Ex. 10. In latitude 40° north, required the duration of twilight at the several times when the sun's declination is 23° 28' south, 5° 50′ south, and 23° 28' north; the twilight being supposed to begin when the sun's centre is 49' below the horizon,* and to end when it is 18° below. Ans. 1 h. 35 m.; 1 h. 29 m.;† and 2 h. 4 m. Ex. 11. Given, two zenith distances of the sun's centre, 65° 20′ and 60° 18', taken at the same place, both being in the forenoon; the interval between the observations, measured by a good time-piece, 1 hour 32 minutes; the sun's declination 20° south, and the approximate latitude of the place 40° 15′ north; to find the time of the last observation, and the correct latitude of the place. cos BZ = cos ZPB.sin BP.sin ZP + cos BP.cos ZP. Hence (AP being = BP), COS AZ cos BZ = (cos ZPA cos ZPB) sin AP.sin ZP; * The refraction of the sun's light, when in the horizon, is 33', and apparent semi-diameter about 16'; hence his upper limb is just visible when the centre is 49' below the horizon. †This is the shortest twilight in latitude 40° N., and occurs twice in the year, viz., in spring and autumn, when the sun's declination is 5° 50' S. consequently (Art. 37, eq. 13), sin (BZ + AZ).sin (BZ — AZ) = sin 1(BPZ + APZ). of which, BPZ-APZ is given from the elapsed time. Hence, sin (BPZ + APZ) = 15° 51' 11". sin (BZ + AZ).sin (BZ — AZ) = Whence APZ = 4° 21′ 11′′, and (by Art. 59) AZP = 175° Ex. 12. Given, the approximate latitude, 39° 26' N.; the Ans. Lat. 39° 29′; watch too fast, 18 min. 57 sec. Ex. 13. At the time when Sirius and Aldebaran were in Ans. Lat. 35° 8′ 42′′ N.; right ascension of mid-heaven, *When the latitude thus found differs considerably from the approxi- Examples of a Mixed Character. Ex. 1. From the top of a cliff near a river, two buoys at anchor being observed, whose distance from each other was known to be 300 yards, their angles of depression below the plane of the observer were found to be 30° and 40° respectively; and the angle at the eye, subtended by the line joining them, was 37°. Required, the distance of each buoy from the observer, and the altitude of the cliff above the level of the water. The observed depressions, each increased by 90°, form two sides; and the angular distance, the base of a spherical triangle, with which the angle opposite the base is found=44°. This is the horizontal angle, subtended by the line joining the buoys. Drawing then a vertical line through the position of the observer to meet the plane of the water; and, from the point where it meets that plane, drawing lines to the buoys; those lines will be to each other as the cotangents of the given angles of depression; and the angle which they make with each other will be 44°, as above found. The construction is this: C Take AB=300, the given distance; on AB describe a segment of a circle ACB, containing an angle of 44°; complete the circle, and bisect the arc AEB in E; make the angles ABF and BAF 30° and 40° respectively; draw FG at right angles to AB; join EG, and produce it to meet the circle in C; join CA, CB; then, since the angle ACB is bisected by the line CG, A G E B = As AC CB AG; BG (3.6): cot BAF: cot ABF. Consequently, C is the point in the plane of the water which is cut by a vertical line passing through the place of the observer. The calculation is easily made. For (Art. 28 and Art. 37, eq. 8), As sin (BAF+ABF) : sin (BAF-ABF) :: AB : BG-AG. Hence AG is known. If we join AE, BE, the side AB, and all the angles of the triangle ABE, are given; whence AE becomes known. Then, in the triangle AGE, the sides AE, AG, and the contained angle, are known; from which the angle AEG ABC is found. In the triangle ABC, we then have the base AB and all the angles, to find AC and BC. Then, from either of these and the angle of depression, the altitude of the cliff may be found. Lastly, with the distances AC, BC, and the angles of depression, the distances from A and B to the place of the observer are determined. = Result: Altitude of cliff, 249; distances, 388 and 498. Ex. 2. Suppose an observer on a frozen lake takes the altitudes and angular distance of two cliffs on the shore, as follows: altitude of first, 50°; of second, 55° 30'; angular distance, 25° 20'. Then, advancing on the ice 500 yards, in the vertical plane which passes through the first cliff, the altitudes are 57° and 59° respectively. Required, the distance of the cliffs from each other, and their respective altitudes above the surface of the lake. Ans. Distance of cliffs, 3087 yards; altitude of first, 2636 yards; of second, 4070 yards. Ex. 3. The crew of a vessel at sea discovering a light in the horizon, which they suppose to be a vessel on fire, sail directly towards it, over 1° of a great circle, when they perceive that it is a fire on a mountain, which is then 1° 30' above the horizon. Required, the distance of the light wh first seen, and the height of the mountain above the le the sea; the earth being considered as a spnere wnose radius is 3968 miles. Answer. Distance, 138.5 miles; height, 2.4 miles. Ex. 4. Given, AB, a horizontal line, 1785 yards in length, running exactly north; D, C, two elevated peaks, eastward from AB, such that the elevation of C above the plane of the horizon, seen from A, is 16° 30', and the elevation of D 20° 40'. But, seen from B, the elevation of C is 18° 25', and the elevation of D 13° 15′. Also, the angle BAC, taken in the oblique plane passing through AB and C, is 38° 16'; the angle CAD, taken in the plane which passes through A, C and D, 87° 20'. The angles at B, also taken in the oblique planes, are ABD 40° 30′, and DBC 94° 16'. Required, the distance and bearing of DC when reduced to the horizontal plane on which AB lies. Result: DC, N. 2° 55' E.; 11,748 yards. THE END. |