the required relations. These formulæ are Napier's Second Analogies. The following are left as exercises :— Smil cos + (A+B) = cos (a + b) co+ cos(A-B) = sin (a+b)sin C cos c sinc SOLUTION OF SPHERICAL TRIANGLES. 11. The right-angled spherical triangle. The solutions of all the cases of right-angled spherical triangles are given in Art. 6. Before proceeding to the solution of numerical examples, it will be well to examine whether any of those solutions are ambiguous. Now if the value of any quantity is to be determined by its cosine, tangent, or cotangent, there will be no ambiguity; for each quantity is supposed to be less than 1802, and therefore the sign will show whether it is greater or less than 90 (Plane Trig. Art. 15.) If, however, a quantity is to be found from its sine, there will in general be two values (Plane Trig. Art. 15,) which will satisfy the equation. It will be sufficient therefore to examine those solutions only in which a quantity is determined by its sine. And first let A and a be given; then to find c, b, and B, we have by (2), (4), and (10) of Art. 6, Hence in each of these the solution is ambiguous, as there is nothing to enable us to determine whether the smallest corresponding angles, or their supplements, are to be taken. The same also follows when B and b are given. Next, let A and c be given, then to find the other parts, we have by (2), (3), and (8), of Art. 6, sin α = sin e sin A, tan b = tan c cos A, cot B = cos c tan A. Hence a only is given by its sine, and in this case there is only one solution; for by (4) of Art. 6, tan a = sin b tan, A, and sin b is always positive, since b is supposed to be less than 180' hence tan a and tan A must have the same sign, that is (Plane Trig. Art. 15), a must be greater or less than 90, as A is greater or less than 90°. Consequently as A is known, there can be no ambiguity with respect to a. If a and c be given there will also be only one solution. Hence in right-angled spherical triangles, the only case which is really ambiguous is when the data are a side and its opposite angle. 1. Given b = EXAMPLES. 78° 20′, A = 37° 25', and B = 90 of a spherical triangle A B C, to find c. Take 90- A for the middle part; then by the first of Napier's rules (Art. 6), = cos A tan b: 9.8999506 This problem, by the preceding discussion, is not ambiguous, and c is acute (Geo. of the Sphere). 2. Given a = 36° 31', A 1=3 37° 25', and B = 90°, to find c. Let c be the middle part; then by the first of Napier's rules, с As the side a and its opposite angle A are given, the problem is ambiguous. Whence c = 75 25′ 42′′, or 104° 34' 18". Scholium.-A quadrantal spherical triangle which has one of its sides equal to 90° may be solved in the same manner by means of the polar triangle. EXERCISES. 1. Given a = 48 24' 16", c = 70 23' 42", and C = 90', to find the other parts. Ans. b = 59' 38′ 27′′, A = 52 32′ 55", B = 66° 20′ 40′′. 50° 30' 30", A 2. Given c = 3. Given a = the other parts. = 47° 54′ 20′′, and B 90', to find b. Ans. b = == 61 4' 56". 148 27' 10", c = 37 10' 20", and C = 90', to find Ans. A = 59 59′ 17", B = 144 3′ 40′′, b = 159 13′ 46′′. 4. Given a = 40 30′ 20′′, A = 47° 54′ 20′′, and B 5. Given a = 98° 20′ 20′′, B = 57° 43′ 12", and C 90', to find the other parts. Ans. b = 57' 26' 40", c = 94 28' 33", A = 97 2' 35". 51° 30', A = 58 35', and B = 90', to find c. 12. OBLIQUE-ANGLED SPHERICAL TRIANGLES. CASE I. (1.) Given two sides (a, b) of a spherical triangle, and one of the opposite angles (A), to find the other parts: (2.) Or given two angles (A, B) and one of the opposite sides (a), to find the other parts. [This is usually divided into two cases, according as there are given two sides and an angle, or two angles and a side, but as both are solved by the same formulæ, it seems better to include both in one general case, as in the analogous one of Plane Trigonometry.] When a, b, and A are given, then by Art. 8, = which will give the angle C. The remaining side c may then be found by Art. 8; or the side c may be found independently of C by Art. 10. Since sin B sin (180B), the solution is sometimes ambiguous, as in the corresponding case of plane triangles (Geo. of the Sphere). If two angles and a side be given, the mode of solution is exactly similar. EXAMPLES. 1. Given a = 115' 20' 10", b = 57° 30' 6", and A to find the other parts. To find the angle B by (a): 9.8744347 B 48 29' 48" = As a is between b and 180° b, the solution is not ambiguous, (Geo. of the Sphere); and since B and b are of the same kind, the value of B is acute, as given above. A somewhat different method of putting down the work to that employed in Plain Trigonometry is adopted in this and some of the subsequent examples. By sin, cosec, etc., are meant log sin b, log cosec a, etc. The value of a is not comprehended between 6 and 180°-b, and hence (Geo. of the Sphere), the solution is ambiguous; whence, B = 46 18' 6", or B' = 180° 46° 18' 6" Consequently to find C and C', we have (Art. 9), cotCtan (A + B) cot C'tan (A + B') cos (ba) cos + (b − a)' By means of these and the formula of Art. 8, each of the triangles ABC, AB'C' may be solved. 51° 30' 6", B = 59' 16' 10", and a = 63° 50′ 30′′, to 3. Given A = As the value of the angle A does not lie between B and 180°-B, the solution is ambiguous (Geo. of the Sphere), and therefore b = = 180° 80° 20' 42" = 99' 39' 18". 80° 20′ 42", or If C and c were also required, the former could be obtained from the formula of Art. 9, and the latter from that of Art. 8. As b has two values, it will be obvious that C and c have also two values; that is, there are two triangles A B C, AB C', that fulfil the conditions. EXERCISES. 1. Given a 80° 5' 4'', b = 109° 49′ 30", and A = 33° 15' 7", to find B. Ans. B = 148 25' 22'3. 34' 15' 2", = = 51° 32' 15", and a = 56° 17′ 12′′, 47° 19' 10", B = = = 42° 15′ 13" 25, and a = 40° 0' 10", = 50° 10' 30". 95' 57' 15", 58 30' 56'. CASE II. (1.) Given two sides (a, b) of a spherical triangle and the included angle (C), to find the other parts: (2.) Or given two angles (A, B) and the included side (c), to find the other parts. When two sides and the included angle are given, A + B and A B can be determined, and thence A and B, by (6) and (7) of Art. 9. The side c may then be found by Art. 8. The side c may also be obtained independently of the angles A and B, by means of (8) and (9) of Art. 9. If two angles (A, B) and the included side (e) be given, by taking the polar triangle, there will then be two given sides and the included angle of a spherical triangle, which may be solved as the preceding. Or the formulæ of Art. 10 may be employed. The sub-cases (1) and (2) constitute one general case for the reason stated in Case I. = sin = Cosec= 9.5388804 10 0074043 cot 10 3166443 = tan (CB) = From these we get (C+ B) 84° 37' 23', and 9.8629290 |
