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This least Sine being thus found out, the Sine of one of the Fig. VI. first Scruples, that is, of part of the whole Quadrant, or of one hundredth part of a Degree, that is, of part of the whole Quadrant is to be found. Therefore by the second Theorem, as 4 is to 8192, so the quantity of the first Part of this Divifion is to the quantity of the first Part of that Divifion, and by the first Theorem; so is the Sine of Part, which you have in

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the Table at the Sine of {part of Degree.

Therefore the Sine of the first Minute, or of the first hundredth part thus form'd by Prob. 1. extract the Sine of the Comple

ment, to wit, of the Ark 89 Deg. {

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then by Prob. 3. find

২ out the Sine of the second Minute, and its Co-fine by Prob. 1. and from thence you will find the Sine of the Sum of 2 Min. and 1 Min. that is 3 Min. by Prob. 3. and its Co-fine by Prob. 1. and from the Sine and Co-fine 2 Min. or from the Sines and Co-fines 3 Min. and 1 Min. you may look for the 4th Sine by Prob. 3. and the Sine of the Complement by Prob. 1. Likewife from the Sines and Co-fines of 2 Min. and 3 Min. or 4 Min. and 1 Min. you may find the 5th Sine and Co-fine by Prob. 3. and 1. &c. even

to {

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or 1 Deg. and from the Sine of 1 Deg. you may by the fame means find all the Sines of the 90 whole Deg. and from the Sines and Co-fines, before found, of 60 fingle Minutes, it will be eafie by 3d Prob. taking the 4th Prob. too, when it will be useful, to extract the single Sines of all the single Minutes interspers'd.

A Deduction of the Tangents and Secants from the Tables of Sines

The Tangents are made thus:

As A C, the Co-fine, is to CB the Sine; fo is A E, the Ra- Fig. VII.

dius, to ED the Tangent.

But the Secants thus:

As AC, the Co-fine, is to A B the Radius; so is A E, the Ra

dius, to AD the Secant.

By this way the whole Canons of Tangents and Secants are extracted from the Canon of Sines.

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Fig. VII. I pass by all Rules of Calculation, for I don't undertake to make new Canons; forasmuch as some of the most excellent Artifts, by their Study and Industry have fav'd me that Trouble, it fufficiently answers my Design, if the Reason of Syntax, whatever it be, be only understood, and the Truth of the Numbers put in the Canon which the Propofitions above abundantly prove.

CHAP. IV.

Of the Affections of Right-lined Triangles, in order to
the Calculating (or Resolving) of them.

I.

Plain Triangle is contained under Three Right Lines,

A and secret Right-a

or Oblique.

2. In all Plain Triangles, Two Angles being given, the third is
also given; and One Angle being given, the Sum of the other
Two is given; because, The Three Angles together are equal
to Two Right Angles. By the 32 Pro. Euclide. Lib. I. and by the
11th of Sect. I. hereof. Therefore,

In a Plain Right-angled Triangle, One of the Acute Angles is the
Complement of the other to 90 Deg.

3. In the Resolution of Plain Triangles, the Angles only being
given, the Sides cannot be found; but only the (Reason, or) Pro-
portion of them: It is therefore neceffary that one of the Sides
be known.

4. In a Right-angled Triangle, Two Terms (besides the Right Angle) will serve to find the third; so that One of the Terms be a Side.

5. In Oblique-angled Plain Triangles there must be Three Things given to find a fourth.

6. In Right-angled Plain Triangles there are Seven Cafes, and Five in Oblique: For the Solution of which, the Four following AXIOMS are fufficient.

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1

In a Right-angled Plain Triangle: The Rectangle made of Radius,
and one of the Sides containing the Right Angle, is equal to
the Rectangle, made of the other containing Side, and the Tan-
gent of the Ange thereunto adjacent.

DEMONSTRATION.

