which are the same as (442) and (470). part, we should have In like manner, if co. B were made the middle cos. B = cotan. h tang. a, cos. B = sin. A cos. b; which are the same as (443) and (471). Thirdly. If a were made the middle part; then, by (473), co. B and b would be the adjacent parts, and co. A and co. h the opposite parts; and, by (474) and (475), we should have (483) (484) (485) which are the same as (455) and (448). In like manner, if b were made the middle part, we should have sin. b = cotan. A tang. a, sin. b = sin. B sin. h; which are the same as (454) and (449). Having, thus, made each part successively the middle part, the ten equations, which we have obtained, must be all the equations included in (474) and (475); and we perceive that they are identical with the ten equations of (472). 7. Corollary. When his less than 90°, the first member of (436), cos. h cos, a cos b, (490) (491) is positive; and therefore the factors of its second (493) member must either be both positive or both negative; that is, the two legs a and b must, by the following Lemma (496), be both greater or both less than 90°. But when his greater than 90°, the first member of (492) is by (496) negative; and therefore one of (494) the factors of the second member must be positive, while the other is negative; that is, of the two legs a and b, one must be less while the other is greater than 90°. These results may be simply expressed as follows: The three sides of a spherical right triangle are (495) either all less than 90°; or else, one is less while the other two are greater than 90°; unless one of them is equal to 90° as in (522). 8. Lemmas The sine and cosecant of an angle, (496) which is greater than 90° and less than 180°, are positive; but its cosine, tangent, cotangent, and secant are negative. 'Demonstrations Let the given angle be 90° + N, N being less than 90°. Then 90° + N and 90° (497) N are supplements of each other, since their sum is equal to 180°, and we have from (195) and (5), = sin. N (498) sin. (90° + N)= sin. (90° - N) = cos. N (499) cos. (90° + N) cos. (90° - N) =(500) tang. (90° + N)=-tang. (90° - N) = -cotan. N (501) cotan. (90° + N)=-cotan. (90° -N) = - tan. N (502) sec. (90° + N)=- sec. (90° -N)=-cosec. N (503) cosec. (90° + N)= cosec. (90°-N)=sec. N all which equations agree with (496). 9. Corollary. The equation (436) cos. h=cos. a cos. b, (504) leads also to the result, that the hypothenuse differs (505) Demonstration. The factors cos. a and cos. b of the second member of the above equation are, by (5), fractions whose numerators are less than their denominators. Their product, neglecting the signs, must then be less than either of them, as cos. a for instance, or cos hcos a; and therefore h must differ less from 90° than a does, as is evident from the following Lemma. (506) 10. Lemmas Of angles less than 180°, the one which differs the least from 90° has the largest sine, tangent, and secant; and the smallest cosine, cotan- (507) gent, and cosecant; no regard being had to the signs. Demonstrations Let the quantity by which an angle differs from 90° be N; and the angle is either 90° (508) + N or 90° - N. But, by (498), (509) sin. (90° + N) = sin. (90° - N). Now the smaller N is, the larger must (90° — N) be, and by (30') the larger the sine of (90° --- N) ; (510) that is, the less the angle differs from 90° the larger is its sine. Again, by (499), cos. (90° + N)=-cos. (90° - N) = - sin. N. Now the smaller Nis, the smaller, by (30'), must its (511) A (512) sine be, since it is less than 90°; and therefore the smaller, neglecting the signs, must the cosine of the given angle be. In the same way, by means of (48'), the proposition (507) might be proved, with regard to the tangents, cotangents, secants, and cosecants. Indeed, it readily appears from the equations (13), (7), and (10) (513) that the sine, tangent, and secant of an angle increase, while the cosine, cosecant, and cotangent diminish. (514) 11. Corollary. When A is less than 90°, the first member of (470) cos. Acos. a sin. B is positive, and therefore the factor cos. a of the sec(515) ond member, being multiplied by the positive factor sin. B (496), must be positive; that is, a must be less than 90°. But, if A is greater than 90°, the first member of (514) is, by (496), negative, and therefore (516) the factor cos. a of the second member must be negative; that is, a must, by (496), be greater than 90°. We may express this result as follows: An angle and its opposite leg in a spherical right (517) triangle must be both less or both greater than 90°, or by (522) both equal to 90°. (515) 12. Corollary. The equation (470) leads also to the result that an angle differs less from Demonstration. Since the second member of (518) is the product of the two fractions cos. a and sin. B, (520) the first member must be less than either of them. Thus, neglecting the sines, cos. A <cos. a; hence, by (507), A differs less from 90° than does a. (521) 13. Corollary. When, in a spherical right triangle, either side is equal to 90°, one of the other two sides (522) is also equal to 90°; and each side is equal to its opposite angle. Demonstration. First. If either of the legs is equal to 90°, the corresponding factor of the second (523) member of (436), cos. h = cos. a cos b, (524) is, by (157), equal to zero; which gives it follows, from (157) and (524), that 0=cos. a cos. b, (528) and therefore either cos. a or cos. b must be zero; that is, either a or b must be equal to 90°. (529) Secondly. When either side is equal to 90°, it fol lows, from (523) and (526), that h=90°. (530) This result, substituted in equation (448), sin. asin. h sin. A, (531) |