Elements of Plane and Spherical Trigonometry

In like manner, the triangles AOH, ADH, right angled

in н and D, give

AH AO sin AOBAO sin c
ADAH Sin AHD

Making these two values of

AD equal, we have

AO sin B sin c = AO sin c sin b.

And therefore, by division,

AO sin c sin B.

Hence, the sines of the angles of a spherical triangle are proportional to the sines of the opposite sides.

21. Draw CE and DF, respectively perpendicular and parallel to OB; then will the angle DCF = EOC = a. But the right angled triangles AOC, ACD, FCD, give

ACAO sin b, DC AC Cos CAO sin b cos c
and FD DC sin a = AO sin a sin b cos c.

NOW OHOE+EHOE + FD,

er ao cos c = oc cos a + FD = AO cos a cos b + FD =AO cos a cos b + Ao sin a sin b cos c.

Therefore, dividing by Ao, we have

cos c cos a cos b + sin a sin b cos c. Similar relations are deducible for the other sides a

and b: hence, generally

cos a = cos b cos c + sin b sin c cos A
cos b = cos a cos c + sin a sin c cOS B
cos c cosa cos b + sin a sin b cos c

(2.)

22. These equations apply equally to the supplemental triangle. Thus, putting for the sides a, b, c,

180° - A', 180° - B', 180° c', &c. and for the angles A, B, C, 180° — a′, 180° - b', &c. we shall have

cos A' COS B' COS C'

sin B' sin c' cos a'.

Here again we have three symmetrical equations applying to any spherical triangle, viz.

cos A= cos a sin в sin c
COS B cos b sin a sin c

COS A COS C(3.) cos c = cos c sin A sin в - COS A COS B

23. Another important relation may be readily deduced. For, substituting for cos b in the third of the equations marked (2) its value in the second; substituting also for cos2 a its value 1-sin2 a (chap. i. 19), and striking out the common factor sin a, we shall have cos c sin a sin c cos a cos B + sin b cos c.

But, equa. (1) gives sin b =

Hence, by substitution,

sin B sin c

sin c

Thus, again, we get three symmetrical equations, cot a sin b = cos b cos c + sin c cot a

cos c cos A + sin A Cot B
cos a cos B + sin в cot c

}

(4.)

cot b sin c = cot c sin a 24. The classes of equations marked 1, 2, 3, 4, comprehend the whole of spherical trigonometry: or, in truth, the equations (2), from which the others may be made to flow, may be regarded as comprehending the whole. They require, however, some modifications to fit them for logarithmic computations, and become simplified in their application to some kinds of triangles. We shall, therefore, now show the pupil how they be

come transformed when they are applied to the prin cipal cases which occur in practice.

SECTION II.

Resolution of Right angled Spherical Triangles.

25. Suppose, in the first diagram in this chapter, the angle A to be right, or the faces OAB, OAC, to be perpendicular to each other. Then, since sin A = 1, the equations marked (1) become

sin B

sin c

Consequently,

sin a

sin b

sin c

cos a = cos b cos c.... (6.)

For the same reason, the first of equa. (3) gives cos a sin B sin c = cos B COS C;

whence cos α =

COS B COS C

sin B sin c

cot в cot c (7.)

....

Upon the same hypothesis, cot A becomes = o, so that the first of equa. (4) becomes

cot a sin b = cos è cos c. Or, dividing by sin b,

cos b

cot a = cos ccot b cos c .... (8.)

sin b

The two last of equa. (3) give also, upon the same

hypothesis,

COS B = sin c cos by

cos c = sin B cos c

And, lastly, from equa. (4) we have

(9.)

From these equations by a few obvious transforma tions, the six usual cases of right angled spherical triangles may be solved, as below.

or, tan side req. = tan hypoth. x cos included angle, cot c = cos a tan B;

or, cot angle req. = cos hypoth. x tan given angle. In this case there can be nothing ambiguous; for, in applying the first form, it is known that the angle and the opposite side are always of the same affection; and in the two latter the rules for the changes of sines in the different quadrants (chap. iv. 9), will determine to which the result belongs.

Case II.

27. Given the hypothenuse a, and one of the sides ; to find the rest.