There is a Right-angled Triangle, wherein a Right-line is drawn parallel to the Cathetus; there is given the Cathetus, that Segment of the Hypothenufe next to the Cathetus, and the alternate Segment of the Base; thence to find the Base, &c. viz. Let 1 b =20.24 . and h = =15 2bathe Bafe. Quære a Then Here And 9 8 X 2+ That is, ccaa bb + 2ba + aa - аа - ее bb 2ba+aa = bb -aa 2ba3+hhaa―at ccaa-hhbb-bbaa +2hhba-2ba3+ hhaa—a+ 10 a++2baaa+ccaa +bbaa-bhaa-2hhba=hhbb II aaaa+40aaa+751aa-9000a=90000 For a Solution of this Equation, let it be made aaaa+baaa+caa — da G Putrea Then . G = 90000 +4rrre+6rree Then 10000 + 4000e + 600ee. +40000 + 12000e+1200ee 7510015020e + 751ee =G=90000 That is, 35100 + 22020e + 2551ee = 90000 45 &c. Fittr 10 +e=2,1 r+e = 12,1=r for a fecond Operation, which being involv'd, and multiply'd into the Coeffici ents, as before, will produce thefe Numbers: +21435,8881 + 7086,24e+ 878,46ee 9000,00€ C. Viz. 93352,2381 + 33829,64 +3081,46ee90000 90000 therefore 12,1 > a, and therefore it must be made rea, which will produce the fame Numbers, only all the fecond Signs must be changed. Thus, 93352,2381 - 33829,64 + 3081,46ec90000 from whence will arife this Equation: +33829,64e3081,46ee3352,2381 Confequently, 10,9784eee 1,08787332 = D In the Oblique-angled Triangle CAD, there is given the Side AD, and the Sum of the Sides AC+ CD; alfo within the Triangle is given the Line AB perpendicular to the Side CA; thence to find the Side CA, &c. Let Suppofe the Line D F Then B parallel to AB; CA being produced to F And 6 BC: CA: DC: CF √ bb + aaa : : s Let AF-e, and FD=y 8 7 a: a te 5 ΙΟ OCD II 12 Per Fig. 11 ss 2sa+aaaa+2ae+ee+yy=□CF+□FD 14, D This Equation being brought out of the Fractions, and into Numbers, will become 2018a+ + 125409a3-2464230,25a2+ 354683074274183922,25; which being divided by 2018, the Co-efficient of the higheft Power of a, will be at +62,145 5a3 - 1221,125a2 + 17575,9697a=135869,138375 &c. And And from hence the Value of a may be found, as in the last Pro blem, due Regard being had to the Signs of every Term This Work of reducing, or preparing Equation, for a Solution by Divifion, hath always been taught both by ancient and modern Writers of Algebra, as a Work fo neceffary to be done, that they do not fo much as give a Hint at the Solution of any adfected Equation without it. Now it very often happens, that, in dividing all the Terms of an Equation, fome of their Quotients will not only run into a long Series, but alfo into imperfect Fractions (as in this Equation above) which renders the Solution both tedious and imperfect. To remedy that Imperfection, I fhall here fhew how this Equation (and confequently any other) may be refolv'd without fuch Divifion, or Reduction. Let b 2018. c = 125409. d 2464230,25 = 35468307. And G274183922,25 Then the precedent Equation will stand thus: baaaa+caaa daa+fa = G Put rea as before. This is plain and eafily conceived. The next Thing will be, how to eftimate the firft Value of r; and, to perform that, let G be divided by b, only fo far as to determine how many Places of whole Numbers there will be in the Quotient; confequently, how many Points there must be (according to the Height of the Equation.) Thus b 2018) G274183922,25(130000 2018 Now from hence one may as eafily guefs at the Value of r, as if all the Terms had been divided. That is, I fuppofer 10, = which being involved, &c. as the Letters above direct, will be Viz. 213489045 +15734402e + 87239,75ee=2741839 &c. fecond Operation, with which you may proceed, as in the laft Problem, and fo on to a third Operation, if Occafion require fuch Exactness. But this may be fufficient to fhew the Method of refolving any adfected Equation, without reducing it; which is not only very exact, but alfo very ready in Practice, as will fully appear in the laft Chapter of this Part, concerning the Periphery and Area of the Circle, &c. wherein you will find a farther Improvement in the Numerical Solution of High Equations than hath hitherto been publifh'd. CHAP. V Practical Problems, and Kules for finding the Superficial Contents, or Area's of Right-lin'd Figures. BE Efore I proceed to the following Problems, it may be conve nient to acquaint the Learner, that the Superficies or Area of any Figure, whether it be Right-lin'd or Circular, is compofed or made up of Squares, either greater or lefs, according to the different Measures by which the Dimenfions of the Figure are taken or mea fur'd. That is, if the Dimenfions are taken in Inches, the Area will be compofed of fquare Inches; if the Dimensions are taken in Feet, the Area will be compofed of Square Feet; if in Yards, the Area will be fquare Yards; and if the Dimenfions are taken by Poles or Perches, (as in Surveying of Land, &c.) then the Area will be fquare Perches, &c. Thefe Things being understood, and the |