GIFT OF 
 
 Miss e.t.x 
 
 University of California • Berkeley 
 
SELECT 
 
 EXERCISES 
 
 FOR 
 YOUNG PROFICIENTS 
 
 IN THE 
 
 MATHEMATICS. 
 
 C ONTAINING 
 
 I. A great variety of Al- 
 gebraical Problems with 
 their Solutions. 
 
 II. A choice Number 
 of Geometrical Problems 
 with their Solutions both 
 Algebraical and Geome- 
 trical. 
 
 III. The Theory of 
 
 Gunnery, independent of 
 the Conic Seftions. 
 
 IV. A new and very 
 comprehenfive Method for 
 finding the Roots of Equa- 
 tions in Numbers. 
 
 V. A fliort account of 
 the Nature and firft Prin- 
 ciples of Fluxions. 
 
 By THOMAS SIMPSON, F. R. S. 
 
 A NEW EDITION. 
 
 CAREFULLY REVISED AND CORRECTED 
 
 By J. H. HEARDING. 
 
 LONDON : 
 PRINTED FOR F. WINGRAVE, IN THE STRAND, 
 
 iSio. 
 
Printed by T. C HANSABW, 
 Petcrbor* Court, 
 rifW-streat, 
 LXJJ.'DON. 
 
TO 
 
 JOHN BACON, 
 
 Of Newtoncap, Efq. F. R. S. 
 
 Sir, 
 
 When. Gentlemen of your 
 Station and Figure become the 
 Patrons of Science it is a Benefit 
 to the Public : their Expeftations 
 of farther Improvements having 
 then the beft Foundation. And 
 All who have the Pleafure of your 
 Acquaintance, and know your At- 
 tachment to polite and ufeful 
 Learning, in which a Knowledge 
 of the Mathematicks may be juftly 
 included, will be fenfible of my 
 Happinefs in being thus permitted 
 to addrefs You. 
 
 Believe 
 
Believe me, Sir, whatever may- 
 be the Fate of thefe Sheets, I fliall, 
 at all Times, confider this Ufe of 
 your Name as a lingular Honour 
 to. 
 
 Sir, 
 
 Tour mq/l obedient, and mojl 
 
 Humble Servant^ 
 
 T. Simpson. 
 
 Woolmdi, Mm U 
 
THE 
 
 PREFACE. 
 
 -L HE enfuing Work, or at leaji the greateft part ofii^ 
 was originally compofed for my own uje m the KoYA.h 
 Academy. And it is upon a prefumption that it may 
 aljo be offervice to others, efpecially thofe employed in a 
 like public way of teaching, that it now appears in the 
 World, 
 
 The work itfelf con/ijls effx * diJlinB parts, or traEts; 
 each of which I fhall here give fome account of. 
 
 The frji part contains a number of Algebraical Pro- 
 blcms, with their Solutions ; defgned as proper exercifes 
 for young beginners. In the courft of thefe Problems 
 and Solutions f whereof the greater part will appear to 
 be new) the art of managing equations, and the various 
 methods of fubflitution^ are taught and illuflrated. 
 
 The fecond part comprehends a variety of Geometrical 
 Problems with their Solutions, both hy Algebra and alfo 
 independent of it, from principles purely Geometrical. 
 In this part the learner will find a large field to exercife 
 his indufiry in : he will moreover have the opportunity of 
 comparing the two methods of folution together, and from 
 thence obferving, thatfometimes the one has theadvantage 
 andfometimes the other ; that, in fome cafes they both 
 proceed upon the very fame properties, and in others, 
 upon quite different ones : and it miy be further remark- 
 id from hence (which will be of fome ufe to know) that, 
 
 * The fixth part, relating to the Dodrine of Annuities, 
 is omitted in this Edition, and added to thi Author*s 
 Trcatifc on that subje&. • 
 
PREFACE. 
 
 zvhe7i quantities are given in magnitude only, the Alge^ 
 braic method generally claims the preference^ in point oj 
 eafe and expedition^ at leajl ; whereas the advantage is 
 ahnoji always on thejide of the geometrical effcBion, 
 when the pofitions of points and lines^ and the quantitief 
 of angles are given. 
 
 There isy however, one particular^ or two, in this party 
 that may bethought tofiandin need cf fome apology , 
 
 In the frjl place, the frequent ufe offymbols, common 
 to the Algebraic notation, may, perhaps, be looked upon as 
 repugnant to the rigour and jlriElnefs of Geometry . But 
 it is not the ufc offymbols (which fome, morefcrupulous 
 than difcerning, have condemned) bui the ideas annexed 
 to them, that render th$ conf deration Geometrical, or , 
 Ungeonjetrical. In pure Geometry regard is always had 
 to the abfolute quantity offome one of the three kinds of 
 e^tenfion, abfiraRedly confidered ; and, whatever fymbols 
 are uftd hert, are to be confidered as exprefjive of the 
 quantities tkemfelves, and not asanymeafures, or numeric 
 cal values of them. Thus by hy(^^, taken in a geome- 
 trical finfe, jve have an idea, not of the produB of two 
 numbers (as in the algebraic notation), but of a jeaJ, 
 r^Bangular ,f pace comprehended under two right-lines, 
 ripr^fented by A and B, and two others equal to the?n, 
 
 B V C 
 
 So, likewife, • — ^^ is not to be underflood here in the 
 
 Light of an algebraic fraBion^ but a^ a right line, which 
 is fourth prap or tional lo three other right. Unes,repre^ 
 fentedby A, B^ and Q.-^The/e diflinBions are abfolutely 
 ne^-^ary t^ 4kefe who would have an accurate idea of 
 tkefubjeS, 
 
 Higfecond particular, above hinted at, relates t0 thp 
 
PREFACE. 
 
 quotations ; wherein I have referred to my own Elements 
 of Geometry *, and not to thefe of Euclid, fo univerfally 
 known andestablifhed. Butfor this there wer e two reafons : 
 Firjl, thofe perfons.for whofe infru£iion thef fleets are, 
 in a more particular manner, defigned, are taught the 
 frf principles oj Geoinztryjrom other Elements than thofe 
 of Euclid : and, fecondly, a number of propoftions are 
 ifed here that are only to be met with in modern authors. 
 In the third part the Theory of Gunnery, or the motion 
 of ProjeBiles, is confdered, exclufive of the conic Sec- 
 tions ; and the praBicalfolutions of the fever al cafes 
 depending on the theory (as well thofe where the objeBis 
 elevated or depreffed as where it is Jituate in the plain of 
 the horizon) are given, at large, by plane Trigonometry, 
 The fourth part exhibits a new, and very comprehen- 
 Jivemethodfor extra&ing the roots of algebraical Equa- 
 tions ; whereby the number fought may be determined, to 
 any propojed degree of exaSness, without the trouble of 
 repeating the operation, as in the cnmrtion way, by con- 
 'verging Series* s. 
 
 The fifth part gives fome account of the Nature of 
 Fluxions, together with the Invejligation of thefunda- 
 mental rules; and may be of ufe, not only to beginners, 
 but alfo tofuch, who, though tolerably well vjerfed in the 
 practice and application of Fluxions, have neverthelejs 
 but an imperfed idea of the frfi principles of this dif- 
 ficult branch offcience. 
 
 May, 1752. 
 
 * The references in this prefent Edition, are made to 
 correfpond with the lajl Edition of Mr* Simpfon's Geo^ 
 metry, printed for F. Wingrave. 
 
ADVERTISEMENT. 
 This new Edition of the Select Exercises of Mr. 
 Simpson, a Book which is now introduced into moji of 
 ourfnncial Mathematical Schools, has been carefully 
 revifed, corre&ed, and improved, throughout. The No- 
 tation is made to agree with the Notation that is generally 
 adopted by the firjl Mathematicians of the age, according 
 to the prefent refnedjiate of Analyfis : Thus, the value of 
 X, in the Solution of Quefl, XCVI, is written ^ [\b-\' 
 
 \^{bb^ 4<23)], and that ofy:=z ^ 
 
 infiead of V {bJ^^^bb-^^a^ and ^ ^ b -\- '^ bb-^a\ 
 as they were printed informer Editions. The like changes 
 will be found to have been jnade in allfmilar expreffions 
 throughout the work. A Jew Notes have been added, 
 wherein different methods of Solution^ to fome of the 
 Algebraic ^ejlions^ are pointed out, and Rules are given 
 Jor folving, with great Jacility, by means oj the Logarith- 
 mic Sines and Cofines, thofe Cafes of Cubic Equations that 
 are ufually termed, with refpeSfto Cardan's Rules, ** The 
 '* impoffible Cafes." 
 
 Pall Mall, March 31, 1810. 
 
 J. H. H. 
 
PART I. 
 
 CONTAINING 
 
 A 
 
 SeleB Number of Algebraical Problems, 
 with their Solutions. 
 
 DESIGNED 
 As proper Exerclfes for young Beginners. 
 
 QUESTION I. 
 
 TJ/'HAT Number is ikat, which being doubled and 16 
 added to the Product^ the Sumjhall ^^188 ? 
 
 Let X reprefent the required number ; 
 then 2x will denote the double thereof; 
 and fo 2;v-|-i6z=:i88, by the quejiion. 
 Therefore 2^=188 — 16 = 172, by tranfpofition, 
 
 AndA;:=-^:zz:86, by divifion, 
 
 QUESTION II. 
 
 To find that Number ^ which being added to 56, the Treble 
 of the required Number Jhall be produced, 
 
 Jf a: be put for the number fought, 
 then 3^: will be the treble thereof : 
 and therefore 3jv=i:^-[-56, by the quejiion* 
 Hence q.x z=z 56, by tranfpofition. 
 
 And X s=: — ^zzz 28, by divifion. 
 
 B QUESTION 
 
t Algebraical Problems, 
 
 QUESTION III. 
 
 The Sum of i^^l. teas ralfed (for a certain PurpofeJ by 
 three different Perfons, A,^, and C ; whereof^ ad- 
 vanced igl. more than A ; and C, 20/. more than B: 
 How much did each contribute ? 
 
 Let A" be the number of pounds advanced by A. 
 Then ;c -j- 15 is the number of pounds advanced by B, 
 and X -\- 35 the number of pounds advanced by C. 
 Therefore 3;!^-}- 50=1: 155, by the quejiion. 
 Whence 3;t:i:z:i05, 
 and Ar=z35. 
 
 From which it alio appears that B contributed ^o/. and 
 C 70/. 
 
 QUESTION iV. 
 
 A Gentleman (by Will) left ^50/. to be divided among four 
 Servants, A, B, C, and D ; whereof B was to have 
 tzvice as much as A ; C as much as A and B ; and D 
 as muck as C and B, How much had each f 
 
 Let X be the number of A's pounds 
 Then 2.x is the number of B's pounds, 
 alfo o^x is the number of C's pounds, 
 and ^x the number of D*s pounds : 
 therefore ha: ==550, by quejiion. 
 
 ceo 
 
 And, confequently, ;v=~-=50. 
 
 From which the icft of the fhares are eafily determined. 
 
 QUESTION V. 
 
 It is required to divide the Number g2 intofour fuck Parts ^ 
 that the firjl may exceed the fecend by 10, the third by 
 18, and the fourth by 24. 
 
 Let X be the firft part. 
 .0 
 Then-^ a— 18 J> will be the other parts. 
 
 And 
 
with their Solutions. 
 
 And 4^ — 52 z-z 92, by the queftien. 
 Hence j^x:=zz 144 ; 
 
 and x:=,^^z=L^6. 
 
 QUESTION VI. 
 
 A certain Sum of Money wasjkared among Jive Perfons, 
 A, B, C, D, and E; whereof "B received toL lefs 
 than A ; C 16L more than B ; D 5/. lefs than C ; 
 and E 15/. more than D. Moreover it appeared that 
 E received as much as both A and B. What was thi 
 whole Sumfhared^ andhsw much did each receive ? 
 
 Let X be the fhare of A. 
 Then 
 
 fx — io^ fB, 
 
 \x+ 6 I will be I C, 
 J ;t-4- ^ [that of | D, 
 La: 4- 16 J IE: 
 
 therefore x-^iGzzz.Qx — 10, by the queftion* 
 
 Whence 26z=zx. 
 From which it appears that 26, 16, 32, 27, and 41 
 (pounds) were the refpeftive (hares ; and that tha 
 whole fum was 143/. 
 
 QUESTION VIL 
 
 To find that Number^ whereof the Double increafed by 
 24, fliall as much exceed 80, as the Number itfelf is 
 below 100. 
 
 
 Let X be the required number. 
 hen 2;« -]- 24 — - 80 nz: 100 — x, by the quefiion, 
 "hence zx-\-xzi=:ioo — 24-J-80, 
 that is 3^:^=1: 156 ; 
 
 id therefore ;*?=— =52. 
 
 B a QUESTION 
 
4 Algebraical Problems, 
 
 QUESTION VIII. 
 
 What two Numbers are fhofe, whereof the Difference is 
 7 and the Sum 33. 
 
 Let X be the lefTer number, 
 then X -\~y will be die greater, 
 
 and2.r4-7 = 33- 
 
 Theretore Qx -==, 33 — 7 rz= 26, 
 
 and x:=z — =13= the leffer. 
 2 
 
 Whence a: -|- 7 zz=: 20 =zz the greater. 
 
 QUESTION IX. 
 
 To divide the number 75 into two fuch Parts, that ^ times 
 the greater may exceed 7 times the lejfer by 15. 
 
 If a; be the greater part, 
 
 then 7 « — a; will be the lefTerl , .7 n- 
 
 1 ' / ^v.- J >by the queltion, 
 
 and 3^ = (75—a:JX 7 + 15/ ^ ^ ^ 
 that IS, 3^ =1=525 — 7^ -j- 15 ; 
 
 therefore 10^=11:540, and a: 1=54. 
 
 From whence the lefler part [j^ — x) Is found r=:2i. 
 
 QUESTION X. 
 
 A, after zuinning 10 Guineas of B, had as much Money 
 as B and 6 Guineas more) and betwixt them both they 
 kadforty Guineas, What inoncy had each atfirji ?' 
 
 Let X be the guineas that A began with ; 
 then 40 — X, are the guineas that B began with : 
 therefore, after play, 
 
 A had, ;i: -j- 10 guiqeas; 
 and B, 30 — x guineas. 
 Whence a: -j- 1 o = 30 — x^S (hy the queJHon.J 
 Therefore 2x z=. 26. 
 AndA:= 13. 
 
 QUESTION 
 
with their SoLU-^iovs. g 
 
 QUESTION XI. 
 
 The Sum of gooL was div' Jed ^m'-^ig fourPerfons, Jo 
 that the jlrjl and Second, between them ^ had 2%ol ; 
 the fir II and third 260/; and the firjt and fourth 
 2LoL How viany Pounds had Each ? 
 
 If X be the number of pounv'^s the fiiii had, 
 
 ( 2d "I r28o — X. 
 
 then the -^ 3:1 \ bad < 260 — x. 
 
 (4rhJ V.220 — x. 
 
 The fum of ail which, 
 being 760 — -2 a:, is =500, by the quefiion. 
 
 VV hence x zzz. — = i 30. 
 
 2 
 
 Therefore the four fhares were 130, 1^0, 130 and 9a 
 
 pounds, refpectiveiy. 
 
 QUESTION XIL 
 
 ^Tis propofed to divide 60 into two fuch parts, that the 
 Difference between the Greater and 64, may he equal ts 
 twice the Difference between tht Leffer and 38. 
 
 If X be the greater part ; 7 
 
 then 60 — X will be the leffer: 
 alfo 64 — X will be the firft mentioned difFerence, 
 and 38 — 6o-[-Af, or (^ — 22,) will be the fecond. 
 Therefore 64 — ^ = 2 X [x — 22,) by the quejiion, 
 that is, 64 — xz=: 2x — 44. 
 Whence io8z=3a^, and '^S-zzzx, 
 
 QUESTION XIII. 
 
 After 34 Gallons had been drawn out of one of two, equals 
 Cafks, and ^o Gallons out of the other^ there remained 
 juji twice as much Liquor in the one as in the Qtha\ 
 What did each Cafk contain when full f 
 
 Let X be the number of gallons fought; 
 then X — 34 will be what remained in the firll cafk, 
 
 B 3 and 
 
6 Algebraical Problems, 
 
 and X — 80, what reraained in the fecond, 
 Hence x — 34= 2 X ('^ — 80), by the qiieftion. 
 Or, a: — 34zii:2a: — 160. 
 Therefore 126 = a:. 
 
 QUESTION XIV. 
 
 A Son a/king his Father how old he was? received the 
 Jollowing Anfwer^ Your Age four Years ago^Jays the 
 Father, was only \ ofmine^ at that Time; but now your 
 '^S^ yjjz^ y oj "^i^^* What was the Age of each ? 
 
 Let X be the age of the fon, 
 .then o^x will be that of the father : 
 
 alfo X — 4 will be the age of the Ton four years before 
 the time in queition ; and 3^: — 4 will be the corref- 
 ponding age of the. father: which, by the quejiion, 
 is equal to 4 times x — 4. Hence we have this equation 
 ^x — i6zrr 3;^ — 4. 
 
 Therefore a; z= 12, and '^x-z=.'^^\ which are the two 
 ages required. 
 
 QUESTION XV. 
 
 What number is that, whofe f exceeds its \ Part by 16 ? 
 Let X be the required number ; then, its \ part 
 
 X X 
 
 beincr — » and its ~ part — > 
 3 4 
 
 XX 
 
 we have =16, by thequeflion. 
 
 3 4. , / 
 
 Hence \x — 2>^-=zi^'i, by reduction* 
 
 that is, a: 13=192. 
 
 QUESTION 
 
with their Solutions. 
 
 QUESTION XVL 
 
 In a Mixture of TVine and Cyder ^ one half of the whoii 
 -}- 2^ Gallons was Wine; and ^ Part — 5 GallcTtS 
 Cyder, How many Gallons were there of each ? 
 
 If^be put for the number of gallons in the Whole 
 mixture, the gallons of wine will be expreiled by 
 
 1" 25, and thofe of cyder by 5 : which together 
 
 . o 
 
 being equal to the whole, we 
 
 X X 
 
 therefore have V-^sA ,5=;x; 
 
 2 ' 3 
 
 X X 
 
 or, 20_^— ~— — .. 
 
 Hence 1 20 zzz 6x — 3^ — 2x ; 
 
 or i2oz=:zx. From which it appears that the mixtur#^^ 
 
 confided of 85 gallons of wine, and 35 of cyder. 
 
 QUESTION XVII. 
 
 In a Lottery conjijiing of 100,000 Tickets^ half the 
 Number of Pri%es added to ~ of the Number of Blanks^ 
 was 35,000 How many Prizes were there in the 
 Lottery ? 
 
 If X be the number of prizes ; 
 then 100000 — X, will be the number of blanks. 
 
 . - _ X , 100000 X 
 
 And 10, 1 1^:35000. 
 
 ^ 3 
 
 Hence 2,x -f- 200000 — ix-zzz 2 10000 : 
 and X = 10000= the number fought. 
 
 B 4 QUESTION 
 
B Alcebraical Problems, 
 
 QUESTION XVIII. 
 
