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Here the triangle which is supposed to undergo a minute variation is oblique, one of the sides being PZ, the co-latitude, or distance from the pole to the zenith, another being Ps, the co-declination of the luminary at the time proposed, and the other zs 90° the distance from the zenith to the horizon. It is required to ascertain the variation in the hour angle P which corresponds to any assigned variation in the opposite side zs.

The differential equation which applies to this inquiry is, obviously, the first of class 1, oblique angled spherical triangles, which, when accommodated to the present notation, becomes

zs

бр

sin s sin PS; or, de=

dzs
sin s sin PS

Now, from the 2d fundamental theorem of spherical triangles, (chap. vi. p. 84), we have

COS PZ = cos zs cos PS + sin zs sin PS cos s. But, in the example before us, cos zs =

0, and

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It may be readily shown, by multiplying together the corresponding values of cos (A + B) and cos (AB), (equa. c, p. 42), and substituting 1-cos2 for sin2, that COS2 PZ = COS (PZ + PS) COS (PZ — ps).

sin2 PS

-

Hence it follows that

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time sun ris. =

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✔[cos (lat. + dec.) cos (lat. — dec.)]

Let it be proposed to ascertain the time occupied by

7

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'.... 9.5085850
COS (LD).. 29° 13′.... 9·9409048

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So that the time required is 3.9727 minutes, or 3m 58. Note. By a similar theorem we may find the time which the sun's rising or setting is affected by horizontal refraction.

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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. > sin right ascen. from solstitial colure.

2. Var. right ascen. —

var. dec. × 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.
var. obliq. x cot obliq. x tan dec.

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* 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.

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Rem. - 10, is log 301·2067.. 2.4788646

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"..7.1641417 Log tan 9' 20"..7.1599080

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=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 = c2, 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

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Substituting, then, for c2 in this equation, its value in each of the three former, there will result, respec tively,

In the 1st case, cos c = 0 = cos 90°

In the 2d

In the 3d

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= cos 60°.

Corollary. In like manner it may be shown, that

when

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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 н the hypothenuse;

log N

Find N so that 2 log P- log B = and make B+ NM.

Then,(log м + log в)= log н.

H.

For, from the nature of logarithms,=N;

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B

M; whence

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