Again, let x be the common distance of the three places from the required place, and L" and "" its longitude and co-latitude ; then from the A (, 1", x), (a', a'", x), (a", 1", x), we get -- cos. 2. cos. 2." cos. (L - L"') = . (1) sin. a. sin. a" COS. cos. (L'- L") = COS. 2 cos. a'. cos. a'" sin, a'. sin. 2" (2) (3) . COS. X COS. COS. cos. (L" — L") = sin. a" sin. 2'" :: sin. a, sin. 2"". cos. (L - L"') + cos. a cos. a' sin. a'. sin. a" cos. (L' – L") + cos. a' cos. 2" And sin. 2. sin. a". cos. (L – L") + cos. a. cos. 9" = sin. a" sin. 2" cos. (L" - L") cos. a" cos. 2" (5) which two equations will give, when reduced, " and L"; and we shall then have more than resolved the problem. By taking the A (d, d' d") we may easily show that 1 d = cot. x tan. sin. d cos. D" 8 being the inclination of x to d. Hence, by reduction, sin..D" cot.?r= tan.d 2ď d ď + tan. 2 tan. tan. 2 2 2 which gives a. Hence 0 is known. cos. a. cos. ď sin. a. sin. ď Hence we have the 2 between x and a, and from the A (2.!" x, ~) we get cos. a" COS. A COS. C cos. (1, 2) = sin, a sin. which gives a", cos. a" and a " being substituted in the equat. (1), gives L- L' and :. L". This latter method of finding a'" and L" is evidently more commodious in practice than the former. 898. Since the star is given, its co-declination (D), and right-ascension (A) are known by the tables. Let H be the hour angle from noon at sunrise, A' the sun's rightascension, and D' his co-declination, and L the co-latitude of the place. Then cos. H = cot. D cot. L and from the right-angled A (90° – D, A'), we have cot. D = tan. 23° 28' x sin. A' .. cos. H= sin. A'. cot. L. tan. 23° 28' But since the star is due south, A' A = H, sin. (A + H) tan. L. cot. 23° 28' cos. H and reducing, we finally get tan. L. cot. 23° 28' tan. H= - {tan. A + cos. A which gives H the hour of the day. Hence A' = A + H the sun's right-ascension is known, which gives the day of the year. 899. Let L be the co-latitude, z the zenith distance, and D the co-declination of the sun at six ; then from the right-angled A (2, L, D) we have .. cos. D COS. Z cos. L which gives D, and therefore by the tables, the day of the year. 900. Let D be the sun's co-declination, 21 the hour 2 from 'noon at rising, and L the co-latitude ; then from the A (right-angled by the question) cot. D = cos. H. cot. L and from the A (90°, L, D) we get cos. 2H = cot. L. cot. D = 2 cot. D. tan.L-1 cot. D + 3 cot. D .:. tan. L = 2 = 2 cot. D or cot. D which gives the latitude L. 901. The perpendiculars will evidently meet at the centre of the earth, being I to its surface. Hence the between them is measured by the arc of the meridian, which is intercepted by the two horizons, or it is the difference of the latitudes of the two places. 902. Let o be the angle which any great circle, passing through the star, makes with the horizon, the altitude of the star being a, and the distance between the intersections of the vertical and great circle with the horizon being x; then from the rightangled A (2, a) we have tan. a 1 sin. x tan. 0 = tan. a, and 0 = a. Q. E. D. 903. For the sun's place, when the aberration in declination = 0, (see Woodhouse, p. 269.) If the unresolved aberration ( 0') were constant, the aberration in right-ascension would plainly be greatest when that in declination = 0. But o óoc sin. o T, and consequently varies, :. &c. 904. Let z, z be the zenith distances of the sun when the coincidences of the extremities of the shadows with the bases of the rods take place; also D the co-declination of the sun, and L the co-latitude of the place. In the first place, since the shadows, although in opposite directions, are in the same straight line, between the first and second shadow, the sun must have passed through 180 degrees of azimuth. In the next place, we have, by the question, 6 = 20 cot. 2, 3 = 20 cot z' 3 2 5 which gives z and z'. Again, let a, 180 – a be the azimuth angles on each side of the meridian, then from the A (2, L, D), (z' L, D,) we get cos. L. cos. z - cos. L. cos. COS. a = sin. L. sin. z sin. L. sin, a which gives sin. z + sin. z' cos. L = cos. D X sin. z . sin. z' + cos. 2. sin. z' cos. D. cos. D 905. Let L be the co-latitude, D the co-declination of the sun, a the azimuth, and in the A (L, D.) let the opposite L, or that in wbich the sun is situated be called 0 ; then sin. D sin. a = sin. A X OC sin, o sin. L .: a = maximum, when 8 = 90°. Or when the vertical circle passing through the sun touch his diurnal circle. But this contact, when the zenith is between the tropics, can evidently take place on one side of the meridian as well as on the other. Consequently, to persons situated within the tropics, there will be two maximum azimuths daily. 906. Let h be the difference between the times of setting, or the distance between the two declination circles which pass through the stars. Also let D, D be their co-declinations, and L the latitude of the place ; then from the right-angled A (L, D), (L, D), we have cos. 0 = tan. L. cot. D cos. (0 + h) = tan. L. being the 2 included by the altitude of the pole and D. tan. D cos. 0 +h = cos. h sin. h . tan. O tan. D cos. A cot. D' and tan. 6 = cot. h tan. D sin. h tan. D which gives 8; and .. tan. L = cos. 6. tan. D also gives L. 907. The moon's greatest declination is evidently = the inclination of her orbit to the equator. Hence, if I be the inclination of her orbit to the ecliptic, the greatest declination is 23° 28' FI= D. Again, let a be her latitude at this epoch, and L the given longitude of her node; then her longitude is L + sin.-? (tan. a cot. I) = 1 (1) Now let M (Fig. 117) be the place of the moon, R the intersection of the equator Rm, with the ecliptic Rm', &c., as in the figure. Then from the A MRm, MRm', we get these three equations, (A being the right ascension of the moon), tan. D = tan. (8 + I'). sin. A (3) sin. A + (5) by supposition, m and n being known quantities. . . |