 To the Compofdidn of a certain ^antity ef Gunpowder^ 
 i of the whole Weight -|- 6lb, oj Saltpetre was necejfary: 
 ^he Sulphur ufed was f of the whole — c^lh, and the 
 Charcoal \ of the xJbhole — 3/^. How many Pounds of 
 tach of the three Ingredients were there taken? 
 
 Let X be the number of pounds in the whole : then 
 
 + 6, pounds of faltpetre. 
 there ^ 
 
 were -^ 5, pounds of fulphur. 
 
 o 
 
 X 
 
 3, pounds of charcoal. 
 
 • XXX 
 
 And therefore 1 1 2 = a:, by the aiuefiion^ 
 
 2 » 3 • 4 ^ . ^ *^ 
 
 Whence izx-^-^x -\-6x — 48 z=: 24A: : 
 
 And confequently;!^ zz:: — = 24. 
 
 Therefore there were taken i81b. of faltpetre, 31b, of 
 fulphur, and 31b. of charcoal. 
 
 QUESTION XIX. 
 
 A Generaly after having lojl a Battle^ found that he had 
 only \ of his Army --j- 3600 lefty fit for Action : \ of 
 his Men '\- 600 being wounded y and the Refl^ which 
 were \ of the whole Army, either flo in, taken Prif oners, 
 or mijing. What was the whole Number of the Army f 
 
 The number fought being denoted by ;t, the num- 
 
 X 
 
 ber of men left unhurt will be 1-3600. 
 
 • X 
 
 And the number of the wounded f- 6co. 
 
 X . • 
 
 To which adding, — i the number of the flaln and 
 
 S 
 prifoners, wc have the number of the whole army ; 
 
 or 
 
with their Solutions. rj 
 
 or ~.4.36oo + -|--f-6oo+y— Af. 
 
 Hence 4=00 (^,v— -.___-)=^— _. 
 
 Therefore 168000 (i=: i^x — ^x) z=z yx : 
 and 24000=3: a:. 
 
 QUESTION XX. 
 
 A Prize of 2,000/. was divided between two Per/ons, A 
 and B ; lohofe Shares therein zoere in proportion as 7 to 
 9. IVhat Was thi Share of Each ? 
 
 Let X denote the (hare of A ; 
 then 2000 — jv, will be that of B. 
 But X : «ooo — X i 17 : 9. by the quejiion. 
 From whence (as the rectangle of the tv o extremes, of 
 any four propoitional numbers, is equal to the rectangle 
 of the two means *) we get this equadon, 
 
 9X.^ = (2ooo— .^JX7; 
 that is, 9;^= 14000 — jx. 
 
 Hence 16:^=14000, 
 
 and x =3 875. 
 
 Therefore the fhare of A was 875/. and that of B,' 
 
 1125/. 
 
 QUESTION XXI. 
 
 To divide /^^into two fuck Parts, that the Greater in^ 
 creafed by 5, may be to the Ltffer increafed by 7, as 4 
 is to 3. 
 
 If X be the greater part, 
 
 44 — x^ will be the leffer, 
 
 and ^ + 5:51 — X : ; /^ i ^, by the gueflion. 
 
 Therefore (by multiplying extremes and means) we have 
 
 3^4-i5:=204~4^: 
 
 TxrL 204 — 15 189 
 Whence xz=z — ^z= — ^ = 27. . 
 
 7 7 
 
 * See Prop, xvi, p. 164, 14th Edit, of Dr. R. Simfon's 
 £uclid. 
 
 QUESTION 
 
ro Algebraical Problems, 
 
 QUESTION XXII. 
 
 Tojindtwo Numbers in the proportion of \ to 2',fo that, 
 12 being added to Each^ the Sumsjhall be in proportion 
 as 5 is to J. 
 
 Let X be the leffer number ; 
 
 then 2.r, will be the greater. 
 
 Hence x-\- it: : 2;^-}- 12 : : ^ : y^ by the qv.ejrion. 
 
 Therefore (sa; -f- 1 2) X .5 = ("^ + ^ 2) X 7 J 
 
 that is io;<f 4" ^o = /-^ "t" ^4- 
 From which 3x1:1124 ; andxz=:8. 
 
 So that the two numbers are 8 and 16. 
 
 QUESTION XXIII. 
 
 A, at Play.JlrJi won ^ Guineas oj B, and had then as 
 much Money as B ; but E, upon winning back his own 
 Money and 5 Guineas more, had 5 times as much 
 Money as A. What Money had each atfirji ? 
 
 If X be the number of A's guineas, at firfl ; then it 
 is plain, from the queflion, that j^-f-^^ ^'^^^^ be the 
 number of B's guineas. 
 
 Whence j;4-io + 5=(x — 5)X5 = 5'^ — 25. 
 that is, 40 = 4;^. 
 
 Therefore x z=. 10= the number of A's guineas ; 
 and a: -J- 10 = 20 = the number of B's guineas. 
 
 QUESTION XXIV. 
 
 A Grocer t with ^6lb, oj fine Tea, at 20 Shillings <t 
 
 Pound, would mix a coarfer Sort, of \\ Shillings, 
 
 Jo as to afford the whole, together, at 18 Shillings, 
 
 per Pound, What ^antity of the latter Sort mujt he 
 
 take? 
 
 Let X be the number of pounds required ; 
 then 1 120 is the value of the fineft fort, 
 and i4Jf that of the coarfeft. 
 
 Moreover, 
 
^ith thdr Solutions. ii 
 
 Moreover, the number of pounds of both forts together 
 
 being ^6-f-'^. it is evident 
 
 that (56-|-a:)XiH, or ioo8-fi8^, is the value of the 
 
 whole mixture. And therefore 
 
 11 204- 14^:1= 10084-18;^. 
 
 Whence ii2rr4x. 
 
 And confequently 28=:.v. 
 
 QUESTION XXV. 
 
 A Farmer would mix Wheat, at 4 Shillings a Bujhel, with 
 Rye, at 2j. (id. a BuJJiel ; fo that the whole Mixture 
 may conjijl of 90 Bujhels, and be afforded at ^s. 2d. 
 a BuJJieL It is required to find how many Bujhels of 
 each Sort mufi be taken. 
 
 If the number of bufhcls of wheat be x, 
 
 Thofe of rye will be 90 — ;ic. 
 
 Moreover, the value of the wheat will be ^Zx, pence, 
 
 and the value of the rye fc,o — ;c)X30, pence. 
 
 Whence 48,\'-f (90— ;v)x 30=90 XS^ {by the queflion.) 
 
 that is, .48X-I-2700 — 30^^=13420 : 
 
 therefore 1 8 J\;r= 720 : 
 
 720 
 and confequently xiz. — —zi^^o. 
 
 QUESTION XXVI. 
 
 A Workman was hired for 40 Days, at 31. 4^. per Bay^ 
 
 for every Day he toorked ; but with this Condition, that 
 
 fer every Day he played, he was to forfeit ts. /^d. : and 
 
 it fo happened, that, upon the whole, he had 3/. 3J. 4^, 
 
 to receive. The quejiion is, to find how many Days of 
 
 the 40 he worked^ and how many he. played. 
 
 Let X be the number of days he worked : 
 then 40 — X will be the number of days he played. 
 Moreover, as he was to receive 40 pence, for every 
 day he worked, and to forfeit i6 pence for every day 
 he played, 
 
 we 
 
12 Algebuaical Problems, 
 
 we have 40A:=zthe number of pence earned by work, 
 and (40 — jv) X 16 z=z the number o\ pence forfeited by play : 
 whence 40;*? — (40 — x) X i6z=:76o, hy the que/iiun ; 
 that is, 40^1: — 640 -[- i6x zziz 760 ; 
 theretore ^6x = 1 400 ; 
 and confequentiy x z= 25. 
 
 From which it is evident that he worked 25 days, and 
 played 15, 
 
 QUESTION XXVII. 
 
 A Bill of 120I. was paid in Guineas and Moidores, and 
 'the ISumbir of Pieces ufed of both Sorts was jufi 100. 
 Mow many wire there of each ? 
 
 If ^ be put for the number of guineas ; 
 
 then too — x will be the number of moidores. 
 
 And fo,2i;t4-(ioo — at) V27 zn: 120X20 1 , ^, n- 
 
 ,*^ y/^ / — ^^ yby theque lion* 
 
 or, 2i.r-|-27oo — 27;czz:2400 / -^ ^ -^ 
 
 Hence 300 = 27;^: — 2 1;^ = 6a:. 
 And confequently x .'' ' — ^50. 
 
 QUESTION XXVIII. 
 
 One bought 30 Pounds of Sugar, of two different Sorts, 
 and paid for the whole 19 Shillings ; the beji Sort ccfl 
 lod per Pound, and the worji yd. How 77idny Pounds 
 were there of each? 
 
 Let X ftand for the number of pounds of the heft fort, 
 and then 30 — x will express the pounds of the other fort. 
 
 Therefore a:X io-j-( 30— x)X7= 19X^2> ^^ ^^^ ^"C/^^^"' 
 
 that is, iQX'\-2io — 7a; = 228. 
 
 Whence 3;c=:228 — 210= 18, 
 
 and a: = 6. Therefore there were 6 pounds of the befl 
 
 foi't, and 24 of the worft. 
 
 QUESTION 
 
with their Solutions. 13 
 
 QUESTION XXIX. 
 
 A Lady gavs a Guinea in Charity^ among a Number of 
 Poor^ conjijiing of Men^ Women and Children : each 
 Alan had i2d. each Woman 6d. and each Child '^d. 
 Moreover there were twice as many Womm as Men, 
 wanting 2; and 3 times as many Children as Women, 
 wanting^. How many Perjons were there relieved"^ 
 
 Let X be the number of men ; 
 then 2x-^2, will be the number of women, 
 and 6x — 10, the number of children. 
 Hence a:X 12 -\-[2x — 2)y^6'\~[6x — io)X3 = 2iX^2; 
 that is, i^x-\-\2x — 1 2 -|- 1 8^ — go •=. 252, 
 or, 42^:^=294. 
 Whence x' — y. 
 
 Therefore there were 7 men, 12 women, and 32 
 children ,* in all 51 perfons. 
 
 QUESTION XXX. 
 
 To find that Number^ which being divided^ either inta 
 three or four equal Parts, the continual Product of all 
 the Parts, in both Cafes, fhall be exa&ly the fame. 
 
 Let X be the required number ; 
 fo fhall the continual product of the three equal part* 
 
 be — X ""^ X — =— ; and that of the four equal 
 3 3 3 -27 ^ 
 
 X X X X x'^ 
 
 parts —x—X— X-7=' 
 
 4 444 256 
 
 x' 
 
 Whence — -z=: — 1 by the que/Hon, 
 256 %y ^ ^ -" 
 
 Therefore 2ixz=:i 256 ; and x =:q — , 
 
 ^27 
 
 QUESTION 
 
14 Algebraical Problems, 
 
 QUESTION XXXI. 
 
 Tojind two Numbers, in the Proportion of o^ to 4, whojt 
 Sum is CO the Sum of their Squares, as 7 to 50. 
 
 Let 2,x denote the lefler number : 
 then ^x will express the greater. 
 And we fhall have 2>x-\-^x : g^t^-f- i6x" : : 7 : 50, 
 or, yx : 2^jf^ : : 7 : ,50, by the qucfiion. 
 
 Therefore 25^:^ X 7 = 7*^ X 5°» 
 or, i^x''-:=z^Qx\ 
 
 whence A-nz: — ^=1:2. So that 6 and 8 are tHi'two num- 
 
 bers that anfwer the queftion. 
 
 QUESTION XXXII. 
 
 To find two Numbers in the Proportion of ^ to 'j \ fo thai 
 the Square of their Sum^ and the Cuke of their Differ- 
 ence, fhall be equal. 
 If 9;t be put for the greater number ; 
 
 then yx will be the lefler ; 
 
 And fo (16^:) *rr: {^x) ^ by the quejlion, 
 
 that is, 2^6x"-z=i^x^. 
 
 Hence ;\;zi=— —-==32. 
 o 
 
 Therefore 288 and 224, are the two numbers fought. 
 QUESTION XXXIII. 
 
 To find two Numbers whofe Difference is 4 and the Dif 
 Jerence of their Squares 120. 
 Let X be the leffer number * ; 
 then a: -{-4, will be the greater: 
 alfo XX will be the fquare of the lefler, 
 and XX -j- ^- + 16 that of the greater. 
 Whence 8a: -|- i6=z: lao, by the quefion. 
 Therefore 8a ziz: 104 : and ^z= 13. "Cf- 
 
 So that 13 and 17 are the two numbers that were to 
 *be found. 
 
 * Or thus: — 1 20 -f- 4 1:11:30, the fum of the two num- 
 bers ; then {\ of 30) 15 + 2, gives 17 and 13 for the num- 
 bers themselves. — J. H» H. 
 
 QUESTION 
 
 I 
 
with their Solutions. i^ 
 
 QUESTION XXXIV. 
 
 To divide loo into two fuch Parts ^ that the Difference of 
 their Squares may be 1,000. 
 
 If X be the greater part, 
 100 — .a:, will bethe lefTer. 
 Therefore^;t — (10b — a:)^=icoo; 
 that is, x^ — 10000 -j- 200;c — x^ z=: 1000. 
 Whence 200.V =zz 1 1000 ; 
 
 and confequently x zzz • = 55 *. 
 
 200 
 
 QUESTION XXXV. 
 To divide 100 into two Parts, fa that the Square of their 
 Difference may exceed the Square of twice the leffer Part 
 by 2000. 
 
 The lefTcr part being denoted by at, 
 the greater will be expreffed by 100 — x, 
 and the difference by 100 — zx. 
 
 Therefore, by the problem, (loo — 2x) '^ = (sa*) '-^ -|" 2000, 
 that is, 1 c 000 — 40o.r -j- \xx = axx -f- 2000, 
 or, 10000 — 2ooo = 4ooA\ 
 
 8000 , o 
 
 Hence x:=: = 20; and 100 — xzzzoo. 
 
 400 
 
 QUESTION XXXVI. 
 
 A and B make a joint Stock of ^00 1, by which' they gain 
 160L whereof A^ for his fliare^ had 0^9.1, more than 
 B. What did each P erf on bring into Stock ? 
 If X be the number of pounds advanced by A ; 
 
 then it will be, as 500 (the whole ftock) is to 160 (the 
 
 whole gain) fo is ;: (the flock of A) to .,the 
 
 ° ^00 
 
 i6x 
 gain of A: whence the gain of B being i6o 
 
 50 
 
 * Or thus : ' — iooo-r-iooc=: lozzzdiff. o? the two 
 Darts; then 50 +5 = 55 and 45, the parts required. 
 J.H.H. 
 
 we 
 
i6 Algebraical Problemj, 
 
 16^ /- i6x . ■ _ _. 
 
 we have =160 h'^s, by the quejtion, 
 
 50 ' 50 * '^ ^ -^ ^ 
 
 Therefore — — = 192, or — = 6 ; 
 
 and confequendy ^ = 50 X ^ = 300. 
 
 QUESTION XXXVII. 
 
 A Sum of Money was divided between two Perfons^ A 
 and B, Jo that the Share of A was to that oj^^ as 5 
 to 3, and exceeded |- ef the whole Su?n by ^oL What 
 was the Sliare of each Per/on f 
 
 Let c^x exprefs the (hare of A, 
 and 2t^ the Ihare of B ; 
 then 8a: will be the whole fum, 
 
 and -2 — will be — thereof. 
 9 9 
 
 Therefore 5Jf — -^ — =50, by the queflion, 
 
 that IS, — =50, or x nr: 90. 
 
 9 
 Hence 4 ^o/. and 270/. are the two fhares required. 
 
 QUESTION XXXVIII. 
 
 A and B began to play together with equal Sums of Money : 
 A firfl won 20 Guineas, but afterwards loji back the 
 Half of all he then had ; and thereupon had only half 
 as much Money as B. What Money did each begin with ? 
 
 Let X be the number of guineas required : 
 then A, after winning 20 guineas had x-\-^o\ the 
 
 X 
 
 half of which, or U 10, is therefore what he had at 
 
 laft : and this deducted from ^x (the whole fum betwixt 
 
 X 
 
 both) leaves ^x — 10=: what B had at laft. 
 
 Therefore 
 
with their Solution!. 17 
 
 X 
 
 Therefore 2^ 10 =.^4- so; 
 
 2 
 
 whence ^x — zozzz^x-^- 40, 
 
 and consequently x = 60. 
 
 QUESTION XXXIX. 
 
 Ji Gentleman left his whole EJlate among his four Sons ; 
 zvhereoj the Eldeji had | wanting 800/. tne Second | 
 and 120/. over ; the Third had half as much as the 
 Eldeji ; and the Youngeji | of what the Second had. 
 What was the whole EJiate? and how muck had Each*? 
 
 htt X be the whole Eftatc ; then 
 
 The Firft had ^—800. 
 2 
 
 X 
 
 The Second (- 1 so. 
 
 4 
 
 The Third -^ — 400 
 4 
 
 The Fourth —4- 80, 
 12 ' 
 
 The Sum of all which is, confequently,equaI to the Whole, 
 
 -_ X t X ^ X . X 
 
 Or, 4 \ 1— 7- — 1000=;!?. 
 
 2 4 4" 
 
 ♦ This queftion may be readily folved thus : 
 If II (=1) reprefent the whole Eftate, it is plain, that 
 the ift (hare will be -fi^ — 800, the 2nd /_. ^ ^ao, 
 the 3rd i\ — 400, and the 4th y^-j- 80, and their Sum 
 ^1 — 1000 muft be=: 41; therefore -^^ (or-|) is==: 1000 
 and •|-*=6coo, and hence their (hares are found 
 = 2200/. 1620/. 1 100/. and 1080/. refpectively. J. H. H. 
 
 But 
 
i6 Algebraical Problems, 
 
 But it is plain (without Reduction) 
 
 that i 1 z=.x. 
 
 244 
 Hence our Equation becomes 
 
 X X 
 
 -7: 1000= o, or -2r = 1000. 
 
 O" o 
 
 Therefore the whole Eftate was 6000/. whereof the 
 
 eldeft Son had 2200/. the Second 1620/. the Third 
 
 iiDo/. and the Youngeft 1080/. 
 
 QUESTION XL. 
 
 One being ajk'd his Age, reply' d; If ^ of my Years be 
 multiply' d by 3, and^ of them be added to theProdud, 
 the A?nount will be 115. What was his Age ? 
 
 If X be the required Number of Years*, 
 
 ZX X 
 
 then X 3 -j =115. hy the Queftion ; 
 
 ^^ ^ 
 
 . 6x , X 
 
 that IS, 1= 115. 
 
 5 3 
 
 Therefore iSx-^-sx^rziSX 115, 
 
 or 23^= 15 X ^^5 ' 
 ConfequentlyAr=^^-^^^t=i5X 5^=i75' 
 
 / Differently. - ? X 3 + 7 = 4| + tV = l^thi 
 ©f his Age, which amounts to 11^; confequent ly, 
 «3 ; 15 : 113: 7Sy the Age required. 
 
 QUESTION 
 
with their Solutions. 
 
 QUESTION XLI. 
 
 ^9 
 
 A Perfon being ajk'd the Hour of the Day, anfweredthus^ 
 If^ of the Number of Hours remaining till Midnight 
 be multiply* d by 4, the ProduB zuill as much exceed 12 
 Hours, as Half the prefent Hour from Noon is below 
 4, What was the Hour after Noon ? 
 
