Multiplying (4) by (7), member by member, we have,

sin b sin c = tan b tan c tan (90%-B) tan (90°— C).

Dividing both members by tan b tan c,

we have,

208 b cos c = tan (90°-B) tan (90°— C');

Formula (6) may be written under the form,

Substituting for cos a, its value, cos b cos c, taken from (1), and reducing, we have,

Changing B, b, and C, in (9), into C, c, and B, we have,

These ten formulas are sufficient for the solution of any

right-angled spherical triangle whatever.

73. The two sides about the right angle, the complements of their opposite angles, and the complement of the hypothenuse, are called Napier's Circular Parts.

If these parts be arranged in their order, as shown in the figure, we see that each part is adjacent to two of the others, and that it is separated from each of two remaining parts by an intervening part. If any part be taken as a middle part, those which are adjacent to it are called adjacent parts, and those which are separated from it, are called opposite parts. Thus, 90° and

90°

B,

C, are adjacent parts to 90°-a; and c and bare opposite parts; and so on, for each of the other parts.

74.

Formulas (1), (2), (5), (9), and (10), of

Art. 72, may be written as follows:

Comparing these formulas with the figure, we see that,

The sine of the middle part is equal to the rectangle of the cosines of the opposite parts.

Formulas (8), (7), (4), (6), and (3), of Art. 72,

Comparing these formulas with the figure, we see that,

The sine of the middle part is equal to the rectangle of the tangents of the adjacent parts.

These two rules are called Napier's rules for Circular Parts, and they are sufficient to solve any right-angled spherical triangle.

75. In applying Napier's rules for circular parts, the part sought will be determined by its sine. Now, the same sine corresponds to two different arcs, supplements of each other; it is, therefore, necessary to discover such relations between the given and required parts, as will serve to point out which of the two arcs is to be taken.

Two parts of a spherical triangle are said to be of the same species, when they are both less than 90°,

or both greater than 90°; and of different species, when one is less and the other greater than 90°.

From Formulas (9) and (10), Art. 72, we have,

since the angles B and C are both less than 180°, their sines must always be positive hence, cos B must have the same sign as cos b, and the cos C must have the same sign as COS C. This can only be the case when B is of the same species as b, and C of the same species as c; that is, the sides about the right angle are always of the same species as their opposite angles.

From Formula (1), we see that when α is less than 90°, or when cos a is positive, the cosines of band с will have the same sign; that is, b and c will be of the same species. When a is greater than 90°, or when cos a is negative, the cosines of b and c will be contrary; that is, b and C will be of different species: hence, when the hypothenuse is less than 90°, the two sides about the right angle, and consequently the two oblique angles, will be of the same species; when the hypothenuse is greater than 90°, the two sides about the right angle, and consequently the two oblique angles, will be of different species.

These two principles enable us to determine the nature of the part sought, in every case, except when an oblique angle and the opposite side are given, to find the remaining parts. In this case, there may be two solutions, one solu tion, or no solution at all.

C'

B

and b are given. Prolong

the sides BA and BC till

they meet in B'. Take

B'A' = BA, B'C' = BC, and join A' and C" by the arc of a great circle: then, because the triangles BAC and B'A'C' have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, the remaining parts will be equal, each to each;

that is, A'C' angle A

=

AC, and the angle A' equal to the hence, the two triangles BAC, B'A'C',

right-angled; they have also one oblique angle and the opposite side, in each, equal.

Now, if b differs more from 90° than B, there will evident

ly be two solutions, the sides

B

are

>B'

A'

including the given angle, in the one case, being supplements of those which include the given angle, in the other case.

If b B, the triangle will be bi-rectangular, and there

=

If 6 differs less from 90° than B, the triangle cannot be constructed, that is, there will be no solution.

SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES.

76. In a right-angled spherical triangle, the right angle 18-always known. If any two of the other parts are given, the remaining parts may be found by Napier's rules for circular parts. Six cases may arise. There may be given,

I. The hypothenuse and one side.

II. The hypothenuse and one oblique angle.
III. The two sides about the right angle.

IV. One side and its adjacent angle.

V. One side and its opposite angle.
VI. The two oblique angles.

In any one of these cases, we select that part which is either adjacent to, or separated from, each of the other given parts, and calling it the middle part, we employ that one of Napier's rules which is applicable. Having determined a third part, the other two may then be found in a similar manner.