TRIGONOMETRY. 



15- 



Tan. 



1 6. Tan. 



17- 



Tan. 



1 8. Tan. 



19. 



Tan. 



= cot. 



= cot. 



A C 



2 ^ 



iA 



cofj (b -^l 

 cof. i 



fin. 



(b + a) 

 i (^ - *) 



= cot. 



fin. 

 cof. 



+ 



h) 



5= cot. 



cot. 



B 



B 



cof.i (. + 

 fin. \ [a — 





lin. i (a + 



cof. \ [ a — ^ ) 

 cof. i (a + c) 



method ; but in the cafe of contiguous parts they admit of 

 other folutions. 



1. Given the hypothenufe B C, 60°, and the angle C, 

 23° 30' ; to find the oppofite leg A B [Jig. 4. ) Since A B is 

 the middle part, C and B C are disjuntt (fee Part) ; the 

 produft of the whole fine, into the cofine of the comple- 

 ment A B, i. e. the fine itfelf of A B, is eqi.al to the pro- 

 duft of the fines of C and E C 



Therefore from the log. fine of C 9.6006997 

 + fineof BC 9.9375306 



20. Tan. i £ = tan. 



\[h - a) 



2Ic 



Tan. 1 



= tan. 



[b + 



22 Tan. \ a = tan. | [c 



23' 



Tai>. 



tan. 



24. Tan. i ^ = tan. 



25. 



Tan. i i = tan. 



26. Cot. I C = tan. 



27. Cot. 



28 Cot. 



29. Cot. 



30. Cot. 



31' 



Cot. 



hi' + b) 

 iia - c) 



M'' + 

 1 (B- A) 



(B + A) 

 (C-B) 

 (C-f- B) 

 (A-C) 

 (A + C) 



We are indebted for many of the above transformations to 

 Gregory's Trigonometry ; others have been derived from 

 Bonnycaftle's treatife on the fame fubjeft. 



We fhall now conclude this article, by illuftrating the fo- 

 lution of fpherical triangles by means of Napier's_/fi;« circu- 

 lar parts. See Part and Circular Part. 



Solution of right-angled fpherical Triangles, by Napier's cir- 

 cular Parts. — If either one or both the fides, including the 

 right angle, come into the queftion, for it among the data, 

 write its complement to a quadrant. Since then by the 

 general rule dehvered under the article Circular Parts, and 

 rendered logarithmic, the whole fine, with the fine comple- 

 ment of the middle part, muft be equal to the fines of the dis- 

 junft parts, and the cotangents of the conjunft parts ; from 

 the fum of thofe data fubtraft the third datum ; the remainder 

 vsrill be fome fine or tangent, the fide or angle correfponding 

 to which, in the table of fines, &c. is the fide or angle fought. 



This umverfal rule being of great fervice in trigonometry, 



we (hall apply it to the various cafes thereof, and illuftrate it 



with examples ; which examples, in the cafe of disjunct or 



feparate parts, will at the fame time illuftrate the common 



8 



Suin 

 Subtraft whole fine 



i9-53823°3 



10.0000000 



Remains fine of A B 9.5382303 The neareil 

 correfponding number to which, in ti:e tables, is 20° 1 2'. 



2. Given the hypothenufe B C, 60°, and the leg A B, 

 20° 12'; to find the oppofite angle C. It is evident from 

 the preceding problem, that from the fum of the whole fine, 

 and the fine of the leg A B, the fine of the hypothenufe 

 B C is to be fubtrafted, the remainder is the fine of the 

 angle C. The example therefore of the former cafe is eafily 

 converted into an example of this. 



3. Given the leg A B, 20° 12', and the oppofite angle 

 C, 23° 30' ; to find the hypothenufe B C. It is evident from 

 the firft cafe, that from the fum of the whole fine, and the 

 fine of A B, is to be fubtrafted the fine of the angle C, 

 and the remainder is the fine of the hypothenufe B C. 



4. Given the hypothenufe B C, 60°, and one leg A B, 

 20° 12' ; to find the other leg. Since B C is a mean part, 

 and A B and A C are disjunft parts, the product of the 

 whole fine, into the cofine of the hypothenufe B C, is equal 

 to the produft of the fines of the complements, i. e. the 

 cofines of the legs A B and A C. 



Therefore from the whole fine 10.0000000 

 Cofine of B C 9.6989700 



Sum 

 Subtraft cofine of A B 



19.6989700 

 9.9724310 



Remains cofine of A C 9.7265390 The cor- 

 refponding number to which, in the tables, is 32° 11'; 

 therefore A C = 57° 49'- 



5. Given the legs A C, 57° 49', and A B, 20° 12' ; to 

 find the hypothenufe B C. It is evident from the preceding 

 cafe, that the whole fine is t > be fubtrafted from the fum of 

 the cofines of the legs A B and A C ; the remainder is the 

 cofine of the hypothenufe B C. The example, therefore, 

 of the preceding cafe is eafily applied to this. 



6. Given the leg AC, 57° 49', and the adjacent angle 

 C, 23° 30' ; to find the oppofite angle B. Since B is a 

 middle part, and A and C are disjunft parts ; the produft 

 of the whole fine, by the cofine of B, is equal to the pro- 

 duft of the fine of C, and the fine of the complement, i. e. 

 the cofine of A C. 



Therefore from the fine of C 9.6006997 

 Cofine A C 9.7265310 



Sum 

 Subtraft whole fine 



19.3272307 

 10.0000000 



Remains cofine of B 9.3272307 The neareft 



number correfponding to which, in the tables, is 12° 15'; 

 therefore B ^^ 77° 45'- 



7. Given the leg AC, 57° 49', and the oppofite angle 

 B, 77° 45'; to find the adjacent angle C. It is evident 

 from the preceding cafe, that the cofine of A C is to be 



fubtrafted 



