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878. The apparent diameter of the sun o inversely as the radius vector of the orbit. Hence since

a. (1 – e)

-e cos. A 0 being measured from

apogee,

the apparent diameter oc1-ecos. 0. Hence the augmentation of the apparent diameter is 1 e cos. A

1

1 e cos. 0 - Ite
al - e) ali+e) a. (1 ~.c)
e. (1 cos. 0)

O vers. O
a (1
whatever be the magnitude of the eccentricity.

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879.

The visible illumined part of the disk ; whole disk :: vers. exterior z of elongation : 2.

See Woodhouse, p. 552, or Vince. But in the case of the moon, the distance of the sun from the earth being incomparably greater than that of the moon, the exterior 2 of elongation may be considered the 2 of elongation itself. Hence visible part : whole disk :: vers. 60° : 2

::1 cos. 60° : 2

:: 1: 4 :, visible part : dark part :: 1:3

880. Let D, A; L, 4 be the declination, right ascension, longitude, and latitude of the star; d (D) d (A) the given aberrations in declination and right ascension. Then, if o, ó (Fig. 114,) be the two places of the star, P, # the poles of the equator and ecliptic, Pon, mom referred to the parallels of the equator and ecliptic, on om will denote the co-declination and the co-latitude ; and we have on = cos. D X dA

(1) o'm=

= cos. a X dL Again, since the aberration is small, the A orm', o'rn may be considered rectilinear. Hence the 2 no'r = {mor = angle of position (P) VOL. II.

2 s

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.

And it is easily shown that

o'm = on x sin. P o'n cos. P

om = on x cos. P F o'n sin. P. But from (1)

o'm

; and on = da, on = dD COS, a

dL =

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And da = dD x cos. P F dA cos. D sin. P See also Woodhouse's chap. on Aberrations.

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881. The surface is expressed generally by

S = f 2xyds s being the arc of the generating curve. Bat ds =

-rdy

y

ydy S = 2tr

(r-yo)

2xr. {C + V (ro - y^)}
Let S = 0, when y=r; then C = 0, and
S = 2# 1 pe-Y?

= 2ro sin, lat. Let lat. = 90 and 30; then the whole surface and the surface between the specified parallels are

:: 28 = 4719 and 28 = 4tro X

x

respectively; so that

28 = 2 X 2S

Q. E. D.

882. Let D, D, z be the co-declinations of the stars and planet, a the given difference of the right-ascensions of the stars, and y the difference between the right-ascensions of the planet and that star whose declination is D.

Also let d, d be the given distance of the stars from the planet then from the A (D, 2, 3) D', d', *) we get

cos. d

cos. D. cos. x cos. Y =

sin. D. sin. 2

cos. ď' - cos. D'. cos. x And cos. (y+a)

sin. D sin. x which, by the Arithmetic of Sines, will give us y and x.

883. The shadow will evidently be a parallelogram deviating from a rectangle by half a right-angle. Its breadth will therefore = breadth of the wall x cos. 450 =

1

x breadth

of the wall.

The datum of the sun's altitude is unnecessary. It will give the length of the shadow.

884.

This is easy, since
merid. alt. = co-lat. = dec.
midnight depress. = co-lat. F dec.

merid. alt. + med. dep.
.. co-lat. =

2 And I dec. =

merid. alt. - mid. dep.

2

885. Since the star is known, his declination is given by the tables. Let D be the co-declination, a its altitude when on the Prime Vertical, and L the co-latitude of the place; then from the right-angled A (D, a, L) we have

cos. D = cos. a cos. L

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886. Let ss' (Fig. 115,) be two successive positions of the sun, P being the pole and z the zenith. Then if Pz the colatitude of the plane be put = L, the co-dec. = D, the hour 4

zPs = H, and azimuth sz P = a, &c., as in the figure, we have

o'n = da. sin zs.

ss' = dH. sin. D. and o'n = ss' x sin. s'sn = dH sin. D cos. zsP

sin. D. cos. zsP i. da = dH X

sin. zs

cos. L

=dH X

cos. zs. cos. D

sin. Ezs Hence, putting zs = 2, we have, by the question, cos. L cos. z. cos. D

= min,

sin. z which gives by the rule,

2 cos. L cos. 2

1

COS. Z

cos. D

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COS. Z

COS. a

887. Let L be the co-latitude, - the zenith distance, « the azimuth, and D the co-declination; then from the A (2, L, D) we have

cos. D - cos. L.

sin. L. sin, z which gives (Woodhouse's Trig.)

L+ .

M) sin.

2 wherein

sin.

2

sin.'M = sin. L. sin. z. cos.

This gives the declination.
Again, the declination being found, it would be easy to find the

right ascension in terms of the obliquity of the ecliptic. But we are required to express it otherwise.

Since the day and hour of observation are given, let the known right-ascension and hour of the sun, be denoted by A, and H; also let A', H' be the right-ascension and meridian distance of the star; then

A - A = H - H'

.. A' = A + H - H' But H' is known from

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888. Let L be the latitude of the place, D the known co-declination of the star; then from the A (L, D, 45°) we have cos. D = cos. L.

COS. 45
.. cos. L = cos. D x 72.

889. Let L denote the co-latitude, D the co-declination of the sun, and a the required azimuth; then from the rightangled A (L, D) we have

tan. D tan, a =

sin. L

890. Since the day and hour are given, the right-ascension (A), and hour 2 (H) of the sun are known, also since the latitude and longitude of the star are given, its right-ascension (A) and declination (90° – D) may be found (see Vince.) Hence the angular distance of the star from the meridian (H') is known from

H' H = A' A, and the latitude of the place (90° – L) is given. Consequently if 90 — z, a denote the required altitude and azimuth, from the A (2, D, L) we have

S

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