713 



SPHERICAL TRIGONOMETRY. 



SPHERICAL TRIGONOMETRY. 



711 



and A', B, c', for angles ; the third has a', b,' e for sides, and A', B', c, 

 for angles. Hence every spherical triangle has another, with one side 

 and its opposite angle remaining unchanged, and all the other parts 

 changed into their supplements. 



Again, if the three circles be taken which have A, B, and c for their 

 poles, the intersections of these new circles are themselves the poles of 

 A B, B c, and c A ; and, of the eight new triangles thus formed, each 

 one has all its angles supplemental to the sides of its corresponding 

 triangle in the first set, and all its sides supplemental to the anyles. 

 Thus there exists a triangle which has the sides A', B', c', and the 

 angles a', U', c' ; which is called the supplemental triangle of that 

 which has a, b, c, for sides, and A, B, c, for angles. Hence if, in any 

 general formula, sides be changed into supplements of angles, and 

 angles into supplements of sides, the result is also a general formula. 



Any two sides of a spherical triangle are together greater than the 

 third, and the sum of the three sides is not so great as 360. Any 

 two angles of a spherical triangle are together less than the third angle 

 increased by 180, and the sum of the three angles is more than two, and 

 leas than six, right angles. And the greater side of a spherical triangle 

 is opposite to the greater angle ; and the sum of two sides is greater 

 than, equal to, or less than, 180, according as the sum of the opposite 

 angles is greater than, equal to, or less than, 180. 



The formula) for the solution of a spherical right-angled triangle are 

 nix in number. Let c be the right angle, and let c be called the 

 fii/pothenust, as distinguished from a and b, which are still called sides. 

 [CIRCULAR PARTS.] 



1, 2. The cosine of the hypothenuse is equal to the product of the 

 cosines of the sides, and of the cotangents of the angles : 



coa e = cos a cos 4 ; cos c = cot A cot B. 



3. The sine of a tide is t,he Bine of the hypothenuse into the sine of 

 the opporitt angle : 



sin a = sin c sin A ; sin b sin e sin B. 



4. The tangent of a side u the tangent of the hypothenuse into the 

 cosine of the included angle : 



tan a ~ tan c cos B ; tan 4 = tan c cos A. 



5. The tangent of a side is the tangent of its opposite angle into the 

 side of the other side : 



tan a = tan A sin b ; tan b = tan B sin a. 



6. The cosine of an angle is the cosine of its opposite side into the 

 side of the other angle : 



cos A = cos a sin B ; cos B = cos b sin A. 



These formulae are sufficient for every case. Name any two out of 

 the five a, b, c, A, B (c being a right angle), and in the preceding six 

 formula), by repetition ten, will be found those two combined with 

 each of the other three. Thus, having given a side a and its adjacent 

 angle B, we find the other parts from 



tail a 

 tan 6 = tan B sin a, tan c = C03 B , cos A = cos a sin B. 



These formula! should be committed to memory : the abbreviation, 

 to called, described in CIRCULAR PARTS, is only an expeditious mode 

 of wasting time. 



When all the angles are oblique, the principal formula) are as follows 

 (in most cases we give only one, those for other sides, &c., being easily 

 supplied) : 



I. 



sin A sin B sm c 



sin a = sin b siu c' 



as the sines of their opposite angles. 



or the sines of sides are to one another 



2. cos c = cos a cos b + sin a sin 4 cos c. 



3. cos -. = 



- c) 



4. tan 



/ 



/ V 



in - - / sin ( - q) sin ( - It) 

 2 ' V sin a sin 4 



o / sin ( o) sin ( 4) 



2 " V" 



sin sin b 



sin * sin ( c) 

 where it = V( sin (* ) "in < i 



A + B c cos 4 (o 4) 



5. tan n = cot o 



sin ( c) ' 

 " ( - c) -f- sin ). 



