Draw any right Line, as GEC; take any Extent of the Compaffes, and place one Foot in E, and draw the Semicircle GBC; erect EF perpendicular to GC, and draw the right Line CB: Begin to divide the Quadrant CB into three equal Parts, at the Points 30, 60, and B, and divide each of these into three more equal Parts, at the o Bol Points 10, 20, 30, 40, &c. then fetting one Foot of the Compaffes in the Point C, extend the other to 10, 30

1401

F

B

70 60 50 40 30

18

30

60

150

140

30

1201

110

20, 30, 40, &c. and transfer it to the Chord-Line CB; fo will this Line be divided in the Points 10, 20, 30, 40, &c.-` the Chords of thofe Degrees. If you continue to divide the Arch CB into ftill fmaller Divifions, and transfer them in the fame Manner, these Divifions, where they interfect the right Line CB, will be the Chords of thofe Degrees; which, in the Figure, are divided but to every fifth Degree.

Of the Line of Sines.

If thro' the Points 10, 20, 30, &c. of the Quadrant CB, Lines are drawn parallel to the Diameter G C, they will cut the Radius BE, and the Part cut off will be the Sine of the Number of Degrees which fuch Parallel paffes thro' in the Periphery of the Circle: Thus Ea is the Sine of 10 Degrees,

becaufe

because the Parallel which cuts the Radius EB in the Point à paffes thro' 10 Degrees in the Periphery. Thus, by drawing Parallels thro' each Degree of the Quadrant, the Radius BE would be divided into a Line of Sines to all the 90 Degrees of the Quadrant.

Of the Lines of Tangents and Secants.

A Perpendicular being raised, by Prob. 3, on the End of the Diameter G C, as CH, is a Tangent-Line; to which, Lines drawn from the Center E, thro' the feveral Degrees of the Quadrant, will divide it into a Line of Tangents to the Radius BE; and the Lines fo drawn from the Center are the Secants of the feveral Degrees thro' which they pafs, which may be transferred to the Line BF, by placing one Foot of the Compaffes in E, and extending the other to 10, 20, 30, 40, &c. on the Tangent-Line, and drawing the respective Arches 'till they interfect the Line E F, which Distances from the Center will be the Secants of their refpective Degrees.

If the Reader takes the Chord of 60 Degrees from his Scale, and proceeds as directed, he may compare the Truth of his Conftruction with the Lines on the Scale.

S

Of Spheric GEOMETRY.

PHERIC GEOMETRY confifts in projecting the Circle of the Sphere on a Plain: In which Projection there are great Circles of three different Sorts, viz. primitive, right, and oblique Circles.

An oblique great Circle is a Circle which cuts the primitive Circle in two oppofite Points, but does not pass thro' the Center, as the Circle Z DN.

An oblique lefs Circle is a Circle parallel to a right Circle, as the Circle k Im, parallel to HOR.

A fpherical Angle is the Meeting of two Circles in a Point; thus, by the Meeting of the primitive Circle H ZRN with the oblique Circle Z DN, is made the Angle R Z D.

The Pole of any Circle is that Point which is ninety Degrees from it every Way.

When any Circle paffes thro' the Pole of another Circle, it interfects it at right Angles, or is perpendicular to it.

Prob. 38. To find the Poles of any right Circle, A B.

The Poles of every right Circle are, in the Periphery of the primitive Circle, ninety Degrees from it; therefore, draw CD perpendicular to A B, paffing thro' the Center O; then C and D are the Poles of the given right Circle A B.

Prob. 39. To lay any Number of Degrees, as 51°. 30′, on the primitive Circle.

Take 51°. 30' from the Line of Chords, and set it on the primitive Circle, as from B to a.

Prob. 40. To measure any Part, as CG, of the primitive Circle.

Take CG in the Compaffes, and, applying it to the Line of Chords, it will reach from the Brafs Center to 46°.36', the Answer.

Prob.

Prob. 41. To lay any Number of Degrees, as 68°, on a right Circle.

This Problem admits of two Cafes:

Cafe 1. If the Degrees are to be laid from the Center towards the Periphery of the primitive Circle, take 34°, the Half of 68°, from the Line of Tangents, or 68° from the Line of Half-Tangents, and lay them from O to E.

Cafe 2. When the Degrees are to be laid from the Periphery towards the Center or Pole of the primitive Circle; As, to lay 43°.24 on the right Circle A B from A towards O, take the Complement of 43°. 24 to 90°, which is 46°. 36', from the Scale of Half-Tangents, or 23°. 18', the Half of it, from the Line of Tangents, and lay it from O to H; then is A H 43°. 24, becaufe OA is 90°. Or, fet one Foot of the Compaffes on 90° in the Line of Half-Tangents, and extend the other 'till you have 43°. 24' between the Compaffes; with this Extent, place one Foot of the Compaffes in A, and the other will fall on H; thus will A H be 43°. 24′, as required.

Prob. 42. To measure any Part of a right Circle.

This, like the laft Problem, admits of two Cafes:

Cafe 1. To measure the Part OE of the right Circle A B, take OE in the Compaffes, and, applying it to the Line of Half-Tangents, it will reach to 68°; or, applying it to the Line of Tangents, it will reach to 34°, which, being doubled, gives 68°, as before.

Cafe 2. To measure a Part, as A H, of the right Circle A B, take OH in the Compaffes, and, applying it to the Line of Half-Tangents, it will reach to 46°. 36, the Complement of which to 90° is 43°. 24', the Measure of AH; or, take A H in the Compaffes, and fetting one Foot on 90° in the Line of Half-Tangents, and turning the other towards the Beginning of the Scale, they will be found to include 43°. 24, the Measure of A H, as before.

Prob. 43. To find the Pole of an oblique Circle, BED.

Draw the Diameter BO D, and the Diameter A O C perpendicular to BOD; then draw a Line from B thro' E, meeting

mitive Circle and its Center. If the other Pole be required, fet off go° on the primitive Circle from F

to I, extend AC towards K, and draw BI meeting A C in K, the other Pole of BED.

From what has been fhewn, of finding the Poles of the feveral Circles, the Learner may perceive, that the Pole of every Circle is ninety Degrees diftant from the Circle itself: Thus, if EH, the Distance between the oblique Circle BED and its Pole, be measured, by Prob. 42, Cafe 2, the Part EO will be found thirty-four Degrees, and OH fifty-fix Degrees, which together are ninety Degrees.

Prob. 44. To lay any Number of Degrees, as 37°, on oblique Circle, BED. See the laft Figure.

Find H, the Pole of the oblique Circle, by Prob. 43; take 37° from the Line of Chords, and fet from B to a; then draw Ha, cutting the oblique Circle in r; thus will Br be 37°, as was required.

Prob. 45. To measure a Part, as Br, of an oblique Circle, BED.

Find H, the Pole of the oblique Circle, by Prob. 43; then draw Hr, cutting the primitive Circle in a; and Ba, being measured on the Line of Chords, will give 37°, the Measure of Br. Again, if it were required to meafure the Part rs of the fame oblique Circle, firft draw Hr, cutting the primitive Circle in a ; then draw H s, cutting the primitive Circle in t'; and the Distance at, being meafured on the Line of Chords, gives 87°, the Measure of rs, required.