 Let X be the required Hour * ; 
 then 12 —•;c, will be the Hours till Midnight, 
 
 and '^ — A - X 4 — ^2 =:4 » by the QuefliQn. 
 
 That IS, 12 = 4 , 
 
 2 9 
 
 whence 36 — 3^; — 24 = 8 — x; 
 
 and confequently ;v = 2. 
 
 QUESTION XLII. 
 
 A Market-Woman bought in a certain Number of Eggs, 
 
 at the Rate of ^ for two Pence ; one half of which Jhe. 
 
 fold out again at 2 a Penny, and the remaining Half 
 
 at <^ a Penny ; and cleared 4 Pence, byfo doing : What 
 
 Number of Eggs hadfhe ? 
 
 Let 2x be the Number fought; 
 then, by the Queflion, 
 
 5:2: : 2a: : -^ the Number of Pence the Eggs cofl ; 
 v5 
 
 * Let the hours from Midnight zzzy ; then, fmce 
 — ^ X 4 = — * 2^^ 12 — y = required time, from the 
 
 nature of the queflion -^^^ 12 is =4 — |(i2 — y\ 
 
 or 2^=20; confequently jv:::r= 10, and 12— •^=is2, 
 ihe hour of the day. J. H. H. 
 
 C a But 
 
to Akgebraical Problems, 
 
 But the Number of Pence They were fold for, again, 
 
 X X 
 
 is —J : Therefore we have this Equation, 
 
 2*3 
 
 X ^^ X 4x 
 
 From whence i^^f+icv — 24jv=:12o, or«vz=:i20. 
 
 QUESTION XLIII. 
 
 jf certain Sum of Money ^ put out at Interejly amounts^ in 
 8 Months, to zgyl. 12s. And, in 15 Months its 
 Amoiint (computtdaccording 10 Jimp le Inttreji) is Qo6/, 
 What is that Sum ? And what the Rate oflnterejt*? 
 
 Let X be the Number of Pounds in the required Sum : 
 Then, the Intereft thereof for 8 Months being 297.6 — x, 
 and for 15 Months 306 — x^ we have, 
 as 8 : 15 : : 297.6 — x : 306 — x : 
 Whence, by multiplying Extremes and Means, 
 we get 2448 — Sx:=. 4464 — i^x. 
 Therefore 7^ = 2016, 
 
 and confequently x nr 288/. the Sum required. 
 For the Rate of Intereflr, it will be, 
 as 288/. X ^5 • 100/. X 12 : : 18/. (the Intereft of 288/* 
 for 1,5 Months) to 5/. the required Intereft of lool, for 
 12 Months. 
 
 * Here, go6/. — 296/. 12^. = 8/. Ss. — intereft for 
 7 months, the intereft for 8 months is, therefore, =9/. 
 12J. or 192J. which taken from 297/. 12s, leaves 288/. 
 for the principal. Then, 288/. : 19 2 j. : : 100/. 
 
 : ^^^ , which V~ V -^ = 5/. the rate of Intereft. 
 3 '^ 2 "^ «o ^ 
 
 J.H.H. 
 
 QUESTION 
 
with their Solutions. 21 
 
 QUESTION XLIV. 
 
 A Waterman finds by Experience^ that he can, with th^ 
 Advantage of a common Tide, row Jrom London to 
 Greenwich^ which is five Miles, in 3 Quarters oj 
 an Hour ; and that, to return to London, againfi an 
 equal Tide, though he rows back along-Jhore, where the 
 Stream is only half as ftrong as in the Middle, takes him 
 a full Hour and Half. ' Tis required to find^ from 
 hence, at what Rate, per Hour, the Tide runs in the 
 Middle where it is Jirongeji *, 
 
 In the firft Place, it will be 
 
 *"' 
 
 3 : 4 : : 5 : 6| = Dirt, row'd per Hour wiih the Tide, 
 6 : 4 : : 5 : 3y z=: Dill, row'd per Hour againft Tide. 
 If now the former of thefe two Diftances (6|.) be put=<7, 
 and the latter (3!) = ^; and x be affumed to expreis 
 the required Diftance run, per Hour, by the Stream in the 
 Middle of the River ; then a — x will be the" r^a/ Effect 
 of his Rowing, per Hour, in going from London, the 
 
 Motion of the Tide being deducted ; and ^-j will b« 
 
 the like Effect in his Return : 
 
 And fo, thefe two Quantities being equal to each other 
 
 we have b 4-— = a — x: 
 ' 2 
 
 Whence 2 ^ -j- ^ = 2 fl — 2 a*; 
 
 * The voung Student will, doubtlefs, obferve that this, 
 like almoft all the other queflions, up to the 6^th, may be 
 readily anfwered without the afTiflance of Algebra, al- 
 though, in fome cafes, not fo elegantly as by the Algebraic 
 Analyfis ; a few more Solutions will be pointed out, but 
 it will be well for him to attempt folving queflions, in 
 general, in any way, that an attentive confideration of 
 theirnature may fuggefl, and we venture to predict that 
 he will find himfelf amply rewarded, in the end, for any 
 additional trouble that may be taken by this manner of 
 proceeding. — See Rev, Mr. Lawfons Differtation on th^ 
 Geometrical Analyfis of thi Jntisnts, J. H. H. 
 
 and 
 
i« Algebraical Problems, 
 
 2.CI — — zb "* 
 
 and confequently x = :z=: 2|. 
 
 3 
 
 QUESTION XLV. 
 
 To divide 36 (a) into 3 fuck Parts, that i of the FirJI, 
 4 of the Second, and * oj the Third, may he equal to 
 each other, 
 
 \{x be put for the firft Part, 
 
 X 
 
 then IS — =^ of the fecond Part, (hy the Ouejlion,) 
 
 And fo ~ z= fecond Part. 
 
 2 
 
 Moreover — being -=.1 of the third Part, 
 
 therefore ^(or 2A')= third Part. 
 2 ^ ^ 
 
 Hence jt 4—^ -i- 2.y — /7. 
 
 and 2x -J- 3^ -[- 4jf :;!::: 2^ . 
 
 Confequently x-=. ^^ - - ^ ^ ==: 2 V 4 = 8. 
 
 9 9 
 
 From which the fecond Part \~~\ 
 
 appears to be :z= 12, and the Third (2a:)z=: 16. 
 
 QUESTION XLVI. 
 
 To divide the Number 90 into 4 fuch Parts, that, if 
 the Jirjl he increafed by 5, the fecond diminijlied by 4, 
 the third multiply d hy 3, and the fourth dividedby 2, 
 the Refult, in each Cafe,fhallhe exaSlly the fame. 
 
 Let X be the fourth, or laft. Part : 
 
 Then, three times the third Part being = — 
 
 th« third Part will be— . 
 
 X 
 
 Moreover, the fecond I^art — 4 being, alfo, z= — 
 
 tlie 
 
with their Solutions. t^ 
 
 the fccond Part will be [- 4 : 
 
 2 ' ^ 
 
 And, the firll Part 4- 5 being 
 
 X 
 
 X 
 
 the firll Part, alone, will be 5. 
 
 And, by adding all the Parts thus found together, 
 
 XX X 
 
 we have^ + g+- + 4 + j—5 = 9o; 
 
 that is, 2a: 4" r — 1 = 90. :, 
 
 Whence 13x11:1:91 X ^ 5 ^"^ a: = 42. 
 
 Therefore the four required Parts are 16, 25, 7, and 42, 
 
 refpeftively. 
 
 QUESTION XLVII. 
 
 Za;^ Workmen A ^^^ B z^i^r*? employed _ together for 50 
 Days, at 5 Jkillings per D^^, ^^zt/z y during which 
 Time A, ^y [pending only Sixpence a Day lefs than 
 B, hadfaved twice as much as B, be/ides the expence qf 
 2 Days over : What did each Perjon expend a Day ? 
 
 Let X be the Pence A fpent per Day ; 
 then 60 — X, will be what he faved/?^r Day, 
 and 54 — X, what B faved. 
 Therefore 3000 — ^ox are A's whole favings, 
 and 2700 — ^ox thofe of B. 
 Hence 3000 — 50Afz=2 x(2700 — ^Qx)'\'<ix ; 
 Or, 3000 — 50JV ==5400 — 98;^ : ^ 
 From which 48;^ z=. 2400, and a: =: ^o. 
 
 QUESTION 
 
*4 Algebraical Problems, 
 
 QUESTION ^LVHL 
 
 Two Pirfons, A and B, have both, the fame Income > 
 A lays by y of his ; but B, by spending 60/. per Ann. 
 more than A, at the End of three Years finds himfelf 
 lOoL in debt. What did Each receive, and expend^ 
 per Annum ? 
 
 Let X be the yearly Income of Each * ; 
 
 then 21 is the Sum expended by A, per Ann, 
 o . 
 
 and ^ + 60, That expended by B. 
 o 
 
 Therefore ^ -f 60 — a:, is what B runs in Debt. 
 
 
 Coofcquently (y + 60 — ;v )x 3 = 100, 
 
 or -_- + 180 — 3a;=ioo; 
 
 that is, 180 — 1! —ICO 
 5 * 
 
 Whence 900 — 3^^ = 500, 
 
 and;r=-4^_i33/. 6j. 8fl^. 
 
 Therefore A expended 106/. 13^. ^d, and B 166/. 
 13J. 4^. per Annum, 
 
 B evidently fpends 60/. befides 4 of his Income each 
 year and confequentiy will fpend - + 180/. in 3 years, 
 which per q. is = 100 + y (his 3 years income); 
 hence I inc. = 8o/. and 4 = ,33/. |^, [he income re. 
 quired. J. H. H. 00 ^> 
 
 QUESTION 
 
with thtir Solutions* *o 
 
 QUESTION XLIX. 
 
 A Grazier bought in as many Sheep, of different Sorts, 
 as coji him 33/. js. 6d, For the firji Sort, which 
 were ^ of the whole, he paid gs. 6d. a-piece ; for the 
 fecond Sort, which were ^ of the whole, he paid 1 u, 
 each; and for the reji, i2j. ^d. each: What Num- 
 ber of Sheep did he buy in all? 
 
 If X be the whole Number of Sheep ; 
 
 then, the Number of the firft Sort being - , and of the 
 
 X • 
 
 fecond Sort - , the Number of the remaining Sort (at 1 2s. 
 
 4 
 
 cj l-^ n u XX 12X AX QX ^X 
 
 6^. each) muft be a; z=, i~:=r'=^ 
 
 34 12 12 
 
 Whence, by the conditions of the Problem, we have 
 
 , . IQiX , 22JV , IIRX 
 
 that IS, ^+_-4--^=:i335. 
 
 Let eich Term of this Equation be now multiply'd by 12, 
 and it will become 
 
 76;^+ ^^-^ + 1*5^ =16020, 
 
 or, 'i^'^xzuz 16020. 
 Therefore a; = 60. 
 
 QUESTION L. 
 
 A Draper, of a Piece of Cloth, /landing him in 3^. 2d, 
 per Yard, fold 4 Part, at 41. per Yard; I at o^s. Sd. 
 per Yard; -f at 31. 6d, per Yard ; and the Remnant 
 at 3^. 4^. a Yard: And his Gain upon the Whole was 
 i^s, 2d, How many yards did the Piece contain ? 
 
 If the Number fought be denoted by x ; 
 then the Number of yards in the Remnant 
 will htx^^^^ ^6ox-.2ox^ 15X- 1 .^^13^^. 
 3 4 5 00 60 
 
 Therefoie, 
 
26 Algebraical Problems, 
 
 Therefore, by the Queftion, we have 
 
 ^ X 48+7 X 44+7 X 42 +-^— X40— 3^^= 1^2, 
 3 4 5 "o 
 
 ' 5 3^ 
 
 , . 4ZX , 26X 
 that IS -A rii-^^^ios: 
 
 6 * 3 
 Whence 12 6x-|- 130A: — iG^xzziz^j^o ' 
 
 And therefore a' = ^^ =:=: ,'^0. 
 
 QUESTION LI. 
 
 A DiJiilUr propofes to mix Foreign Brandy ^Jlanding him 
 in 8 Shillings a Gallon, with Britijh Spirits of 3 
 Shillings per Gallon, in fuch Proportion that he may 
 gain 30 per Cent by felling out the Compound at gs. 
 a Gallon. What is that Proportion ? 
 
 Suppofe, that, with a Gallons of Brandy, he mixes 
 X Gallons of Spirits ; then, the Brandy, ftanding him 
 in 8^ \^Sbillings) and the Spirits in 3.T (Shillings), the 
 ■ true value of the whole Mixture will be 8^-4-3^ : 
 But the Value of ^ -|- a: Gallons, at 9 Shillings per 
 Gallon, is 9*2 -j- 9^ : Therefore, by laying out 8a -J- ^x, 
 he gains a-\'6x : And fo 
 
 we have %a-^^o^x \ a -\-6x : : 100 : 30, by the Qjuefiion, 
 Confequently 100a -|- 600^ =21240^ -\- gox ; 
 
 whence 510=^ = 140^, 
 14a • 
 
 From which it appears that, to every 51 Gallons of 
 Brandy, there muft be taken 14 Gallons of Spirits. 
 
 QUESTION LII. 
 
 To find two Numbers in the Proportion of /^ to ^^from 
 which two other (required J Nu7nbers, in the proportion 
 of 6 to y being, refpe£lively, deduEled^ the Remainders 
 fhall be in the proportion of 2 to 3, and their Sum 
 equal to 20. 
 
 Ixt 4AC and ^x be the 2 firft Numbers, 
 
 aiMl 
 
with their Solutions. ^ 
 
 and 6y and j y, the other 2 Numbers. 
 
 Then ^x — 67 : 5;^ — z;/ : : 2 : 3 1 , , ^j^^ Ouejlion, 
 
 And 9^- — 13^ = 20 / ^ ^ ^ 
 
 From which Proportion, by multiplying Extremes and 
 Means, we have i2x — 18^ = iox — i4y, and 
 therefore x zzri^y \ which fubftituted in the above 
 Equation gives i8jy — - i3y= 20 ; 
 
 whence ):= =1:4 ; and x (= ^y) :z=8. 
 
 5 
 Thereiore the 2 firft Numbers are 32 and 40 ; and the 
 other Two, 24 and 28. 
 
 QUESTION LIU. 
 
 A Farmer fold, at one time, 30 BuPiels of Wheat and 
 40 of Barley, and for the whole received 13/. lOi ; 
 and, at another time, he fold 50 Bujhels of Wheat and 
 30 of Barley, at the fame Prices as before, and for tht 
 whole received lyl. The Qjiejlion is, to find what 
 each Sort of Grain xoas jold at per BufJiel. 
 
 Let X and y exprefs the Numbers of Shillings, re- 
 fpecStively, that the Wheat and Barley were fold at per 
 Bufhel ; and then, from the Conditions of the Queftion, 
 wc ihall have the two following Equations, viz. 
 
 3o;v-f 40>'zzz27o, 
 
 5o^-f3oyz=34o.^ 
 From 4 times the Second of which Equations let 3 
 times the Firft be fubtraQpd, and there will remain 
 iiox = 550. 
 
 Therefore a: =^ = 5: 
 
 QUESTION LIV. 
 A Farmery with 28 Bifhels of Barley, at 2j. 4c/. per 
 Bufhel, would mix Rye, at 3^. per Bujliel, and Wheats 
 at 4J. per Bufhel ; fo that the whole Mixture may con- 
 fiji- of 100 Bufiels, and be worth ^s, \d, a Bufhel'. 
 How many Bufiels of Rye, and how many of Wheat 
 muji he mingle with the Barley ? 
 
 Let X be th« Number of Bufheh of Rye, and y 
 
 thofe 
 
«S Algebraical Problems, 
 
 thofe of the Wheat : Then, the value of the Barley 
 being 784 (Pence), *of the Rye 36a: (Pence), and of thic 
 Wheat 48> (Pence), we have 
 
 784 + 36^ + 48y = 40oo-t ^ Utc ^u.Jlzcn. 
 
 and 2 5-J- A-f- v:z= 100 J ^ "v -/ 
 Jrom ihe fiift of which Equations, take 36 times the 
 fecond, and there refuhs, 
 784 — 36 X 28 4- 1 £?jr z= 400, 
 that is, — gg^ -^ 1 g V — 400. 
 Therefore 12;' z=: 624, 
 
 and confequendv v = -^ = ^ - • 
 
 Whence x (= 100 — 28 — >)= 20. 
 
 QUESTION LV. , 
 
 A end B, wdrking iogeiker on the famt Work^ can earn 
 40 Shillings in 6Da\s ; A and C together can earn 
 ^ Skillings in 9 Days ; and Band C, 80 Shillings in 
 15 Days: 'Tis required to jind what each Perjon, 
 (done, can earn per Day. 
 
 Let jr, y and z exprcfs the Numbers of Shillings im 
 the three required V«i]ues, refpectivelv. 
 
 Then < 9* -j" 9-^ 54 f h '^" Qjuefiion. 
 
 And < X -j- 2;= 6 > ^^ Divifiou. 
 
 Hence^ — z — \. byfuhtraBi n^ tie zd Equation Jrom i/f. 
 
 And 2y z=:6, I'y adding the two lafi, 
 
 Confequently > zzn 3 : From which 
 
 we have^ (=6f — >}=3f=3J.8^. and 2(=5^— ^^^ 
 
 =Iz2y=z2J. \d. 
 
 QUESTION LVI. 
 ToJindthreeNmnbersJotkat^tkeFirfl, ^oftheStcond, 
 and ^ of the Third, Jkall together be equal to 62 ; alfo | 
 of the FirJI, 1 of the Second, and ^ of the Third, 
 equal to 47 ; and, lajily, I of the Firfi, '^ of ike Second, 
 and ^ of the Third, equal to 38. 
 
 Put a = 62, ^ =47, and f = 38 ; and let the three re- 
 quired 
 
vnth iheir SoLUTIOKS. ^ 
 
 quired Jhxahtnhe^aiatedhjr x^y-mtizt ic^efihrdf ; 
 then the CoaHaom of tlie'^Dbfefli triU Iw exprefM 
 in the Hues follofwmg E^uj tkm f ' -- 
 
 3 4 i 
 
 4 i t5 
 
 Wiiich, ckv'd of riacdoQS, Isecome 
 
 1 2x 4- 8^ + ^ === *4^ 
 
 licnr (m ocdbr to extenBoaiCe z) let the Second of tfae^ 
 £^ii^OBS be taken from the Doable of the TiA, and 
 i^ tbe Treble of the Third Erom the Qcintaple of the 
 Seoood; ami there re&iks 
 
 W^lxMice, bjr drdmlwg tfe leeoiid of Tbele kam die 
 Trd^ of Ae Fonmtr, and dMding b^ c, there cones ott 
 
 j:=7M — «40*4-i8of=«4. 
 FiQM vibch^ (=4& — 6o^ — 4x} k^lb f9aad=:6b, 
 
 iadz[=>— ix— 4j^X4]=i»o. 
 
