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

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

we have,

Dividing both members by tan b tan c,

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

Formula (6) may be written under the form,

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

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.

90-B

B

If we take any three of the five parts, as shown in the figure, they will either be adjacent to each other, or one of them will be separated from each of the other two, by an intervening part. In the first case, the one lying between the other two parts, is called the middle part, and the other two, adjacent parts. In the second case, the one separated from both the other parts, is called the middle part, and the other two, opposite parts. Thus, if 90°-a, is the middle part, 90° – B, and 90° — C, are adjacent parts; and b and c, are opposite parts; and similarly, for each of the other parts, taken as a middle part.

74. Let us now consider, in succession, each of the five parts as a middle part, when the other two parts are opposite. Beginning with the hypothenuse, we have, from formulas (1), (2), (5), (9), and (10), Art. 72,

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.

Let us now take the same middle parts, and the other parts adjacent. Formulas (8), (7), (4), (6), and (3), Art. 72, give

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 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,

sin C =
COS B
cos b

and

sin B =

cos C

COB C

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 e; 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 C 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 с 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 speries; 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.

to determine the nature

These two principles enable us 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.

B'A' = BA, B'C' = BC, and join A' 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' = AC, and the angle A'
angle A hence, the two triangles BAC,
right-angled; they have also
one oblique angle and the op-
posite side, in each, equal.

Now, if b differs more from 90° than B, there will evidently be two solutions, the sides

B

C

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 will be but a single solution.

If b 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 is 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.