the sun in rising from the horizon on the 25th of May, 1816, in latitude 50° 12′ N. It appears from the Nautical Almanac, that the sun's declination on the given day is 20° 59′ N., while its apparent semidiameter is 15′ 48′′. Hence, diam. ÷ 15 = 2·1078. Log cos (L+D) . .71° 11′. cos (LD).. 29° 13′. .... 9.5085850 9.9409048 So that the time required is 3·9727 minutes, or 3m 583. Note. By a similar theorem we may find the time which the sun's rising or setting is affected by horizontal refraction. Example VI. To determine the annual variation of the declination and right ascension of a fixed star, on account of the precession of the equinoxes. Here, if A (in the spherical triangle ABC) be the pole of the ecliptic, в the pole of the equinoctial, and c the place of the star; the sides AB or C, and Ac or b, must be regarded as invariable. The differential equations applicable to this question are the 1st and 2d of class 1, oblique spherical triangles; from which the following are at once deduced. 1. Var. dec. preces. equinox x sin obliq. eclip. x sin right ascen. from solstitial colure. 2. Var. right ascen. = var, dec. x cot ang. of posit. cos dec. Example VII. To determine the variation in right ascension and declination, occasioned by any variation in the obliquity of the ecliptic. In this example, the hypothenuse and the opposite angle of a right angled spherical triangle are assumed as constant. The 2d and 6th equations, class 3, right angled spherical triangles, give 1. var. dec. = var. obliq. x sin right ascen. 2. var. var. obliq. tan obliq. * For more on this curious subject the reader may consult Cagnoli's Trigonometry, chap. xix, and xxi., and Lalande's Astronomy, vol. iii. pp. 588-604. Conseq. arc 301′′·2067 = 5′ 1′′·2067. 2. By Hutton's tables. 3. By Borda's tables. Log tan 5′ 2′′ . 7·1655821|Log tan 9′ 30′′..7.1646031 Log tan 5' 1" ..7.1641417 Log tan 9′ 20′′..7·1599080 Difference.... 14404 Difference.... 46951 Given log tan..7.1644398 Given log tan..7.1644398 Log tan 5'1"..71641417 Log tan 9′ 20′′.. 7·1599080 = 9' 29"-652 = sexiges. 5' 1"-2032. Hence it appears that in this part of the tables Hutton's has the advantage of Borda's in point of accuracy. Borda, however, gives a rule to approximate more nearly to the truth; while in other parts of his tables the decimal division supplies great facilities in the use of proportional parts. PROBLEM II. It is required to demonstrate that if a, b, and c, represent the three sides of a plane triangle, then will a2 + b2 = €2, indicate a right angled triangle, a2 + ab + b2=c2, one whose angle 'c is 120°, ab + b2 - c2, one whose angle c is 60°. a2 It appears from equa. 11, chap. iv. that a2 + b2 - c2 COS C= 2ab Substituting, then, for c2 in this equation, its value in each of the three former, there will result, respec = cos 120° cos c = + = cos 60°. Corollary. In like manner it may be shown, that and the two va 1 a2±—ab + b2 = c2, cos c = ± ; n 2n lues of c supplements to each other. PROBLEM III. Required a commodious logarithmic method of finding the hypothenuse of a right angled plane triangle, when the base and perpendicular are given in large numbers. B denoting the base, P the perpendicular, and и the hypothenuse; Find N so that 2 log P log B = log N and make в + NM. Then,(log M + log в) = log H. For, from the nature of logarithms, 1⁄2 (log M + log E) = § log MB = } log (2a B х в B) = log √ (P2 + B2) = log н; as it ought to be. Scholium. The following formulæ, the first three for right angled triangles, will often be of use, and may be easily demonstrated by the student. Let a and b be any two quantities, of which a is the b greater. Find x, z, &c. so, that tan x = √ sin z = sec y log√(a2b2) = log a + log sin y= log b + log tan y log√(a2 = b2) = [log (a + b) + log (a - b)]. log (a2 + b2)=loga + log seculogb+ log cosecu. log √ (a + b) loga + log sec x = log a log √ (a - b) = log a + log cos z = log 2 + log cos ty. log a + log 2 + log sin by. log (a+b)"= 土)]. m == [loga + log cost + log tan (45° n PROBLEM IV. To investigate a method of resolving quadratic equations, by means of trigonometrical tables. 1. Let x px = q2 be the equation proposed to be resolved. Suppose AB, and BC, the perpendicular and base of a right angled triangle ABC, to be respectively equal to 9 and p: then D в с CA = CD = CE = √(AB2 + BC2) = √ (q2 + £p2) = √(x2±px + ‡p2) = x ± {p. E |