 QUESTION LVU. 
 
 r« dUdif /^ Nmmher 90 /^a^/ 2>/<^ filr^ Parts, fo ikat^ 
 du JUM^^ At frj Part + 40 fbj ; tk^. trehk cj 
 ike Secmd+tofcJ ; aMdiktQwMdrm^ 1^ tU Tlird 
 -^ 10 fdj^ WMj kt all epudu ^me arnHker*, 
 
 Let X, y, and z Fq>re{eiit the three r e qnued Faiti 
 rdy et liwly i then, frtmi the ConAaons of AeFroblri, 
 we Ih^hare 
 
 ftr + ^ = 4x4-dL 
 
 * Tiis qaeftion. a^ be.cicgaBtij £9hred by the afift- 
 aace of one wfaowM ni imm ji , J. H. HL 
 
 Ncnr 
 
30 ' Algebjiaical Problems, 
 
 Now, in order to exterminate y and z, let 12 times the 
 firft Equation, 4 times the Second, and 3 times the 
 Third, be added all together ; and you will have 
 26^-j-i2^-j-i2z-J-7/'= 12^-^ 12^ -J- 122 -|- 4c 4"3^« 
 Therefore ^6x :z= 1 2^ -|- 4c -j- 3" — jK 
 , i2fl-|-4c+3^ — jb 
 
 Whence^ 
 
 ^--.illll ^j=:3o,andz(=r(Z— a:— >)=25. 
 
 QUESTION LVIII. 
 
 To find three Numbers, Jo that the Firji with half the other 
 Tzvo, the Second with ^ of the other TwOy and the 
 Third with ^ of the other Two, may be the fame, and 
 amount to ^1 in each Cafe. 
 
 Put fl = 5i, and let x, y and z denote the three re- 
 quired Numbers ; then, by the queftion, 
 , jv + z 
 ' 2 
 
 y^ ^=''' 
 
 z -+-- — *~^z=:a : 
 / 4 
 Which, cleared of Fraflions, become 
 
 2X -\-y -[- Z = 2d!, 
 
 ^ + 3:)' + 2 = 3«, 
 
 '^'+JV + 4;2 = 4^- 
 From whence, by taking the Second from the Third, 
 and the Firft from the Double of the Second, there rc- 
 
 fuhs 2j/-}-32 = ^, 
 
 and 5>' 4- -2 = 4^- 
 
 And, by deducing the former of Thefe, from the Treble 
 
 of the latter, we have lyyzniia. 
 
 Therefore y=z~-z=z^s» 
 
 »(=4^ — 57)='^ = 39. 
 
 and^(=:3a — 3;^ — 2}=^ = i^. 
 
 QUESTION 
 
with their Solutions, 
 
 5» 
 
 QUESTION LIX. 
 
 A certain Sum of Money was divided between three Per- 
 fons^ A, B and C ; Jo that, A's Share exceeded ^ of 
 the Shares of B and C by 30/; alfo the Share of& 
 exceeded^ of the Shares of A and C by 30/; and the 
 Share ofC likewife exceeded ~ of the Shares of K and 
 B, ^y 30/. The Qiieflion is, to fnd the Share of each 
 Perfon, 
 
 Let «zz=*3o ; and let x, y, and z beafTumedtoexprefs 
 the three required Numbers ; then by the Condition! of 
 the Problem, 
 
 3^ + 3z _ 
 
 y 8 "— ' 
 
 2a: + 2;;_^ 
 
 9 
 
 Whence, by Reduction, 
 
 /Of — 4;/— 42=7^!, 
 _3;t_^8y— 3^ = 8^, 
 
 — 2a: — 2^-j-9zii=9'2. 
 
 Now, to get rid of y (which, becaufe of the even Coef- 
 ficients, is the eafieft to be exterminated) let the Double 
 of the firft Equation and the Quadruple of the Third 
 be, fucceflively, added to the Second ; by means where- 
 of we have 
 
 \\X 1123=22^, 
 
 — 11^+33^ = 44^. 
 
 Moreover, by adding thefe two laft Equations together, 
 we have 222 =: ^^a. 
 Therefore z = 35 = gci ; 
 whence x (= 2a +z)=rr 5a = 1^0, 
 
 and;^[=« + |x(A;+x)]=:4<z = i2o. 
 
 QUESTION 
 
Algebraical Problims, 
 
 QUESTION LX. 
 
 Ifh and B together can perform a Piece of Work in 8 
 Days : A and C together in 9 Days ; and B and C in 
 10 Days : How many Days will it take each Per/on, 
 alone, to perform the fame Work ? 
 
 Let the whole Work be reprefented by a, and let x^ 
 y, and z fland for the Parts thereof performed by A, B, 
 and C in one Day, refpeftively. 
 
 will 
 
 
 tgx -\-gz:z=:a J> by the Quejiion. 
 
 I ^1 
 
 And 
 
 "T^ — -- y by Divifion* 
 
 Whence j— z(=|-^)~. 
 
 A 1 a . a 82a 
 
 And2y = — >A = 
 
 ^ 10 ' 72 720 
 
 Confequently >=— : From which x fz=:| — ^^ it 
 
 foundr=-^^, 
 720 
 
 \ 10 / 720 
 
 Now, the Part of the VVork faj performed by each Per- 
 fbn in one lingle Day being thus afTigned, the Number 
 of Days it will take any one of them to do the Whole, 
 will be found by dividing the whole by the afligned Part* 
 
 Thus,lifV=^^=^=i7 5^ is the Number of 
 
 Dayt in which 15, alone, can do the Whole. And, in 
 
 like 
 
u)ilk their SOLUTIONS. 33 
 
 like manner, the Number of Days in which A, or C, can 
 do the Whole, appears to be 
 
 i4^,or23~» rerpe6iively, 
 49 31 
 
 QUESTION LXL 
 
 i/"A, B and C can, together^ finifti a piece of Work in 9 
 Days ; A, B and D together, in 10 Days ; A, C and 
 D together, in 11 Days ; andB, C and u in 12 Days: 
 In how long Time can they all Four, together, Jinifh iti 
 
 Here, denoting the given Numbers by a, b, c, and d, 
 and putting u, x, y, and 2, for the Parts of the whole 
 Work fg) done by Each in one Day, refpectively, we 
 fhajl, by the Queflion, have thefe Equations, 
 
 iz. 
 
 {ax[u-\-x-^y)~=.g'\ 
 by^[u-\-x-\-z]—g \ 
 
 or<( y by Divifion, 
 
 The Sum of all which, divided by 3, 
 
 give. u + x + y + ^ = jx{^+-J+j-+^)- 
 Therefore, feeing the Work done by all the Four, in 
 
 one Day,isexprefredby-ix i^ + jr^^^'^^)'^^^ 
 Tv'hole Work g, divided hereby, will confequently 
 
 D give 
 
34 Algebraical Problems, 
 
 3^ 3 
 
 -ve 
 
 a^ h ^ c^ d a^ b ^ c ^ d 
 
 ?,^bcd '^7^7 r 1 XT , 
 
 L~7~\ jn — r-j-i — ; — =7 ~^ » fo^' the Is umber 
 
 ^i:df-|-^c^-|-a/>^-[-^^6- ^ 2289 
 
 of Days required. 
 
 QUESTION LXir. 
 
 To find that Number whcfe square Root is to its Cuhi 
 Root, in the Proportion of ^ to 2. 
 
 Let x^ exprefs the required Number : 
 Then jt^ will be its fquare Root, 
 and x' its Cube Root.. 
 Now x^ : x"" \ : ^ \ 2,by the Outftion, 
 Therefore ix^ :z=z S^\ or 2 a: zzi: 5 ; 
 
 or, laftly, x — -^ — 2.5. 
 
 Whence ^''z^i: 244.140625 :r^ the Number fought. 
 
 QUESTION LXIII. 
 
 To find Two Numbers in the Proportion cf'>, to 5 ; where- 
 of the fifth Power of theFirjlfhall be to the third Power 
 of the Second as 972 to 125. 
 
 If 3;^: be put for the firft Number, 
 then ^x will exprefs tlie Second ; and we fiiall 
 
 Useful Theorems. 
 
 I. — Half the Sum of the Sum and Diffdjence of two 
 Numbers is the greater Number, and half the Difference 
 of their Sum and Difference is the leflfer. 
 
 II. — The fquare Root of the fquare of the Difference 
 of two Numbers, increafed by 4 times their ProduEl is 
 the Sum of the two Numbers, and the fquare Root of the 
 Square of the Sum of two Numbers diminified by 4 times 
 their ProduSi is the Difference of the two Numbers. 
 
 J. H. H. 
 
 have 
 
with their Solutions. %$ 
 
 have (3^^)^ : [^x)^ : : 972 : 125, by the QueJliOHi 
 that is, 2j^^x^ : 125;^^ : : 972 : 125 : 
 Hence n^'^x^ X » 25 i— : 1 25^-^ X 972» 
 Or, 243x^:^1:972. 
 
 Therefore ;i;^ zn -^ zzi: 4 ; 
 243 
 
 and .V izr: -y/ 4 zzz: 2 : So that 6 and 10 are the tw® 
 
 Numbers that were to be found. 
 
 QUESTION LXIV. 
 
 To find three Numbers in the Ratio of 1, I-, and ~ ; 
 whereof the S'am oj ail the Squares fliall be 549. 
 
 L-Jt X denote the firft Number. 
 
 ^x 
 
 then it will be | : i : : ;^ : :i=i, the fecond Number, 
 
 3 
 
 X 
 
 Andi \\\ \ X \ — zizthe third Number. 
 ~ 2 
 
 Hence x~A^[ — )'+(— ")'==<549> hy the Q^ueflion^ 
 
 that IS, X' -\ .^ =549' 
 
 9 4 
 
 or, 36x^ + 16^^4-9.^^3=1:36 X 549 • 
 
 From which x' = ' ^^ ^ '^^^ zn: 36 X 9, 
 01 
 
 and a;:i±:6 X 3=1^- Therefore 18, 12 and 9, are the 
 three required Numbers. 
 
 Otherwfe, 
 
 By reducing the given Fractions |, ^, and |, to th© 
 fame Denomination, they will appear to be in the Pro- 
 portion of 6, 4 and 3. If, therefore, the firft of the 
 three Numbers fought be denoted by 6x, the other 
 Two will be exprefled by 4X, and gx, refpe6lively : 
 And fo we fhall have I 
 
 g6c^' ■\-i6x'' -\-gx'' =11:549. 
 Whence a:i~:3, and the Numbers fought, as before^ 
 
 D 2 QUESTION 
 
^G Algebraical Problems, 
 
 QUESTION LXV. 
 
 Having given the Difference of two Numbers •=zi6^ and 
 their Produci :=zz 720 ; to Jind the Numbers *". 
 
 Let the Lefler of them be denoted by a: ; then the 
 Greater will beA:-J-6; and fo, by the Queflion, 
 we ihali.have xx-\-^xz=i'j'^o. 
 
 But in order to the Kefolution of this Equation (in 
 which both the firft and fecond Powers of x are in- 
 volvedj let Half the Coefficient of ^, which (in this 
 Cafe) is 3, betaken and fquared, and let that Square be 
 added to both Sides of the Equation : By which Means 
 it becomes ;\r.3c -|- 6.V -|- 9 zzi 729. 
 
 Whereof the former Part being now a compleat 
 Square, its Root may, therefore, be extrafted ; and 
 will be exprefled by the faid Plalf Coefficient joined 
 to X with its proper Sign; that is, byx-f-3 (as 
 may be very eafily proved by the Multiplication of 
 a: -[-3 into itfelf; whence ata: -}- 6-v -|- 9» the very 
 Quantity above, is produced). Kence it is evident, 
 that A:-f-3 = '\/(729)=27 (for equal Quantities have 
 equal fquare Roots); and confequently .^1=24, 
 
 QUESTION LXVI. 
 
 The Sum of Two Numbers being given zz=. 60, and the 
 Sum of their Squares z=: 1872 ; /<? find the Numbers t. 
 
 Let X be the Greater of them ; 
 then 60 — X will be the LelTer : 
 
 * By Theorem IL on p. 34, the Sum of the two 
 * Numbers is z=:'\/(36-)-4 X 72o)zi=54, and, by Theo- 
 rem L T (54 + 6)/^'^ 30 and 24 are the two Numbers 
 required. 
 
 + From the fquare of the Sum. of the two Numbers 
 3600, fubtra£fing the Sum of their Squares 1872, is got 
 twice the Product of the two Numbers zzi 1728 ; 4 times 
 their Produft is therefore zzi: 3456. Then by Theorem IL 
 y^(3Goo — 3456), or 12, is the difference of the two 
 Numbers, and by Theorem L i(6o42i2), or 36 and 24 
 are the Numbers themfelves. J. H. H. 
 
 And 
 
with their Solutions. g7 
 
 And therefore x''-\'{6o — .s;)* 1^=1872 ; 
 Or x''-\- 2^600 — 120x-[-a:*= 1872 : 
 Whence 2jv'— i2o;t zm-^- 1728, 
 
 and a;' — 6qx z=z — 864 : 
 From which, by compleating the Square fas in the lajl 
 ProblemJ^Q get x"- — 6o.t -|- 900 (m: — 864-i-90o):rr36: 
 And confequentl y,by taking the Root,;v — 30311 y^(36)z=6. 
 Therefore ;f zzz: 6 -(^30 = 36 ; and 60 — xz=z2.^'. Which 
 are the two Numbers that were to be found. 
 
 But, to folve the Problem in a more general manner 
 (by Letters) put the Sum of the two Numbers z=:^, the 
 Sum of their Squares ■=. b, and the greater Number r=: x, 
 as before. 
 
 Then will A^^-j-a* — 2ax-\-x''z=zbt by the Quejiion, 
 Hence 2x'' — 2ax-z=:b — ^% 
 
 b a" 
 
 and x^ — <ixz=i . 
 
 2 2 
 
 Where, Half the Coefficient of the fecondTerm being — ♦ 
 the compleated Square will therefore be 
 
 ,___ \£j^^t ^jL.£y 1. ^. 
 
 ' 4 V 2 2 ' 4 J 2 4 
 From which, by extrafting the Root, 
 
 webavcA: z=. J \ ) ; 
 
 2 ^ ^ 1 4 / 
 
 And therefore Arr^-i/ ( — )M : Which \{ a be 
 
 . ^ \ 2 4 / ' 2 
 
 taken zz: 60, and ^3=1872, will come out ::=: 36, the 
 
 very fame as before, 
 
 QUESTION LXVII. 
 
 Ta divide the Number 60 fa) into two fuch Parts^ that 
 their ProduB may be 864 (b) *.. 
 
 If X be put for the greater Part, the LefTer will be de- 
 
 * By Theorem II. on p. 34, \/ {60'' — 4 X 864)^=12 
 is the Difference of the two Numbers, and, by Theorem I. 
 on the fame page { (6oH7i2), viz. 36 and 24 are the 
 lumbers fought. J. H. H. 
 
 D g noted 
 
3^ Algebraical Problems, 
 
 noted by a — x -, and we fhall therefore have 
 
 ax — XX izz: b ; by the Conditions of the Queftion. 
 
 This Equ;ttion, by changing the Signs ot all its Terms 
 
 (in order to have the highell Power oi x affirmative) be, 
 
 comes XX — ax zzz: — b. 
 
 Whence by compleating the 'Square, 
 
 , . aa aa 
 
 we have xx—axA zzz — b-\ ; 
 
 4 4 
 
 and confequently x -uny'i ^h\ 
 
 Therefore x-^iy/ f b j-\ '=36, the greater Part ; 
 
 and ^ — ;vzzr 24, the LefTer. 
 
 QUESTION LXVIII. 
 
 To divide the. Number 60 (a) into two Part9,fo that the 
 Square of the Greater multiply' d by the Leffer, added 
 to the Square of the Lejfer multiply' d by the Greater, 
 may amount to 51840 (b). 
 
 Vix be the greater Part ; then a — x will be the Lefler • 
 and ^' X f ^ — x)-\-[a — xj X ^' ^= ^» 
 that is, ax'' — x^ -\- a^ x — 2gx'- -\-x^ -^zb, 
 
 or, a" X — ax^ z=zb : 
 
 Whence x^ — 6lxz=z — — ; 
 a 
 
 and, confequently, xz=i \/ f j-} r=: 36. 
 
 QUESTION LXIX*. 
 
 The Sum of two Numbers being given nn 20 fa) and 
 the Sum of their Cubes :=z 2240 fbj ; to determine the 
 Numbers, 
 
 Let X be the Greater ; then a — x will be the Lefler ; 
 
 * Let 2z z=2 Difference of the two Numbers. 
 Then lo-j-zzn: greater Number, arid 10 — zzrzthe 
 leflfer, by 1 heorem 1. ; therefore the Sum of their Cubes 
 is {10 -\- %y '\- {10 — z)\ or 2ooo-|-6oz% which, per q. 
 isi:r:2240, confequently Z:=:\/^z=ziy ^o -{- zz=z 12, 
 thegrcater Number, and JO — »=^i, the leffer. J. H.H. 
 
 an4 
 
with their Solutions. 39 
 
 and therefore A^-{-(^ — xYz=ib^ 
 
 that is, x'^ -\- a^ — 3^^ x -f- '.^ax'^ — x'^ z=z b : 
 
 Whence '^ax^ — <^a^ x z=z b — ^^, 
 
 b a" 
 and ;t^ — ax ■= — , 
 
 3^ 3 
 Therefore, by compleating the Square, 
 
 , aa f b ^^ _\^^ \ ^ ^^ 
 
 ^^ 4 I 3« 3 4>' '^a 72' 
 
 And, by extraftlno^ the Root, x — — zzizt/ (-^ .) : 
 
 2 "^ V3^ 12/ 
 
 Confequently x z=: -- -4- 1/ ( ' Izzz^ -4- 
 
 2 ' ^ ^3^ 12/ 2 ^ 
 
 /(37t"~33|]=i2. 
 
 QUESTION LXX. 
 
 To divide the Number 240 (a) irdotwo fuch Farts, that 
 the greater Fart divided by the Lejfer^ may be to the 
 lejfer Fart divided by the Greater^ in the Proportion 
 0/ i^j to 75 for of m to nj. 
 
 If the greater Part be denoted by x, the LefTer will be 
 exprelTed hy a — x", and we (hall have 
 
 X a — X 
 
 a — - X X 
 
 nx ?n\{a 
 
 Hence 
 
 a — X 
 
 and zz=:frt — x^f". 
 
 m 
 
 From which, by extraBing the fquare Root, on boih 
 Sides, X v^(— jzi=:a — x. 
 
 Whence, putting x/l-^ Jizzz—' , 
 
 we have bxz=zca — ex \ and confequently xzzzT—t — 
 
 b-\- c. 
 
 But 
 
 D4 
 
40 Algebraical Problems, 
 
 But, in the Cafe above propofed, y^^— J being =; 
 
 ^(tJ) = ^i^)=j ' ''' ^'''' ^ = 5,c = 7;^n 
 therefore x == - ^ =: 7 X 20 =;■ 140. 
 
 QUESTION LXXI. 
 
 Two Wcrkmtn A and B wen employ d, by the Day, at 
 different Rates ; A, at- the End of a certain Number of 
 Days, had g& Shillings to receive ; but B, who play* d 
 6 of thofe Days, received, only, 54 Shillings : But^ 
 had B worked the whole Time, and A play* a 6 Days, 
 They would have received exa&ly alike, 'Tis propofed 
 to find the Number of Days They were employed ; 
 and what Each had a Day, 
 
 Let X be the Number of Days that A work'd ; 
 then X — 6 will be the Days that B work'd. 
 
