9. Given, a = 89° 21′ 37′′, b=97°18′ 39′′, c = 86° 53′ 46", to find A, B, and C. Ans. { A = 88° 57′ 20′′, B = 97° 21′ 26", C = 86° 47′ 17". 10. Given, a = 31° 26′41′′, c = 43° 22′ 13′′, and the angle A=12° 16', to find the other parts. Ambiguous; b angle B=157° 3′ 44", or 4° 58′ 30′′; C=16° 14′ 27′′, or 163° 45′ 33′′. Ans. { = 73° 7' 34", or 12° 17′ 40′′; 11. In a triangle, ABC, we have the angle A=56° 18' 40", B = 39° 10′ 38′′; AD, one of the segments of the base, is 32° 54' 16". The point D falls upon the base AB, and the angle C is obtuse. Required the sides of the triangle and the angle C. Ambiguous; C=135° 25', or 135° 57'; c=122° 29', or Ans. 12. Given, A = 80° 10′ 10′′, B = 58° 48′ 36′′, C = 91° 52' 42", to find a, b, and c. Ans. a=79° 38′22′′, b = 58° 39′16′′, c = 86° 12′50′′. SECTION V. APPLICATIONS OF SPHERICAL TRIGONOMETRY TO ASTRONOMY AND GEOGRAPHY. SPHERICAL TRIGONOMETRY APPLIED TO ASTRONOMY. SPHERICAL TRIGONOMETRY becomes a Science of incalculable importance in its connection with geography, navigation, and astronomy; for neither of these subjects can be understood without it; and to stimulate the student to a study of the science, we here attempt to give him a glimpse at some of its points of application. co.latitude. Qeq is a portion of the equator, and the dotted, curved line, mS'S, parallel to the equator, is the parallel of the sun's declination at some particular time; and in this figure the sun's declination is supposed to be north. By the revolution of the earth on its axis, the sun is apparently brought from the horizon, at S, to the meridian, at m; and from thence it is carried down on the same curve, on the other side of the meridian; and this apparent motion of the sun (or of any other celestial body,) makes angles at the pole P, which are in direct proportion to their times of description. The apparent straight line, Ze, is what is denominated, in astronomy, the prime vertical; that is, the east and west line through the zenith, passing through the east and west points in the horizon. When the latitude of the place is north, and the declination is also north, as is represented in this figure, the sun rises and sets on the horizon to the north of the east and west points, and the distance is measured by the arc, cS, on the horizon. This arc can be found by means of the right-angled spherical triangle cqS, right-angled at q. Sq is the sun's declination, and the angle Scq is equal to the co.latitude of the place; for the angle Pch is the latitude, and the angle Seq is its complement. The side cq, a portion of the equator, measures the angle cPq, the time of the sun's rising or setting beforo or after six o'clock, apparent time. Thus we perceive that this little triangle, cSq, is a very important one. When the sun is exactly east or west, it can be deter mined by the triangle ZPS'; the side PZ is known, being the co.latitude; the angle PZS' is a right angle, and the side PS' is the sun's polar distance. Here, then, are the hypotenuse and side of a right-angled spherical triangle given, from which the other parts can be computed. The angle ZPS' is the time from noon, and the side ZS' is the sun's zenith distance at that time. The following problems are given, to illustrate the important applications that can be made of the right angled triangle cqS. PRACTICAL PROBLEMS. 1. At what time will the sun rise and set in Lat. 48° N., when its declination is 21° N.? In this problem, we must make qS=21°, Ph=48°=the angle Pch. Then the angle Scq = 42°. It is required to find the arc cq, and convert it into time at the rate of four minutes to a degree. This will give the apparent time after six o'clock that the sun sets, and the apparent time before six o'clock that the sun rises, (no allowance being made for refraction). Making cq the middle part, we have Sun rises A. M., 4* 19m 4", apparent time. From this we derive the following rule for finding the apparent time of sunrise and sunset, assuming that the declination under. goes no change in the interval between these instants, which we may do without much error. RULE. To the logarithmic tangent of the sun's declination, add the logarithmic tangent of the latitude of the observer; and, after rejecting ten from the result, find from the tables the arc of which this is the logarithmic sine, and convert it into time at the rate of 4 minutes to a degree. This time, added to 6 o'clock, will give the time of sunset, and, subtracted from 6 o'clock, will give the time of sunrise, when the latitude însă declination are both north or both south; but when one is north, and the other south, the addition gives the time of sunrise, and the subtraction the time of sunset. 2. At what time will the sun set when its declination is 23° 12′ N., and the latitude of the place is 42° 40′ N.? Ans. 7h 33m 4, apparent time. 3. What will be the time of sunset for places whose latitude is 42° 40′ N., when the sun's declination is 15° 21' south? Ans. 5h 1m 23s, apparent time. 4. What will be the time of sunrise and sunset for places whose latitude is 52° 30' N., when the sun's declination is 18° 42' south? Ans. { Rises 7h 44m 423, } apparent time. 5. What will be the time of sunset and of sunrise at St. Petersburgh, in lat. 59° 56′, north, when the sun's declination is 23° 24', north? What will be its amplitude at these instants? Also, at what hours will it be due east and west, and what will be its altitude at such times? Sun sets at 9o 13m 30′ P.M. } } Sun rises at 2* 46m 30 Α.Μ. Ans. Sun sets N. of west apparent 52° 26′ 18" Sun is east at 6h 58m 28 Α.Μ. Sun is west at 5* 1m 58o P.M. Alt. when east and west is 27° 18' 57". ON THE APPLICATION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. One of the most important problems in navigation and astronomy, is the determination of the formula fo |