Arts. 10, 21). A spherical triangle can be solved graphically by drawing (Art. 24) upon any sphere a triangle that satisfies the given conditions, and then measuring the required parts of the triangle. The sides and angles (see Art. 11. c) can be measured with a thin, flexible, brass ruler, on which a length equal to a quadrant or a semicircle of the sphere is graduated from 0° to 90° or 180° respectively.

Small slated globes can be obtained fitting into hemispherical cups, whose rims are graduated from 0° to 180° in both directions. With such a globe, cup, and a pair of compasses, the constructions discussed in Art. 24 and the measurements referred to in this article are easily made.

If the student has the means at hand, it is advisable for him to solve some of the numerical problems graphically.

N.B. Questions and exercises on Chapter II. will be found at p. 102.

CHAPTER III.

RELATIONS BETWEEN THE SIDES AND ANGLES OF SPHERICAL TRIANGLES.

35. In this chapter some relations between the sides and angles of any spherical triangle (whether right-angled or oblique) will be derived. In the next chapter these relations will be used in the solution of practical numerical problems. The first two general relations (namely, the Law of Sines and the Law of Cosines), which are by far the most important, can be derived in various ways. In a short course it may be best to deduce these laws by means of the properties of right-angled triangles as set forth in Art. 26; and, accordingly, this method is adopted here. These laws are also derived directly from geometry in Note A at the end of the book. It may be stated here that the geometrical derivation will strengthen the student's understanding of the subject, and will show more clearly the correspondence (Art. 14) between the parts of a spherical triangle and the parts of a triedral angle.

36. The Law of Sines and the Law of Cosines deduced by means of the relations of right-angled triangles.

A. Derivation of the Law of Sines.

Let ABC (Figs. 39, 40) be any spherical triangle. From B draw the arc BD at right angles to AC to meet AC, or AC produced, in D.

Hence, in both figures, sin a sin C sin c sin A.

=

Similarly, by drawing an arc from C at right angles to AB, it

In words: In a spherical triangle the sines of the sides are proportional to the sines of the opposite angles. (Compare Plane Trigonometry, Art. 54, I.)

B. Derivation of the Law of Cosines.

cos BC= cos CD cos DB

= cos (b― AD) cos DB, or cos (AD – b) cos DB
= cos b cos AD cos DB + sin b sin AD cos DB.

(a)

cos a = cos b cos c + sin b sin c cos A.

(2)

Similarly, or by taking the sides in turn,

cos b = cos c cos a + sin c sin a cos B,

cos c = cos a cos b + sin a sin b cos C.

In words: In a spherical triangle the cosine of any side is equal to the product of the cosines of the other two sides plus the product of the sines of these two sides and the cosine of their included angle. (Compare Plane Trigonometry, Art. 54, II.)

NOTE 1. The law of cosines, (2), is the fundamental and the most important relation in spherical trigonometry. For, as shown in Note A, it can be deduced directly; the law of sines, (1), can be deduced from it; all other relations follow from these; and the relations for right triangles, Art. 26, can be deduced from the relations for triangles in general, on letting C be a right angle. The formulas for cos a, cos b, cos c, were known to the Arabian astronomer Al Battani in the ninth century. (See Plane Trigonometry, p. 166.)

C. The Law of Cosines for the angles. Relation (2) holds for all triangles, and, accordingly, for A'B'C', the polar triangle of ABC. (See Fig. 14, Art. 16.) That is,

cos a' cos b' cos c' + sin b' sin c' cos A'.

.. cos (180° — A) :
= cos (180° — B) cos (180° — C)

+ sin (180° — B) sin (180° — C) cos (180° — a). [Art. 16. d.]
cos A = (— cos B) (— cos C') + sin B sin C (— cos a).

Relation (3) can also be derived by means of right-angled triangles.

NOTE 2. From (2),

-

cos A =

cos a
cos b cos c
sin b sin c

The denominator in the second member is always positive. If a differs more from 90° than does b, then cos a is numerically greater than cos b, and, accordingly, greater than cos b cos c; hence cos A and cos a have the same sign, and thus, a and A are in the same quadrant.

Similarly, a and A are in the same quadrant when a differs more from 90° than does c.

it can be shown in a similar way that if A differs more from 90° than does B or C, then a and A are in the same quadrant.

Ex. 1. Derive cos b and cos c by means of right triangles.
Ex. 2. Derive cos A and cos B by means of right triangles.
Ex. 3. Derive cos C from cos c by means of the polar triangle.

37. Formulas for the half-angles and the half-sides.

[Compare the method and results of this article with those of

a-b+c=2(sb), and a+b-c=2(s—c).