 Moreover — , will be the Watres of A per Day ; 
 
 X o i V 
 
 and _ the Wages of B per Day. 
 
 Therefore -^^i- v x, is what B would have earn'd, had( 
 X — 6^ 
 
 he work'd the whole Time : 
 
 And — X ("^ — ^)» ^^^t ^ would have earn'd had he 
 play'd 6 Days. Which two Values being equal, by the 
 Queftion, we have-^-L-z=-^-^^ ^. 
 
 X O X 
 
 Whence, by Reduftion, 54^:' 1:11:96 X {^ — 6)% 
 
 Or, -~= (a: — 6)': Therefore, by extraQing the fquare 
 
 Koot on both Sides, ^ :=:X—6; and confequentlyA::r24, 
 
 4 
 From which it is evident, that A had 4 Shillings, and B 
 
 3 Shillings a Day, 
 
 V QUESTION 
 
7 with their Solution s, 44 
 
 QUESTION LXXII. 
 
 From two Places, at the Dijiance of '^10 (a) Miles, two 
 Fer/ofis, A, and B,Jet out, at the fame Time, in order 
 to meet each other ; A travelled 8 fbj Miles a Day 
 more than B ; and the Number of Days in which They 
 met was equal to Half the Number of Miles B went in 
 a Day : 'Tis required to find how far Each travelled 
 to meet the oth&r. 
 
 Let X be the Number of Days in which They met ; 
 
 Qu;H:{r.+.}-"''-l^=Nu,nberofMi.es{ ^ } 
 
 went a Day. 
 
 Therefore, by multiplying each of Thefe by (x) the 
 
 Number of Days, we have ^xx, and ^x\-bx, for 
 
 the whole Number of Miles travelled by B and A, 
 
 refpe£lively : 
 
 And confequently a^xx '\-hx-zzza. 
 
 -T . hoc . hb a . bb 
 
 jtlence xx A -f- -7^^=-^ + ~r~ » 
 
 ^ 4 ^64 4^64 
 
 Therefore 2xx:=: 128 z=: the Miles travelled by B ; and 
 ^xx 4" ^^ = 192 31= Thofe travelled by A. 
 
 QUESTION LXXIII. 
 
 Two Meffengers^ A and B, were dif patched at the fame 
 Time, to a Place, at the - Dijiance of 90 (a) Miles; 
 the Former of whom, by riding one Mile an Hour 
 more than the Other, arrived at the End of his Journey 
 one Hour before him : The Quefiion is, tofnd at what 
 Rate Each travelled ^qi Hour, 
 
 If X be the Miles that A rode per Hour ; 
 Hien X — 1 will be the Miles which B rode per Hour : 
 
 Moreover -^willbethe Number of Hours in which A 
 
 performed 
 
Algebraical Problems,. 
 
 a 
 
 performed the whole Journey ; and ^ will be the 
 
 X — 1 
 
 Number of Hours wherein B performed it. 
 And therefore -- rz:: i, hy the Quejlion, 
 
 X X 1 
 
 whence, by reduftion ax — a m: ax — x'' -\- x. 
 
 Or x"- — X :=: a : 
 
 Therefore x'- — ^-|-4^^^~t~4(^y compleating the Square) 
 
 and confequently x-=z \/[a 4-1)+ i = 9T + i==; lo. 
 
 QUESTION LXXIV. 
 
 To find two Numbers^fo that their Sum multiply d hy the 
 Greater may produce lOO times the Lejfer, and being 
 multiply d by the Lejftr may produce 64 times the 
 Greater. 
 
 Let X denote the greater Number, and y the Lefler. 
 Then [x -\- y) y^ x z=z ioo>', 
 
 And(^4-jv)X7=64^:- 
 Now, the firft o\ thefe Equations being multiply 'd by^, 
 and the Second by x^ they become both alike ; 
 and fo we have loojy^' •==. 64A;* : 
 Thirefore, by taking the fquare Root, io^z=8jv, 
 
 and confequently y==L— : Which Value fubftituted in 
 5 
 
 , ^ . ' . / . AX\ AX ^ 
 
 thefecond Equation, gives ('^ + — JX — =04.-^; 
 
 Whence ;c = 2^-^^1=144 - ; ^n^y=<^5^' 
 9 ^9 9 
 
 O.UESTION 
 
with their Solutions. -43 
 
 QUESTION LXXV. 
 
 To find three Numbers, fo that their continual ProduB 
 divided, by the Sum of each two of them, mav qajte 
 given Numbers ; or f which is the fame Tiling) to de^ 
 termine the Values of x, y and z, in the underwritten 
 Equations. 
 
 xyz xyz xyz, 
 
 200, — \ — z=. 150, — ^j — z=: 120. 
 
 x-\-y 'x-\-z' ' y~\~^' 
 
 Pirft, by Multiplication, 
 
 xyzz=:iS.Qox-\-20oy=:ziS^^ ~\~ i^oz^ir: 120V-|- 120Z : 
 Therefore by Red.. tlion, 
 
 ^cx-\-2Qoy:=zt.^oz\^^f x-\- ^y:=.-y,z, 
 200X -j- 807^=1: 120Z J XqX -^-^y^ri'^z. 
 Hence ;>^ ~|- 4^)/ z=:sx-^2.y, and ihcrefore y -zziz ix : 
 Which, fubflituted in 2,'^:=zzx -^/^y, gives "^z' — Cfx; 
 
 and z z=z %x. 
 And, by subftituting for both y and z, in the Equation 
 
 -^, — = 200, we ffet ^ , ^^ — 200 ; 
 X -\-y x-f-2x 
 
 that is, 2x^=2 200. 
 
 Confcquentiy x •=. 10, yz=z 20, and z ziz 30. 
 
 QUESTION LXXVI. 
 
 To find the Ratio of two Numbers, whofe. Redangk is 
 equal to the Square of their Difference, 
 
 Let the lefTer Number be to the Greater as 1 is to x ; 
 then, if the faid leffer Number be denoted by z, the 
 Greater will be exprefled by xz, and we fliall haye 
 xz X 2;zzz {xz — 2)", or xz^ z=ix^ z' — ^.xz"- -f- z' (by the 
 Quejiion), Whence, dividing the whole by z^, there rc- 
 
 fultS X -ZULX^ — 2X-\- w 
 
 Therefore x'^ — 3^'= — 1 ; 
 
 and confequently x:=, ■ - ■ - = 2.618 Be, 
 
 Hence 
 
44 Algebraical Problems, 
 
 Hence the Ratio of any two Numbers, whofe Reftangle 
 is equal to the Square of their Difference, muft be that 
 of 2.618 &c. to Unity, 
 
 QUESTION LXXVII. 
 
 To find two Numbers, whofe ProduB is 300 fa) ; fo thai, 
 
 ^ ^10 fbj be added to the Lffer, and 8 (c) fubtraBed 
 
 from the Greater, the ProduB of the Sum and Re- 
 
 mainderfhall, alfo, be equal to 300 (a). 
 
 Let the greater Number be denoted by x, and the 
 
 Leffer by y ; 
 
 then will { f^{Z7) X + *)= a } ^>' '^' Prolkm. 
 By the laft of which xy-\-bx — cy — cbz=:z a. 
 Prom whence, the fir'ft Equation being fubtrafted, there 
 *refts bx — cy — cb zzz o : 
 
 Therefore bxzzzcb -^-cy, and x =: — "[ : Which fub- 
 ilituted in the firft Equation, gives • ^ ' - z=:a. 
 From which we have )/'-j-^^' = — : 
 
 And, by compleating the Square, jy'-J-^;' -J ■= \ — •- 
 
 Whence J z= -/( -^4 — —)—-= 15 ; 
 ^ndx[z=z^j=:2o. 
 
 QUESTION LXXVIII. 
 
 To divide 100 i?ito two fuck Parts, that their Difference 
 may be to their Sum, as their Re^an^le to the Dif- 
 ference of their Squares, 
 
 Let a reprefent half the given Number, and x half 
 the Difference of its two Parts ; then, the Greater of 
 
 there 
 
with their Solutions. 45 
 
 tliem being exprefled hy a-^-x^ and the Lefler by a — x, 
 it will be 2x : 2a : : ( a-\'x)y^[a—x)\ [a-{-xY—[a — xY; 
 that is, by Redu£lion, x : a \ : aa — xx ; ^ax : 
 Hence /^ax^ z=:ay<^[aa — xx). 
 Or ^x^zuza* — x\ 
 
 and therefore xzzr.\/{ — J=z 22.36. 
 
 So that, the greater Part 1S72.36 ; and the Lefler 27.64. 
 
 QUESTION LXXIX. 
 
 To divide the Number 60 (a) into two fuch Parts, that 
 their ProduB may be to the Sum of their Squares, in 
 the Ratio 0/2 (m) to 5 (n)^ 
 
 Let X be the greater Part : 
 Then, the lefler Part will be 13 — x, 
 the Prod u6l ax — xx, 
 
 and the Sum of the Squares a^ — 2aX'\-ix^, 
 Therefore m : n : \ ax — xx \ a' — 2ax-\-2x^; 
 and fo, 2mx' — 2max '^ma^z=znax — ?z^^ 
 Whence, 2mx''\-nx^ — 2max — ?iaxzz=: — ma^ ; 
 that is, {2?}i -\- n]xx' — (2?^ -|~^)X^'^' = — ^^'5 
 
 Or, a;^ — ax-:=z i — • 
 
 2ZW -f- w 
 
 Therefore ^^—^a;4 z=( r— )=: — X "i , 
 
 '4 V4 2m-\-nJ 4 ^^71-^-2711 
 
 . a jn — 2m\ , a 
 and;;<^=-^ \/\ — \ H--=z:40. 
 
 2 ^ V/2-J-2?«/ ' 2 ^ 
 
 QUESTION LXXX. 
 
 To find tzuo Numbers whofe PrcduBJhallbe 320 (a)^ and 
 the Difference of their Cubes to the Cube of their Dif- 
 ference, as 61 fnj is to Unity, 
 
 Let X be the greater Number, and jy the Lefler ; 
 then will xy z=zcf, 
 and x^ — y^ ', [x — yY : : n : i, by the Queflifin : 
 
 "" Whick 
 
4^ Algebraical ProrlemSp 
 
 Which, by a6^ually involving x — y, 
 
 becomes ;t^ — y^ : x^ — Z^^'y -\-'^^y* — y^ ' - n: i. 
 
 Prom uijence (by fubtrafting the Confequents front 
 
 their Antecedents) v 
 
 we get '^x-y — ^xy^ : x^ — Q^y -\-3^y' — ^ • ^ ^' — i • i; 
 
 Or, which is the fame, S^yy<.{^~yj • {^ — yY- '- ^ — i • i» 
 
 Whence, dividing by x — y^ 
 
 we have ^xy : {x — yY : : n — 1:1; 
 
 Or, 3a : [X — yy~ : : n — 1:1 [hecdiuk xyzn^aj. 
 
 Prom whence (x — y)' z=:—^ — ; 
 
 n — 1 
 
 and confequcnily x — y zzzi'^f—^ ); which putzz:^: 
 
 Thea from the Equaions Aryzzr ^, and a: — v — ^, thp 
 
 Value of X will be found =z— — ~^ ^ ' — 20 : and 
 
 2 
 
 That of y ^/(^Mlifizi^ ^ ,5^ y^^^ p,,^^ 6^^ 
 
 The fame' ether wife, 
 
 - Let % denote the Half Sum, and x the Half Difference, 
 of the two Numbers ; then the Greater will be expreffed 
 by z+a:, and the Leffer by % — a;*, and we {hall therefore 
 
 that IS ^6^^;,_^2;,^__,8«A'3. 
 The laft of which, divided by 2.r, 
 
 fives '^^z' -\- X- znz \nx'' . 
 rom whence, the Treble of the firft being deduced, 
 we have \x^ zzz Ajix'^ — 3<z ; and confequently x zr: 
 
 ^ -^" Z=2. 
 
 Hence % [= ^(a -j- xx )]r=: 18 ; therefore z-^-x-zzi 20, 
 
 By Theorem I. on p. 34. 
 
 and 
 
with their Solutions. 47 
 
 snd % — jv zz- 16 ; which are the two Numbers that were 
 to be found. 
 
 QUESTION LXXXI. 
 
 A Farmer received yl. 41. for a certain Quantity of 
 Wheat, and an equal Sum, at a Price lefs by is, 6d, 
 a Eujhel, for a Quantity oj Barley, which exceeded 
 That of the Wheat by 16 BuJJiels : How many Bujhels 
 were there @f Each ? 
 
 Put a ■=: the total Value of each Sort of Grain, 
 b z=z the Difference of the Quantities, 
 cz=: the Difference of the Prices per Bufliel, 
 
 and xz=z the Number of BuQiels of the Wheat : 
 Then, dividing the whole Price by the Number of 
 
 Bufhels, wehave-^ for the Price of the Wheat /)er 
 
 Bufhel : And, in the fame manner, the Price of the 
 
 Barley per Bufhel will appear to be — -tTa' 
 
 Therefore r- -zzic, by the Queftion, 
 
 X x-\-I? '^ 
 
 Whence (by Redufclion) ax '\-ab — ^x z=z ex"" -\- hex ; 
 
 „ ab , . 
 
 Or, — z=ix'' -\- bx, 
 c * 
 
 Confequejitly x = \/{~-\', — ^)— ^ ^^^*' ^^^ Num- 
 ber of Bufhels of Wheat ; and ;v + i^ = 4^j the Num- 
 ber of Bufliels of Barley. 
 
 QUESTION 
 
4^ Algebraical Problems, 
 
 QUESTION LXXXII. 
 
 One bought two Pieces of Cloth of different Sorts, zohcreof 
 the Finer Cofl 4 Shillings a Yard more than the Other; 
 fo that, for the Fineji, he paid 0^60 Shillings ; zuhereas 
 the Coarffl^ which exceeded the^FineJl hfio Yards^ 
 cojl him, only, 320 Shillings : How many Yards were 
 there of each Piece ? 
 
 Let X be the Number of Yards of the Finefl, 
 and^, the Number of Shillings a Yard : 
 Then jv -|- io» will be the length of the coarfeft Piece, 
 and^ — 4, its Price ^^r Yard. 
 
 Hence {(^._^7o)=(^^°^)^3,<,}^;' the Quejlion. 
 
 By the laft Equation xy -\- icy — 4^1=: 360 ; 
 from whence deducing xy zzz 360, 
 we have 10^ — 4^ = o. 
 
 Therefore 10^' ziz: ^x ; and y = — . Which, being fub- 
 
 5 
 
 2x' 
 
 ilituted in the firft Equation, we get — z=:^6o: Whence 
 
 5 
 X comes out zzz 1/900 = 30. 
 
 QUESTION LXXXIII. 
 
 There are two Numbers, whofe ReBangle is equal to the 
 Difference of their Squares ; and the Sum of their 
 Squares is alfo equal to the Difference of their Cubes : 
 
 ' What are thofe Numbers ? 
 
 Let X denote the lefler Number, and let the Greater be 
 in Proportion thereto as y is to Unity ; Or, which is the 
 fame Thing, let the sfreater Number be denoted by xy. 
 
 Then {"^"Jr^fy'Tf \hytheProbUm. 
 \x^y^-\-x^-=.x^y^ — x^ } ^ 
 
 Therefore k'^rf^ ^ , \ h Divifion, 
 
 From 
 
with their Solutions. 49 
 
 From the firfl of which Equations, jy* — ^ = 1 ; 
 whence jv* — jy-|~4=-4' ^"^ confequently ^z= — ^ — 
 
 But, bythcfecond Equation x z=z'^-^--^--^z=z'^--^ — (be- 
 
 yyy—i ^y 
 
 caufe yy z=.y + 1)= 1- ""^ ='^ + —1 i — = -' + 
 
 g 4/c 2 
 
 -i- (by multiplying both the Numeratoi; and De- 
 
 4 
 
 nominator by -y/^ — 1) = ^!--^. Therefore the two 
 Quantities fought are \\/ ^, and ' ' ^--^. 
 
 QUESTION LXXXIV. 
 
 h fets out from London for York, at the fame time as B 
 Jets out from York for London ; and theRate^at which 
 They travel is fuch, that A, 9 Hours after their meet- 
 ing arrives at Yorky and B at London^ in j6 Hours 
 after. The Quefiion is, to find in what Time each 
 Traveller perforins his Journey. 
 
 Let X denote the Number of Hours travelled by Each 
 before the Time of their Meeting on the Road. 
 Then, fmce A (by the former Part of the Quefiion) 
 goes over the very fame Ground in 9 (a) Hours, as B 
 travelled in x Hours, we have, thereFore, as ^ : ;v : : at : 
 
 XX 
 
 — , the Time wherein B travels a Diftance equal to 
 
 That gone over hy A, in x Hours : But (by the latter 
 Part of the Quejtion) B, in 16 (b) Hours, travels the 
 very fame Diftance as A in a; Hours, Hence it is evi- 
 
 XX 
 
 dent that and h are equal to each Other : 
 
 a ^ 
 
 And confequcntly that x i^z: \/(^/^)=z: 12. 
 
 Therefore A performs the Journey in 21 Hours, and B 
 
 in 28 Hours. 
 
 QUESTION 
 
50 Al&ebraigal Problems, 
 
 QUESTION LXXXV, 
 
 Tke Sum of igoL was divided bitzceen three Perfons^ 
 lohojejhares were in Geometrical Proportion^ and the 
 greateji of them exceeded the leaji by ^o/. What were 
 
 • all the fever al Shares ? 
 
 If X be put for the Lcaft of them, then the Greaieft 
 will be;^-j-5o; the Sum of which two, fubtrafted from 
 (190) the Whole, leaves 140 — 2x for the Mean Share : 
 Therefore, x : 140 — 2x : : 140 — 2.x : ^-(-50, by the 
 Q^uejiion, 
 
 And confequently (140 — 2xYz=zx X^^~l~5^) » 
 that is, 19600 — 560A -J-4A;'— rA'-[-^0^. 
 Whence 3^?' — 6iqx zz=. — 19600, 
 
 I , 610A: 10600 
 
 and ;c' ^zzn ■. 
 
 3.3 
 Hence, by compleatmo- the Square, 
 
 ... 6ioa: 93025 / 1C-600 Q8o?5\ 34225, 
 
 39V "^ 9 / 9 ' 
 
 And, by extra6iing the Root, x — ' — ^zzi+ — . 
 
 ^ 8 ^ ^ 
 
 Therefore a: z=z '^—^ =40: And fo the other twa 
 
 3 
 Shares are 60 and 90 Pounds. 
 
 QUESTION LXXXVI. 
 
 Two Notes, one of 120L payable in 6 Months^ and the 
 Other of i^ol, 'payable in 9 Months, were dif counted 
 for 8/. loj. ^^. What Rate of Inttrejl were They df- 
 counted at ? 
 
