sin B tan B tang) da. ; la B Let be the latitude of the place; h, h' the hour angles, and, the declinations of the sun corresponding to the common zenith distance b. Then cot c=tan 4, cot a=tan d, -dad, and therefore dв and de cotemporary increments of h-hand -; hence the last formula becomes tan tan or if - be seconds of arc, h=h′ (in seconds of time) = 2 tan 7 ____ tan § 818 tan h). 8—¿ sin h tan h 30 and subtracting this quantity from the middle time between the instants shown by a chronometer, at which, on a given day, the sun has equal altitudes, gives the time by the chronometer when the sun is on the meridian.* PROPOSITION XIII. (635) The error in taking the altitude of a star being given, to find the corresponding error in the hour angle. As in the preceding proposition, let в represent the pole of the equinoctial, a the zenith, and c the observed place of the Then e will be the co-latitude, a the star's co-declination, and ¿ its co-altitude; the sides e and a will be constant quantities. It is shown (626) that dв: db:: rad: sin A. sin c, hence Now when the errors are very small, dв and db are the corresponding errors of the hour angle and zenith distance; therefore, radius being unity, the error in time= A being the star's azimuth. error in altitude COR. Hence, the error in time, from a given error in altitude, is less the greater sin a is, or the nearer the body is to the prime vertical, at the time of observation. PROPOSITION XIV. (636) The error in the altitude of any tower, or other object, is to the error committed in taking the angle of elevation, as double the height of the observed object is to the sine of double the angle of elevation. This proposition is designed to illustrate the scholium (629). From the first set of equations (616) we have da= sin c sin b da, that is, da da :: sin b: sin c, and considering the triangle as plane (629), da da But, sin 2A=2 sin a cos A, rad=1 (228), hence 1 = 2 Sin A da 2b sin a COS A sin 2A da sin 2A Again, sin a : a :: rad=1: b (110), consequently b sin a=a, and by substitution da 2a 2a: sin 2A which was to be shown. = C that is da da :: : COR. Because da= in altitude will be a minimum, when sin 2a is a maximum, that is sine of 90°. Hence if da the error in the observed angle be l', the sine or arc of which is 0002909, and the observed angle be 45°, we have 2a da da=- =2a x 00029090005818 a = the note page 77. a 1718.8. See CHAP. XIV. MISCELLANEOUS PROPOSITIONS, ETC. Of the French Division of the Circle. (637) The modern French writers on Trigonometry divide the circumference of the circle into 400* equal parts or degrees, each degree into 100 equal parts or minutes, each minute into 100 equal parts or seconds, &c. which degrees, minutes, &c. they write in the usual manner, thus 126° 80′ 64′′, &c. A French degree is therefore less than an English degree, in the ratio of 90 to 100, or of 9 to 10; a French minute is less than an English minute in the ratio of 90 × 60 to 100 × 100, or of 27 to 50; and a French second is less than an English second in the ratio of 90 x 60 x 60 to 100 x 100 x 100 or of 81 to 250. Hence, if n any number of degrees, to turn English degrees into French, we have 9: 10::n: =n+ and to turn 10n 9 n 9 French degrees into English, 10:9:: n : PROPOSITION I. (638) To turn French degrees, minutes, &c. into English. RULE. Consider the degrees as a whole number, after which place the minutes and seconds + as decimals; of this mixed decimal deducted from itself will give the English degrees corresponding to the French. EXAMPLE I. The latitude of Paris is 54° 26' 36" in the Eléments de Géométrie, par A. M. Legendre, 6th ed. page 328. to Borda's Trigonometrical Tables (Paris, An. IX.), page 18, et seq. Preface The minutes and seconds, if under 10, must have a cipher prefixed; thus 27° 7′ 35′′ must be written 27° 07′ 35′′ or 27°0735; 45° 18′ 4′′ 45° 18' 04" or 15° 1804, &c. = French division of the circle, what is the corresponding latitude in the English division? 54° 26′ 36′′ 1 10 54°.2636 French degrees. = 5·42636 48 83724 English degrees. 50 23440 60 14.06400 Answer. 48° 50′ 14′′ English. 2. What number of degrees, &c. in the English division of the circle will correspond to 74° 4' 8" in the French division? 74° 4' 8"-74° 04′ 08′′=74°.0408 10 = 7·40408 66 .63672 60 38 20320 60 12.19200 Answer. 66° 38′ 12′′ English. PROPOSITION II. (639) To turn English degrees, minutes, &c. into French. RULE. Reduce the minutes and seconds, &c. to the decimal of a degree, and annex it to the given number of degrees; this mixed decimal increased by of itself will give the French degrees corresponding to the English. EXAMPLE. The latitude of Greenwich Observatory is 51° 28′ 39′′ N., according to the English division of the circle, what is the corresponding latitude by the French division? 51° 28′ 39′′= 51°4775 ᄒ = 5 .7197 57 1972 57° 19′ 72′′. Answer. 57° 19' 72" North, by the French division of the circle. On the 12th March, 1840, the moon's distance from the sun, at midnight, was 116° 20′ 38′′ by the Nautical Almanac, what was the distance according to the French division of the circle? 116° 20′ 38′′= 116° 343889 10 = 12 927099 129 270988=129° 27′ 10′′. Answer. 129° 27′ 10′′. (640) A table of arcs differing by 10 degrees according to the French division of the circle, or by 9 degrees by the English division, with the corresponding natural and loga rithmic sines. The latitude of the Observatories of Paris and Pekin are 54° 26′ 36′′ N., and 44° 33′ 73′′ N., and their difference of longitude 126° 80′ 56′′, according to the French division of the circle, what is their distance?* SOLUTION BY THE FRENCH DIVISION OF THE CIRCLE.† Here we have two sides a and b of a spherical triangle given, and the included angle c, to find the third side c. By the formulæ (434) tan = COS C. tan b and COS C = cos b cos (a-), where a=100°-54° 26′ 36′′-45° 73′ 64′′, b=100°-44° 33′ 73′′=55° 66′ 27′′, and c=126° 80′ 56′′. * Legendre's Geometry, page 403. + By Borda's Tables, Paris, An. IX. |