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FOr finding this arch of the Aequator, there must be given the angle of Inclination of the Circle of position with the Meridian, & the height of the Pole above that Circle, both which may be found by the 9 Chap∣ter:
but •••• to our present purpose, we are not tied unto such a ••••dious calcu∣lation, because the Pole of the Circle may be supposed, and then the angle of Inclination may be found at one Operation, and this arch of the Ae∣quator at another.
And first, the Elevation of the Pole above any Circle of position being given together with the latitude of the place or Countrey, the angle of In∣clination may be found, by this analogy.
As the sine of the Pole of the place, is in proportion to Radius: So is the sine of the Pole of the Circle, to the sine of the angle of Inclination.
For Example. In the Rectangle Spherical Triangle of the 9th Chapter L M C right angled at L. Let M C the Elevation of the Pole of the place be 45. And the Pole of the circle LM 42, hence to find the angle LCM. I say,
| As the sine of M C | 45. | 9. 8494850 |
| Is to the Radius | 90. | 10. 0000000 |
| So is the sine of L M | 42. | 9. 8255108 |
| To the sine of L C M | 71. 13 | 9. 9760258 |
Then to find the arch of the Aequator, the proportion is: As the Radius to the sine of the Complement of the Pole of the place: So is the tangent of the angle of Inclination, to the tangent of the arch of the Aequator. For Example, In the triangle A D F of the afore-said Diagram, Let there be given the side A D the Comple. of the Pole of the place 45. The angle of Inclinat. DAF 71. 9. Hence to find the arch of the Aequator. DF, I say,
| As the Radius | 90 | 10. 0000000 |
| To the sine of AD | 45 | 9. 8494850 |
| So is the tangent of D A F | 71. 13 | 10. 4662••85 |
| To the tangent of D F | 64. 20 | 10. 3157275 |
Which is the arch of the Aequator sought.