 Let X denote the Intereftof One Pound for rs Months : 
 Then the Amount of 1/, in 6 Months being 1 -| — , and; 
 
 * Allowing the Intcreft to be proportional to the time, 
 agreeable to the nature difimple IntereJL 
 
with their Solutions. ^i 
 
 in 9 Months i -}-— the Prefcnt Value of the Bill, due 
 
 at the End of 6 Months, will therefore be ■ ; , — : and 
 
 Thatoi the Bill, due at the End of o Months, -^^^. 
 
 i+-lx 
 
 Whence, wc have _i^4-— ^^ ( = 120+150 — 
 
 8*5) = 261*5 fh ^^^^ QueJHon,) 
 
 Which, by Reduftion, becomes 120 -j-9oa'-}- 1 50 + 
 
 , ^ 26i-AX(8+ 10^ + qa;') 
 
 or, 270 + i65;t=zz '^"^- \. ' ^. ■ 
 
 o 
 
 Therefore (^70+ ^65^)X B^ .^ ^^^ _^g 
 251-5 ^ ' « * 
 
 , . 2160 , 1Q20JV , 
 
 that IS, —p 1 ==3^'+ 10^ + 8 : 
 
 From which we have 
 , , 2500 iq6 
 
 ' 3X^23 -^ 3X523 
 
 Whence, by compleating the Square, &c: 
 
 X is found =v/r-^+f-^^^yi--^^ — 
 
 L 1569 ' V^ 1569 / J 1569 
 v!li3iXi569j4-i29jx 1295)— « 295. 
 
 1.569 -^°-°S°93- 
 
 Which multiply'd by 100, gives 5*093/. or /. 5 : 1 : lof, 
 learly, for the Rate per Cent, at which the Notes were 
 
 1 
 discounted 
 
 QUESTION LXXXVII. 
 
 A and B ia/ie, in Trade, 5940/. per Annum, each ; but 
 
 A, whefe Profits are 2 per Ct^nt. greater than Thoje of 
 
 B, clears looL per Annum more than B : What are 
 the Profits oj Each, per Cent ? And What do They 
 clear per Annum ? 
 
 Letrzizrioo (Pounds), 
 
 b (zr: 594o)zz:: the whole Sum taken by Each, 
 d[z=. iqqI.)zz=, the given Difference of their Gains, 
 X z=z What A gains per Cent. 
 X — a{=.x — 2)z=: What B gains j&(?r C^«/, 
 
 E 2 Now 
 
52 Algebraical Problems, 
 
 Now, " fmce A, in taking c-^-x Pounds, gains x Pounds, 
 
 DX 
 
 It will be c+ ^ : ^ •• : ^ : —. — , the whole Gain of A : 
 c-\-x 
 
 'And, in the fame manner, we have, 
 
 c+x — a \x — a\ ',b\- ^^"^"^^ the whole Gain of B: 
 c-j- X — a 
 
 Therefore .--p--~~Ki^)_ j^ i,y the dueJHon. 
 
 Hence abc :=z=.[x'' -\- 9.cx — ax — ac -\- cc))^d ; 
 
 and confequently x^-\- [q.c — «)x x =. —-. — |- ac — cc: 
 
 Whence, by compleating the Square, 
 
 /2C — ay abc aa 
 
 , , r /abc , aa\ 2c — a 
 
 and thererore x-=,\/[ r~-)-— zn: lo. 
 
 ^ V ^ 4^ 2 
 
 From which it appears that A gained lo per Cent, and 
 cleared 540/. per Annum; and that B gained 8 per Cent, 
 and cleared 440/. per Annum, 
 
 QUESTION LXXXVIII. 
 
 Of four Numbers in Geometrical ProgreJJion, there is 
 given the Sum ofihe two Leaf -zzz 20 faj and the Sum 
 of the two Greatef :=z 45 (bj ; to find the Numbers, 
 
 Let X denote the firft Number, and^ the Third ; 
 then the Second being expreffedby a — ^, and the Fourth 
 by b — y^ the four Terms of the Progreffion, placed in 
 Order, will ftand thus, a;, a — x, >:, b—^y^ 
 Whence, by the Nature of Proportionals, 
 
 ,vehave /' X(*->)=:(<^-Ar)X7. 
 \xy^=[a — x)\ 
 
 bx 
 From the firft of which Equationsjy is z=. — : And, by 
 
 bx^ 
 fubftituting in the Second, we have zzr f ^ — xY : 
 
 Whence, by extrafting the Square Root on both Sides, 
 
 and confequently \a;z=^ -r (1 -j-v^" )=^- 
 
 Therefore 8, 12, 18, and 27, are the four Numbers 
 that were to be found. 
 
 QUESTION 
 
with tkeir So LVriom. ^3 
 
 QUESTION LXX:^IX. 
 
 , The Sum ofyool. fa) was divided among four Perfonshy 
 B, C andD ; whofe Shares were in Geometrical Pro^ 
 grejjion ; and the Difference between the Create]}, and 
 Leaji was to the Difference between the two Means, as 
 37 f'^J io 12 fnj : What were all the fever al Shares ? 
 
 Let the Share of A, or the firfl; Term of the Progref- 
 fion, be denoted by x, and let the Common Ratio b« 
 That of 1 to^ ; 
 
 and xy^ — x : xy* — xy \ : m : n j ^ ^ -^ 
 From which Proportion, we have 
 
 ^3 — J ___.|^^ — ij^-J.^ orjy'-j-jv-J- 1=— (by dividing 
 
 n n 
 
 the whole by ^ — 1 ). 
 
 H=nce^ is found ^"^+^(^'»ZL2.^^3^L!iHL7 
 =- : Whence x (= — ; \ — r-; — 3^ is given = 
 
 ■ ^l^^Ti A = 108. Therefore the four Shares 
 
 27 + 36 + 48 + ^4 
 
 are 108, 144, 192, and 256/. 
 
 QUESTION XC. 
 
 Of four Numbers in Arithmetical Progreffion, the Sum oj 
 the Squares of the two Means is given r= 400 faj ; and 
 the Sum of the Squares of the two Extremes z=: /{6^ 
 fb) : To determine the Numbers *. 
 
 If X be put for the lefler Extreme, and y for the 
 Common Difference ; then the four Numbers will be 
 exprefled by, 
 Xi X -f-jy, X -\- 2jy, and ^ + 3)' refpeftively ; and we 
 
 * Let the Progreflion be denoted by ;\? — 2ty*^ — 7* 
 X'\-y, and x -\- 3jy. Then the Sum of the Squares of the 
 Means will be 2 [x^ -\- y*)z=z J^oo , and that of the two 
 Extremes 2(^'-j-9jv^) z= 464; their Difference gives 
 2 X 8jy^ = 64, or y z=. */ \ = 2 ; therefore a'zz=400 — 
 4 = 196, ;v =z= v^ 196 zzz: 14, and the Numbers are 
 '^ 12^ i6, and2o. J. H. H. 
 
 • (hall 
 
 lis 
 
^4 Algebraical Problems, 
 
 (hail therefore have S i^ + y)[ + i^ + ^yY = ^ \ or 
 
 \V hence, by Subtraftion, 4}/*=:^ — a : 
 And therefore yz=: — = 4. 
 
 But, by the firll Equation, x^ -{- ^xy z=z j2lL : 
 
 From which (looking upon y as known) wc get 
 
 ^ \2 J^J 2 
 
 Therefore 8, 12, . 16, and 20, are the four Numbers 
 that were to be found. 
 
 QUESTION XCI. 
 
 The Sum of four Numhtrs, in Arithmetical ProgreJJion, 
 being given =56 fbj aiid the Sum of their Squares nr 
 864 fcj : to find the Numbers. 
 
 If Half the Sum of the two middle Numbers be de- 
 noted by a, and Half their Difference by .r, the Num- 
 bers themfelves will be exprcfTed hy a — a:, and a-^x: 
 And we fhall have 
 
 r («— 3^) -\-[a—x) -\-{a-\-x] 4-(^+3^) ==:^ Xh the Nature 
 \ {a--^xy+{a—xY+{a-^xy-\-[a-\-^xy==:c fofthe^aestion 
 Hence> by Reduftion, 4a z= b, and ^aa ~|- 2Qx.r — r. 
 
 Therefore ^ = --= 1 4 ; and Xz=:z\/f — 1 — 2. 
 
 4 ^20 ^J — 
 
 From which the Nambers themfelves are given ; being 
 
 8, 12, 16 and 20. 
 
 By the fame way of proceeding the Problem may be re- 
 folved, when the Progreffion is fuppofed to confift of 
 6, 8, 10, or any other, even, Number fnj of Terms *. 
 
 For the Sum of the Squares of the two Means (a — x 
 
 and 
 
 * In the Solution of this Problem, in its generalized 
 
 n 
 State, it is neceflary to find the Sum of -« terms of the 
 
 Series of Squares 1 +9+ 25 + 49 + ^^' whofe Roots 
 arc the Scries of odd Numbers 1, 3, 5, 7, &c.; which 
 
 may 
 
^wiih their SoLVriOi^S. ^j 
 
 and a-\-x)heing m2 \(aa-^xx) ; and the Sum of the 
 Squares of the two Terms {a — ^x and a -|- 3:1) next ad- 
 jacent to them zm 2 x(«^-{- 9-'»^«^) ; 3lfo the Sum of the 
 Squares of the two Terms {a — ^x and a -f- ^x) next ad- 
 jacent to thefe laft being zz: ^Y^' aa -j- 23 xx)^ &c, &c. it 
 follows that 
 
 Which Equation, by puttmg 1 --(- 9~|- 25 + 49 + <^<:. 
 (continued to - Terms) ^zzfi is reduced to naa -\- 2fxx 
 
 Whence x -= ^I — —j. — V, from which, as a is given 
 
 zzz-^, the Value of x will alfo be known. 
 n 
 
 QUESTION XCII. 
 
 Having the Sum (h) and the Sum of the Squares (c) of 
 Jive Numbers^ in Arithmetical Progrejfion ; to deter ^ 
 ?mne the ProgreJJion. 
 
 Let a denote the middle Number, and X the Com- 
 mon Difference ; then the five Terms of the Progreflion 
 will be, a — 2^, a — x, «, a^^-x, and a^2x \ whence we 
 have ^a = b, and 5^^ -j- loxjv = c, by the Conditions of 
 
 the Queftion. From which a is found z=z~^ , and ;^(— - 
 
 5 
 
 After the fame Manner the Problem may be iavefiigatf d^ 
 •when any od^ Number fn) of Terms is propounded ; 
 Tor the Sum of the Squares of the 2 Terms [a — x and 
 a-\-x) adjacent to the middle One being z=-:2\[aa'^xx)i 
 and That of the two Terms (a — 2x and a -(~ 9.x) next 
 
 may be eafily done by the Corol. at p. 210 of Mr; Sim- 
 fon's Algebra, or by pp. 94 and 95 of his EJfays, and will 
 
 be found to be =:— — - — , the exprefTion that is sbovc 
 
 reprefented by^. J. H. H. 
 
 E4 
 
56 Algebraical Problems, 
 
 to Thefe z=:zx(aa-\- ^xx), &c, ^c, it is evident there- 
 fore that 
 
 flfl -j- 2 X'Cia ■\- xx)-\- 2 X (^« + 4^^)+ 2 X (^^ + 9^a:)4- 
 &c, :zzz c. Which Equation, by putting i -|-4-|-9 -j- i6 -{- 
 
 &c. (to Terms)=y, becomes naa -f- zjxx z=z c. 
 
 Hence x z=z \/l T"^)'* ^^^"^ which, as a is given= 
 
 *- , all the Terms of the Progreflion will be known. 
 
 By a like Procefs, if, inftead of the Sum of the Squares, 
 the Sum of the Cubes, or Biquadrates, be given, the 
 Problem may be refolved. 
 
 QUESTION XCIII. 
 
 J Traveller, bound to a certain Place at the Dijlancc 
 ' of i^o Miles, goes 26 Miles the Jirji Day, 24 Miles 
 
 the Jecond Day, 22 Miles the Third ; and fo on, in 
 . Arithmetical ProgreJJion, decreafing 2 Miles every Day, 
 
 In how many Days will he arrive at the End of his 
 
 Journey f 
 
 Put^= 140, the whole, given, Diftance, 
 
 f = 26, the Part thereof travelled in the firfl: Day, 
 dz=z 2, the Common Difference by which each 
 Day's Journey dccreafes, 
 and xzzz. the Number of Days wherein the whole 
 Journey is perform'd. 
 Then, from the Nature of the Queftion, ft is evident, 
 that, the laft Day's Journey will tall fhort of the Firft, 
 by a: — 1 times the Common Difference V^y' ; and is, 
 therefore, truly expreffed by c — [x — i)x d: 
 But it is well known that the Sura of any Arithmetical 
 Progreffion is equal to the Sum of the firil and laft Terms 
 multiply'd by Half the Number of the Terms : 
 
 Hence we have[c-|--c— (^— IJX^JX J for the whole Dif- 
 tance travelled ; and confcquently this Equation 
 
 [ac-(^-i)X^]xf = ^- 
 
 Whence icx — dx'-\- dxzzz 2 b, 
 
 and 
 
with their Solutions. ryj 
 
 d zb 
 
 , tc-X-d 2b 
 and XX — - X ^ = T * 
 
 From which a-z=:4- k/\ ' — -^-r-^ — -r H J— =7- 
 
 QUESTION XCIV. 
 
 Jifter A, w'^<? travelled at the Rate of \ Miles an Hour^ 
 had been Jet out txoo Hours and three Qiiarters^ B Jet 
 out to overtake hiin ; and in order thereto went a\ 
 Miles the firji hour^ /\^ the Second, 5 the Third ; and Jo 
 on, gaining a Quarter of a Mile every Hoar, Hozo 
 many Hours did it take him to come up with A? 
 
 Put^z±r4, the Dirtance travelled by A, per Hour. 
 (^=11, the Diftance gone by A before B fet out, 
 c=:^i, the Miles travelled by B in the firft Hour, 
 d -— I of a Mile, the Common Excefs, 
 and X z=z the Number of Hours required. 
 Then it is evident (from the preceding Problem) that the 
 Diftance travelled by B, in the laft Hour, will hec-i- 
 
 {x — i)X^5 and that [2<:-|-(^— i)X^]X^^'il^ exprefs 
 
 the whole Diftance travelled by B in ;t Hours. 
 
 But the Diftance of A in x Hours being ax, the whole 
 
 Diftance travelled by A will therefore be ax-^b; which 
 
 being equal to That of B (by the Que/lionJ we thence 
 
 have, 
 
 [2c4.{^--iix^3xj=«^+^/ 
 
 and confequently 2cx -\- dx^ — dx = 2ax -J- 2b ; 
 
 2c — 2a — d 2^ 
 
 oixx-^- ^— -X^=-J.• 
 Which (by making ^£llilZ^_/) gives 
 
 Af=/(^+-^j-^ = 8, the Number of Hours 
 required. 
 
 QUESTION 
 
 y 
 
58 Algebraical Problems, 
 
 o 
 
 QUESTION XCV. 
 
 To find jour Numbers in Arithmetical Progrejficn, which 
 being increafed by 2, 4, 8 and 15, refpedively^ the 
 Sumsjhall be in Geometrical ProgreJJion, 
 
 Let X denote the leaft Number, and y the Common 
 Difference. Then the four Nutnbers will be expreffed 
 by x,^ x-^-y.x -f- 2;/, and x-\'<^y \ 
 and the lour fpecified Suras, 
 
 by X'\- 2, x-\-y -|- 4, ^ -j- ajy -J- 8, and;c-|- 3jv -J- 15- 
 Whence, by the Nature ot" Geometrical Proportionals, 
 
 " \(^+'')X(^+37+'5J=(^+>-f4)X(^+2>'+8) 
 h ' /jV'+4>' = 2'^ ' \ hy Multiplication, 
 ^ ^ ^^ l^y'-\- ^oy'\- 2z=z5x ] andTranJpofition, 
 Hence 5j)/' + 2q>' = 4;^' + 20^ + 4 ; 
 thereiore>'^=i:4, and jyzir: 2 : 
 
 From which a: m 6 ; and fo the other three Numbers are 
 8, 10, and 12 refpe6iive]y : 
 
 For 6, 8, 10 and 12 are in Arithmetical Progreflion ; 
 and 6-|-2, 8-1-4. 10 -|- 8, and 12 -|-* 15, are in Geome- 
 trical Progreflion. 
 
 QUESTION XCVI. 
 
 The Reclangle (a) and the Sum of the Cubes (b) of two 
 Numbers being given ; to deter?nine the Numbers. 
 
 Let X denote the greater, and y the leffcr Nurnber : 
 Then will V^Jj^^^—iJ h ^^^^ Ouejlion. 
 
 Whence, by Involution, ^-^Jy3:zz:a^ 
 and;c*-4-2A:y4-/ = <^^- 
 
 Let the Quadruple of the former of thcfe two Equations 
 be fubtrafted from the Latter, and you will have x^ — 
 
 and, by extra£ling the fquare Root, on both Sides, 
 x^ — y^z=z^(b^ — 4«^)- Which, addedtothefirft Equation^ 
 . gives <2.x^-=.b-\- y/{b* — 4^^). 
 Confequcntly x :=:^[\b-\- \\/[bb — 4^^)] ; 
 
 &y (=- )=:{7[i^+V(^"-4-^ • 
 
 • The 
 
 an 
 
with tkdr Solutions. 59 
 
 Hhtfame otherwife. 
 
 Since xy z=:ci, we have^ z=- ; and therefore 
 
 x^ -\ — ^z= K by the Oueftion. 
 
 Whence x^'\-a^z=. bx^^ 
 
 or, x^ — bx^z=. — ^^ 
 
 Therefore, by complcating the Square, 
 
 ,.__ Bx^ +-{=-- ^^' )=^i=ifl; 
 ' 4V 4 '' 4 
 
 and, by extra£lin^ the Root, x^ — --z=:-^ — -• 
 
 Hence ;t'= i /^ + 1 /(/^^ — 4^^] ; 
 and confequently the Values of x and jv, ^/z^ very fame 
 as before. 
 
 From either of the above Solutions, a General Theorem, 
 for the Refolution of Cubic Equations (according to the 
 Manner oi Cardan) is very eafily deduciblc. 
 For, by putting zz=:^-J-^» and involving each Side to 
 the third Power, we have z^ z=ix^ -|- o^x^y -|- o^xy"" -|-^^= 
 \x^ -\-y^-\-2f^yy^[x '\-y)z=:x^ +7^ 4" S^X^^ ( ^7 fubflitu- 
 ing for ;\^j)/and;\^ -|-_y, their Equals, d^ and z). From 
 whence, by Tranfpofitlon, 
 z' — ^az[z=.x^-\- y^]z=zb. 
 But it appears, from above, that z(zzi x -j-^') 
 
 Which Value, therefore, is the true Root of the 
 Equation z^ — ^az z=: b. 
 
 From which the Root of the Equation z' -\- cz z= b 
 (where the fecond Term is pofitive) will alfo be given, 
 by affuming — 3^ zzz: ^ , and fubftituting for a : Vv^hence 
 % is found 
 
 bb .71, 
 
 =='^t+nf+5!')]-'^[^+/(:+4i')]. 
 
 QUESTION 
 
6o Algebraical Problems, 
 
 QUESTION XCVII. 
 
