:: cot the perpendicular arc CD, we shall have cos AC : cos BC::cos AD: cos BD. · This proportion produces this subsequent one, cos AC +cos BC : cos BCcos AC::cos AD+cos BD : cos BDcos AD, and from this we easily deduce B this third proportion (19 Pl. Trig.) AC+BC BC-AC. BD-AD cot : tang cotAD+BD : tang 2 But AD+DB=AB, when the perpendicular falls within the triangle; by substituting tangents for cotangents, we have therefore the following theorem of such frequent use, and which may be announced as follows: 84. In any oblique-angled spherical triangle, on the base of which we have let fall a perpendicular arc, and which lies within the triangle, the tangent of the half base is to the tangent of half the sum of the other two sides, as the tangent of the half difference of these sides is to the tangent of half the difference of the segments of the base. Observe that if the arc falls without B E That is to say, in any right-angled spherical triangle, radius is to the cosine of one of the sides of the right-angle, as the sine of the oblique angle opposite to the other side is to the cosine of the other oblique angle. 86. And consequently if we let fall an arc A B perpendicularly on the base of an oblique-angled triangle, the sines of the angles al the summit will be proportional lo the cosines of the angles of the base. In the oblique-an C AS B D 87. Let there now be the triangle ABC, right-angled at A, and let there be drawn the tangents BP a and BQ: the latter to the hypothenuse BC, and the first to the side BA. If we draw the secants EP and EQ, F whose extremities are joined by the line PQ, we shall have the triangles PEQ and BPQ, włose planes are perpendi B cular to that of the base BAE. Their GH P intersection BQ will consequently be perpendicular to the same plane; and the triangle_BPQ, rightangled at P, will be similar to the triangle CFE. Thus we shall have FE: GE :: BQ : BP or, R: cos B :: tang BC : tang AC Therefore in any right-angled spherical triangle, radius is to the cosine of one of the oblique angles, as the tangent of the hypothenuse is to the tangent of the side opposite to the other angle. 88. If the trian C gle ABC is obliqueangled, we may let fall on its base the perpendicular arc CD, which will A A give B B R: cos ACD :: tang AC : tang CD and R: cos BCD ; : tang BC : tang CD Whence we infer, cos ACD :cos BCD :: tang BC : tang AC That is, if we let fall an arc perpendicularly on the base of an oblique-angled spherical triangle, the cosines of the angles at the summit will always be reciprocally proportional to the tangents of the adjacent sides. 89. And as in the complemental triangle CDE, we have R: cos D :: tang CD : tang DE D It is evident that we also have E R:sin AB :: cot AC :cot B::tang B: tang AC Therefore, in any right-angled triangle, radius is to the sine of one of the sides of the right-angle, as the tangent of the oblique angle opposite to the other side is to the tangent of this last side. 90. The same triangle CDE gives R:sin CE:: tang C : tang DE A B therefore, R: cos BC ;: tang C: cot B : : Whence it follows that in every right-angled spherical triangle, radius is to the cosine of the hypothenuse, as the tangent of one of the oblique angles is to the colongent of the other angle. APPLICATION OF THE PRECEDING PRINCI PLES AND PROPORTIONS. 91. By means of the propositions which we have just demonstrated, we can very easily solve any spherical triangle, provided three of its parts are given. We shall commence with right-angled triangles. And first, as the right-angle is always given, it is sufficient to know two of the other five parts, that make up these triangles. Again the number of combinations of m quantities, taken two m. (m-1) by two, is expressed generally by (Page 111). It is evi. 2 dent, therefore, 1“, that in this point of view, right-angled spherical triangles present ten of these combinations. But, as for each combination we have three quantities to determine, it is clear 2dly, that all the possible varieties in the solution of right-angled spherical triangles are 30 in number. The following Table contains them all, the angle A is supposed E to be the right-angle; and the other two angles are denoted indifferently by B or C. B 92. The construction of this G Table is wholly founded upon two propositions already demonstrated. We shall place them once more F under the eye of the student. Prop. 1. In every right-angled spherical triangle, radius is to the sine of the hypothenuse, as the sine of one of the oblique angles is to the sine of the side opposite to it (81). Prop. 2. In every right-angled spherical triangle, radius is to the sine of one of the sides of the right-angle, as the tangent of the oblique angle opposite to the other side is to the tangent of that side (89). Sometimes these proportions are applied immediately to the triangle ABC, and sometimes we must have recourse to one of the com plemental triangles CDE, BFG, in order afterwards to transport the results to the triangle ABC, as will be shewn in several examples. TABLE For the solution of all the possible cases of a spherical triangle ABC, right-angled at A. a Given. B BC с AC с AC AC Required Tie part required is less than 30o. с R: cos BC::tang B: cot C If BC and B are of the same species B R: cos BC::tang C : cot B If BC and C are of the same species B R; cos AC::sin C:cos B If AC < 90° c. cos AC :cos B::R:'sin C Ambiguous B sin BC: sin AC::R: sin B If AC < 90° с tang BC : tang AC::R: cos C If BC and AC are of the same species B cos AB: cos C::R: sin B Ambiguous < с R: sin AC::cot AB : cot C If AB < 90° с isin BC: Sin AB::R: sin C If AB < 90° AB с АВ AB AC АВ BC B с 90° To facilitate the application of this Table, let the hypothenuse BC=81° 13, and the angle B=37° 19'; required the side AC, opposite to the angle B. To solve this case, we must employ the first proportion in the Table, and say R: sin BC:: sin B : sin AC But log sin BC (814 13') 9.994877 log sin B (37° 19') = 9.782630 19.777507 9.777507 = 10. = 10. = 1 Therefore AC-36° 48', or 143° 12', which is its supplement. To decide which of these values is the right, we must call to mind that the side AC must be of the same species as the angle B oppa to it (74.) By way of reminding the calculator of the conclitions upon which depend the results which he seeks for, they are inserted in the last column of the Table. 93. Supposing still the same hypothenuse BC and the same angle B to be given, and that the adjacent side AB were required; it would be easy in the first place to find the side AC, as before, and afterwards to employ the proportion tang B : tang AC::R: sin AB. But this process introduces two proportions into a computation which may be managed by only one. For in the complemental triangle CDE we have R: sin DE : : tang D : tang CE, and therefore transporting this proportion to the triangle ABC, it will become R: cos B : ; cot AB : cot BC, or perhaps better, R: cos B :: tang BC: tang AB, But log radius taken negatively log cos B (37° 19') 9.900529 log tan BC (81° 13') = 10.811042 log tang AB = 10.711571 This last logarithm corresponds to 799 or 101°; but as BC and B are of the same kind, we must take the first value: Consequently the side AB is 79', or calculating as far as seconds, 79° 0 20". N. B. When the logarithm of radias is to be subtracted, we shall simply write it 10; and in the aggregation of the three terms which form the proportion, the unit surmounted with the negative sign must be considered as nega. tive and subtracted. To find the angle C, we must have recourse to the triangle CDE, in which we have sin CE : R:; tang DE : tang C, whence we obtain by substitution, cos BC:R:: cot B : tang C R: : cos BC:: tang B:cot C. By logarithms, we obtain the angle C as follows: neg. log of radius = 10. = 9.065935 which gives C=83° 22', and not 96° 38', since 'BC and B are of the same kind. 94. If instead of the hypothenuse being given, we had known the adjacent side AB, with the same angle B, and had desired the |