-46 



= 177-73 

 * a= 82-61 



cos. JA, 16 



2-2498 

 1-9170 



2)199652 



9 -9826 



.'. A = 32 15' 



In the first of these logarithmic operations there 

 enters a log. sine, and in the second a log. cosine : each, 

 therefore, requires a correction, as noticed at page 1038 ; 

 10 should be added to the right-hand member of each of 

 the fonnuliB (I), (II), when expressed logarithmically. 

 Now two tens have been added to the sum of the four 

 logs, above, on account of the two complements ; so that, 

 after dividing this sum by 2, for the square root, one ten, 

 in excess, affects the result, which therefore supplies the 

 10 that ought to have been added, and thus the proper 

 correction is provided for. 



We shall now exhibit the work of the preceding 

 example without logarithms. The most convenient 

 formula for this purpose is, perhaps, that immediately 

 derived from the expression for cos. A at page 1046, by 

 first adding and then subtracting 1 ; we thus get 



coe. A = 



(a + b+cXb + c-a) 

 2be 



o= 9512 

 6 - 162 34 

 96 



a -f-6 + c = 355-46 . . . 355'40 

 6 + c a = 165 -22 reversed => 22561 



6 = 162-34 

 c= 98 



12-I872 

 140106 



15909-32 X 2 = 

 8457 = cos. A, 32 15' 



31818,64)58729(1-8457 

 31819 



The contracted method, so often re- 

 commended in these pages, is used 

 throughout this operation. 



26910 

 25455 



1455 

 1273 



182 

 159 



23 



It may be remarked here, that in computing, as in this 

 specimen, the first figure in the quotient will always be 

 1, followed by decimals ; no that there will be no occasion 

 to take any notice of decimal points, either in the 

 dividend or divisor : it is sufficient that we know that 

 the first figure of the quotient is always an integer (unit), 

 and the following figures decimals. But there is no 

 question that, in general, the operation by logarithms 

 would be preferred. It is worthy of notice, however, 

 that whenever a, 6, c are all divisible by the same 

 number, we may divide accordingly, and use the quotients: 



thus, instead of the above values, we might have taken 

 o=47 '56, 6 = 81 '17, and c = 49; these being the halves 

 of those values. 



2. Given a = 174-1, 6 = 232, and c = 345, to find the 

 angle A. Ans. A =27 4'. 



3. Given o = 95 -12, b = 162-34, and c = 98, to find 

 the angle B. Ans. B=114 24'. 



4. Given a = 3388, 6 = 2065, and c = 1637, to find 

 the angle A. Ans. 132 7'. 



5. Given a = 112, 6 = 112-6, and c = 120, to find all 

 the angles. Ans. A=57 28', B = 57 57', 0=64 35'. 



6. Given a = 698, 6 = 352, and c = 467, to find all 

 the angles. Ans. A = 116 12', B = 26 54', C = 36 54'. 



MISCELLANEOUS QUESTIONS REQUIRING THE SOLUTION 

 OF PLANE TRIANGLES. 1. Being on one side of a river, 

 and wishing to know the distance of a tree (C Fig. 10, 

 ante) on the other side, I measured a length of 500 yards 

 along the side of the river, and set up a rod at each end 

 (A and B) : the angle at A, subtended by B C, was then 

 measured, and found to be 74 14'; and the angle at B, 

 subtended by A C, was found to be 49 23': required the 

 distance of the tree from each station. 



Here the two angles at A and B are given, as also tho 

 interjacent side A B ; the third angle at C is therefore 

 known ; namely, C = 180 (74 14' + 49 23 ) = 56 23'. 

 As sin. C, 56 23' Arith. Comp. . -0795 

 sin. B, 49 23' ... 9-8803 



A B=500 .... 2-6990 



AC=455-8 .... 2-6588 



As sin. C, 56 23' Arith. Comp. . -0705 

 sin. A, 74 14' ... 9 -9833 

 AB = 500 .... 2-6990 



B 0=577-8 . 27618 



2. From the top of a mountain 3 miles high, the angle 

 of depression of the visible horizon that is, of the circle 

 where sky and sea appeared to 

 meet was found to be 2" 13' : it 

 is required from this to determine 

 the diameter of the earth. 



Let A (Fig. 11) be the centre of 

 the earth, C tho summit of the 

 mountain B C, and E C D the angle 

 of depression of the remotest visible 

 point of the surface of tho sea, 

 below the horizontal line C E. 



The angles A C D, D C E, make up a right angle ; so 



do the angles A C D and A, because D is a right angle 



(Euc. 18 of HI.); therefore the angle A = the angle 



BCD. No w by right-angled triangles A C = A D sec. A. 



.'. AC-AB=BC=AD(see. A-l); that is, 



A D (sec. 2 13'i - 1)=3, /.AD- gec . y^.j. 



By help of a table of natural secants this expression 

 for the semi-diameter AD is very readily calculated. 

 To convert it into a form adapted to logarithms, we have, 



by putting -- j for sec. A, 



AD = 



BC 



B C cos. A 

 sec. A 1 1 cos. A 



But it was shown at page 1046 that 1 cos. A= 2 sin. 2 JA, 

 BCcos. A . n AD BCcos. A 



** *-* ^ s . a |~~i~ 4 A J ^~t 



2 sin. 2 JA sin. 2 JA 



.'. log. diameter log. 3 -|- log- s - 2 13'J 2 log. sin. 

 1 6'J 4- 10, 



the 10 being added because there are two subtractivo 

 log. sines namely, log. sin. 1 6'J -f- log. sin. 1 6'J, and 

 only cue additive trigonometrical quantity namely, log. 

 cos. 2 13'i. 



log. 3 .... -477121 



log. cos. 2 13' J .... 9-999672 



log. sin. 1 6' Arith. Comp. 1-71185!) 



Repeated .... 1-711859 



log. diam., 7952 J 



3-900511 