 The Difference of two Ni.mLers being given :z=zj{, and the 
 Sum of their Cubes = 1240 ; to determine the Numbers i 
 
 Let X denote their Half Sum, and d their Half Dif- 
 ference, then the Greater being a: -|- d^, and the Lefler 
 X — f/, we have 
 {x -|- dy + (^ — dy = 2340, 
 that is Q.x^ ^6d^x = 2240, 
 
 I ; 2240 
 
 or, x^ + <^d'x = —^' 
 
 2240, 
 2 
 then it will become x'^-\-cx:=zb : 
 From whence by the general Theorem, in the 
 Problem, x zzz 
 
 2240 
 Put cz=z^d'{z=. 12} and bzzz — —{'=: 1120) 
 
 ^[r+v'(f+ll]-^[fM|5?])]=.o. 
 
 Therefore 12 (z== a: + ^) ^nd 8 (=^ — dj arc the two 
 Numbers that were to be found. 
 
 QUESTION XCVIII. 
 
 *Tis propofed to divide the Number 24 (^a)into two fuck 
 Parts, ^that the Difference of their Cubes may be 
 3584 (2b J. 
 
 Let a-^-x exprefs the greater Part, and a—x the Lefler ; 
 then will [a -^ xY — [a ^ xf z=i %b ; 
 that is, 6a^x-\- 2^'= ^b ; 
 or x^'\- ^a^x z=zb. • 
 
 Then will x'^-{'CXz=. b, 
 
 andA:i=;r — =4, by the Theorem in the preceding 
 
 Page, Whence 8 and 16 are the two required Numbers. 
 
 QUESTION 
 
with their Solutions. 6i 
 
 QUESTION XCIX. 
 
 The Sum of the Squares of two Nuvibers beirTg given = 
 208 fgj and the Sum of their Cubes :=. 2240 fh) ; t 
 Jind the bJ umbers. 
 
 Let the greater Number be denoted by x -\-y^ and the 
 Leffer by x — y : 
 
 Then will {^^::j:^Jj^^ ]l>ytheau.Jlion. 
 
 From the hrft of thefe Equations, multiply 'd by 3;^, let 
 
 the fecond be fubtrafted ; 
 
 whence you will have /^x^ z=z ^x — h ; 
 
 and confequently x^ — \- ■- z=z o ; 
 
 4 4^ 
 
 that is, in Numbers, x^ — 156;^ -j- 500^=0 ; 
 the Roots of v/hirh Equation (by Seel. i£ of my 
 Treatife of Algebra) * will be found to be 10, 4, and — 14; 
 
 whereof 
 
 * Cubic Equations that have three pofTible Roots, in 
 the rcfolution of which Cardan's rule fails and the method 
 oj divifors is fometimes but of little ufe, may be more 
 dire6tly folved by means of the tables of logarithmic 
 Sines and Cofines, 
 
 The general form to which fuch Equations belong 
 \%x^ — ax-=.'j^b. 
 
 Therefore, suppoling the radius to be 1, in the tabic of 
 logarithmic Sines take the arc correfponding to log. 
 
 3^ — log. a y/— , or to log. 1680 — log. 156 y'aoS in 
 
 the Equation above-mentioned, i.e, to 1 '873,1530, which 
 is 48° 18' 22''-8, its t=:i6° 6' f,e. whole log. fine 
 
 = T-443,o282; then will T -443, 0282 -}-i ^og- — 
 
 (i-i59,03i7), viz. 0*602,0599, be the log. of one of the 
 values of x, or of one of the Roots of the Equation, nameily, 
 4. — But the Equation has befides this tw6 other Roots, 
 which may be obtained as follows : 
 
 To the arc already found 48"^ 18' 22"'8 add a circle, 
 or 360°, the Sum is 408° 18' 22"-8, its 1=136° 6' f-^ 
 
 whereof 
 
62 Algebraical Problems, 
 
 whereof the Firft, only, is for our Purpofe : By Mean* 
 of which we get y {■=: \/{jg — x^)z=z 2. 
 Therefore 12 and 8 are the two Numbers that fulfil the 
 Conditions of the Problem. 
 
 Note. 
 
 whereof the log. fine is 1 '840,9684, then adding the ^ 
 
 log. — , as before, the Sum is 1*000,0001, the log. of 10, 
 
 3 
 another Value of x, or Root of the propofed Equation ; 
 and that both thefe Roots mull be affirmative is evident, 
 fmce the 3rd part of each of the arcs found falls within the 
 two firft quadrants of the circle. — Laftly, 48° 18' 22"*8-|-j2 
 Circles, or 720°, is ==768'' 18' 22"-8, >vhofe f z= 
 2^6° 6' f'*6i correfponding to the Ipg. fine T -987,0964, 
 
 which increafed by \ log. 2— is 1 '146, 1281 zin log. of 14, 
 
 o 
 another value of x, or the third Root of the Equation pro- 
 pofed, and which muft be confidered as negative, becaufe 
 the arc 256° 6' y"'6 extends beyond the femicircle. — It 
 may not be amifs juft to obferve that, when the Root 4 
 was difcovered, the other two remaining Roots might have 
 been found by dividing the given Equation by x — 4, and 
 then refolving 'the quadratic that would refult, or that, 
 having found the two firft values of Jt, 4 and 10, their 
 Sum, 14, with the negative Sign prefixed, muft have been 
 equal to the third Root. 
 
 Jn exa?nple of the other cafe contained in the general 
 Jorm x'^ — qxzzzL^ b^ namely^ when the fgn efb is affirmA^ 
 'tive, or zvhcn., injteqd of working by the log. Sines, it be- 
 comes necejfary to have recourfe to the log. Cofnes. 
 
 Let the Equation propofed be x^ — 1 17;^ z=: 324. 
 
 Here, log. 3^-— log. a v^ — , or log. 972— log. 117 v/i 56, 
 
 3 ' . 
 
 gives 1 '822,9181 = log. cos. 48° i8'22"'8, its ^-zzz 
 i6°6'7"'6, to which the correfponding log. cos. is 
 T*982, 6189 ; therefore T'982,6i89 4- 1*096, 5623 
 
 [i. e. the \ log. i-, or of 156,)=: 1*079,1812, the log. 
 
 of 12, which is a value of x^ or one of the Roots of the 
 propofed Equation. Again, 16** 6' 7'''6-|- i^o*' (= f 
 
 of 
 
with their Solutions. 63 
 
 Note. The Refolution of the above Equation, by the 
 Theorem referred to in the preceding Examples, is im- 
 poffible; becaufe the fquare Root of a negative Quantity- 
 is to beextrafted ; as, upon Trial, will be found. :• 
 
 QUESTION C. 
 
 The StimfaJ and the continual ProduB fb J of four Nuin- 
 bers, in Geometrical ProgreJJion^ being given ; to deter- 
 mine ike Numbers : 
 
 If the lefler of the two Means be reprefentcd by x — -y, 
 and the greater by x -\-y^ we fhall have 
 
 V"^ ^^ . . ^. , ^— ->' XQ^ueftion, 
 
 From the fecond of thefc Equations, by putting cz=i\/b^ 
 and extrafting the Square Root, there comes out 
 [x—y) X {x-^y]:=zxx—yy=LC. 
 Moreover, irom the Firft, we get 
 
 {x-\-y) ; or, q,cx-\-\x — yy-\-[x-\-yYz=:ac (by Subftitution) 
 that is, 2CX -}- 2x^ -\- 6A7* -zz-i-ac ; 
 
 of 360°, or the whole Circle) is ziz 136^6' f'^^ whofe log.. 
 COS. ;z=T.857,68o2, and confequently this -j- 1*096, 
 56231Z1: 0*954,242^, the log. of 9, a value of .v, or a 
 Root of the propofed Equation, and this Root, it is 
 plain, mufl be negative, becaufe the arc, whofe log. cos. is 
 ufed, falls in the fecond quadrant of the circle. — LaRly, 
 16" 6' f'^ +240' [z=.\ of 72o°):rr: 2^6" ^' f'^ and the 
 corrofpondincp log. cos. zzi: T'38o,5590 ; therefore 
 i''3^0'559i^ + i"096,<5623zzzo-477,i2i3, the log. of 3, 
 the remaining value of x, or Root of the propofed Equa- 
 tion, and which muft alio be confidered as negative, fmce 
 the arc correfponding to the cos. ufed falls in the third 
 quadrant of the circle. 
 
 This method of the logarithmic fines and cofines may 
 also be eafily extended to the refolution of other Equa- 
 tions of a higher order than the third, as the Reader may 
 find by turning to Problem, xxxiv. SeB. xviii. of Mr. 
 Simpfens Algebra^ the Prob. from which th? preceding 
 methods have been deduced. J, H. H, 
 
 or 
 
64 Algebraical Problems, 
 
 or 2CJf-}-aA''~|- 6x\[xx — c)zz:zac (becaufe y):=:zxx — c). 
 
 X nc 
 
 Hence jf3 zzn-Tr'' From which, by the Theorem 
 
 2 8 
 
 in Page 59, the Value of x may be found. 
 
 QUESTION CI. 
 
 The Compound Inttreji of a certain Sum of Money ^ put- 
 out for 4 Years^ amounted to 344/. tVo > ^^^ thefimple 
 Inter eft thereof^ for the fame Time^ and at the fame 
 Rate^ would have been only. 320/. What, was the Sum 
 put out ? and what the Rate of In^erefi ? 
 
 Put a ziz 344*8i, and b z= 320 ; and let x denote the 
 Inttreft of i/./for one Year. 
 
 Therefore, fmce the Simple Intereft of 1/. for 4 Year 
 is 4a:, and the Compound Intereft (1 -j-^)* — 1, or 4^:4- 
 
 we have, as ^x -^ &x^~ -\- 4x^ -^^ x^ : /^x : : a : b,hy the 
 Nature of the Queftion : 
 
 Aa — a5 
 and confequently 6Ar+4 x^-]-x^:= -. = 0.3101 25. 
 
 From the Refolution of which Equation, x will be 
 found :=:'o^ : Therefore the Rale of Intereft was 5 per 
 Cent, and the Sum put out 1600/. 
 
 • QUESTION CII. 
 
 The Sum (sj and the Producl (p) of any two Numbers be- 
 ing given J to find the Sum of the Squares ^ Cubes, Bi- 
 quadrates, £Sc. ofthofe Numbers. 
 
 Let the two Numbers be denoted by x and y : 
 Then 
 
 ^+>!_T"^ IbytheQueft 
 
 and xy z=Lp J ^ ^ 
 
 The former of which Equations, fquared, 
 given x^-^ 2xy ^y y^ z=:i s^ ; from whence, the Double 
 of the Latter being fubtrafted, 
 we get x' -j-/=nz i"* — 2/?, thefum of the Squares, 
 Let this Equation be multiply 'd by x ^yzzzs^ 
 and there arifes x^ -j- xy'' -^yx- -^y^ z=: s^ — nsp^ 
 or, x^ -\- xy\(x -\- y)-]^ y^ =:s^ — 2j/?, 
 that is, x^ -}"/'X"^"T~y = -^' — ^^P (becaufe xyz=zpi and 
 x+y=zs). 
 
 Therefore 
 
with their Solutions. 65 
 
 Therefore ^^-j-^^zrzJ' — '^sp, the Sitm of the Cubes, 
 Again, multiply this laft Equation by x-\-y:=:S, 
 and you wilJ have x^ -{- xy y^[x^ -^ y'']-\- y"^ z=i s^ — 3i'/>, 
 or x'^-^-pX^s^ — 2/>)-j-jv4rr:J'^ — '^s''p (becaufe A:'-|-y'=-^' — 
 2/>j. W Hence x'^'\-y'^-=.s^ — ^s'p -|-2/>% the Sum of the 
 fourth Powers. 
 
 Multiply, again, by x-\-y::=zs, and you will have 
 x'''\-xy X (x^ '\- y') -^ y^ :=: s' — j{s^ p-\'2sp\ 
 
 or, x^'{-px U^ — ?>^P) -\-y^ = J^ — ^J'p 4- ^^p\ 
 
 and therelore x^ +^' = s^ — ^s^p -[- 5-f/'% thefum of the 
 ffth Powers. 
 
 From whence the Law of Continuation is manifeft ; be- 
 ing fuch, that the Sum of the next fuperior Powers will, 
 always^ be obtained by multiplying the Sum of the 
 Powers laft found by j, and fubtratiing the Sum of the 
 preceding Ones drawn into/?, from theProduft. 
 So that the Sum of the «*'* Powers (expreffed in a ge- 
 neral Manner) 
 
 n , Yi -111. ^ ^ — 2 , ;?-o n — 4 * 
 
 oxx -[-jy", svillberz:J —ns pJ^nY^—^s p — 
 
 n — 4 n — ,5 n — 6/ , n — ^ n — 6 n — 7 «— 8 
 
 "X— X —' P +-X— X —X -fs 
 
 p^—&c. 
 
 QUESTION cm. 
 
 The Sum faj and the Sum of the Squares fbj of four 
 NumberSyin Geometrical Frogreffon^ being given ; to 
 find the Numbers, 
 
 If X and y be affumed to reprefent the two middle 
 Numbers, then, from the Nature of continued Pro- 
 portionals, the two extremes will be expreffed by 
 
 XX , yy Air 
 — and -^ : And fo 
 
 x^ y* 
 
 we (hall have 1-a:-4-v-| =«, 
 
 y ^ ^ ■' * X 
 
 F Pwt 
 
66 Algebraical Problems, 
 
 Put X -\-y z=. «, and xy z=iz: 
 
 Then, from the firft Equation, U<- -a — u, 
 
 y * X 
 and therefore x^ -|->^ = ■^>'X(^ — ^)* 
 But, becaufe the Sum of ihe two Means [x and y) Is ex- 
 prelfcd by m, and their Reftangle by z, it is evident, 
 from Que/iion 102, that the Sum of their Squares (a:'4" 
 y^) will he exhibited by m* — 22, and the Sum of their 
 Cubes (x^ -f->') by u^ — 3^2. 
 
 And, in like manner, the Sum of the two Extremes 
 /x* y* \ 
 / 1 J being denoted hy a — u, and their Reftangle 
 
 by z, the Sum of their Squares will be — (/? — uy — 2z, 
 
 and the Sum of their Cubes z=i:(^ — w)' — {^a-^^u) y^z {but 
 
 this Laft is of no Ufe in the prefent Cafe.) 
 
 Hence, by fubflituting thefe feveral Values in our 
 
 fecond, and \di\ Equations, 
 
 we get (a — i/f — 2z-\-u^ — azzzr^, 
 
 and u^ — 3MZ zzz 2;X(^ — «) I 
 
 that is, Irv Reduftion, 
 
 2tt* — 2fi«-j- ^j' — I? z=: 4Z, 
 
 and ?/3zz:(2w -j-«)Xz : 
 
 And, confequently, {2u -[-fl)X(2«' — 2au-\'a^ — ^)=4«'i 
 
 Whence, by Reduftion, u^ -4 =zz ; 
 
 ^ * a 2 
 
 ind therefore u z=: x/i— 1 j : 
 
 V 2 4<2tf / 2a 
 
 From which the feveral Values of 2, a*, y will alfo be- 
 come known. 
 
 QUESTION CIV. 
 
 The Sum fa) and the Sum of the Cubes (c) of four 
 Numbers in continued Geometrical Proportion being 
 given : to determine the Numbers, 
 
 Let the Notation of the preceding Problem be retained ; 
 then our two Equations (in this Cafe J 
 
 will 
 
with their Solutions. 67 
 
 win be — 4-A:+y 4-— zzifz, 
 jy ' ' -^ ' a: 
 
 But, it appears,yr^w ThencCy that the Sum of the Cubes; 
 (x^ "i"^0 of the two Means is zzr u^ — 3"zy and that the 
 
 Sum of the Cubes ( — + <— j of the two Extremes is =: 
 ^y X / 
 
 [a — uY — (3^ — 3")X 2:; Therefore our laft Equation, by 
 
 fubflitutingthefe Values, becomes u^-\~[a — uf — ^azz=zc\ 
 
 And, it appears, by the laft Problem^ 
 
 that the firft Equation (by a like Subftitution) 
 
 is reduced to z^^ zzi:(2« -j- ^]X z. 
 
 Hence by exterminating % out of the two Equations 
 
 thus derived, and putting, -^ z=id. we obtain 
 
 3 3^ 
 (2u -{■ '2)X(«* — au'\- d)z=i u^ ; 
 or u^ — au"^ — [aa — 9^d)y^u -\- ad-zzzo. 
 From whence, the Value of u being found, the reft of 
 the Quantities will be known. 
 
 In the fame Manner the Problem may be refolved, 
 when (inftead of the Sum of the Cubes) the Sum of the 
 4th, 5th, or 6th, &c. Powers is given. 
 Eor the Sum of the w*" Powers of the two Means (or 
 
 jv"-}-^") being, univerfally , r= m" — nu^'**z 4" ^ X 
 
 w"~-*z* ^c, (See ^ejiion 102. J 
 
 and the Sum of the n^ Powers of the two Extremes 
 
 —-(a -^ uf^nxi^ + uy-'Z'\-nX-—-u'"'^z\&c. (fince 
 
 the Sum of the Roots is here zzz fl — u) ; 
 
 we therefore have m"-}-(^ — m)"~7?zX ["""'+ (^ — ")""']+ 
 
 1 " 
 
 Equation, by writing -r-— inftead of its Equal z, be- 
 
 F « comes 
 
68 Algebraical PRdBLEMS, 
 
 comes u''-X-(d—uY "7— X [«*""'+( ^ — w) "•**1 + 
 
 X 
 
 Whence u may be found. 
 
 QUESTION CV. 
 
 Having pven the Sum fa) and the Sum of the Squares 
 (b) of Jive Numbers in Geometrical Progrejfion ; t9 
 determine the ProgreJJion, 
 
 Let Xy z, and y denote the three middle Numbers, 
 
 XX 
 
 taken in Order : Then — will be the firfl; Number, 
 z ' 
 
 yy 
 
 and — the Laft ; and we fhall have 
 
 z * 
 
 Put u z=.x-\'y ; then from the firfl Equation, 
 
 _^ , 21 __ 
 
 z % 
 
 Therefore, feeing the Sum of the two Extremes is ex- 
 prefled by a — u — 2, and their Reftangle by z' (from the 
 Nature of Proportionals) the Sum of their Squares will 
 be z={a — u — z)* — 2z' fby Quejhon 102J. 
 Moreover, the Sum [x^y) of the two Terms adjacent 
 to the middle one being zrrw, and their Re6Ungle — 
 z\ the Sum of their Squares {x'-^y^) will therefore be 
 rz: w* — 22* fby the Same). 
 And fo, by fubttituting thefe Values above, 
 we get [a — m — z)^ — tz^'-^u^ — 2z* -|- z'= b, 
 
 tt* — 2Z* . 
 
 and ^=z[a — u — z). 
 
 z ^ 
 
 Whence 
 
with their Solutions. 6^ 
 
 Whence a* — zau — 2flz-{- iiu^-^ 2uz — 22*zz=^, 
 
 and az — «' — «/z-j- 2'= o. 
 
 The Double of which laii Etjuation, added to the Former, 
 
 , , ah 
 gives a* — 2<2M rz: ; whence u zn ^ . 
 
 From which, and the Equation az — u" — wz -j- z^-=i O, 
 the value of z will alfo become known. 
 
 QUESTION CVI. 
 
 The Sum (a) and the Sum of the Cubes (h) of five 
 Numbers^ in continued Geometrical Proportion being 
 given; to find the Numbei s. 
 
