directly to the solution of spherical triangles. This is especially the case when two sides and the included angle, or two angles and the included side, are given and one or more of the remaining parts required. If, for instance, A, B, and c are given, we may from are the first two of equations (11) obtain the values of C and a by Prob. I. Chap. V. Then from the next two we may obtain c and C, and from the last cos C, and thus a third value of C. If the work is correct, these three values of C will all agree.

Example. Two of the face-angles of a trihedral angle are a = 132° 46′.7 and b = 59° 50′.1, and the included edge-angle is C 56° 28'.4. Find the remaining parts.

=

The computation of A, B, and c may be effected by the equations (8), as follows:

NOTE. We may omit the "log" from the designation of the logarithms of the trigonometric functions whenever no uncertainty will thus arise.

In this computation the numbers in parentheses show the order in which the lines may be written. Lines (15), (23), and (31) are the addition or subtraction logarithms, from the tables for finding the logarithm of the sum or difference of two numbers which are given by their logarithms. The student can equally well find the numbers, add them together, and take the logarithm of their sum.

The agreement of the two values of sin C with each other and with cos C shows the correctness of the calculation.

107. The following transformation, similar to that of Prob. IV. Chap. V., will often render the work convenient. In the second and last of equations (11) let us put

k sin K = sin A cos c;

k cos K =

cos A.

By substitution these equations will then become

Ksin

sin C cos a = k cos K sin B+ k sin K cos B

= k sin (KB);

cos C = k cos K cos B + k sin K sin B

= k cos (KB).

(a)

(b)

To apply these equations we compute k and K from (a), and then sin C cos a and cos from (b). We complete the work by computing sin C sin a from the first equation of (11).

We may also transform the fourth equation by computing h and Д from the equations

h sin Hsin B cos c;

h cos H =

cos B.

We shall then have

sin C cos bh sin (H — A);

cos Ch cos (H — A).

To transform the equations (8) on the same plan, we may com

pute k, K, h and H from

k sin K = sin a cos C;

k cos K = cos a;

I sin H = sin b cos C';

h cos H = cos b.

(c)

We then have, in the same way as before,

sin c cos A = k sin (b — K);

(d)

cos C = k cos (b − K) = h cos (a - H).

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

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 cpposite side,

prove

cos b + cos c = 2 cos ta 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

A

Therefore the pole of BC is on both BA and CA, and there

fore 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 which, when any two parts are given, any вone of the three remaining parts may be

с

a

found; hence between every combination of three parts out of the five there must be an equation. The number of combinations of 3

in 5 being

5.4

1.2

=

10, there must be ten such equations.