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6. Set 24° 20', the Plane's Declination, from the Scale of Chords, on the Primitive Circle from H to c, from D to e, and from O to f.

7. Then a Ruler laid from Z, to the Point c, e, f, will give the Points W, B, E for the West, South, and East Points of the Horizon of London.

8. Lay a Ruler from W to Z, and it will cut the Primitive in b; then lay off 38° 28' (the Co-latitude of London) from the Chords from b to d.

9. Take the Tangent of 54° 00' and set from 2 to K, and draw thro' Ka Line parallel to HO, infinitely extended, and thereon set off the Half-Tangent of 24° 20', from K to M; then with the Tangent of 65° 40' ( set the other way from K, on the fame Line extended) you find the Center whereon to draw the Circle Z B M, which shall be the Meridian of London.

10. A Ruler laid from W to d will cut the Meridian in P for the North Pole of the World.

11. Thro' P and 2.draw the Right Line N 2 T, which shall be the Meridian of the Place fought after, and the Axis of the World ; and, being produced, will meet the Meridian of London in S, the South Pole of the World.

12. Cross N T at Right Angles with V X, and a Ruler laid from 0 to V will give the Point Æ, where the Equinoctial interfects the Meridian of the enquired Place.

13. Thro' V, Æ, and X, draw the Equinoctial Circle, as hath been taught; which also will pass thro' W and E, the East and West Points of the Hori. zon of London.

14. Divide P S into two equal Parts in E, and draw E G at Right Angles thro' the fame ; extending it infinitely make P E Radius, and lay off the Tangents of 15°, 30°, 45°, &c. both ways from E on

the

the Line E G ; and thus you will have the Centers of the Hour Circles to be projected ; which when done your Projection is finished.

15. Now to find the Requisites, take QP in your Compasses, and measure it on a Scale of Half-Tangents,

and

you will find it equal to 72° 34' whose Complement PN, or QÆ, is 179 26'; and that is the Latitude in which this Reclining Plane will be an Horizontal Dial.

16. Lastly, lay a Ruler from P to a, and it will cut the Primitive in h; then Ih measured on the Chords will be found 14° 41', equal Æ a, the Difference of Longitude required.

Therefore a Plane declining 24° 20', and Reclining North 360 oo' in the Latitude of London, will be an Horizontal Plane in the Latitude 17° 26', and Difference of Longitude from London 14° 41', which is about Caffena in Negroland.

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CH A P. VII.

Theorems for the Explication of the Doc

trine of Spherical Triangles, and the Manner of their Solution.

THEOREM I.

A

NY two Great Circles of the Sphere mutually bisect each other.

Demonstration. For since one Great Circle

ABC, hath the Center

D in common with the other Great Circle AFC; therefore the common Section

AC, shall be a Diameter of each ; and so will cut them into two equal Parts. 2. E, D.

THEOREM II.

If from the Pole B, of any Great Circle AFC, be drawn a Right Line B D, to the Center thereof, the faid Line will be perpendicular to the Plane of that Circle.

Demonstration. Let be drawn the Diameters

EF, GH, n the Circle

A ECF;

Then

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then because in the Triangles BD F, BDE, the Sides

BD, BF, are equal to the Sides

BD, DE, and the Base

BF= BE; therefore shall the Angle

BDF=BDE, and so each is a right one ; thus it is demonstrated that Angles

BDG, BDH, are right ones, and so is the Line

- BD, perpendicular to the Plane of the Circle

AECF, by Euclid 11. Prop. 4.

2. E. D.

THEOREM III.

If a great Circle E B F, be described about the Pole A ; then the Arch B F, intercepted between AB, A F, is the Measure of the Angle B AF, or BD F.

Demonstration. By Theorem 2, the Angles

ADB, ADE are right ones ; because the Arches: A B, AF, are Quadrants ; Consequently the Angle B DF, ( whole Measure is the Arch

BF, ) is equal to the Inclination of the Planes A B C, AFC; and so equal to the Spherical Angle B AF, or BCF,

2. E, D.

COROLLARY I. Because AB, AF, are Quadrants, A shall be the Pole of the Circle passing thro the Points B, F; for A is at Right Angles to the Plane FD B, by Euclid 11. Prob. 14.

COROLLARY II. The Vertical Angles are equal ; being each equal to the Inclination of the Circles; also the adjoining VOL. II.

I a

An

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