' 2 cos 4 (a + b) 



Un 



c sin 4 (a - 4) 

 = * 2 sin 4 (o + &) 



2 



A B 



s 



C. cot a sin c = cos I cos B + sin B cot A. 



7. tan = sin 4 c V sin a sin 4 -=- sin 4 (<* -~ b}, 



gives sin 4 < = "in i ( a *) "T cos * 



8. sio 9 = sin 4 c V si" " m ' -r cos 4 (< 4), 



gives cos 4 c cos 4 (a 4) os * 



The formula) (5) which are called Napier's Analogies, may be demon- 

 strated more easily than in the usual way, as follows. First 



tan (4 A + 4 B) . tan 4 c = 



tan 4 A tan 4 c + tan 4 B tan \ c 

 1 ^p tan 4 A tan 4 B 



, . .. sin (s - c) 



from (4) tau 4 A tan 4 B = "^~, , c. 



tan (4 A + tau 4 B) . tan 4 c 



sin s 



sin (s 4) + sin (s a) 

 sin s 4! sin (s c)' 



2 siu 1 c cos i (a 4) 

 [TRIGONOMETRY.] tan (4 A + 4 B) . tan 4 c = f 



2 sin 4 c cos 4 (a + 4) 



since It a 4 = c. Hence the first of (5) easily follows, and the 

 second in a similar manner. 



The formula (6) is not easily remembered, except by the following : 

 Write the sides in any successive pairs, as ab, be, ca, or ac, cb, ba. : 

 change the last three into the corresponding angles giving ab, be, c A, 

 or ac, CB, BA; remembering the formula cos -;- sin = cot make the 

 middle terms cosines, those on the right and left sines, and those on 

 the extreme right and left cotangents. We have then 



cot a sin b = cos 4 cos c + sin c cot A, 



which is a case of the formula in question. 



We now proceed to the differeut cases of triangles, observing that 

 these may be taken in pairs, owing to the property of the supple- 

 mental triangle. Thus, suppose it granted that we can solve the case 

 of finding the three angles when the sides are given, it follows that we 

 can solve that of finding the three sides from the three angles. For, if 

 A, B, and c be given, find the anyles of the triangle whose sides are A', 

 B', and c'. If a', b', and c? be these angles, then a, b, c are the sides of 

 the original triangle. Nor is it worth while to separate the several 

 cases, since it generally happens that out of each pair one is of much 

 more frequent occurrence than the other. 



Case 1. Given the three sides, to find the three angles. If one 

 angle only be wanted, one of the formula) (3) answers as well as any- 

 thing. If all three angles be wanted, the shortest way is to calculate 

 M from 



M = V {sin (s a) sm(sb) sin (s c) -f- siu s} 

 and then the angles from 



' tan i c = 



Supplement. Given the three angles, to find the three sides. Make 

 the supplements of the given angles sides, calculate the three angles, 

 and the supplements of the last three angles will be the sides 

 required. 



Cote 2. -Given two sides (a and 4), and the included angle o, to find 

 the remaining parts. If all the parts be wanted, calculate 4 (A + B) 

 and 4 (A B) from (5), and then find A and B by addition and sub- 

 traction : lastly, find c from one of 



sin c sin c 



sin 4 , sin c = sin a 

 sin B' 



' siu A ' 



or from both, which will be a verification. But if the remaining side 

 only be wanted, use the formula (7) or (S), which gives this third side 

 at once, by means of the subsidiary angle S. Or from the extremity 

 of the shorter side given (say o), let fall a perpendicular arc z on ft, 

 dividing 4 into x and y. Then 



tan x = tan a cos c, sin z = sin a sin c 

 y = b x, cos c = cos z cos y. 



Supplement. Given a side (c) and the two adjacent angles (A and B) ; 

 required the remaining parts. Make A' and B' the sides of a triangle, 

 and c' the included angle ; find c' the remaining side, and a' and 4' the 

 remaining angles. Then o is the remaining angle of the original 

 triangle, and a and 4 are the remaining sides. To find the remaining 

 sides alone, the following formula may be used : 



+ 6 c cos 4 (A - B) 



tan ~T" =tan 2'cos4(A + B) 



tan 



a 4 



c sin 4 (A - B) 

 tan ' 



Case S. Given two sides (a and 4) first both less than a right angle, 

 and an angle opposite to one of them (A) ; required the remaining 

 parts. This case may afford no solution at all, or may give two 

 solutions ; it is therefore sometimes called the ambig*uous case. The 

 formula (6) may be used by the usual introduction of a subsidiary 

 angle ; but we should recommend a person who is not well practised 

 in the subject to prefer the following simple method: From the 

 extremity of 4 which is not adjacent to the given angle, drop a 

 perpendicular 2 on the side c, and let x be the segment adjacent to 

 A. Let a'z and 4"z be the angles made by a and 4 with z. And first 

 calculate sin 4 sin A ; if this be greater than unity, the triangle does 



