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:s BC = 61° 461 = 11° 46'. The Sun's Place. Then
As R:ct B :: 1 AC:s BA= 59° 39', the Right Ascension. This Problem being the Reverse of the two laft, needs no Example in Numbers.
PROBLEM VII. Given the Sun's Place I 1° 46' and greatest Declination 23° 29'; to find the Angle of the Sun's Position, or that Angle which the Ecliptic makes with the Meridian, viz. AC B.
Practice. Use the Analogy in the Synopsis Case 2, Var. 3 ; saying,
As the Co-sine of B C = 610 461 9.6749134 Is to Radius So is the Co-tangent of B = 23° 29' 10.3620437 To the Tang. of the Ang. C-780 23' 10.6871243
Or thus ; R:csBC::tB:ctC=789 23'.
Note, If the Sun's Place and present Declination be given, that is, BC and AC; then the Angle C is found thus ;
R:ct BC::+AC:csC = 789 23!
PROBLEM VIII. Given the Latitude of the Place, and the Sun's Declination, to find its Amplitude of Rising and Setting,
PraEtice. Suppose at Chichester it be required to find the Sun's Amplitude, May the ist, 1735, when the Declinanation is 20l 34'; by Prob. 4. To answer this,
In the Scheme annexed, there are formed two Right-angled Triangles A B C, and CP H, by either of which this Problem may be satisfied. For
H Н in the Triangle A BC, besides the Right-angle at 4, there is given the Side A C, which is the present
S Declination, and the
N Angle at B the Complement of the Latitude ; to find the Amplitude BC.
Also in the Triangle CPH, Right-angled at H, there is given the Side HP the Latitude, and CP the Complement of Declination, to find HC, the Complement of the Amplitude.
In the Triangle ABC, the Analogy is this,
9.7994951 Is to the Sine of AC = 20° 34' 9.5456896 So is Radius
10.0000000 To the Sine of the Amplit. BC=33° 52' 9.7461945
In the Triangle C P H, this is the Analogy ;
As csPH:R::cs PC:cs HC= 56° 08'. Hence we see on that Day the Sun will rise and set 33° 52' from the East and West Points of the Hori
zon towards the North ; or 5608' from the North towards the East or West Points of the Horizon ; that is on the Points N E by E, and NW by W, very nearly
Given the Latitude and present Declination, to find the Ascensional Difference ; and so the true Time of the Sun's Rising and Seting, and the Length of Days and Nights.
PraEtice. Example. Let the Latitude be that of Chichester, viz. 500 56', and the Sun's Declination 20° 34', as on May 12th, 1735 ; I demand the Ascensional Difference ?
In the foregoing Scheme, by either of the two Right-angled Triangles ABC, or C P H, this Problem allo is to be answered.
For in the Triangle A B C, the Side AC, and Angle B, are given to find A B, the Ascensional Difference.
Also in the Triangle C P H, there is given HP, and CP, to the Angle at P the Complement of the Ascensional Difference.
'Tis most directly found by the Triangle A B C, whose Analogy runs thus,
10.0000000 Is to Co-tangent of B = 399 04' 10.0905978 So is the Tangent of A C = 20° 34' 9.5742761
To the Sine of AB = 279 327 9.6648739
The Analogy for finding the same in the Triangle CPH, is this; As t PC:R:: PH: CP= 62@ 28'.
Or varied thus ; As R:ctPC::tPH:CP. Now the Ascensional Difference A B being found 27€ 321, this Motion in the Equinoctial is performed by the Sun in 1 Hour, 50 Minutes, 8 Seconds of Time ; and this being the Time the Sun riseth before, and fetteth after, 6 a-Clock, 'tis evident the Sun will then rise at 4 H. 9 M. 52'' in the Morning ; and set at 7 H. 50' 8" at Night.
Consequently the Length of that Day will be 15 H. 40' 16W; and the Length of the Night 8 H. 19'44", as was required.
Having given the Right Ascension, and Ascensional Difference ; to find the Oblique Ascension, and Defcenfion.
Practice. 1. If the Sun be in Northern Signs, fubftract the Ascensional Difference from the Right Ascension, the Remainder is the Oblique Ascension.
Thus on May 12th, 1735. The Ale. Diffe. is 27 32
. 59 The Difference is the Oblique Ascension = 32 07
2. Also to the Right Ascension
59 39 Add the Ascensional Difference
27 32 The Sum is the Oblique Descension = 87 11
3, But if the Sun be in the Southern Signs, the Sum of the Right Ascension and Ascensional Difference gives the Oblique Ascension ; and their Difference, the Oblique Descension,
The Latitude and Declination given, to find the Time (Morning and Evening ) when the sun will be due East and West.
PraEtice. In the adjacent Diagram, there are formed two Right-angled Æ
P Triangles A B C and CP Z, in either of which this Problem may be answered. For in the Triangle ABC, H there is given the Side A C, the Declination ; and the Angle at B the Latitude ; to find the
S Side A B, the Time
N before and after six aClock, wherein the Sun will be in the Azimuth of East and West Z N, in the Point C.
Also in the Triangle C P Z, there is known, the Side C P the Co-Declination ; the Side Z P the Co-latitude; to find the Angle P, the Time from Noon, &c.
Example. I demand what Time the Sun will be due East or West on May the 12th, 1735, when the Declination of the Sun will be 20 34', at Chichester, whose Latitude is 50° 56' ?