We then have, in the same way as before,

sin e cos Ak sin (b-K);

sin c cos B = h sin (a·

H);

(d)

cos Ck cos (b-K) = h cos (a - H).

We compute the same example as before by these formulæ, as

follows:

1. Transform the equations (6), (7), (9), and (10) in the same

way that we have transformed (8) and (11).

2. From the values of A, B, and c, which we have obtained in the last example, find those of a, b, and C with which we started. 3. If m be the arc joining the vertex A to the opposite side,

prove

cos bcos c = 2 costa cos m.

CHAPTER II.

RIGHT AND QUADRANTAL TRIANGLES.

Fundamental Definitions and Theorems.

108. Def. A right spherical triangle is one which has a right angle.

Def. A quadrantal spherical triangle is one which has a side equal to a quadrant.

Def. A trirectangular triangle is one which has three right angles.

Def. A birectangular triangle is one which has two right angles.

Def. A biquadrantal triangle is one which has two sides equal to a quadrant.

THEOREM I. Every birectangular triangle is also biquad

rantul.

Proof. Let ABC be a spherical triangle in which angle B = angle C 90°. Then:

Because angle B is a right angle, the pole of the great circle BC is on the great circle BA. Because angle C is a right angle, this pole is on the great circle CA. (Geom.)

B

Therefore the pole of BC is on both BA and CA, and therefore at their point of intersection A.

Because A is the pole of BC, AB and AC are quadrants.

Q.E.D. THEOREM II. Conversely, Every biquadrantal triangle is also birectangular.

Proof. Because every point of the polar circle of the point A is a quadrant distant from A, and because AB and AC are quadrants, this polar circle must pass through both B and C.

But only one great circle can pass through these points.

Therefore BC is the polar circle of A, and A the pole of BC. Therefore the great circles AB and AC intersect BC at right angles. Q.E.D.

Cor. Every trirectangular triangle has three quadrants for its sides; and,

Conversely, Every triangle having three quadrants for its sides is trirectangular.

THEOREM III. In a birectangular triangle the oblique angle is equal to its opposite side.

Proof. Because the plane of the great circle BC intersects the planes of AB and of AC at right angles, the arc BC measures the dihedral angle between the planes AB and BC.

But the angle A is equal to this same dihedral angle.

Therefore BC= angle A.

THEOREM IV. The polar triangle of a right triangle is a quadrantal triangle.

This follows at once from the fact that the angles of the one triangle are the supplements of the sides of the other.

EXERCISE.

Let the student translate the preceding definitions and theorems into those relating to the face- and edge-angles of a trihedral angle, and, which is the same thing, into those relating to the angles between three lines emanating from a point and the angles between their planes.

109. Formula for right triangles. Since in a right triangle one of the parts, the right angle, is known in advance, if two other parts be given the remaining three parts may

be found.

An equation must therefore exist by

a

A

which, when any two parts are given, any Bone of the three remaining parts may be found; hence between every combination of three parts out of the five there must be an equation. The number of combinations of 3

To find these equations let C be the right angle, and therefore c the hypothenuse. We seek for those equations in the sets (6) to (12) of the last chapter, in which the angle Centers, and in which the equation contains only three different parts. We then suppose

sin C = 1;

The set from which the required equations are taken, the number in the set, and the result are shown as follows:

From (6),,

66

(7),,

These ten equations will be found to include all combinations of three out of the five parts a, b, c, A, B. From each of them we may determine any one part in terms of the other two; for example, the first equation gives not only

Properly speaking, only six of these equations are really distinct, as the other four can be derived from them by a mere interchange of letters between corresponding parts. For instance, since the same relation must hold between each oblique angle and its opposite side, the second equation may be derived from the first.

The equations which are thus related are connected by braces in the formulæ above.

110. Napier's rules. The six preceding formula, which may be found difficult to remember, have been included by Napier in two precepts of remarkable simplicity,

and easily remembered.

B'

Let us take for the five parts the sides a and b as before, and, instead of the other three parts, the complements of the oblique angles and of the hy pothenuse. The fact that the complements are understood is indicated by accenting the letters in the diagram. We suppose

a

a.

B' = 90° B; c′ = 90°. A' = 90° c; Omitting the right angle, the five parts a, b, A', c', B' form a continuous series, B' being followed in regular order by a. Now if we select any three of these parts, one of two cases must occur. Either

(1) The three parts all adjoin each other, as B', a, b; a, b, A', etc., or

(2) Two of the parts adjoin each other and the third is separated from each of them by the remaining intervening parts.

The middle part of the three in the first case, or the separated part in the second, is called the middle part.

In the first case the extreme parts of the three are called adjacent parts.

In the second case the adjoining parts are called opposite parts. 111. Napier's rules are:

I. The sine of the middle part equals the product of the tangents of the adjacent parts.

II. The sine of the middle part equals the product of the cosines of the opposite parts.

The concurrence of the vowel a in tangent and adjacent, and of the vowel o in cosine and opposite, will help in remembering the relations.

Examples. 1. Let the parts be the hypothenuse and the two adjacent angles, or c, A and B.

The middle part is c', and A' and B' are adjacent parts.