3. cos 671° + i sin 674°, cos 1574° + i sin 1574°, cos 2474° + i sin 2471°, cos 3374° + i sin 3374°. 4. sin 404 cos3 0 sin 0 - 4 cos 0 sin3 0. 2. I. The cosine of the middle part = the product of the cotangents of the adjacent parts. II. The cosine of the middle part the product of the sines of the opposite parts. EXERCISE XXXII. 24. A = 175° 57′ 10′′, B = 135° 42′ 50′′, C. 135° 34' 7". 25. C 104° 41′ 39′′, a = 104° 53′ 2′′, b = 133° 39′ 48′′. 26. a 90°; b and B are indeterminate. 27. a = A = 60°, b= 90°, B= 90. 28. The triangle is impossible. 29. b = 130° 41′ 42′′, c = 71° 27′ 43′′, A = 112° 57′ 2′′. 30. a = 26° 3′ 51′′, A = 35°, B = 65° 46′ 7′′. 31. Impossible. EXERCISE XXXIII. 1. cos A cot a tan b, sin B = csc a sin 1 b, cos h = cos a sec 1 b. 4. Tetrahedron, 70° 31′ 46"; octahedron, 109° 28′ 14"; icosahedron, 138° 11′ 36′′; cube, 90°; dodecahedron, 116° 33′ 44′′.. cos A = cos B csc x sin (C − x) ; cos B = cos C csc x sin (A — x). EXERCISE XLI. 4. 2066.5 square miles. EXERCISE XLII. 1. If x denotes the angle required, sin x = cos 18° sec 9°, x = 148° 42′. 2. cos x = cos A cos B. 3. Let w = the inclination of the edge c to the plane of a and b. Then it is easily shown that V= abc sin l sin w. Now, conceive a sphere constructed having for centre the vertex of the trihedral angle whose edges are a, b, c. The spherical triangle, whose vertices are the points where a, b, c meet the surface of this sphere, has for its sides l, m, n; and w is equal to the perpendicular arc from the side to the opposite vertex. Let L, M, N denote the angles of this triangle. Then, by means of [39] and [48], we find that sin w = sin m sin N = 2 sin m sin N cos N where V = 2 abc √sin s sin (s — 1) sin (s — m) sin (s — n). 4. (i.) 9,976,500 square miles; (ii.) 13,316,560 square miles. 5. Let m = longitude of point where the ship crosses the equator, B = her course at the equator, d = distance sailed. Then tan m = B= 6. Let karc of the parallel between the places, x = difference required; then sink = sin 1⁄2 d sec l. x = 90°( √2 − 1). 7. tan + (m— m') = √sec s sec (s — d) sin (s — 1) sin (s — l'); where 2 s = 1+ l'+d, and m and m' are the longitudes of the places. t 15 of sunset= o'clock P.M.; cos a = sin d sec l. For longest day at Boston: time of sunrise, 4 hrs. 26 min. 50 sec. A.M.; time of sunset, 7 hrs. 33 min. 10 sec. P.M. Azimuth of sun at these times, 57° 25′ 15′′; length of day, 15 hrs. 6 min. 20 sec.; for shortest day, times of sunrise and sunset are 7 hrs. 33 min. 10 sec. A. M. and 4 hrs. 26 min. 50 sec. P.M.; azimuth of sun, 122° 34′ 45′′; length of day, 8 hrs. 53 min. 40 sec. 12. The problem is impossible when cot d <tanl; that is, for places in the frigid zone. 13. For the northern hemisphere and positive declination, sin h = sin l sin d, cot a = cos l tan d. Example: h = 17° 14′ 35′′, a = 73° 51′ 34′′ E. 14. The farther the place from the equator, the greater the sun's altitude at 6 A.M. in summer. At the equator it is 0°. At the north pole it is equal to the sun's declination. At a given place, the sun's altitude at 6 A.M. is a maximum on the longest day of the year, and then sin h = sin l sin e (where e 23° 27′). 15. cost = cot l tan d. Times of bearing due east and due west are t t 12 o'clock A.M., and 15 o'clock P.M., respectively. 15 Example: 6 hrs. 58 min. A.M. and 5 hrs. 2 min. P.M. 16. When the days and nights are equal, d = 0°, cos t = 0, t = 90°; that is, sun is everywhere due east at 6 A.M., and due west at 6 P.M. Since I and d must both be less than 90°, cos t cannot be negative, therefore t cannot be greater than 90°. As d increases, t decreases; that is, the times in question both approach noon. If l<d, then cos t>1; therefore this case is impossible. If l=d, then cost = 1, and t=0°; that is, the times both coincide with noon. The explanation of this result is, that for d= the sun at noon is in the zenith, and south of the prime vertical at every other time. And if l>d, the diurnal circle of the sun and the prime vertical of the place meet in two points which separate further and further as 1 increases. At the pole the prime vertical is indeterminate; but near the pole, t = 90°, and the sun is always east at 6 A.M. 17. sin l = sin d csc h. 18. 11° 50′ 35′′. 19. The bearing of the wall, reckoned from the north point of the horizon, is given by the equation cotx= cos l tand; whence, for the given case, x = 75° 12′ 38′′. 20. 55° 45′ 6′′ N. 21. 63° 23′ 41′′ N. or S. 22. (i.) cost = tan l cot p; (ii.) t = z; (iii.) the result is indeterminate. 31. 3 hrs. 59 min. 27 sec. P.M. 27. t = 45°42′, l = 67° 58′ 54′′. |