2. Given a 88° 12' 20", b = 124° 7' 17", C= 50° 2' 1";

=

find A = 63° 15′ 11′′, B = 132° 17′ 59′′, c — 59° 4′ 18′′.

3. Given a = a 120° 55′ 35′′, b = 88° 12′ 20′′, C= 47° 42' 1"; find A129° 58' 3", B = 63° 15' 9", c=55° 52′ 40′′.

4. Given b 63° 15' 12", c = 47° 42' 1", A= 59° 4' 25"; find B = 88° 12′ 24′′, C=55° 52′ 42′′, a = 50° 1′ 40′′.

5. Given b 69° 25' 11", c=109° 46' 19", A 54° 54' 42"; find B = 56° 11' 57", C=123° 21′ 12′′, a = 67° 13'.

§ 57. CASE II.

Given the side c and the two adjacent angles A and B.

The sides a and b may be found by the third and fourth of Napier's Analogies,

and then the angle C may be found by [44], by Napier's second Analogy, or by one of Gauss's equations, as, for instance, the second, which gives

If the angle C alone is wanted, the best way is to decompose the triangle into two right triangles, and then apply Napier's Rules, as in Case I., when the side c alone is desired.

Let (Fig. 46) ABD=x, ▲ CBD=y, BD=p; then,

Rule I.,

It is clear that C may be computed from the equations

cot xtan A cos c,

cos C′ = cos A cse x sin (B — x).

EXAMPLE. Given A=35° 46′ 15′′, B=115° 9' 7", c=51° 2';

1. What are the formulas for computing A when B, C, and a are given; and for computing B when A, C, and b are given? 2. Given A= 26° 58′ 46′′, B=39° 45′ 10′′, c=154° 46′ 48′′; find a = 37° 14′ 10′′, b 121° 28' 10", C=161° 22' 11". 3. Given A-128° 41′ 49′′, B=107° 33′ 20′′, c= find a — 125° 41′ 44′′, b= 82° 47' 34", C=127° 22'.

= 124° 12′ 31";

4. Given B=153° 17' 6", find b 152° 43′ 51′′, c = 88°

= 86° 15′ 15′′;

C=78° 43' 36", a=
12′ 21′′, A = 78° 15′ 48".

5. Given A=125° 41' 44",
find a c=
= 128° 41′ 46′′, c = 107° 33′ 20′′, B = 55° 47′ 40′′.

C= 82° 47′ 35", b = 52° 37' 57";

§ 58. CASE III.

Given two sides a and b, and the angle A opposite to a.

The angle B is found from [44], whence we have sin B sin A sin b csc a.

When B has been found, Cand e may be found from the fourth and the second of Napier's Analogies, from which we obtain

The third and first of Napier's Analogies may also be used.

NOTE 1. Since B is determined from its sine, the problem in general has two solutions; and, moreover, in case sin B>1, the problem is impossible. By geometric construction it may be shown, as in the corresponding case in Plane Trigonometry, under what conditions the problem really has two solutions, one solution, and no solution. But in practical applications a general knowledge of the shape of the triangle is known beforehand; so that it is easy to see, without special investigation, which solution (if any) corresponds to the circumstances of the question.

It can be shown that there are two solutions, when A and a are alike in kind and sin b>sin a> sin A sin b; no solution when A and a are unlike in kind (including the case in which either A or a is 90°) and sin b is greater than or equal to sin a, or when sin a <sin A sin b; and one solution in every other case.

NOTE 2. The side c or the angle C may be computed, without first finding B, by means of the formulas

tan m = cos A tan b, and cos (c — m) = cos a sec b cos m,
cotx = tan A cos b, and cos (C― x) = cot a tan b cos x.

These formulas may be obtained by resolution of the triangle into right triangles, and applying Napier's Rules; m is equal to that part of the side c included between the vertex A and the foot of the perpendicular from C, and x is equal to the corresponding portion of the angle C.

NOTE 3. After the two values of B have been obtained, the number of solutions may readily be determined by § 48 — I. If log sin B is positive, there will be no solution.

Given a = 57° 36′, b = 31° 12′, A = 104° 25′ 30′′.

EXAMPLE.

1. Given a = 73° 49′ 38′′, b = 120° 53′ 35′′, A = 88° 52′ 42′′; find B116° 42' 30", c == 120° 57′ 27′′, C=116° 47′ 4".

2. Given a = 150° 57′ 5′′, b= 134° 15′ 54′′, A = 144° 22′ 42′′; find B1 = 120° 47′ 45′′, c1 = 55° 42′ 8′′,

B2 59° 12' 15", c2 = 23° 57' 17.4",

2

=

C1 = 97° 42' 55.4";
C2-29° 8' 39".

3. Given a = 79° 0′ 54.5", b=82° 17' 4", A = 82° 9' 25.8";

find B = 90°, c= 45° 12′ 19′′, C= 45° 44′.

4. Given a = 30° 52′ 36.6", b=31° 9' 16", A= 87° 34' 12"; show that the triangle is impossible.

§ 59. CASE IV.

Given two angles A and B, and the side a opposite to one of them.

The side b is found from [44], whence

sin b sin a sin B csc A.

The values of c and C may then be found by means of Napier's Analogies, the fourth and second of which give

NOTE 1. In this case the conditions for one, two, or no solutions can be deduced directly by the theory of polar triangles from the corresponding conditions of Case III. There are two solutions, when A and a are alike in kind and sin B>sin A> sin a sin B; no solution when A and a are unlike in kind (including the case in which either A or a is 90°) and sin B is greater than or equal to sin A, or when sin A <sin a sin B ; and one solution in every other case.

NOTE 2. By proceeding as indicated in Case III., Note 2, formulas for computing c or C, independent of the side b, may be found; viz.:

tan m = tan a cos B, and sin (c — m) = cot A tan B sin m,
cot x = cos a tan B, and sin (C — x)
x) = cos A sec B sin x.

In these formulas m = BD, x = ▲ BCD, D being the foot of the perpendicular from the vertex C.

NOTE 3. As in Case III., only those values of b can be retained which are greater or less than a, according as B is greater or less than A. If log sin b is positive, the triangle is impossible.

EXERCISE XXXVIII.

1. Given A= 110° 10′, B=133° 18', a = 147° 5' 32"; find b=155° 5' 18", c=33° 1′ 36", C=70° 20′ 40′′.