Line) and one Foot in the given Point A, cross the given Line in b and c; then with any Diftance more than half the Distance between b and c, and one Foot in & draw the Arch d, and with the fame Extent, and one Foot Fig. 2. in, cross the said Arch at d; then lay a Scale from the Point A to the Croffing of the Arches at d, fo draw the Line AE, which is the Perpendicular required.

PROB. III. How to raise a Perpendicular upon the End of a given Line.

With your Compaffes at any Extent, and one Foot in the given Point A, make a Mark at any conveFig. 3. nient Diftance above the Line, as at b; then keeping one Foot in the Mark b, with the fame Extent cross the given Line at c, and turning your Compaffes about, make the Arch d; then lay a Scale from the Croffing at c to the Mark at b, and make a Mark where it croffes the Arch d; fo a Line drawn from that Interfection to the Point A is the Perpendicular required.

PROB. IV. How to let fall a Perpendicular upon the End of a given Line, from any given Point over the End of the faid Line.

Draw a Line from the given Point A to interfect the given Line at any convenient Diftance, as at b; Fig. 4. then divide the Line A b into two equal Parts at e; and upon c, with the Extent c A, or cb, crofs the given Line in d; then a Line drawn from the given Point A to the Interfection at d, is the Perpendicular required.

PROB. V. To draw a Line parallel to another Line at any given Diflance.

Take in your Compaffes the given Diftance, and with one Foot in the given Line, towards each End Fig. 5. of it, draw the two little Arches a and b: a Line drawn from the Extremity of thefe Arches is the

Parallel required.

PROB.

PROB. VI. To bring any three Points (not fituate in a right Line) into the Circumference of a Circle.

Suppose the three Points be ABC; first take more than half the Distance A B, and with one Foot in A, fweep an Arch; then with the fame Diftance, and one Foot in B, draw another Arch to cross the aforefaid Arch in the Points d and e; then with more than half the Distance BC, and one Foot in B, draw the Arch gb, and with the fame Diftance, and one Foot in C, draw an Arch to cross

the aforefaid Arch in the Points g and h; then Fig. 6. draw the Lines a e d, and a gh, and where these Lines crois, as at a, is the Center of the Circle required.

SECT. II.

The Construction of Sines, &c.

EVERY great Circle is (fuppofed to be) divided into 360 equal Parts, called Degrees, whereof the Half, or Semicircle, contains 180 Degrees, and the Quarter or Quadrant is 90 Degrees, and upon one Quadrant is projected the Sines, Tangents, &c.

The Radius is the Semidiameter of a Circle, upon which the Projection is made; as AK, or A 90, or AS, is the Radius of the projected Diagram, and is commonly fuppofed to contain 10000 equal Parts.

A Sine is a Perpendicular let fall from the given Degree to the Bafe or Semidiameter of the Circle; as the Line g 30, is the Sine of Degrees, and the Line b 80, is the Sine of 80 Degrees, &c. the Sine of go Degrees is equal to the Radius.

A Tangent, or Touch-line, is a Perpendicular erected upon the End of the Semidiameter, juft fo as to touch the Periphery of the Circle: Thus the Line K is a Tangent Line. The Tangent of any De

B 2

Fig. 7.

gree

gree is the Distance from the Beginning or Foot of the Tangent Line to that Part of it where the Line, drawn from the Center over the given Degree, cuts the Tangent Line: Thus the Part of the Tangent Line k p is the Tangent of 50 Degrees. The Tangent of 45 is equal to the Radius.

A Secant is a Line drawn from the Center through the given Degree 'till it interfect the Tangent Line: Thus the Line Ap is the Secant of 50 Degrees. The Secants begin at the Radius,, and proceed to Infinite. The Secant of o Degrees being equal to the Radius.

A Chord is the nearest Distance in a ftrait Line between any Quantity of Degrees; or from o Degrees to the Degree whofe Chord is required: Thus the Line S. 90 is a Line of Chords, and the Distance S. 30 upon that Line is the Chord of 30 Degrees, &c. the Chord of 60 Degrees is equal to the Radius; and hence it is, that the Chord of 60, commonly called the Sweep of 60, is generally taken in the Compaffes to draw any Great Circle, or Arch of a Circle, whofe Quantity in Degrees is to be measured.

