190 



THE POPULAR EDUCATOR. 



Let A B c be the triangle (Pigs. 12 and 13). From c drop c p 

 perpendicular to A B, or, as in Fig. 13, to A B, produced either 

 way; 



C 



C 



Fig. 12. 



Fig. 13. 



then sin. A = C , and sin. B = ; 

 AC CB 



sin. A 

 sin. B 



CP 

 AC 

 c P 

 c B 

 a 



CT . ., , 

 Similarly, - = 



c B _ a 

 AC ~ fr ' 



sin. A b 



sin. B 



sin. C c sin. C 

 If A or B be a right angle, there is no need to drop the 

 perpendicular above referred to. The sine of the right angle 

 will of course be unity, but the above reasoning will hold good, 

 and the result be the same. 



This statement of ratios between sides and sines of opposite 

 angles is called the rule of sines, and may be thus written : 



sin. A sin. B sin. C 



__ ______ _____ (66) 



2. The sum of any two sides is to their difference as the 

 tangent of half the sum of ^the opposite angles is to the tangent 

 f half their difference. 



By the last proposition, -" = 8in " A . 

 b sin. B 

 Then, componendo et dividendo 



a + b _sin. A + sin. B 

 a b sin. A - sin. B' 



Whence, by (47), i 



Oi 



This may be written differently; for (A + B) (180 C) ; 



.-. tan. g (A + B) = tan. (90 - _r C) = cot. J C ; 

 . a + b _ cot. J C 

 ** a - 6 



= _ ...... (67) 



tan. 

 Whence, 



(A-B) 



a b 



= cot. i (A B) cot. i C 



= tan. |(A - B) tan. 



(68) 



a+b 



3. The sum of any two sides is to the third side as the cosine 

 ef half the difference of the opposite angles is to the cosine of half 

 their sum. 



Since A + B = 180 - C, 



. a sin. A 



sin. (A -f B) = sin. C ; 

 sin. B 



' sin. (A + B)' 



and = 

 c 



sin. (A + B) 



Adding these equations, and using (41) and (60), we get 

 a + b __ sin. A + sin. B _ 2 sin. ^ (A + B) cos. % (A - B) . 

 c ' sin. (A + B) 2~dnT (A + B) cos. ~(AT+ B) ; 

 L+j! = cos. 4 (A - B) .... 



c cos. 4 (A + B) 



Similarly (by subtracting the second from the first equation 

 above, instead of adding them together), we find that 



The difference of any two sides is to the third side as the sine 

 of half the difference of the opposite angles is to the sine of half 

 their sum ; 



or a -^ = sin, k (A - B) (?0 



c sin. t (A + B) 



4. The square of any one side = the sum of the squares of the 

 other sides less twice the rectangle contained by them multiplied 

 into the cosine of the opposite angle (i.e., the angle included 

 between the sides last mentioned). 



If the opposite angle be a right angle, the "contained 

 rectangle," being multiplied by cos. 90, i.e., by 0, disappears, 

 and leaves only that part of the statement which concerns the 

 squares of the sides, which is proved in Euclid I. 47. 



If the opposite angle be acute (Fig. 12), by Euclid 11. 13, 



BC 2 = AC 2 + AB 2 2 AB AP. 



But since cos. A = , AP = AC cos. A - 

 AC 



.'. BC 2 = AC 2 + AB 2 2AB AC COS. A ; 

 or writing a, b, and c for BC, AC, and AB, 



a* = 6 2 + c" - 2bc cos. A. 

 If the opposite angle be obtuse (Fig. 13), by Euc. II. 12, 



BC* = AC 2 + AB 2 -f 2AB . AP ; 



but AP = AC cos. A = AC cos. (180 A) = AC cos. A 

 (A being in 2nd quadrant) ; 



therefore, as before, a 2 = 6 2 + c 2 - 2bc cos. A") 



Similarly, b' 2 - a? + c 2 - 2ac cos. B ...(71) 

 and, c 2 = a 2 + b- 2ab cos. C j 



5. Sines and cosines of angles in terms of sides. 

 by transposition, 



Cos. A = 



From (71), 



2bc 



Cos. B = v I 



2ac 



Cos. C = 



(72) 



2ab 



Since sin. 2 A= 1 cos. 2 A, 

 Sin. 2 A = (1 + cos. A) (1 - cos. A) 



_ - 

 ~ 



2bc 



.. _ 



2&c 



2bc 



2bc ) 



a 2 - (b 2 - 2bc + c")\ 

 2bc / 



_ / 

 \ 



2&c \ 2bc 



(b + c-a) (a + b-c) (a + c-b) 

 4& 2 c 2 



This expression for sin. 2 A (and therefore sin. A by extracting 

 the root) is in better form for calculation than (72), as it con- 

 sists entirely of factors. It can, however, ba further simplified 

 by taking 



s = semiperimeter of triangle ; 



then 2s = a + b -f- c, 



and 2(s a.) = b -j- c - a, 



2(sb) = a + cb, 



2(sc) = a + b-c; 



therefore, extracting the root, and simplifying, 



2Vs(s a) (s-b)(s-c) 

 _ > ?_\ ' ^ 



be 



(73) 



The expressions for sin. B and sin. C are similar, but the deno- 

 minator is ac in the one case and ah in the other. 



6. Sines, cosines, and tangents of the semi-angles. 



By (62), 1 + cos. A = 2 cos. 2 \ A. 



4s (s - cttb 



But by the preceding calculations, 1 -f- cos. A = ; -&-J. 



8v0 



, , . 2s (s - a) 

 .: 2 cos. 2 ^ A = '; 



be 



Similarly, 



and 



/ s (s - b) 



s (s- c) 

 ab 



(74) 



Deducing in a similar manner from other values of 1 - cos. A, 

 etc., we get expressions for the sines of the semi-angles, and by 

 dividing the latter by the corresponding expressions for the 

 cosines (74), we get the tangents, as under 



