CASE IV. 184. Given two angles and the included side. Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the angles A and B, and the included side c; to solve the triangle. To find a and b. By means of Napier's analogies (188) and (189), we have tan (a + b) cos = cos (AB) tanc, log tan (a+b)= log cos (A-B)-log cos (A+B) +log tanc, +log tanc, log tan (a-blog sin (A-B)-log sin (A+B) which determine (244) (245) (a + b) and (a - b), and thence a and b. To find C. We use equation (239), as in the first two cases; but (147) or (149) may be employed, as in the last case. This case is analogous to Case III., and gives rise to no ambiguity. EXAMPLES. 1. Given, in an oblique-angled spherical triangle A B C, the angles A and B equal to 119° 15′ and 70° 39', and the side c equal to 52° 39′ 4′′; to solve the triangle. Ans. Sides a and b, 112° 22′ 58′′ and 89° 16′ 53′′; angle C, 48° 36'. 2. In an oblique-angled spherical triangle, given two angles equal to 130° 3′ 11′′ and 31° 34′ 26′′, and the included side equal to 38° 30'; to find the other parts. CASE V. 185. Given the three sides. Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the sides a, b, and c; to solve the triangle. To find A, B, and C, we have, by (164), (165), and (166), or, by logarithms, log sin (s—b)+log sin (s—c)—log sin blog sin c 2 log sin (s-c)+log sin (s—a)—log sin c-log sin a (246) 2 (247) log sin C log sin (s-a)+log sin (s—b)—log sin a-log sin b 2 (248) A, B, and C can also be determined by formulæ (167), (168), and (169) for the cosine of half an angle, and by formulæ (170), (171), and (172) for the tangent of half an angle. Since the half-angles must be less than 90°, there is no ambiguity in determining the angles by any of these formulæ. EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the side a equal to 70°, the side b equal to 38°, and the side c equal to 40°; to find the angles. a, 2) 19.919372 ar.co.log sin 0.027014 ar.co.log sin 0.027014 2) 18.810094 2) 18.850476 4, log sin 9.959686B, log sin 9.405047 C, log sin 9.425238 = | B= A 65° 41' 33".7 B 14° 43′ 18" C-15° 26'21".7 21′′.7 Ans. A, 131° 23′ 7′′; B, 29° 26' 36"; C, 30° 52′ 43′′. 2. Given, in an oblique-angled spherical triangle, the sides equal to 112° 22′ 59′′, 89° 16′ 53′′, and 52° 39′ 4′′; to solve the triangle. CASE VI. 186. Given the three angles. Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the angles A, B, and C ; to solve the triangle. To find a, b, and c, we have, by (175), log cos S+log cos (S-A)-log sin B-log sin C log sina = , (249) log cos S+log cos (S—B)—log sin C—log sin A log sin b= log cos S+log cos (S—C')—log sin A-log sin B log sinc= (251) 2 a, b, and c can also be determined by formulæ (176) for the cosine of half an angle, and by formulæ (177) for the tangent of half an angle. Here a, b, and c are determined without ambiguity. EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the angle A equal to 120° 43' 37", the angle B equal to 109° 55' 42", and the angle C equal to 116° 38' 33"; to find the sides. Ans. a, 115° 13' 26"; b, 98° 21' 39'; c, 109° 50′ 20′′. 2. Given the angles in an oblique-angled spherical triangle equal to 131° 23′ 7′′, 29° 26' 36", and 30° 52′ 43"; to solve the triangle. BOOK VI. APPLICATIONS OF SPHERICAL TRIGONOMETRY TO ASTRONOMY AND GEOGRAPHY. 187. The CELESTIAL SPHERE is the spherical concave surrounding the earth, in which all the heavenly bodies appear to be situated. 188. The ZENITH is that pole of the horizon which is directly overhead. 189. The ALTITUDE of a heavenly body is its distance above the horizon, measured on the arc of a great circle passing through that body and the zenith. 190. The DECLINATION of a heavenly body is its distance north or south of the celestial equator, measured on a meridian. 191. The altitude of the celestial pole is equal to the latitude of the place where the observer is located. For the distance from the zenith to the celestial equator is the latitude of the place, and the distance from the zenith to the pole is its complement; but the distance from the zenith to the pole is also the complement of the altitude of the pole; hence the latitude of the place and the altitude of the pole are equal. 192. To find the time of the RIS ING AND SETTING OF THE SUN at any place, the sun's declination and the latitude of the place being given. Let Prepresent the celestial north pole, EAQ the celestial equator, HAO the rational horizon, S the place of the sun's rising, S' the posi A P S P S tion of the sun at 6 o'clock, PE P' the meridian of the given place, PBP the meridian passing through S, and PAP the meridian 90° distant from PE P', passing through S'. From the time of the sun's rising to 6 o'clock, it will pass over SS, the arc of a small circle, corresponding to B A, the arc of a great circle. The length of BA, expressed in time (Art. 147), will then give the amount to be taken from or added to 6 o'clock, to give the time of the sun's rising or setting. BS is the sun's declination, P O is the latitude of the place (Art. 191), and QO, which measures the angle BAS, is its complement; hence, in the right-angled spherical triangle A B S, there are known the side BS and the angle BA S, from which, by Art. 175, or, log sin BA sin BA tan BS cot BAS, log tan sun's decl. + log tan lat. of place. After reducing the arc BA to time, at the rate of 15° to an hour, or 4m. to a degree, it must be added to 6 o'clock for the time of the sun's setting, and subtracted for its rising, when the declination and latitude are both north or both south; but subtracted for its setting and added for its rising, when one is north and the other south. The preceding reasoning rests upon the assumption that the sun's declination does not change between sunrise and sunset, which, although not strictly true, is accurate enough for our present purpose. The time obtained is apparent time, and a correction must be applied if we wish to find mean time, or that indicated by the clock. Another correction is necessary for refraction. Neither of these corrections has, however, been applied to the answers that follow. EXAMPLES. 1. Required the time of the sun's rising and setting in Edinburgh, latitude 55° 57' N., when the sun's declination is 23° 28' N. Ans. Rises, 3h. 20m. 7s.; sets, 8h. 39m. 53s. 2. What is the time of the sun's rising and setting in latitude 60° 3′ N., when the sun's declination is 23° 28′ S.? Ans. Rises, 9h. 15m. 33s.; sets, 2h. 44m. 27s. |