129. Species of the Parts. Two parts of a spherical triangle are of the same species when both are less than 90° or both greater than 90°. Two parts of a spherical triangle are of different species when one part is less than 90° and the other part greater than 90°. We shall, at present, consider those triangles only whose parts do not exceed 180°. Let it be remembered that the sine is positive from 0° to 180°, and that the co-sine, the tangent, and the co-tangent are positive from 0° to 90°, and negative from 90° to 180°. Hence, if the co-sines, tangents, or co-tangents of two parts have like signs, these parts will be of the same species; if they have unlike signs, these parts will be of different species. COS P sin P cos B = Art. 128, 7, 8. Since neither P nor B exceeds 180°, sin P and sin B are both positive; hence, cos B and cos b have like signs, so also have cos P and cos p. Therefore, B and bare of the same species; so also are P and p. cos h and sin B Hence, The sides adjacent to the right angle are of the same species as their opposite angles. = cos b cos p. Art. 128, 9. If h< 90°, cos h is positive; hence, cos b cos p is positive; .. cost and cos p have like signs; ... b and p are of the same species; ... B and P are of the same species. Hence, If the hypotenuse is less than 90°, the two sides adjacent to the right angle are of the same species; so also are their opposite angles. If h 90°, cos h is negative; hence, cos b cos p is negative; .. cos b and cos p have unlike signs; .. b and p are of different species; .. B and P are of different species. Hence, If the hypotenuse is greater than 90°, the two sides adjacent to the right angle are of different species; so also are their opposite angles. Let us now investigate the case in which a side adjacent to the right angle and its opposite angle are given. = = H B' H' = Let p and P be given. Produce the sides PH and PB till they meet in P'. The angles P and P' are equal, since each is the angle included by the plane of the arcs PHP' and PBP'. Take P'H': PH b and P'B' PB h. The two triangles, PHB and P'H B', have the two sides PH and PB and the included angle P of the one, equal to P'H' and P'B' and the included angle P' of the other; hence, they are equal in all their corresponding parts; .. H' = H, B' = B, and H'B' HB. But H is a right angle; .. H' is a right angle. Hence, either triangle, PHB or PH'B', will answer to the given conditions. Since P'H' and PH are equal, and P'H' and PH' are supplements of each other, PH and PH' are supplements of each other. In like manner it can be shown that PB and PB' are supplements of each other. When, therefore, a side adjacent to the right angle and an opposite angle are given, there are apparently two solutions. The conditions of the problem, however, may be such as to render the two solutions possible, reduce them to one, or render any solution impossible. Let us now proceed to investigate these conditions. 1. When P< 90° and p < P. We have from Napier's principles, = sin b=tan p tan (90° — P), or sin b tan p cot P. Since P 90° and p < P, tan p < tan P; but we have tan P cot P = 1; .. tan p cot P<1; hence, sin b1; then b 90° or b> 90°; hence, b may be either of the supplementary arcs PH or PH' which have the same sine equal to tan p cot P. cos B If b90°, since p< 90°, h< 90°; if b > 90°, since p90°, h> 90°. Hence, Hence, if P < 90° and p< P, either triangle, PHB or PH'B', will satisfy the conditions, and there will be two solutions. = sin (90°-B) = PH' PB' Since p P, tan p cot P= tan P cot P fore, sin b1; b 90°, or PH 90°. From Napier's principles, we have sin (90°— h): =cos b cos p, or cos h Since b == 90°, cos b 0; h 90°, or PB = 90°. cos h Since h 90°, cot h ... cos B 0; .. B P P 180° PH: = = tan p tan (90°-h), which reduces to tan p cot h. B' =cos b cos p. 0; .. cos b cos p= 0; hence, 90°; ... PH': 90°; ... PB' B H 1; there 0; .. tan p cot h PH. = PB. Hence, if P < 90° and p P, b 90°, B: 90°, the two triangles reduce to the bi-rectangular triangle PHB, and there is but one solution. We have sin b before. = 5. When P> 90° and p 3. When P< 90° and p > P. As before, we have sin b Then, if p > P, taħ p> tan P; but tan P cot P=1; .. tan p cot P>1; ... sin b> 1, which is impossible. Hence, if P < 90° and p > P, no solution is possible. = = = 4. When P> 90° and p > P. We have sin b tan p cot P, as before. tan p and cot P are 1; both negative, and tan p < tan P, numerically; but tan P cot P = .. tan p cot P < 1; hence, sin b < 1; b 90°, or b> 90°; hence, b may be either of the supplementary arcs PH or PH' which have the common sine equal to tan p cot P. If b 90°, since p > 90°, h> 90°; if b> 90°, < since p 90°, h < 90°. = P. tan p cot P, as P B' h P H p h Hence, if P > 90° and p > P, either triangle, PHB or PH'B' will satisfy the conditions, and there will be two solutions. B = II B = p II 90°, • ... sin b tan P cot P 1 ... b cos b=0; ... cos h =cos b cos p=0; ... h=90°. .'. cot h=0; ... cos B = tan p cot h=0; ... B=90°. 90°. P 90°, P, b h Hence, if P > 90° and p 90°, B90°, the two triangles reduce to the bi-rectangular PHB, and there is but one solution. sin b 6. When P> 90° and p < P. As before, we have sin b tan p cot P. Since p and P are of the same species, and since P> 90°, p > 90°; hence, tan p, cot P are both negative, and tan p > tan P, numerically; but since tan P cot P 1, tan p cot P > 1; ... sin b > 1, which is impossible. Hence, if P 90° and p < P, there is no solution. tan p = sin B sin b cot P 7. When P P 90' and P 90° and . cos p = 0; cos b cos p= 0; .. h=90'. tan p cot P= = ∞ X 0; ... sin b is indeterminate. 0 sin B is indeterminate. 0 sin b 0 cos h p p = 90°. = Hence, if P90, then p 90°, h 90', b and B are indeterminate; the triangle is bi-rectangular, and there is an infinite number of solutions. Hence, the following results: pP, Two solutions. = > P. Two solutions. P, One solution. €90°. |