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AC = Ao sin Aoc = Ao sin b AD = AC sin ACD = Ao sin b sin c. In like manner, the triangles Aoh, ADH, right angle in H and D, give AH = Ao sin Aob = Ao sinc AD = AH sin AHD = Ao sinc sin B. Making these two values of A AD equal, we have Ao sin B sin c = Ao sin C sin b. And therefore, by division, sin B sin c in 5 of sing" In a similar manner it may be shown that

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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 -Ac = Ao sin b, DC = Accos c = Ao sin b cosc and FD = Dc sin a = Ao sin a sin b cos c. Now on = of + EH = or + FD, or Ao cos c = oc cos a + FD = Ao cos a cos b + FP = 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 or ces a = cos b cos c + sin b’sin c cos A cos b = cos a cos c + sin a sin c cos B F (2.) cos c = cosa cos b + sin a sin b cos C 22. These equations apply equally to the supplemental triangle, Thus, putting for the sides a, b, c,

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180°– A', 180°-B', 180° – c', &c, and for the angles A, B, C, 180° — as, 180° — bo., &c. we shall have – cos A" = cos B' cos c – sin B'sin c' cosa". Here again we have three symmetrical equations applying to any spherical triangle, viz. COSA E COS a Sin B Sin C – COS B COS C cos B = cos b sin A sin C — cos A cos # (3.) COS C = COS C Sin A Sin B - 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 coso a its value 1 — sin” 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 cos c sinc
COs c sin a = sin c CoS a COS B + —HF-
Dividing by sin c we have,
cos c -

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Therefore cot c sin a = cos a cos B + sin B cot c. Thus, again, we get three symmetrical equations, cot a sin b = cos b cos c + sin C cot A cot b sin c = cos c cos A + sin A cot : (4.) cot c sin a = cos a cos B + sin B 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 become transformed when they are applied to the principal cases which occur in practice.

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 * I sin B sin c

i = j = . Consequently,

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Upon the same hypothesis, cot A becomes = o, so that the first of equa. (4) becomes cot a sin b = cos & cos c. Or, dividing by sin b, cos &

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The two last of equa, (3) give also, upon the same hypothesis, cos B = sin c cos b cos c = sin B cos c s ' ' ' ' (9.) And, lastly, from equa. (4) we have cot B = cot b sinc cot c = cot c sin b s ' ' ' ' (10.) 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 G and an angle B ; to find the rest; sin b = sin a sin B; B A or, sin side req. = sin opp. angle c x sin hypoth. tan c = tan a coS B;

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 . 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 5; to find the rest. sin b _ sin given side.

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Case III. 28. Given the two sides including the right angle, mamely, b and c; to find the rest. cosa = cos b cos c; or, cos hypoth. = rectangle cosines of the sides. _ tanb _ tan c. tan B = H. . . . tan c = Hij;

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Case IV.

29. Given a side b, and its opposite angle B; to find the rest. * _ sin given side To sin opp. angle' sin c = tan b cot B; or, sin side req. = tan given side x cot opp. angle. cos given angle.

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s & cos given side

Case W.

30. Given a side c, and its adjacent angle B; to find the rest. tan b = tan B sin c; or, tan side req. = tan opp. angle x sin given side. tan a = *; or, tan hypoth. = tan given side. cos B cos given angle or, cot a = cos B cot c : that is, cot hypoth. = cos given angle x tan given side. Cos C = cos cisin 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 c = cot B cot C ; or, cos hypoth. = rectangle cot's given angles.

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Note 1: Here the rule of the signs (chap. iv. 9) serves all along to determine the kind or affection of the unknown parts.

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