Fig. IV. Fig. V. Fig. VI. CHA P. III. Some Short Problems, to make Canons of Sines, Tangents PROB. I. The Sine of an Ark being given, how to find out the Sine of the BC being given, to find A C. Ecause the Triangle A CB is a Rectangle (by the Definition of a Sine) and the Sides AC, B C, are of the fame Power as the Hypotenuse; that is, the Radius A B: Therefore, if the Square of the Sine B C be fubftracted 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 likewife the Sine of the R Q and A Q being given, to find B. O or RO. As A B is to BO, fo is :: BO: to B G; therefore B O will be the Square Side of the Plain from the Radius A B and B G, the half Verfed Sine. For QB, the Verfed Sine of the Ark BR, is given; becaufe 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- RQ, QA, and S T, T A, being given, to find S P. BT and CP together make SP, the Sine of the Sum of Two PROB. H PROB. IV. The fame being given, to find the Sine of the Difference. RQ, QA, and S P, P A, are given, to find S T. As A Q is to QR, fo is AP to P O, from whence you may find OS. As AR is to A Q, fo is OS to ST. To these join the Theorems. Theorem I. The leaft Sines are, almoft, in the fame Ratio as their This Theorem will be prov'd to be true hereafter in the' continued Bifections; but the leaft Arks are about one of the first Scruples, or lefs, and are almoft in the fame Ratio as their Sines, because they are, almoft, contiguous amongst themselves, and almoft of the fame 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 firft Section, is to the Num ber of the fecond, (reciprocally) as one Part of the Second Section is to one Part of the firft Section. Divide the fame Line, firft into Four, then into Three Parts; then it will be as 4 to 3, fo Part to reciprocally; the Reafon is, because 3 in makes 1, and 4 in makes 1; and because the Products are equal, the Multiplicands will be reciprocally proportionable. The Conftruction 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 1ooooo 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. F 2 Then Fig. VI. Fig. VI. Then bifect the Quadrant, and look for the Sine of the Bifegment by Prob. 2. and its Co-fine by Prob. 1. then bifect this Bifegment again, and look for the Sine of the fecond Bifegment by Prob. 2. and the Co-fine by Prob. 1. then bifect this fecond Bifeg ment, and look for its Sine or Co-fine by Prob. 2. and 1. and then bifect the third Bifegment, &c. and continue bifecting 13 times, till you find a Sine of Part of the whole Quadrant, as 'tis here fet in the Table. Now we come to the leaft Arks, where the Truth of the firft Theorem is illuftrated: For as the Ark of the Quadrant is double to the Ark, so is its Sine almoft to that Sine. T I This leaft Sine being thus found out, the Sine of one of the Fig. VI. firft 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 $400 is to 8192, fo the quantity of the firft Part of this Divifion is to the quantity of the firft Part of that Divifion, and by the first Theorem; fo is the Sine of Part, which you have in the Table at the Sine of {part of Degree. 8192 Therefore the Sine of the firft Minute, part thus form'd by Prob. 1. extract the ment, to wit, of the Ark 89 Deg. { ៖ or of the first hundredth then by Prob. 3. find c. even out the Sine of the fecond Minute, and its Co-fine by Prob. 1. 100 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 A C, the Co-fine, is to A B the Radius; fo is A E, the Radius, to AD the Secant. By this way the whole Canons of Tangents and Secants are extracted from the Canon of Sines: I pafs Fig. VII. I pafs by all Rules of Calculation, for I don't undertake to make new Canons; forafmuch as fome of the most excellent Artifts, by their Study and Industry have fav'd me that Trouble, it fufficiently afwers my Defign, if the Reafon 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 Refolving) of them. I. A Plain Triangle is contained under Three Right Lines, and is either Right-angled, or Oblique. 2. In all Plain Triangles, Two Angles being given, the third is alfo given; and One Angle being given, the Sum of the other Two is given; becaufe, 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 3. In the Refolution of Plain Triangles, the Angles only being given, the Sides cannot be found; but only the (Reafon, or) Proportion of them: It is therefore neceffary that one of the Sides be known. 4. In a Right-angled Triangle, Two Terms (befides the Right Angle) will ferve to find the third; fo 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. |