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From the Angle C, let fall the Perpendicular CL on the Side E P; then in the Right-angled Triangle CIP, find L P thus ;

As Radius Is to the Co-line of L P C = 15° 50' 9.9832019 So is the Tangent of C P = 44° 161 9.9888816 To the Tangent of L P = 43

43° 10 9.9720835

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Then L P + P E = LE = 65° 39'; then say,

As the Co. sine of L P = 43° 10' C. A. 0.1370530 Is to the Co-fine of CP= 44° 16' 9.8547428 So is the Co-sine of L E = 66° 39' 9.5980754 To the Co-fine of C E = 67° 08' 9.5898712 Which Complement is CD= 22° 52' the Latitude North.

Secondly ; Then to find the Longitude, or Angle at E, says

As the Sine of CE = 67° 08' Co. Ar. 0.0355463 Is to the Sine of CP E= 164° 10' 9.4359080 So is the Sine of C P = 449 161

9.8438547 To the Sine of CEP = 11° 57' 9.3153090

;

;

This being the Complement of its Longitude to Cancer os, = DO; therefore subftract it from 90°, there will remain 780 03' from the Vernal Equinox or Sign r, that is in I 18° 03!.

Note ; This is 30" more than the Tables give it ; but in these Operations, which are intended only for Examples, we do not regard such small Differences from the Truth of Things ; and the fame is to be remembered in other places.

PRO

PROBLEM XXIII.

Given the Latitude of the Place, the Sun's Right Ascension, and the Altitude of a known fixed Star; to find the Hour of the Night.

Pračlice.

Admit the Latitude be 50° 56', the Sun's Right Ascension 2280 45' ; and the Altitude of the Bull's Eye, or Aldebaran 32° 30' towards the East of the Meridian ; thence to find the Hour of the Night.

The Right Ascenfion of Aldebaran is

P 65° 8', and his Decli

Æ nation 15° 55'.

Now by constructing the Diagram there will be formed the H

А.

0 Triangle C ZP, in which all the Sides are given ; ZP = 399 04' = the Co-Latit. . CZ = 57° 30' = the S Co-Altitude ; and CP

N = 740 05' the CoDeclination ; to find the Angle APÆ the Difference between the Right Ascension of Medium Cæli & Aldebaran.

Now

Now CP - ZP = AM 35° 01' ; Then

Add (the Sine of Z P= 39° 04' Co. Ar. 0.2005049

the Sine of C P=749 05' Co. Ar. 0.0169777 toge

CZ + AM the Sine of

46° 15' 9.8587561 ther.

CZ-AM the Sine of

= 11° 30' 9.2996553

2

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The Sum of all is

19.3759040 of wh. is the Sine of CPZ=29° 10'= 9.6879520

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Wherefore the Double of 29° 10' gives 58 20=CPZ;
which fub. from the R. Ascen, of Aldeb. 64 08
there will remain the R. Af. of Med. Cæli 6 48
add thereto

360 oo
and from that Sum

366 48
subftract the Sun's Right Ascension
the Remainder is

138 03

228 45

Which reduced to Time, makes 9 Hours 12' 12" the Hour of the Night, as required.

Note, When the Angle C P Z is equal to the Stars Right Ascension ; then the Sun's Right Ascension fubducted from 360° ( the Cor Cæli then ) leaves the Degrees which reduced, gives the Time of Night required.

If the Angle CPZ be greater than the Star's Right Afcenfion, the Difference subducted from 360°, leaves the Right Ascension of Medium Cæli, froni which take the Sun's Right Afcenfion, the Remainder reduced to Time, gives the Hour fought.

PRO

PROBLEM XXIV.

Given the Obliquity of the Ecliptic, and the Right Ascension of Medium Cæli, or Mid-Heaven ; to find the Culminating Point in the Ecliptic.

Practice. Admit the Right Ascension of Medium

P Cæli be 1409 ; then Æ that's 400 short of the Equinox = Æ — ; whence in the Rightangled Triangle Æ - H

M E, there is given the Side 2 = 40 oo'; and the Angle at = 23° 29'; to find the

S Side E, the Point E being that which is fought in the Ecliptic. Wherefore say thus ; As the Radius

10.0000000 To the Co-fine of Æ - E= 23° 29' 9.9624527 So is the Co-tangent of Æ = 40° 00' 10.0761865 the Co-tangent of E= 42° 28' 10.0386392

S That is in the Division of the Ecliptic i

28 Which fubftracted from

6. There remains the Distance

4 - 17 32 from the first Point of Aries, viz. 1 17° 32' the Point of the Ecliptic culminating, or on the Meřidian.

PRO

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00

PROBLEM XXV.

Given the Obliquity of the Ecliptic, and Right Alcension of Medium Cæli, to find the Meridian Angle.

Practice.
In the Scheme to the last Problem, in the Triangle
Æ = E, there is given, the Side Æ = 400 00' ;
and the Angle at ^ = 23° 29'; to find the Angle
ÆEA, made by the Intersection of the Ecliptic and
Meridian,

Ibe Analogy is ;
As the Radius

10.0000000 To the Sine of = 23° 29'

9.6004090 So is Co-sine of Æ= 40' oo 9.8842540 To the Co-line of E= 72° 14' 9.4846630

PROBLEM XXVI.

Given the same things as in the two last, to find the Declination of the Culminating Point.

In the aforesaid Triangle Æ - E there is as before specified, the Side Æ =; and the Angle at , to find the Side Æ E, which is the Declination of the Culminating Point.

The Analogy is thus.
As Radius

10.0000000 To the Tangent of = 23° 29' 9.6379563 So is the Sine of Æ = 40° oo' 9.8080675 To the Tangent of Æ E= 150 36? 9.4460238

VOL. II.

L1

Note i

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