The Sine-Complement of any Arch, is the Sine of the Complement of that Arch to 90: Thus the Sine complement of 30 Degrees is the Sine of 60, and the Sine-complement of 70 is the Sine of 20, &c. And in the Diagram the Line W 40 (equal to the Distance Af upon the Base) is the Sine-complement of 40 Degrees, and fo in others.

A verfed Sine is a Segment of the Bafe, contained between the Sine of the Degree and the End of the Bafe where the Tangent Line begins; thus the Segment of the Bafe e K is the verfed Sine of 50, &c.

From this Projection is reduced the following Axioms for the Solution of all the Cafes in Plane Trigonometry; and which (if well underftood) are the Ground of the whole Art of Navigation, fo far as it depends upon Plane Triangles, and is folved by a Canon; which I fhall first fhew in the following Cafes, and then proceed to fhew how it may be done without.

AXIOM

AXIOM I.

In all right-angled Plane Triangles, if one Side be made Radius, the other Sides will be Sines, Sine-complements, Tangents or Secants, as is evident from the Diagram; for,

Suppofe in the Diagram the Triangle Ap K, here the Side A K is the Radius, and the Angle at A being 50 Degrees, the Side p K is the Tangent of 50, and the Hypotenufe A p is the Secant of 50; and what Proportion the Radius hath to the Sine A K, the fame Proportion hath the Tangent of 50 to the Side K p, and the fame Proportion hath the Secant of the Hypotenufe A p.

Again, if you will make the Hypotenufe Radius, and fuppofe the Angle at A be 40 Degrees; then in the Diagram it is reprefented by the Triangle Af 40, and then the Side f 40 is the Sine of the Angle at A, and the Side Af equal to W 40 is the Sine-complement of the Angle at A, and then what Proportion the Hypotenufe hath to the Radius, the fame Proportion will the Sine 40 f have to the Sine of 40, and the fame Proportion will the Side Aƒ have to the Sine-complement of 40, &c. and from this Proportion proceeds the second Axiom.

AXIOM II.

In all Plane Triangles the Sides are proportionable to the Sines of their oppofite Angles, and the contrary; as in the - Diagram in the Triangle above-mentioned, Aƒ 40, it is demonftrated, As the Radius or Sine of 90 to the Hypotenufe, or Side oppofite; fo is the Sine of 40 to the Side oppofite to the Angle at A, &c.

This Proportion commonly called Oppofite Sides Oppofite Angles, holds true alfo in Oblique Plane Triangles; only observe, that where you have an Obtuse Angle, viz. more than 90 Degrees, the Sine of it is found by fubtracting the Obtufe Angle from 180 Degrees, the Sine of the Remainder is the Sine of the Obtufe Angle required.

B 3

AXIOM

AXIOM III.

In all Triangles, as the Sum of the Legs of any Angle is to their Difference, fo is the Tangent of half the Sum of the other two Angles, to the Tangent of half their Difference; and therefore,

When there are given two Sides, and an Angle included, to find the other Angles, the Proportion is,

As the Sum of the Sides, to the Difference of the Sides, fo is the Tangent of half the Sum of the unknown Angles, to the Tangent of half their Difference; which half Difference added to the half Sum, is the greater Angle, and fubtracted leaves the leffer.

AXIOM IV.

In all Triangles, as the Bafe or greater Side, to the Sum of the other two Sides; fo the Difference of the Sides to the Difference of the Segments of the Bafe, which Difference fubtracted from the whole Bafe, the Perpendicular falls in the Middle of the Remainder; and fo the Oblique Triangle is reduced to two Right-angled ones, and may be. wrought after the fame Manner.

By thefe Axioms are all the following Cafes of Plane Triangles folved; in which obferve, In Right-angled Triangles, the two Sides including the Right-angle are called Legs or Sides, or fometimes Bafe or Perpendicular; and the flope Line is called the Hypotenuse,