in this case C, or c>180°. If neither of the values of B makes A B and a b of the same signs, the problem is

impossible.

This case is known as the ambiguous case, and is like the analogous ambiguity in Plane Trigonometry (Art. 116), though it is somewhat more complex. For a complete discussion of the Ambiguous Case, the student is referred to Todhunter's Spherical Trigonometry, pp. 53–58; McCollend and Preston's Spherical Trigonometry, pp. 137-143; Serret's Trigonometry, pp. 191–195, etc.

Ex. 1. Given a=42° 45', b=47° 15', A=56° 30'; find B, C, c.

Since both values of B are such that A- B, A' — B', and ab, are all negative, there are two solutions, by the above Rule.

(1) When B = 64° 26' 4".

log sin (a - b) =8.5939483-|log sin (A + B) =9.9395560 colog sin (a+b)=0.1505150 colog sin (A-B)=1.1599832log cot (A — B)=1.1589413- log tan (a - b) =8.5942832

(2) When B' = 115° 33' 56".

colog sin (a+b)=0.1505150

log sin (a - b) =8.5939483-|log sin (A + B') =9.9989581 colog sin (A-B')=0.3072223– log tan (a - b) = 8.5942832–

log cot (A-B')=0.2467784

log tan =8.9912417

C' 2

C'

= 5°35′ 504".

2

4° 32′ 474".

B' = 115° 35' 56", C'11° 11' 40", c' 9° 5' 34".

Otherwise thus: Let fall the perpendicular CD; denote AD by m, the angle ACD by 6, and CD by p. Then we have

tan m cot b;

=

cos A =

D

cos a = cos (c — m) cosp; cos b = cos m cos p.

The required parts are given by (1), (2), (3), (4), (5).

Ex. 2. Given a=73° 49′38′′, b=120° 53′ 35′′, A=88°52′ 42′′: find B, C, c.

Ans. B 116° 44′ 48", C = 116° 44' 48", c = 120° 55′ 35′′.

=

212. Case IV.

Given two angles, A, B, and the side a opposite one of them; to find b, c, C.

This case reduces, by aid of the polar triangle, to the preceding case, and gives rise to the same ambiguities. Hence the same remarks made in Art. 211 apply in this case also, and the direct solution may be obtained in the same way as in Case III. Thus,

The side b is found from the formula

Then c and C are found from Napier's Analogies.

(1)

Ex. 1. Given A=66° 7′20′′, B=52° 50′20′′, a=59° 28'27";

find b, c, C.

Solution.

By (1) we find b = 48° 39′ 16′′, or 131° 20' 44".

log cos (A+B) = 9.7057190

log cos(a - b)=9.9980612 colog cos(A-B)= 0.0029244 colog cos(a+b)=0.2314530 log tan (a+b)=10.1397643 log cot (A + B)=9.7704854

The second value of b is inadmissible (see Rule of Art. 211), and therefore there is only one solution.

Ans. b = 48° 39′ 16′′, c = 70° 23′ 41′′, C = 90°.

Ex. 2. Given A = 110° 10', B = 133° 18', a =

find b, c, C.

Ans. b 155° 5' 18", c = 33° 1' 45",

=

: 147° 5' 32" ;

Given the three sides, a, b, c; to find the

The angles may be computed by any of the formulæ of Art. 195; but since an angle near 90° cannot be accurately determined by its sine, nor one near 0° by its cosine (Art. 151), neither of the first six formulæ can be used with advantage in all cases. The formulæ for the tangents however are accurate in all parts of the quadrant, and are therefore to be preferred for the solution of a triangle in which all three sides or all three angles are given.

Since the part under the radical is a symmetric function of the sides, it is in the formulæ for determining all three angles A, B, C, and when once calculated, it may be utilized in the calculation of each angle. For convenience in com

putation, denote this term by tan 7.

Then

B

s-c 16° 13'.

tanr=

sin(s-a) 9.8906049 sin (s-b)=9.7013681 sin(s-c)= 9.4460251

=

tan =9.6302185 tan =9.8194553 tan =10.0747983

tanr 9.5208234

C

2

2

Ex. 2. Given a = 100°, b = 37° 18', c = 62° 46'; find A,

B, C.

Ans. A 176° 15′ 46".56, B=2° 17′ 55′′.08, C-3° 22' 25".46.

-

214. Case VI. Given the three angles, A, B, C; to find the sides.

As in Art. 213, the formulæ for the tangents are to be preferred.

Putting tan R =

cos S

cos (SA) cos (SB) cos (S - C)*

we have, from (7), (8), (9) of Art. 196,