French division of the circle, what is the corresponding latitude in the English division? 54°.26′.36" 54°2636 French degrees. = 542636 48 83724 English degrees. 60 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 740408 66.63672 60 38.20320 60 12.19200 Answer. 66°.38'.12" English. PROPOSITION II. (E) 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'.40" N. according to the English division of the circle, what is the corresponding latitude by the French division? 51°.28.40" 51.477777, &c. + = 5 719753, &c. 57 197530=57°.19′.75′′.3. Answer. 570.19'.75".3 North, by the French division of the circle. On the 22d March, 1813, the moon's distance from the sun, at midnight, was 114°.13'.21" by the Nautical Almanac, what was the distance according to the French division of the circle? (F) 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 logarithmical 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? * (G) 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 (S. 191.) tang = cos b cos c. tang b and cos c= cos (a-p) where a=(100°-54°.26.36")=45°.73'.64" cos o b≈(100°—44°.33′.73′′)=55°.66′.27′′, and c=126°.80′.56′′. log cos c log cos 126°.80'.56′′=log cos 200° - 126°.80′.56′′= log cos 73°.19.44". * Legendre's Geometry, page 403. + By Borda's Tables, Paris, An. IX log cos 73°.19.44" 9.6114358 log tang blog tang 55°.66'.27"=10.0776713 responding in the tables is 28°.94.23". But because the cosine of c is negative (K. 100.), it is plain, by the formula above, that tang is also negative, hence 28°.94".23" and a-9= 45°.73.64"+28°.94′.23′′=74°.67′.87′′. log cos (a-4)=log cos 74°.67.87" 9.5880938 b =log cos 55°.66'.27" 9.8071949 log cos 19.3952887 log cos log cos 28°.94'.23"= 9·9534823 log cos c = 9.4418064. The corresponding arc is 82°.16'.05" the distance between Paris and Pekin 821605 seconds, or 821.605 myriametres; a myriametre being an arc of 10 minutes, and a metre an arc of of a second.* (H) SOLUTION BY THE ENGLISH DIVISION OF THE CIRCLE.† Lat of Paris 54°.26'.36" French=48°.50′.14′′064 English. (C. 360.) Lat of Pekin=44°.33.73" French 39°.54.12" 852 English. Diff. Long=126°.80.56′′ French=114°.7.30" 144 English. Hence a = (90° 48°.50.14" 064) 41°.9'.46" 064; b = (90° 39°.54.12"852) 50°.5.47" 148. and c 114.7.30"-144. log cos c log cos 1140.7.30"-144 9.6114358 log tang log tang 26°.2'.53" 9.6891071 Because tang is negative, a—=41°.9′.46′′+26°.2′.53′′— 67°.12′.39". = =5130740 toises = * According to the French mathematicians of the whole terrestrial meridian (viz. 100°)=5130740 toises in length, the len millionth part of which is the metre; therefore the metre of a second = 3 078444 French feet; and because 107 French feet are equal to 114 English feet nearly, the French metre 3.28 English feet. If a French toise=65945 English fect, a metre will be = 3-280852 English feet. Tables Portatives, par Callet, édition Stéréotype, An. 3. The length of a degree, by the trigonometrical survey of England and Wales, is 364950 feet*; hence, the distance between Paris and Pekin is 26986025 English feet. = 73°.56.40" English 82°.16′.05" French as above (E. 361.) PROPOSITION III. (Plate IV. Fig. 7.) (I) If from the sum of the three angles of any spherical triangle two right angles be deducted, the remainder will be to two right angles, as the area of the triangle is to one-fourth of the surface of the sphere. Let ABC be the triangle; complete the circle ABED, and produce AC, BC to meet in the point F, cutting the circle ABED in the points E and D. Then because BD and CF are each of them a semicircle (C. 133.) BC=DF, and for the same reason AC=EF. The angle ACB 2. dce= 2 dfe (N. and Q. 135.) therefore the triangle ABC is equal to the triangle DEF (D. 141.); let the surface of each of these triangles be represented by m, and let w, x, and y represent the surfaces of the triangles in which they are situated. Now it is plain, that if any great circle of the sphere (as 1, 2, 3.) be divided into any number of equal parts, and through the points of division (1, 2, 3.) great circles be drawn so as to pass through the poles (B and D) of the divided great circle, the surface of the sphere will be divided into as many equal parts, or lunula, as there are parts in the divided great circlet; therefore, the whole circumference of the sphere, will be to any arc of this great circle, as the whole surface of the sphere, is to the surface of the lune comprehended between two great circles, whose greatest distance is measured by the aforesaid arc. Vol. II. part ii. page 113. + This may be farther illustrated by considering the manner in which the several meridians, on a terrestrial globe, divide the equator and the surface of the globe. surface of the sphere: m+y. surface of the sphere: m+w. surface of the sphere: m+x. By Composition 180°: ZA+B+C:: surface of the sphere: 3m+y+w+ x; and, by Division, 180°: (ZA+B+≤c)−180°:: surface of the sphere: 2m. or, 180°: (A+B+c)-180°:: surface of thesphere: m. Consequently (LA+LB+ 2c)-180°: 180°::m:surface of the sphere. Q.E.D. (K) COROLLARY. The measure of the surface of a spherical triangle is the difference between the sum of its three angles and two right angles. For if s=4 of the surface of the sphere, 180° x m=sx (A+B+c-180°). But the whole surface of the sphere is equal to four times the area of one of its great circles*, and the area of a great circle circumference x radius=180° x radius, if therefore radius=1, the surface of one fourth of the sphere may be expressed by 180°; therefore m=(A+B+c)-180°. PROPOSITION IV. (L) The sum of the three angles of every spherical triangle being greater than 180° (T. 137.) the sum of the three observed angles of a triangle, on the surface of the earth+, ought to exceed 180°; which excess may be found by the following rule. From the logarithm of the area of the triangle taken as a plane one, in feet, subtract the constant logarithm 9.3267737, and the remainder is the logarithm of the excess above 180° in seconds nearly.‡ Put of the surface of the sphere=s, then + The sides of the triangles which connect the successive stations of a trigono. metrical survey, may be considered as formed by arcs of great circles, whose radii are equal to the radius of the sphere; and each observed angle, being formed by the tangents of two of these arcs, will be the measure of the spherical angle contained between them, at each successive station (R. 135.) Trigonometrical Survey of England and Wales, vol. i. page 138. |