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 cos a sin в sin c B -COS B COS C 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 в + sin b cos c. But, equa. (1) gives sin b = Hence, by substitution, sin B sin c sin c Therefore cot c sin a = cos a cos B + sin в cot c. (4.) cot c sin a = cos a cos B + sin в cot C 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 principal 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 1 sin B sin c Consequently, sin 6 } (5.) sin B sin a... sin c = sin c sin a Also, since cos A is then = o, we have from equa. (2) (6.) .... For the same reason, the first of equa. (3) gives cos a sin B sin c = cos B cos c; Upon the same hypothesis, cot A becomes = 0, so that the first of equa. (4) becomes cot a sin b = cos 6 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 csin B cos c And, lastly, from equa. (4) we have cot B cot b sin c cot c = cot c sin b .... (9.). From these equations by a few obvious transformations, the six usual cases of right angled spherical triangles may be solved, as below. Case I. 26. Given the hypothenuse a and an angle в; sin b or, sin side req. x sin hypoth. or, tan side req. to find the rest; sin a sin B; or, cot angle req. sin opp. angle B tanc tan a cos B; C tan hypoth. x cos included angle, cot c = cos a tan B; 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. or, cos angle req. = tan given side x cot hypoth. Case III. 28. Given the two sides including the right angle, namely, b and c; to find the rest. cos a = cos b cos c; rectangle cosines of the sides. or, cos hypoth, Case IV. 29. Given a side b, and its opposite angle B; to find the rest. 30. Given a side c, and its adjacent angle в; to find that is, cot hypoth. = cos given angle x tan given side. cos c = cos c sin B; or, cos angle req.= cos opp. side x sin given angle. Case VI. 31. Given the two oblique angles B and c; to find the rest. cos ccot B cot c; or, cos hypoth. rectangle cot's given angles. Note 1. Here the rule of the signs (chap. iv. 9) serves all along to determine the kind or affection of the unknown parts. Note 2. In working by the logarithms, the student must observe, that when the resulting log. is the log. of a quotient, 10 must be added to the index: when it is the log. of a product, 10 must be subtracted from the index. This is done in conformity with the rule (chap. iv. 17), to make the terms homogeneous by multiplying or dividing by the powers of radius. SECTION III. Resolution of Oblique angled Spherical Triangles. 32. This may be effected by means of four general cases; each comprehending two or more problems. Case I. Given three of these four things, viz. two sides b, c, and their opposite angles B, C; to find the fourth. This case comprehends two problems, in one of which the unknown quantity is an angle, in the other a side. They are both solved by means of equa. (1) of this chapter from which we have 33. Of these four things, viz. the three sides a, b, c, and an angle, any three being given, to find the fourth. This case comprises three problems. 1. When the three sides are given, to find an angle. Here from equa. (2) we have |