SOltmON OP TRIANGLES.] 



NAVIGATION. 



1045 



But C B D, the half sum of A and B, added to 



ABD, gives B, the greater, 

 .'. ABD = 4 (B-A), /. sin. ABD . 

 = sin. (B = A) ) 



But sin. A B D', greater than 90, is the same as > ... (2) 

 the sine of an angle as much less than 90 

 .'. sin. A B D' = cos. $ (B - A) 



These relations being established, let us refer to Case 

 L, which gives 



J AIX _ AC + CB = sin. ABD' 

 AB" AB sin. AD B 



. AD _ AC - CB _ sin. ABD 

 AB " ~AB " sin. ADB 



(3) 

 (4) 



Hence, dividing (3) by (4), we have 



AC + CB _ sin. ABD' sin. ADB 



(5) 



AC CB sin. ABD sin. AD B 

 that is, by substituting for these sines their values in (1) 

 and (2), 



AC-r-CB = cos. (B A) sin. \ (B + A) 



AC CB sin. i (B A) cos. 



tan. i(B+A) 



i(B + A) 

 (page 1038) ; 



= 



~ tan. ~(B A) 



so that in any plane triangle, the sum of two fides it to 

 difference, as the tangent of half the sum of the oppo- 

 >ite anyles it to the tangent of half their difference. The 

 following, therefore, is the rule : 



RULE. As the sum of the two given sides 

 Is to their difference, 

 So is the tangent of half the sum of the oppo- 



site angles 

 To the tangent of half their difference. 



a J- 6 : a - 6 : : tan. ^ : tan. ~. 

 2 



The half difference of the unknown angles thus becomes 

 known ; and as their half sum is also known for the 

 whole sum is ISO 9 minus the given angle (Euc. 32 I.) 

 the two angles themselves are readily discovered ; the 

 greater of the two is the half sum increased by the half 

 lili. rence, and the less, the half sum diminished by tlio 

 half difference. 



NOTE. Instead of using the tangent of half the sum 

 of the opposite angles, we may use the cotangent of half 

 th'e included angle, as is obvious. 



EXAMPLES. 



1. In the triangle ABC (Fig. 9), are given 6 = 47, 

 e = 85, and the angle A = 52 40' : required the remain- 

 ing parts, 



^ (C + B) = 90 26" 20 7 - 63 40', and the formula 

 for finding J (C B) is 



c + b : e b : : tan. 1(C + B) : tan. J(C B) 

 As c + b = 132 Arith. Comp. . 7 '8794 

 : e 6- 38 . . . 1-5798 

 : :tn. i(C + B), 6340 / . . . 10-3054 



;tan. $(C B), 30 11' . 



9-7646 



.'. C = 93 51', the sum of i(C + B) 



and $ (C B) 

 B = 33 2!X, the difference. 



As all the angles of the triangle are now known, the 

 remaining aide a may be found by Case I., thus I- 

 To find the side a, 



At sin. B, 33 29' Arith. Comp. . -2583 

 : sin. A, 52 40* . . . 9-9004 

 :: 6 = 47 1-6721 



67-74 



1-8308 



It may be observed here that the third side of the 

 triangle may always be determined independently of 

 Case I., and in a manner somewhat more easily, by 

 employing a relation furnished by the foregoing investi- 

 gation, thus : the equations (3) and (4), after putting for 

 the sines the values previously deduced, are 

 AC + CB_cos. |(B A) AC CB_sin. (B A) 

 AB ~cos. |(B + A) ; - AB~ ~6iu. |(B^f~A) 

 which furnish the proportions 



cos. *(B A) : cos. i(B + A) : : b + a : c . . . . (0) 

 sin. f (B A) : sin. |(B + A) : : 6 a : c . . . . (7) 

 The advantage of using either of these proportions 

 instead of the rule in Case I. is, that all the three loga- 

 rithms employed occur at the same openings of the table 

 as the logarithms in the process for finding the angles, 

 and they may, therefore, be readily taken out at the 

 same time : one of the three that for the sum or differ- 

 ence of the sides has only to be repeated. 



For the particular triangle solved above, the second 

 of these proportions is 



sin. ^(C B) : sin. (C + B) : : c 6 : a ; 

 and in order to show the facilities gained by employing 

 it, we shall re-work the example just given by its aid. 



A e + 6 = 132 Arith. Comp. 78794 To find the Bide a. 



: e b 38 . . 15798 .... 15798 



: :tan. l(C-t-B), 6J W . .103054 . . . sin. 9 9j: 



: tan. |(C B), 30" 11' . . 9 7646 Arith. Comp. sin. -2986 



.'. C = 9*>il' = 67-74 . .1-8308 

 B = 33' 2J' 



It may assist the memory in computing the side by 

 this method, to observe that the three terms employed 

 in the proportion for the side are merely those in the 

 proportion for the angles reversed, with sine in the place 

 of tangent. 



Another use, too, may be made of the proportions (6) 

 and (7) namely, in the case where two angles and the 

 interjacent side are given to find the remaining sides, as 

 in the following example. 



2. Given A = 41 13', B = 62 &, and e = 310, to 

 find a and b. 



Inverting the proportions (G) and (7), we have 

 sin. 4(B + A) : sin. i(B A) : : c : b a 

 cos. l(B -j- A) : cos. |(B A) : : c : 6 4- a. 



Hence the work will stand as follows : 



Aiin. i(B + A), 51" 41' Arith. Comp. -1054 Arith. Comp. cos. -2076 

 : Bin. I(B A), 10 28' . . .92593 . . . cos. 9 9927 

 : : e = 319 . . 2-4914 . 2-4914 



,b a 71-8 . 



'or nearly 71*7 



(S Eiample 4, paje 1044.) 



1-8561 6+0 = 491-7 .2-6917 

 b a= 71-7 



42(1 , 210 =o 

 6634,281-7 = 6 



3. Given a . 512, c = 907, and B = 49 10', to find 

 the remaining parts of the triangle. 



Ans. A = 34 6', C = 9G 44', 6 = 691. 



4. Given a = 95-12, c = 98, and B = 114 24', to 

 find the remaining parts. 



Ans. A =. 32 15', C = 33 21', 6 = 162 34. 



5. Given o = 112, 6 = 120, and C = 57 57', to find 

 the remaining parts. 



Ans. A = 57 28' B, = 64 35', e = 112 6. 



6. Given 6 = 154-3, e = 365, and A = 57 12", to find 

 the remaining parts. 



Ans. B = 24 45', C = 98 3', a = 310. 



HI. WTien the given parts are the three sides. 

 Investigation uf the Rules. 



Returning to the diagrams of Case I., page 1043, wa 

 have by right-angled triangles, 



B D = e cos. B, and C D = 5 cos. C. 



If the angle C of the triangle be acute, the cosine of 

 it will be positive ; if it be obtuse, as in the second of 

 the diagrams referred to, it will be negative. In the 

 equation above, the angle C is AC D, which is acute in 

 the second diagram ; but if we replace it by the angle C 



