2. Given A=113° 39' 21", В=123° 40′ 18", a = 65° 39'46"; find b=124° 7'20", c=159° 50′ 14", C=159° 43′ 34′′.

3. Given A=100°2′11.3", B=98°30′28′′, a=95°20′38.7"; find b=90°, c=147° 41′ 43", C=148°5′33′′.

4. Given A = 24° 33' 9", B=38°0′ 12′′, a = 65° 20′ 13′′; show that the triangle is impossible.

§ 60. CASE V.

Given the three sides, a, b, and c.

The angles are computed by means of Formulas [47], and

the corresponding formulas for the angles B and C.

The formulas for the tangent are in general to be preferred. If we multiply the equation

tan + A = Vese s csc (s-a) sin (s - b) sin (s - c)

by the equation

1 = sin(s-a), and put

sin (s-a)

Veses sin (s - a) sin (s - b) sin (s - c) = tanr,

and also make analogous changes in the equations for tan B and tan C, we obtain

tan+A=tanr csc (s - a),

tan+B=tan r csc (s - b),

tan+C=tanr csc (s - c),

which are the most convenient formulas to employ when all

three angles have to be computed.

s = 81° 46′ 50′′ s-a= 30° 52′18′′ s-b= 43° 59′ 32′′ 6° 55′ 0′′

2)19.21669

log tan A = 9.60835

A = 22°5′ 20′′

1. Given a=120°55′35′′, b=59° 4'25", c=106°10′22′′;

find A=116° 44'50", B=63° 15'18", C =91° 7' 22".

2. Given a=50°12′4′′, b=116° 44′48′′, c=129°11'42"; find A = 59° 4' 28", B = 94° 23′ 12′′, C=120° 4' 52".

3. Given a=131° 35'4", b=108°30′14′′, c=84° 46′34′′; find A = 132° 14′ 21′′, В=110°10′40′′, C=99° 42′ 24′′.

4. Given a=20°16′38′′, b=56°19′40", c=66°20′ 44′′; find A = 20° 9' 54", B = 55° 52' 31", C=114° 20′ 17′′..

§ 61. CASE VI.

Given the three angles, A, B, and C.

The sides are computed by means of Formulas [48]. The formulas for the tangents are in general to be preferred.

If we multiply the equation

tana =

by the equation

cos S sec (S - A) sec (S - B) sec (S-C) =tan R, and also make analogous changes in the equations for tan b and tan c, we obtain

tana = tan R cos (S - A),
tan+b=tan R cos (S - B),
tan+ c = tan R cos (S-C),

which are the most convenient formulas to use in case all

three angles have to be computed.

In Example 1, after we find the values of S, S-A, S – B, S-C, we write the formula for tanta with the algebraic sign written above each function as follows:

NOTE. Here the effect, as regards algebraic sign, of three negative factors, is cancelled by the negative sign belonging to the whole product.

In Example 2, after we find the values of S, S-A, S - B, S-C, we write the formula for tan R with the algebraic sign written above each function as follows:

B = 55°52′32′′

C = 114° 20′14′′

2 S = 190° 22′ 42′′

log cos S = 8.95638 (n)

log sec (S - A) = 0.58768

log sec (S - B) = 0.11143

log sec (S - C) = 0.02472

log tan2R = 9.68021 log tan R = 9.84010

S = 95° 11′21′′
S-A= 75° 1′25′′
S-B= 39° 18′ 49′′
S-C=- 19°8′53′′

log tana = 9.25242
log tan b = 9.72867
log tanc = 9.81538

α = 10°8′18.9"

EXERCISE XL.

1. Given A=130°, B=110°, C=80°;

find

α =

139° 21' 22", b=126° 57' 52", c=56° 51' 48".

2. Given A = 59° 55' 10", B = 85° 36' 50", C = 59° 55' 10"; find a = 51° 17′31′′, b=64° 2' 47", c=51° 17' 31".

3. Given A= 102° 14′ 12′′, B = 54° 32′ 24", C=89°5′ 46′′;

find a = 104° 25'9", b = 53° 49′ 25′′, c= 97° 44′ 19′′.

4. Given A = 4° 23' 35", B = 8°28′20′′, C=172° 17' 56"; find a = 31° 9′14′′, b = 84° 18' 28", c=115° 10'.

§ 62. AREA OF A SPHERICAL TRIANGLE.

I. When the three angles, A, B, C, are given.
Let

R = radius of sphere,

E= the spherical excess = A+B+C-180°,
F= area of triangle.

Three planes passed through the centre of a sphere, each perpendicular to the other two planes, divide the surface of the sphere into eight tri-rectangular triangles.

It is convenient to divide each of these eight triangles into 90 equal parts, and to call these parts spherical degrees. The surface of every sphere, therefore, contains 720 spherical degrees.

Since in spherical degrees, the ABC = E, and the entire surface of the sphere is equal to 720 spherical degrees,

ABC: surface of the sphere = E: 720;

II. When the three sides, a, b, c, are given.

A formula for computing the area is deduced as follows:

=

cos(a+b)-cosc. (a)

cos(A+B) - cos (90°-+C) cos (90°C) _ cos(a+b)-cosc
cos(A+B) + cos (90°-C) cos(a+b)+cosc

Now, in § 31, the division of [23] by [22] gives

in which for A and B we may substitute any other two angular magnitudes, as for example, (A+B) and (90-C), or (a+b) and c.

If we use in place of A and B the values(A+B) and

(90° C), the first side of equation (b) becomes

cos(A+B) - cos (90° - + C);
cos(A+B) + cos (90°-+C)

and the second side becomes

or,

-tan(A+B+90°-C)tan(A+B-90°++C);

-tan(A+B-C+180°)tan(A+B+C-180°).

If we remember that E=A+B+C-180°, and observe

that

tan+(A+B-C+180°)=tan(360°-2C+A+B+C-180°)

= tan(360°-2C+E)

= tan [90°-1 (2C-E)]

=cot+ (2 C-E),

it will be evident that equation (b) may be written

cos(A+B)-cos(90°)=-cot (2C-E) tan+ E. (c) cos(A+B)+cos(90°-C)