Ans. { a = 23° 57′ 13′′, angle B=91° 26' 44, and C⇒ 102° 5' 52". 2. Given, A = 81° 38′ 17′′, B = 70° 9′ 38′′, and C64° 46' 32", to find the sides a, b, c. 3. Given, the three sides, a = 93° 27′ 34′′, b = 100° 4′ 26", and c = 96° 14' 50", to find the angles A, B, and C. A 94° 39' 4", B = 100° 32′ 19", and C-96° 58' 35". Ans. SA { = = 4. Given, two sides, b= 84° 16', c 81° 12′, and the angle C80° 28', to find the other parts. Ans. The result is ambiguous, for we may consider the angle B as acute or obtuse. If the angle B is acute, then A-97° 13' 45', B=83° 11' 24′′, and a = 96° 13′ 33′′. If B is obtuse, then Α 21° 16' 43", B · 96° 48′ 36′′, and a = 21° 19' 29". = 5. Given, one side, c=64° 26', and the angles adjacent, A=49°, and B= 52°, to find the other parts. Ans. b = 45° 56' 46", a 43° 29′ 49", and C-98° 28' 4". {%= 6. Given, the three sides, a = 90°, b= 90°, c = 90°, to find the angles A, B, and C. Ans. A 90°, B = 90°, and C = 90°. = 7. Given, the two sides, a = 77° 25' 11", c = 128° 13' 47", and the angle = 131° 11' 12", to find the other parts. Ans. b. 84° 29' 20", A=69° 13' 59" and B = 72° 28' 42". 8. Given, the three sides, a = 68° 34' 13", b = 59° 21' 18", and c = 112° 16' 32", to find the angles A, B, and C. Ans.{ A = 45° 26' 38", 55". = B 41° 11' 30', C = 134° 53′ 9. Given, a = 89° 21′ 37′′, b=97° 18′ 39′′, c= 86° 53′ 46", to find A, B, and C. A Ans. { ▲ =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. Ans. Ambiguous; b = 73° 7′ 34′′, or 12° 17' 40"; angle B=157° 3′ 44′′, or 4° 58′ 30′′; C=16° 14′ 27′′, or 163° 45′ 33′′. 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 c=122° 29', or Ans. 12. Given, A = 80° 10' 10", B58° 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. Let the lines in the annexed figure represent circles in the heavens above and around us. Let Z be the zenith, or the point just overhead, Hch the horizon, PZH the meridian in the heavens, and P the pole of the celestial equator; Ph is the latitude of the observer, and PZ is the H m Ꮓ h co.latitude. Qcq 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, Zc, 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 before or after six o'clock, apparent time. this little triangle, cSq, is a very important one. Thus we perceive that 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 R sin.cq tan.21° tan.48° tan.21° = 9.584177 Sun rises A. M., 4h 19m 4', apparent time. From this we derive the following rule for finding the apparent time of sunrise and sunset, assuming that the declination undergoes 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, |