In the Right-angled Plain Triangle B E D, draw the Arch FE; Fig. VIII.

then is BE Radius, and D the Tangent of the Angle at B: Make
CA parallel to DE, then are the Triangles ABC and BDE
like, because of their Right Angles at A and E, and their common
Angle at B. Therefore,

As BA: BE::AC:E D.

And, BA in ED is equal to BE in AC.
That is, BA in 1 B is equal to Radius in A C.

Which which was to be demonftrated.

ΑΧΙΟME II.

In all Plain Triangles: The Sides are proportional to the Sines of
their opposite Angles.

DEMONSTRATION.

In the Plain Oblique Triangle C B D, extend BC to F, making Fig. IX.

B F equal to D C, and describe the Arches F G and CH; then
are the Perpendiculars F E and C A, the Sines of the Angles at
D and B; and the Triangles BEF and BAC are like, because
of their Right Angles at E and A, and their common Angle at
B. Therefore,

As BC:CA::BF:FE.

That is,

As BC:SD: : DC (equal to B F): s B.

Which was to be demonstrated.

Fig. X.

AXIOME III.

In all Plain Triangles: As the balf Sum of the Sides, is to their half Difference, so is the Tangent of the half Sum of their oppoSite Angles, to the Tangent of their half Difference.

Otherwise,

In every Plain Triangle: As the Sum of the Two Sides, is to their
Difference; so is the Tangent of the half Sum of the Opposite
Angles, to the Tangent of half their Difference.

DEMONSTRAΤΙΟΝ.

In the Oblique angular Triangle ABC, let the known Sides be B A and BC; and the Angle ABC, comprehended by them: Where it is Obtuse in the superior, but Acute in the inferior, Diagram.

Continue the Side A B to H; so that H B may be equal to BC, and join C and H, - and make BI equal to A B: Also, from the Points B and I, draw the Right Lines BD and IG, parallel to the Side A C.

Then shall the exterior Angle CBH be equal to the Two interior and opposite Angles (by the 32th of the ist Euclid.) For the Angle CBD is equal to the Angle ACB; and the Angle DBH, to the Angle C A B.

(Moreover, from the Point B, let fall a. Perpendicular BE, which shall biset C H at the Point E; then making BE the Radius, upon B, describe the Arch MEL. Therefore shall CE be the Tangent of half the Sum of the oppofite Angles: And DE (to which FE is equal) the Tangent of half their Difference.

Now, because AC, BD, and IG, are Parallel, therefore shall CD, DG, FH, be Equal: As also, D F and GH: Therefore I say,

By Equality of Proportion:

As A H, the Sum of the Two Sides,
Is to I H, their Difference;

So is C E, the Tangent of half the oppofite Angles,

To DE, the Tangent of half their Difference.

Otherwise, Otherwise,

As the Sum of the Two Sides,

Is to the greater Side doubled;

So is the Tangent of half the Sum of the opposite Angles,

To the Sum of the Tangents of the half Sum, and half Diffe

rence of the Angles.

Otherwise,

As the Sum of the Two Sides,

Is to the leffer Side doubled;

So is the Tangent of half the Sum of the oppofite Angles,

To the Difference of the Tangents of the half Sum, and the half Difference of the Angles.

ΑΧΙΟME IV.

In all Plain Triangles: As the Base, is to the Sum of the other
Sides; so is the Difference of those Sides, to the Difference of the
Segments of the Bafe.

DEMONSTRATION.

Fig. VIII.

In the Triangle BCD, let fall the Perpendicular CA; extend Fig. XI. BC to F, and draw FG and DH.

Then is B F equal to the Sum of the Sides CD and CB; and HB equal to the Difference of those Sides; and GB is equal to the Difference between A B and AD, the Segments of D B, the Base: And the Triangles HDB and BGF are like; because of their equal Angles at D and F; the Arch H G being the double Measure of them both: And their common Angle at B.

Therefore,

As BD:BF::HB:G B.

That_is,

As DB: FB (the Sum of the Sides) :: HB (the Difference of the Sides C Band CD) : GB (the Difference of the Segments of the Base.)