45 6, 9, 11 12 = t 4AA 45A aa aa 12, That is 13 aa = $s. 15 ssa - aaa = 45A The Value of a, in this Æquation, may be found as in the third Example, Page 241; that is, by making r te=a, &c. it will be found that a = 60., PROBLEM XVIII. There is an Oblique-angled plain Triangle, wherein a Perpendicu lar is let fall from the Vertical Angle upon the Bafe; the Sum of each Segment of the Base, when added to its adjacent or next Side, and the Area of the Triangle, are given, to find the Perpendicular and each Side. { 5+5== 800 } Quære y, b, e, and a Let 2 letus600 3 A the Area = 141750 2A 5X2-a 6 y te 2. 쯤 Per Fig. {1 321 aa= ble 2 2 у 8 ee + aa = uu I у 9 b =% II 9 bb = zz — 2zy + yy Іо 12UU = SS-2se ee 7, 11 13/27 – 2zy = aa 8, 12 14 SS — 25e = aa 13 + 15 zz — aa = 2zy 14 E 16 SS aa - 2se Having found the Value of a ZZ-aa 15 • 27 17 =y from the 24th Step, e and y will 22 be easily found by these two Steps, and b, 4, by the gth and 16 - 25 18 Toth Step. 25 SS аа + =yte 25 6, 19 SS aa 2A 6, 19 20 + 22 25 ZSS-zaa 4za 20 X 2221 zz - aa + 42A 21 Х S22 saa + zss Tag: 22 X a 23 zzsa - saaa + zssa — Zaaa= = 42 As 23, Num. 241900000a aaa = 243000000 Here a 300 found as in the last Problem. ZZS PROBLEM XIX. There is a Right-angled Triangle, wherein a Right-line is drawn parallel to the Cathetus; there is given the Cathetus, tbat Segment of the Hypothenuse next to the Cathetus, and the alternate Segment of the Base; thence to find the Base, &c. viz. · Let! 11b = 20 •Ć - 24 . and b=15 Then 216+a=the Base. Quære a h e: ca 3 a 5 16ta 예 ccaa 5 Qaee 4 6, 71bb 718 8 x 9 + bb + 2ba+aa 9 ccaa=hhbb -bbaa + 2hhba- 2ba3 + bhaa - ar 10 a++2baaa toccaa tóbaahhaa-2hhba = hhbb 11 aaaa+40aaa+7510a-goooa=90000 That is, For a Solution of this Æquation, let it be made aaaa+baaa + caa da =G € = 751 . Putr teza d=9000 G r+ 4rrre torree at G = 90000 Then =G=90000 « = 8,63 te=e 20 45 &c. + 10000 + 40008 + 600ee Then + 40000 + 12000€ + 12000e 90000 90002 And te = 2,1 1. Divisor = 10 1,52 First re to 2. Divisor = 10,7 1,07 ti=2,1 rte = 12,1 =rfor a second Operation, which being involv’d, and multiply'd into the Coefficients, as before, will produce these Numbers: + 21435,8881 +.7086,248 + 878,46ee- = C. 108900,0000 – 9000,000 Viz. 93352,2381 +- 33829,646 + 3081,46ee =90000 Here, because 93352,2381 > 90000 therefore 12,1 > a, and therefore it must be made rea, which will produce the farne Numbers, only all the second Signs muft be changed. Thus, 93352,2381 — 33829,64€ 3081,46 = 90000 from whence will arise this Æquation : + 33829,646 — 3081,46ee = 3352,2381 Consequently, 10,97844 -e = 1,08787332 = D Operation 10,9784) 1,08787332 (0,0999 = 1 -=,0999 9792 1. Divifor 10,88 108673 Last * = 12,1 2. Divifor 10,879 97911 1= 0,0999 3. Divifor 10,8785 1076232 ré= 12,00018 979065 &c. PROBLEM XX. In the Oblique-angled Triangle CAD, there is given the Side AD, and the Sum of the Sides AC+CD; also within the Triangle is given the Livre A B perpendicular to the Siuc CA; thence to Let find the Side C A, &c. C sa аа D Suppose the Line DF parallel to AB; C A being produced to F But 7 BC= bb + aa. Let AF =e, and FD=y 6, 7 8 v bb + aa :a::s-a:ate 8 9 =ate V:bb + aa 10 ss2sa taa=OCD Per Fig. 11 ss.-2sa taa=aa + 2aetety=0CF+OFD aa 12 ss—2sa=2aetee tyy But 13 dd=etyy=AFT OFD 12 13 14 s$_2så - dd=2ae Let 15 2x=55-diel 14, 15 16 -sa=al 16 a 17 II X sa X 9 =oate sa taa 17 + a 18 =ate bb traa xx - 2xsat2xaatssaa--2sait.at 18 ? =cate 2 saaatat bb + aa 19, 20 21 2x5a + 2xaa + ssaa — 2sa3 + at 20 aa ssaa aa This Æquation being brought out of the Fractions, and into Numbers, will become — 2018a+ + 125409a} – 2464230,250+ 354683072 = 274183922,25; which being divided by 2018, the Co-eficient of the highest Power of a, will be - 04 +62,145 6a— 1221,125a? + 17575,9697a = 135869,138 375 &c. And And from hence the Value of a may be found, as in the last Problem, due Regard being had to the Signs of every Term. This Work of reducing, or preparing Æquation, for a Solution by Division, hath always been taught both by ancient and modern Writers of Algebra, as a Work fo neceflary to be done, that they do not so much as give a Hint at the Solution of ariy adfceled Equas tion without it. Now it very often happens, that, in dividing all the Terms of an Æquation, some of their Quotients will not only run into a long Series, but also into imperfect Fractions (as in this Æquation above) which renders the solution both tedious and imperfect. To remedy that Imperfection, I shall here thew how this Æquation (and consequently any other) may be refolu'd without such Division, or Reduction. Let b=2018. (= c = 125409. d = 2464230,25 f = 35468307. And G = 274183922,25 Then the precedent Æquation will stand thus : - baaaa + caaa das + fa =G Putrteña as before. brt 4brrre — Obrrie = bat Then will + cri + 3crré + 3créé =+ cal daa This is plain and easily conceived. The next Thing will be, how to estimate the first Value of r; and, to perform that, let G be divided by b, only so far as to determine how many Places of whole Numbers there will be in the Quotient ; consequently, how many Points there must be (according to the Height of the Æquation.) Thus b = 2018) G = 274183922,25(130000 2018 7238, &c. Now from hence one may as easily guess at the Value of r, as if all the Terms had been divided. That is, I supposer = 10; which being involved, &c. as the Letters above direct, will be |