Since both values of B are such that B- A, B'— A, and b all positive, there are two solutions.

(1) When B = 64° 26' 4". ·

log sin (b-a)= 8.59395-10 colog sin (b+a)= 0.15051 log cot (B-A)=11.15894–10 log tan C 9.90340-10

C=38° 40.8',

C=77° 21.6'.

log sin (B+4)=9.93956-10 colog sin (B-A)=1.15998 log tan (b-a)=8.59428-10 log tan c=9.69382-10 c=26° 17.65',

c=52°35.3'.

Proceeding in the same manner with the other value of B, viz., B' = 115° 33′ 56′′, we find C" = 11° 11.7', and c' = 9° 5.6'.

154. Exercise XXV.

1. Given b= 99° 40', c = 64° 20', B = 95° 40'; C' = 65° 29', a = = 100° 42'.

2. Given a = 109° 20', C = 82°1'8", A = 107° 40'; find C = 90°, ́ B = 113° 33', b = 114° 48'.

3. Given a = 99° 40′ 48", b = 64° 23' 15", A = 95° 38' 4"; find B=65° 33' C 97° 26' c = 100° 49'.

4. Given a=46° 20.75', b=65° 18.25', A=40° 10.5'; find B=54° 6', C=116° 55', c = 90° 32', B' = 125° 54', C" = 24° 13, c' = 27° 23'. 5. Given a = : 30° 40', c = 67° 24', A = 50° 18.5'; find C, B, b.

155. IV.

given.

When two angles and the side opposite to one of them are

This case reduces, by passing to the polar triangle, to the preceding case, and gives rise to the same ambiguities. Hence, the same remarks made in Art. 153 apply in this case also.

(1) Given A=66° 7' 20", B=52° 50′ 20′′, a=59° 28' 27"; find b, c, C.

By [119], tan 1⁄2 c = cos 1⁄2 ( A + B) tan 1⁄2 (a + b) ÷ cos † (A – B).

(ƒ)

COMPUTATION

By (ƒ), we find b = 48° 39.3', or = 131° 20.7' (= b').

The second value of b is inadmissible, since AB is positive, and

a — b' is negative, and hence there is only one solution.

1. Given B=116°, C = 80°, c = 84°; find b

A = 79° 18'.

=

=

C=45°',

C=90°.

2. Given A= 62°, C 101° 58' 24", a 64° 30'; find c = 90°, B = 63° 47', b 66° 30'.

= =

3. Given B 73°, C= 81° 20', b= 122° 40'; show that the triangle

is impossible.

4. Given A=117° 54.4', B = 45° 8.6', a = 76° 37.5'; find b = 51°18', C = 41° 5', C = 36° 39'.

5. Given A 130° 10', C 59° 13', c = 48° 25'; find B, a, b.

=

=

157. V. When the three sides are given.

The angles are found by means of the formulas in Art. 143. If all the angles are to be computed, the formulas for the tangent are not only the most accurate in general, but the most convenient, since only four different logarithms are required. If but one angle is required, the formula for the cosine will be found to involve the least work, though it will lack precision if the angle is near 0°. If all the angles are required, it is convenient to modify the formulas for tangents as follows:

Whence, logr = [log sin (sa) + log sin (s—b)

+ log sin (sc) + colog sin s];

log tan1⁄2 A = log r + colog sin (s — a);

log tan Blog r+colog sin (s — b);

log tan Clog r+colog sin (sc).

Having found the logr by the first of these formulas, we then determine the angles in a very simple manner, as indicated by the last three formulas.

(1) Given a = 112° 6' 42", b = 127° 39′ 12", c = 71° 12' 42"; find A, B, C.

158. Exercise XXVII.

1. Given a = = 38°,

C= 83° 55'.

b

=

: 42°, c = 51°; find A= 51° 58', B= 58° 53',

2. Given a = = 126°, b = 152°, c = 75°; find A = 142° 24'; B = 159° 16',

3. Given a=101° 31′1′′, b = 36° 16′ 29′′, c = 112° 38'; find A78° 19', B= 36° 15', C= 112° 43'.

=

4. Given a = · 71° 15', b = 40° 35', c = 39° 10′; find A= 130° 35′ 55′′, B = 31° 26' 32", C= 30° 25' 34".

5. Given a 60° 13', b = 70° 24', c= 80° 27'; find A, B, C.

159. VI. When the three angles are given.

The sides are computed by means of formulas [108] to [116]. The formulas for the tangents are in general to be preferred, and they may be modified as in Case V.

(1) Given A

=

150°, B = 131°, C=115°; find the side c. Let us find c by the formula for cosine [115], viz. :

1. Given A= 74°, B=82°, C=67°; find a 68° 46', b = 73° 48', c = 63° 12'.

2. Given A= 91° 10', B = 85° 40', C= 78° 30'; find b = 85° 49'.

3. Given A=154° 4.2', B=87° 27.1', C=76° 32.4'; find a=156° 12.2', b=112° 48.6', c = 63° 48.9'.

4. Given A=70°, B=132°, C= 94° 25.89';

b = 137° 56.3', c = 116°.

find a = 57° 54',

5. Given A= 95° 37', B = 113° 40', C = 78° 50'; find a, b, c.

161. Quadrantal Triangles. A spherical triangle is called quadrantal when one of its sides is equal to a quadrant. The polar triangle of a quadrantal triangle is a right spherical triangle.

Hence to solve a quadrantal triangle we may solve its polar triangle as in Chap. V, and take the supplements of the results.

AREA OF SPHERICAL TRIANGLES

162. The Spherical Excess (E) of a spherical triangle is the amount by which the sum of the three angles exceeds 180° or π. Thus,

It is shown in Geometry that the areas of two spherical triangles are to each other as their spherical excesses, and also that the area of a trirectangular triangle is 2, and its spherical excess 90°. Hence if K is the area of any triangle and E° its spherical excess, we have

Example. On a sphere whose radius is 15 ft., a triangle has A=78° 3', B= 90° 50.4', C=100° 21.6'. Find K.

78° 3' + 90° 50.4' +100° 21.6' = 269° 15'

=269.25°.

(15) sq. ft. 350.5 sq. ft.

=

163. Lhuiller's Theorem. This elegant formula, by which the spherical excess is expressed in terms of the three sides, may be deduced as follows:

tan E

=

sin (A+B+C − π)

cos (A+B+C-T)

sin (A+B) sin (C)
cos (A+B) + cos (π-C)

sin (A+B)-cos C

cos (A+B) + sin C

sinA cos B+cos A sin B- cos C

cos A cos B - sin 1⁄2 A sin 1 B + sin C

Substituting from [99] to [107], and reducing,

tan E = √tans tan (s - a) tan (s - 6) tan (s — c) . [124] 1 1

4

2

We find E by (a), Art. 162, or [124], according as the three angles or sides are given; in all other cases we solve the triangle so far as to find the angles or sides, and then apply (a) or [124]. Having