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MATHEMATICS. SPHERICAL TRIGONOMETRY. [SPHERICAL TRIAXOI.IW. 



(6). To prort tht Funnula CoUn. A tin. C Cotan. a gin. 

 4 Co*. Coo*. 6. 



This formula i* uieil in certain propositions : e.g., it is 

 employed in the astronomical problem of finding the 

 aberration in decimation. 



m formula (2) we have 



Con. A sin. ft sin. e cos. a COB. ft cos. e. 

 And from formula (3) we have 



Cos. e ooe. a cos. 6 + " n - *"- 6 cos. C. 

 /. Cos. A sin. 6 sin. e cos. a cot. a cos. 1 ft sin. a 

 sin. ft cos. ft cos. 0. 



.*. Cos. A sin. 6 sin. e cos. a sin. 1 6 sin. a sin. 6 cos. 

 ft oos. C. 



.' . Cos. a sin. c cos. a sin. ft sin. a cos. b cos. C. 

 Also from formula (1) we have 



sin. C sin. a 



Sin. c = 



sin. A 



.'.Cos A 



sin. C sin. a 



= cos. a sin. 6 sin. a cos. ft cos. C. 



sin. A 



.'. Cotan. A sin. C sin. a = cos. a sin. ft sin. a cos. 



ft cos. C. 



.'. Cotan. A sin. C = cotan. a sin. 6 cos. Ccos. ft. 



Q. E. D. 



THE SOLVTION OF RlfiHT-ANOLED SPHERICAL TRIANGLES. 



There are as many as six different cases of right- 

 fig. S. angled spherical triangles, as will 

 appear from the following conside- 

 rations : 



Let A B C be the triangle, having 

 a right angle at C. Then using the 

 ordinary notation, all possible cases 

 are the following : 



(1). Given the base 'and perpen- 

 dicular, i.e., given a and b. 



(2). Given the hypothenuse and 

 another side, i.e., given c and a, or 

 c and 6. 



(3). Given the base or perpen- 

 dicular, and an adjacent angle, i.e., 

 given n and B, or 6 and A. 



(4). Given the base or perpendicular and an opposite 

 anyle, i.e., given a and A, or 6 and B. 



(5). Given the hypothenuse and an angle, i.e,, given 

 c and A, or c and B. 



(6). Given the two angles, i.e., A and B. 

 If these cases be compared with those on p. 641, for 

 plane triangles, it will be seen that the third case of plane 

 triangles diverges into two cases, viz. , the third and fourth 

 of spherical triangles, while the sixth case is peculiar to 

 spherical triangles.' Both of these differences are due to 

 the circumstance, that in -(.he spherical triangle, A -(- B 

 + C is not known, whereas in the plane triangle A + B + 

 C-180" 



(6). Jo inrestigatr. the Formulas on which the Solution of 

 Jtiyht-anyled Spherical Triangles depends. 



sin. c sin. 6 sin. a 



From the general formula . , , = ~- r 



sin. C sin. B sin. A 



Since 90 and .'. fin. C = J, wo have 

 sin. ft sin, a 

 nin. It sin. A 



sin-e - 



.". sin. 6 = sin. c sin. 1? 

 Similarly sin. a = sin. c sin. A 



(17) 

 (18) 



cos. c cos. a co. 6 

 Again, since cos. C - - - and cos. 



sin. < MIL 



0, we have 



.'. cos. c cos. a cos. ft 



(19) 



cos. C + cog. A cos. B 



Again, since cos. 

 cot. C-0 



.'. oos. c cotan. A cptap B 



sin. A win H 



and 



(20) 



cos. A + cos. B cos. 

 sin. B sin. C 



Again, since cos. a 



cos. C and sin. C 1. 



.'. cos. A = cos. a sin. B 

 Similarly cos. B cos. 6 sin. A 





and 



Again, since cos. A cos. a sin. B and sin. B = 



(21) 

 sin. ft 





.'. oos. A - 



sin. c 



cos. a and cos. a -* - . by (2) 



.'.cos. A--, 

 sin. c 



cos. ft 



.". cos. A = tan. 6 cotan. c 

 Similarly cos. B = tan. a cotan. e 



(23) 

 (24) 



sin. ft <-".s li 

 cos. ft sin H 



Again from (1) sin. a sin. c sin. A 

 from (2) and this 



But by (4) cos. B = cos. I sin. A. 



.". sin. a = tan. 6 cotan. li 

 Similarly sin. 6 tan. a cotan. A 



sin. 6 



Mil. I! ' 



sin. A cos. ft 



sin. A 



COS. U 



(25) 

 (26) 



(7). Napier's Eule for the Solution of Right-angled 

 Spherical Triangles. 



The formulas given in the last article can bo included 

 in a single rule, which is very easily enunciated and re- 

 membered. It is generally called Napier's Rule, having 

 been invented by Napier, who, as wo have already 

 stated, was the inventor of logarithms. Leaving out C, 

 which is 90, there are three sides and two angles in the 

 triangle viz. , a. b. c. A. B. We will call the base, the 

 perpendicular, the complements of the hypothenuse and 

 the angles, circular parts ; if we fix on any of these and 

 call it the middle part, then of the remaining four, two 

 will be adjacent, and the other two opposite : then it 

 will be found that all the formulas of the last article 

 are included in the following rule. "The sine of the 

 middle part equals the product of the tangents of the 

 adjacent parts, and also equals the product of the 

 cosines of the opposite parts ; 



Or, Sin. mid. = tan. ad. = cos. op. 

 Thus, if 90 A be the middle part, then 90 c and 

 6 are the adjacent, and 90 B and a are the opposite 

 part. 



The rule gives us sin. (90 A) = tan. (90 c) tan. 6 

 = cos. (90 B) cos. a 



Or Cos. A = tan. ft cotan. c = cos. o sin. B (a) 

 Similarly, if 90 B is the middle part, then 90 c 

 and n are the adjacent, and 90 A and 6 the opposite 

 parts. The rule gives us 



Cos. B = tan. o cotan. c = cos. 6 sin. A (b) 



If a be the middle part then 90 B and 6 are the 

 adjacent, and 90 c and 90 A the opposite part, 

 then the rule gives us 



Sin. a = tan. 6 cotan. B = sin. c sin. A (c) 



If 6 be the middle part then 90 A and a are the 

 adjacent parts, and 90 c and 90 A the opposite 

 parts, then the rule gives us 



Sin. ft = tan. a cotan. A = sin. c sin. B (d) 



Finally, if 90 c be. the middle part, then 90 A 

 and 90 B are the adjacent, and a and A the opposite 

 parts, then the rule gives us 



Cos. e cotan. A cotan. B = cos. a cos. 6 (e) 



If the five formulas of the present article be com- 

 pared with those of Article (6) tliry will l>o found iden- 

 tical. Hence Napier's Rule, as was stated, comprises 

 all the formulas of Art. 6. It is a question whether, as 

 a matter of practice, Napier's Rule is really inon 

 veuient than the disconnected formulas of Article (6). 



(8). To explain the Mitln"l f K<I!H(!II in the cases of 



Itiijht-angteil Spherical Triangles. 

 All the formulas of Art. 6, to which we refer in the 



