Fig. II.

XXIII. An Equilateral Triangle, is that whose Sides are all Equal, and whose Angles contain (each of them) 60 Deg. So the Triangle EBK hath its Sides BE, BK and E K, all of them equal; and the Angles EBK, EK B and KE B equal also, and each of them equal to 60 Deg. and confequently all of them equal to 180 Deg.

XXIV. A Triangle is either Right Angled or Obliqué Angled.
XXV. A Right-angled Triangle, is that which hath one Right

Angle. And Auch is the Triangle CAB, Right-angled at A.

XXVI. An Oblique-angled Triangle is that which bath all its Angles Oblique. And such is the Triangle BCD.

XXVII. An Oblique-angled Triangle, is either Acute-angled, or Obtufe-angled.

XXVIII. An Oblique Acute-angled Triangle is that which hath all its Three Angles Acute. And fuch are the Triangles DBE, and EB K.

XXIX. An Oblique Obtuse-angled Triangle is that which hath One Obtuse, and Two Acute Angles. And such is the Triangle DBC, whose Angle DBC is Obtuse, and the Angles BDC and BCD Acute.

Fig. III.

(

Forafmuch

CHAP. II.

Of Right Lines, applied to a Circle.

as the Ratio or Proportion of an Arch Line to a Right Line, is as yet unknown, yet it is absolutely neceffary that Right Lines be applied to a Circle, for the Calculation of Triangles, wherein Arch Lines come in Competition: For the Angles of Plain (or Right-lined) Triangles are measured by Arches of Circles.

Now, the Right Lines applied (or relating) to a Circle, are
Chords, Sines, Tangents, Secants and Verfed Sines.

1. A Chord, or Subtenfe, is a Right Line, joining the Extremities of an Ark, as AC is the Chord of the Arks ABC and AD C. 2. A Right Sine, which is singly called a Sine, is a Right Line, drawn from one end of an Ark, perpendicular to the Diameter drawn through to the other End: Or, it is half the Chord of twice the Ark; so A Eis the Right Sine of the Arks A B and A D. The Radius (or Sine of 90 Deg.) is called the Whole

Sine,

Sine, and is the greatest of all Sines: For the Sine of an Ark Fig. III. greater than a Quadrant, is less than the Radius; so FG is the whole Sine or Radius.

3. A Versed Sine is the Segment of the Radius between the Ark and its Right Sine; fo E B is the Verfed Sine of the Ark A B, and of the Ark AG D.

4. The Secant of an Ark, is a Right Line, drawn from the Center through one end of an Ark, till it meet with the Tangent : That is, a Right Line touching the Circle at the nearest end of that Diameter which cuts the other end of the Ark. FMis the Secant, and BM the Tangent, of the Ark A B, or of A D.

5. The Difference of an Ark from a Quadrant, (or 90 Deg.) whether it be Greater or Less, is called the Complement of that Ark; so G A is the Complement of the Arks A Band AG D, and HA is the Sine of that Complement: GI the Tangent of that Complement: and F I the Secant of that Complement. All which (for Brevity) we write Co-Sine, Co-Tangent, Co-Secant of the Ark.

6. The Difference of an Ark from a Semicircle (or 180 Deg.) is called its Supplement; so the Ark AB is the Supplement of the Ark DG A, to a Semicircle.

7. That Part of the Radius which is between the Centre and its Right Sine, is equal to the Co-Sine. As FE is equal to H A, and FO is equal to the Co-Sine of the Ark DS.

8. If an Ark be Greater or Less than a Quadrant, the Sum or Difference, accordingly, of the Radius and Co-Sine, is equal to the Verfed Sine: For FD and HA together, are equal to D E, the Verfed Sine of the Ark DGA; and FB less by HA (or E F) is equal to E B, which is the Versed Sine of the Ark A B.

CHAP. III.

Some Short Problems, to make Canons of Sines, Tangents
and Secants.

PROB. I.

The Sine of an Ark being given, how to find out the Sine of the
Complement.

BC being given, to find AC.

Fig. IV. B

Ecause the Triangle ACB is a Rectangle (by the Definition of a Sine) and the Sides AC, BC, are of the fame Power as the Hypotenuse; that is, the Radius AB: Therefore, if the Square of the Sine BC be substracted from the Square of the Radius A B, there remains the Square of AB, whose Side is the Right Line A C, the Sine fought for.

PROB. II.

The Sine of an Ark being given, and likewise the Sine of the
Complement, to find the Sine of half the Ark.

R Q and A Q being given, to find B. O or RO.

As AB: is to BO, so is :: BO: to BG; therefore B O will be the Square Side of the Plain from the Radius A B and BG, the half Verfed Sine. For QB, the Verfed Sine of the Ark BR, is given; because A Q, the Sine of the Complement, and A B, the Radius, are given by the Suppofition.

PROB. III..

The Sines of Two Arks, and the Sines of the Complements be
ing given, to find the Sine of the Sum.

RQ, QA, and ST, TA, being given, to find S P.

As AR is to RQ, so is AT to T.G, or C P.
As AR is to AQ, so is ST to SC.

BT and CP together make SP, the Sine of the Sum of Two

Arks.

PROB.

7

PROB. IV.

The Same being given, to find the Sine of the Difference.

RQ, QA, and SP, PA, are given, to find ST.

As A Qis to QR, so is AP to P O, from whence you may find OS. As AR is to AQ, so is OS to ST.

To these join the Theorems.

Theorem I. The leaft Sines are, almost, in the fame Ratio as their
Arks.

This Theorem will be prov'd to be true hereafter in the' continued Bisections; but the least Arks are about one of the first Scruples, or less, and are almost in the same Ratio as their Sines, because they are, almost, contiguous amongst themselves, and almost of the same Quantity, as it appears, tho' not to every Search, yet to a very profound one.

Theorem II. If the fame Line be cut into unequal Parts in Number, the Number of the Parts of the first Section, is to the Num ber of the Second, (reciprocally) as one Part of the second Section is to one Part of the first Section.

Divide the fame Line, first into Four, then into Three Parts; then it will be as 4 to 3, fo Part to reciprocally; the Reason is, because 3 in makes 1, and 4 in makes I; and because the Products are equal, the Multiplicands will be reciprocally proportionable.

The Construction of the Canon of Sines.

The Sine of the whole Quadrant is call'd the Radius, for it is the Semidiameter of a Circle. Set in the Canon a Radius of rooooo Parts, or 100000.00 for Neceffity of Calculation; but for the better making of the Canon take a Radius 100000.00000 Parts; for by that Means, the Errors which oftentimes happen amongst the Right Hand Figures may fafely be corrected without any Prejudice to the Canon.

F2

Then

Fig. VI. Fig. VI.

Then bisect the Quadrant, and look for the Sine of the Biseg-
ment by Prob. 2. and its Co-fine by Prob. 1. then bisect this Bifeg-
ment again, and look for the Sine of the second Bifegment by
Prob. 2. and the Co-fine by Prob. 1. then bisect this second Bifeg-
ment, and look for its Sine or Co-fine by Prob. 2. and 1. and
then bisect the third Bifegment, &c. and continue bisecting 13
times, till you find a Sine of Part of the whole Quadrant,
as 'tis here set in the Table. Now we come to the leaft Arks,
where the Truth of the first Theorem is illustrated: For as the
Ark of the Quadrant
Sine almost to that Sine.

I

!

is double to the Ark, so is its