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Proposition 10. To find the Declination, right Ascension and Ascen∣sional difference of the Planets or fixed Stars.
Suppose the place of any Planet be given in Longitude and Latitude: As suppose the Moon were the Planet pro∣posed her true place given, let be Leo, 20 d. 49 m. 53 s. and her Latitude 00 d. 22 m. 52 s. North (as at the time of that solar Eclipse, which will happen in July 1684.) which being given with the greatest obliquity of the Ecliptique, 23 d. 31 m. 30 s. we shall enquire for the Moons declination thus.
| d. | m. | s. | |
| As the Radius, | 90 | 00 | 00 |
| Is to the s. of the Moons Longitude ab. Ariete, | 110 | 49 | 53 |
| So is the Tangent of the greatest obliq. | 23 | 31 | 30 |
| To the Tangent of an Arch. | 22 | 8 | 24 |
| Then from the Radius, | 90 | 00 | 00 |
| Substract the Moons latitude (because its North) | 00 | 22 | 57 |
| Rests distance of the Moon from the Pole | 89 | 37 | 8 |
| From which Substract the first Arch | 22 | 8 | 24 |
| Rests a second Arch | 67 | 28 | 44 |
| Then I say again; | |||
| As the Cosine of the first Arch | 22 | 8 | 24 |
| Is to the Cosine of the second Arch | 67 | 28 | 44 |
| So is the Cosine of the greatest obliq. | 23 | 31 | 30 |
| To the sine of the Moons declination required | 22 | 16 | 50 |
| 2. To find the Moons right ascension, I say, | |||
| As the Cosine of the Moons declination | 22 | 16 | 50 |
| Is to the Cosine of the Moons longitude, ab Ariete | 110 | 49 | 53 |
| So is the sine of her distance from the Pole | 89 | 37 | 8 |
| To the sine of an Arch, viz. | 22 | 36 | 2 |
Which added to 90 degrees, because the Moon is in the second Quadrant of the Ecliptique, (and to 180 deg. when in the third Quadrant, &c.) the Aggregate 112 deg. 36 min. 2 sec. is the right ascension of the Moon, as was required.
Lastly, Her ascensional difference is to be found accor∣ding to the third Proposition aforesaid.