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2A tan, d'

1 - A .. sin. 22 +

sin. 2z = 1+tan.

1+tan.ď which being resolved gives z, and therefore z' =z+ d'. Hence is known the 2 between (z, d); for

sin. (2, d) : sin, a :: sin. z: sin. d. But sin. L: sin. (2, d) :: sin. z: sin. 15° :: sin. L: sin. a :: sin. z. sin. a' : sin. d. sin. 15°

sin. a X sin. z. sin. z' i. sin. L =

sin. d x sin, 15° which gives

90° - L.

849. Let the sun's apparent diameter be D, his co-declination be x, and the latitude of the place be 90

L. Also let 6 be the hour angle corresponding to the time of rising; then (Fig. 110,)

(= -0') : ss' :: 1: sin. x
ss' : D(= ns) :: 1 : sin. (nsP)

D
..0 =

sin. x x sin, ns P

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D

v (sin." x – cos. L) and O is therefore a minimum when x = 90°.

850. Let L, L' be the latitudes of the places, D the co-declination of the sun, and H, H' the hour angles measured from noon at the time of setting; then by means of the spherical right-angled A, two of whose sides are

L, D; and L', D; we get, (by Napier's Rules,)

tan. L = cos. H x tan. D,

and tan. L' = cos. H' x tan. D. nt H - H' = difference of longitude =h a given quantity.

tan, L
tan. D

= cos. (H' + h) = cos. H' cos. h -- sin. H' sin, h

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1 :. tan.: D = tan.L' +

x (cos. h tan. L' tan. L)

sin. h
tan. L' + tan. L - 2 cos. h tan. L. tan. L'

sin." h which will give the co-declination D.

851. On any two days in the year, which are equally distant from the time of the equinox, it may easily be shewn that the time of daylight on the one day is equal to the duration of darkness on the other; and the same is true for all other days equally remote from that epoch. Hence the truth of the problem is manifest.

852.

Since the declination of the sun when in either tropic is 23° 27' 30', it is evident that to people living under the equinoctial, he must rise 23° 27' 30" from the east towards the north, when he is in the northern tropic.

853. Let O denote the obliquity of the ecliptic, D the sun's declination, and A bis right ascension at the required time; then from the right-angled spherical A whose legs are D and A, we easily derive

tan. D - tan. O x sin. A..... (1)
d.D

= tan. O x d.A cos. A

cos. D But, by the question, d.D= d.A

:: cos. A x tan. O = 1 + tan.' D = 1 + tan.20 x sin. O which gives cos.' A + cot. O cos. A = 1+ cot. O

cot. O N. (4+ 5 cot.' O) .. cos. A =

2 and therefore A is known.

Again, if I denote the longitude of the sun, or its distance from Aries, we have by the same A

tan. A tan. 1 =

cos. O ?

which gives the sun's place in the ecliptic.

854. Let pe ø, be the distances of the earth and saturn from the sun, and x that of the moon from the earth. Then when saturn shines with a full orb, the exterior angle of elongation being a right angle, his distance from the earth is

R= v(- 89).

Also when the moon is in the full, her distance from the sun is

r = n(p2 — xo).

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Again, the intensity of the light which bodies receive from a luminary cc

(see Wood's Optics.) (dist.)? from that luminary Hence the real intensities upon the moon and saturn are mede

sured by

and

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1

1 p2-32

P These intensities, in the progress of the light reflected to the earth are diminished in the ratio of

1

1 to

ce and the apparent brightness cc magnitude of the disk x intensity. Hence, by the question,

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2

19

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Х

2 $? () r.(p2-x2)
s and m being the radii of the disks of saturn and the moon.

ma
- 9*** = $? (2 – p)

3 so whence by the solution of a quadratic 22 and 1. x may be found.

2

.

855. Let D, D, be the given co-declinations, A, A', the right ascensions, « the common azimuth, z the given zenith distance, and z' the one not given, and L the co-latitude of the place. Then from the spherical A, whose sides are,

2, L, D; Z'LD we have

cos.L

(1) sin, z. sin. L

cos. D

COS. Z.

COS. a =

.

COS. O

cos. D'- cos. a'. cos. L

sin. =' . sin. L

.

.

(2)

Also

cos. (< — 2) - cos. D x cos. D' cos. (A' – A) =

sin. D. sin. Do :. cos. (z' – 2) = cos. D. cos. D' + sin. D. sin. D cos. (A'-A) and this gives (see Woodhouse's Trigonometry,)

D + D'

D+D + M) sin.

M)....(3) 2

2 wherein sin.' M = sin. D. sin. D'cos.

A' – A

2

sin.2

= sin.

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Hence is found z z, and therefore z wbich being substituted in equation (2) will, by means of equating (1) and (2), give the value of L in terms of D, D, and A'

A.

856. Let sn = h (Fig. 110,) the horizontal refraction; then o o measures the required acceleration.

But

h = sš' x sin. ss'n = ss.cos. Latitude,

=oo.cos, dec. cos, Latitude, whence co' is known.

a.

857. Let a be the given 2 subtended by the diameter of the sun at the mean distance a, and 0 the angular distance from perigee, then at the distance

(1 )

it e cos. A the apparent diameter is

1+ e cos. O x α =

xám le

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a

858. The distance (z) of the pole of the ecliptic from the zenith of the place = the inclination of the plane of the ecliptic to the horizon. Hence, if L denote the co-latitude of the plane, O the obliquity, then it is evident, upon inspecting the A whose sides are z, 0, and L, that z is least, when the pole of the ecliptic is in the meridian. Hence in this case there are situated in the circumference of the meridian the poles of the equator and ecliptic, which, therefore intersect in the pole of the meridian, (see Creswell's Spherics ;) that is, Aries rises in the east or west point of the compass when the ecliptic is least inclined to the horizon.

659. Let the given sum of the sun's co-altitude (3), and co-declination (D) be

m = 2 + D.

Also let H be the hour 2 from noon, and L the co-latitude of the place; then cos. H =

cos. D. cos. L sin. D. sin. L

COS. Z

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