 Retaining the Notation of the laft Problem, and pro- 
 ceeding in the fame Manner, we have 
 
 and az — u" — u% -J- z' z=z o fas before). 
 
 The firft of which Equations, by Reduftion, becomes 
 
 From whence the other Equation, multiply'd by 32, bf- 
 ing fubtra6^cd, there remains 
 
 b a' 
 
 therefore w'+ 2 MZ — 2* — ay^(u-\- 21=3: — , 
 
 ^ 3^ 3 
 But, by the fecond Equation, m'-|- 2mz — z^=: az -\- uz ; 
 
 whence, by Subftitution, az ^ mz — rtX(""i"^)= — ■"" 
 
 «' , . b a^ 
 
 — ; that IS, uz — az = ; 
 
 3 3« 3 
 
 . aa b 
 or az — uzzzzd (by puttmg =:aj. 
 
 From which, the fecond Equation being fubtrafted, 
 there rcfults 
 
 u^—-z^z=:zd: Wherein let ( ) the Value of u 
 
 (found from the former Equation) be now fubftituted, 
 
 F ^ and 
 
7© Algebraical Problems, 
 
 and we fliall have ^ ^J 2'=fl? ; and confequently 
 
 z 
 
 z* — {aa — ^)Xz*-f- 2adz — d' — n. 
 
 Whence z will be found. 
 
 QUESTION CVII. 
 
 The Sum f'aj, the Sum of the Squares (b)^ and the Sum of 
 the Qvbes(c), of any three Numbers being given; to 
 determine the Numbers. 
 
 Let them be denoted by x, y and 2 ; then 
 fince/^^+-^v4%'=^V'^ ^'"' ^^ Tranfpofition, 
 
 have 
 
 x'-X-y'zzzb — z' 
 x'-\-y':=ic^ z'. 
 
 X' 
 
 Now by multiplying together the two firfl of thefe 
 
 Equation «, 
 
 we })ave;t'-{-A:*)'-|- xy'^-\-y^ z=.ah — b% — flz*-|-z% 
 
 And, by cubing of the first, we also have 
 
 •^'+ 3-^> + 3^>*+y' =: ^3 — 3^*2; -|-3fl2*— s^ : Which, 
 
 deduced from the J reble of the Foimer, 
 
 leaves 2a;^-J- ^y^-zzz^.ab — a^-\- ^a^z — 3//Z — 6az*-\-^z^ : 
 
 And this, being z=2r — 2z^(by the third Equation) we 
 
 therefore have 
 
 6z^ — 6tf2'4~(3^* — 3^)X^=^^ — 3«^+2c, 
 
 r , , - I ^' — b a^~riab'4-2c 
 and confequently 2' — az^-\ X ^= i: — • 
 
 From whence (wlien ^2, ^ and c are^expreflfed in Num- 
 bers) three diffeient Roots, or Values of z, may be 
 found, answering all the Conditions of the Problem. 
 
 Thus, for Example, let «=:9, /'=:29, and ^==-99; 
 then our Equation will become z^ — 92" -\-26z — 24^11:0. 
 
 And (by tjither of the two first Methods explained 
 in Set!. 12. of my Treatise of Algebra) the three 
 Roots, in this Cafe, will be found to be 2, 3 and 4. 
 
 Whick 
 
with their Solutions. 71 
 
 Which Numbers are, therefore, the true Values of x, y 
 and 2, in the Equations ^+>'~t~^ =9' •^'+j)'*+~'= ^9* 
 and x^'\-y^'^z^ =99* 
 
 QUESTION CVIII. 
 
 The Suvi fa), the Sum of the Squares fbj, the Sum of the 
 Cubes fc), and the Sum of the Biquadrates (d)^ of 
 any four Numbers being given; to determine the 
 Numbers. 
 
 Let the four Numbers be denoted by ^, y^ 2 and u ; 
 and put A-zzza — «, Bizz^ — «', Cz=.c — w^ and Tizzzd — u^. 
 From whence, by the Conditions of the Problem, 
 
 Now> if the Second of thefe Equations, be fubtrafled 
 
 from the Square of the Firll, 
 
 we (hall have 2xy 4-2X2 J^^yz-z^A^ — B. 
 
 And if, in like Manner, the fourth Equation be fub- 
 
 tracted from the Square of the Second, we fhall have 
 
 Moreover, if from the Square of the former of thefe lafl. 
 Equations, the Double of the Latter be dedufted, there 
 will come out 
 SAT^+S^^jfz-l-S^^jtjyrnA^— 2A^B— B*4.2D ; 
 
 or, 8a:7zx(^4-7-^2)=A^— 2A'B— B^-f 2D ; 
 
 A^—gA^B— B"4-2D ,, 
 whence xyz z=z rr^ ' — — ( becaufe x ^y 
 
 Again, by multiplying the firfl and fifth Equations, 
 into each other, we get 2x*y'^2x'z-^2y''x~\.2y^z^2z*x 
 4-2z'^-|-6 A-jv^-zz A'— A B . 
 
 And, by multiplying the Firff and Third together, 
 there arifes x^ -^y^ ^z^ -^ x'y+x^z-^y^x+y'z^z^x ^ 
 2*;'=AB : 
 
 The 
 F4 
 
7« Algebraical Problems, 
 
 TheDouble of which laft taken from the Precedent, leave 
 Sxy2 — 2x^ — 2^3— 22';=:A3 — 3AB : And this, added to 
 
 gives 6xy^=h^ — 3AB+2C. 
 
 A^— 3AB4-2C, . A^--2A*B— B^-L2D 
 Hence 5 ^ i=xy.)=: ^^ 
 
 /^/>. alcove) ; 
 
 and confeq-ientlv, bv Reduftion, A^ — 6A'B-|-8AC+ 
 
 3B^--6D=o.' 
 
 In which Equation let the feveral Values of A, B, C 
 
 and D, be now fubftituted, and then (divi-iing the Whole 
 
 by 24) we, at length, have m*-^ am '-j-— X "' — 
 
 6 ^ ' 24 
 
 Whofe four Roots (found by any of the known Methods) 
 anfwer all the Conditions oi the Problem. 
 
 QUESTION CIX. 
 
 To find the leofi whole Number^ which being divided by 
 x^Jhall produce a Remainder ofj ; But, being divided 
 By 28, the Reinainderjhall be 13. 
 
 Let 1 9^-]- 7 denote the Number fought; where >r 
 according to the Oueftion, mufl be a whole Number. 
 And, b}^ the Queftion, it hkewife appears that igx-^- 
 7 — 13, or, its Equal, 19^ — 6, mull be divifible by 28 
 (wiihout a Remainder). 
 
 But it is plain that 28^ is divifible by 28 : Therefore 
 [gx -|- 6) the Difference bei ween i^x — 6 and sSa:, muft 
 alio be divifible by the fame Number 28. For it is well 
 known that, whatever Number, or Quantity, mcafurej 
 the Whole, and one Part, of Another (w^ithout a Re- 
 mainder) muft do the ftime by the remaining Part. 
 Hence(3 8x-[- 12) the Double of gx-^-G, being divifible 
 by 28, if the fame be fuhtratled from 19;^ — 6 (in order 
 to ger a: without a Coefficient) the Remainder,*- — -18, 
 will, JliUy be divifible by the fame Number ; and con- 
 
 fequently 
 
with their Solutions. 73 
 
 fequently a: — 18, either, equal ro Nothing, or to fome 
 Muhipleof i 8. But, as the leall Valine o' x is required, 
 X — {8 muR bez^io: And theretore 19^ + 7 =349* 
 the Number required. 
 
 OUESTION ex. 
 
 A cert, in Per [on boud^h^ as many Gtefe and Durks, to- 
 gether, as co/i hnn 14 "shillings : for ih^ G^f'/e he paid 
 2.S. id. a-piece ; and for the Ducks is. 3J. What 
 Number haa he of Each ? 
 
 Let X denote the Number of the Geefe, and y That 
 
 of the Ducks ; 
 
 fo fhall i^x -(- 157 = 16^, by the Quejiion ; 
 
 , , p 168 — 26a: -iix — 3 
 and theretore y=. zzz 1 1 — x — -. 
 
 Which being a whole Number, by the Nature of the 
 Problem, ha: — 3 muft, therefore, be (exaftly) divifiblc 
 
 But it is plain that \^x is divifible by 15 ; and that its 
 Excefs above i\x — 3, which is 4^4" 3' "^"^> likewife, 
 be divifible by the fame Number. Let the laft Expref- 
 fion (4^: -\- 3) be now multiply'd by 3, and the preced- 
 ing One (ha;— -3) fubtra£>ed from the Product (in order 
 to get X without a Coefficient) whence you will have 
 ^-j-ia; which ht'mg^ Jiill, divifible by 15, it is plain 
 that ^ muft cither be 3, or 3 added to fomc Muhipleof 
 15, as 18, 33, 48, i^c. But it is apparent, from the 
 Nature of the Queftion, that all thefe Numbers, except 
 the Firft, are too large. Therefore there were 3 Geefe, 
 and 6 Ducks; which laft Number (the Value of x being 
 known) is found dire6lly from the Equation exhibited 
 above. 
 
 QUESTION 
 
74 Algebraical ProblExMi 
 
 QUESTION CXI. 
 
 One having, at Play, won a certain Number of Guineas, 
 not exceeding loo, and being ajked to tell the Nmnber^ 
 made this Reply : ** If the Number of Guineas I have 
 *' won be divided by g, there will remain 6 \ but, ij 
 " the Number of Shillings contained in them be divided 
 " h 39» ih^^^ ^i^f' rernain 12." The (lueJHon is, to 
 find what Number of Guineas he was a Winner of 
 
 Let9A:--|-6 denote the Number fought; where, ac- 
 cording to the Oueftion, x muft be a whole Number. 
 Then, the Number of Shillings being 189A-I-54, it alfo 
 
 , i8qa:+126 — 12 . „ , , ■ 
 appears that — ' , or us Equal 4^: -j- 2 + 
 
 + 39 
 1 2 
 must be a whole Number; and therefore 
 
 iiA- -f- 12, divifible by 13. 
 
 Let the Number 12 (lor the fake of Brevity) be de- 
 noted by n : 
 
 Then, i<^x, andiuv-j-w being, both, divifible by 13, 
 their Difference 2x — n muft alfo be divifible by the fame 
 Number; and fo, likewife, (2^: — «)X5» ^''" ^** Equal 
 IOa: — p^n. And, if this be fubtrafted tiom 1 ix -|-«, the 
 Remainder x -\- 6n [or x -^^ 72) will, fill, be divifible by 
 
 72 7 
 
 1%. But ^^ — is =.5 + -^: Therefore x-X-j muft be 
 
 13 13 
 
 divifible by 13 ; and confequently the Value o{ x, either, 
 equal to 6, or 6 added to fome Multiple of 13 : But, 
 as the Value of gx '\- 6 is not to exceed 100 (by the 
 Question), That of x cannot be greater than 6 : And 
 therefore the Number fought can be no other than 60. 
 
 QUESTION 
 
with their Solutions. 7^ 
 
 QUESTION CXII. 
 
 A Per/on, in exchange for a Number of Pieces of Foreign- 
 Gold, valued at lys. 4^'. each, received a certain 
 . Number of Guineas (not exceeding ^o) and one Shil- 
 ling over. What was the Sum exchanged ? 
 
 If X be put for the Number of Pieces of Foreign- 
 Gold, and y for the Number of Guineas ; tht n it is 
 plain, from the Queftion, that ^2xz=.^'^y-\-o^', and 
 confequently that 
 
 Here iiy-^-n (fuppodug wz=:3) muft be divifible by 
 52; as muft, alfo, iis Quinruple doJ 'k' 5^' And, if 
 from this laft, 52V be fubtra6ted, and the Remainder be 
 multiply 'd by 4, we (hall have i2y-\- jon ; which muft 
 be, ftill, divifible by the fame Number; and fo like- 
 wife its Excefs (jy-f-i9«) above iiy-^n. But -^ 2= 
 
 ^z=z 1 -f- ; therefore ^-l-<5 is, either, equal to 52, or 
 
 to fome Multiple of it ; and confequently y equal, 
 either, to 47, 99, 151, &c, 
 
 Bi t, as, the Value of y, by the Queftion, is not greater 
 than 50, all the Numbers, after the Firft, are too large. 
 Hen c it appears that he received 47 Guineas and one 
 Shilling, in exchange for 57 Foreign- Pieces ; amount, 
 ing in Value to 49/. 8j. Sterhng. 
 
 QUESTION CXIII. 
 
 One laid out 10 Shillings in 20 Fowls, of three different 
 Sorts, viz. Chickens, Pigeons, and Larks ,^ The 
 Chickens coji him 12 d, the Pigeons ^d, and the Larks 
 id, a-piece. How many had he of Each ? 
 
 Let x, y, and z denote the Numbers of the three 
 feveral Sorts, rcfpeftively. 
 
 Then 
 
y6 Algebraical Problems, 
 
 Then will/ ''+'^,+ ^ = ^° Xhy the Quefiion. 
 
 And, by fubtrafting the former of thefe Equations frbm 
 the Latter, we have i ia: -}- 3>' = loo ; and therefore^ j:^ 
 
 = 33 — 3^" . Now, 2JC — 1 being di- 
 
 vifible by 3, it is evident that (^-|-i) its Difference from 
 3;^, muff likewife be divifible by 3 ; and, consequently, 
 that X muft either be 2, or 2 increafed by fome Mul- 
 tiple of 3 ; that is, equal to fome one of the Numbers 2, 
 5, 8, 11, 14, ^c. But, as neither y nor z can be greater 
 than iH (by the Queftion) fo all the foregoing Numbers; 
 below and above 8, give the Value of ^ either too gre^f, 
 or too fmall. 
 
 But, when x is taken 8, y will come out =13 4, and 2 
 rzr 8 J which are the three Numbers required. 
 
 QUESTION CXIV. 
 
 To determine all the fever al Ways whereby it h pqffibl^t,§ 
 pay 60I. in Guifnas and Moidores, only. 
 
 Let X denote the Number of Guineas, and y the 
 Number of Moidores. 
 Then will 2i;c-l-27^ z=: 1200 ; or jjv-f-gjy rz= 400, ^^y 
 
 the Problem ; and therefore xz=.- =57 — y — 
 
 — ^^^^—. From whence, as 2y ■ — 1 is divisible by 7, it will 
 
 7 . . 
 
 appear, by Reafoning as m the preceding Examples, 
 
 that >' + 3 niufl be divisible by the same Number ; and 
 consequently that the leaft Value of 7 is =4, 
 
 and the correfponding Value of ^ (=57 — y — — ) = 
 
 Now, having found the leafl Value of y, and the 
 Greateft of :v, the reft of the Anfwers will be obtained, 
 by adding 7 (the Coefficerit of x in the above Equation) 
 to the lalt Value of y, continually, and fubtrafting 9 
 (the Coefficient of y) from the lall Value of x. By 
 
 means 
 
with their Solutions. 77 
 
 means of ^vhich we get the 6 following Solutions, being 
 all the Queftion admits of. 
 ^;/2 U==5^^ 43' 34' 25, 16, or, 7, 
 \y= 4, 11, 18, 25, 32, or, 39. 
 
 QUESTION CXV. 
 
 To Jindhow many Ways itis poJfibU to pay 20/. in Half- 
 Guineas^ and Half-Crowns^ without any other Sort 
 cf Coin, 
 
 If X be the Number of Half-Guineas, and y the Nnm- 
 ber of Half-Crowns; we Ihall have 2.ix\-5y:=.%QQ ; 
 
 and therefore^ z= zzr: 160— 4^ ■ . Whence 
 
 it appears, at one View, that ;t is a Multiple of 5 : And 
 therefore, the feveral, required, Values of x being ex- 
 preffed by 5, 10, 15, 20, 25, 30, and 35, Thofe of y^ 
 anfwering thereto, muft be 139, ii8, 97, j6, 55, 
 34, and 13, refpeftively. So that there are 7 Anfwers 
 in this Caie. 
 
 QUESTION CXVI. 
 
 A Reckoning of^o Shillings was f pent by a Company of 
 twenty Per sons ^ con/i/iing of Officers^ Sailors, and 
 Marines : Each Officer paid 2/. 6^. each Sailor i2d, 
 and each Marine 8^. How many Perfons were there 
 cfeach Denomination, 
 
 Let the three required Numbers be denoted by x^ y^ 
 and z, refpeftively ; 
 
 fo (hall /^ +H-^= 20 \},y the dueflion. 
 
 And, by fubtra6ling 8 times the former of thefc 
 Equations from the Latter, we fhall have 2 2v^ -J- 4^=80 
 
 and therefore yz=2o . 
 
 But, y being a whole Number, it is plain that x muft 
 be an even Number, and, alfo, lefg than 4 ; and there- 
 fore 
 
7^ Algebraical Problems, 
 
 fore can he no other than 2. From whence y is given 
 =r 9, and z = 9, likewife. 
 
 QUESTION CXVII. 
 
 To find a Number, which, being divided by 28, Jhall 
 produce a Remainder of \c^ ; but, being divided by 19, 
 the Remainder Jhall be 15 ; and^ being divided by 15 ^ 
 the Remainder Jhall be \i. 
 
 Let 2 8;t-|- 19 denote the Number fouorht : where x, 
 accoiding to the firft Condition oF the Pohlem, mufl 
 be a whole Number. And, by the fecond Condition, 
 it appears that 28;c +19 — 15 ^^^ be divifible by ig. 
 Whence (following the Method obferved in the preced- 
 ing Examples) ihe leaft Value of x is found m: 8 : And 
 To 19Z-I-8 (where z denotes any whole Number) is a 
 general Value of x, anfwering the two firll Conditions. 
 
 Let this be, therefore, fubflituted inftead of x; and 
 our affumed Expreffion will become 5322 -\~ 243. 
 From whence, as 532Z-I-232 is divifible by 15, the 
 leafl Value of z will be found 1^=14. And i5«-(-i4, 
 will be a general Value of z : Which, fubftitutt d in 
 532Z-I-243, gives 7980W-I-7691 for a general An- 
 fwer to the Problem; where n may be, either, equally 
 Nothing, or any whole Number. 
 
 QUESTION CXVIII. 
 
 To find three Numbers, in the proportion of f^, 7, and g; 
 which being, jeverally, divided hv 11, 13, and 15, 
 there JJiall remain 1, 2, and 3, refpeBively. 
 
 Let ^x, jx, and 9A' denote the three required Num- 
 bers : Then, by the QueOion, ^x — 1, 'jx — 2, and gx — 3, 
 mud be, refpeBively, divifible by 11, 13, and 15 
 (without leaving any Remainder). 
 
 But it will be found (by proceeding as in the foregoing 
 Problems) that the leaft Value of x, to anfwer the firit 
 of thefe Conditions, will be =9 : Therefore 9-j- iiz 
 
 (wliere 
 
with their Solutions. 79 
 
 (where z denotes any whole Number) will be a general- 
 Value oS X, anfwering the fame Condition. 
 
 Let this Value be, therefore, fubftituted in the fecond 
 and third Expreffions ; which, by that means, will be- 
 come 77z-(-6i, and 99^ + 7^- And then, as the 
 former of Thefe is divifible by 13, the leaft Value of 2, 
 to fulfil this Condition, will (alfo) be found 1^=9. 
 
 Let, therefore, 9-}~*3^^ (which is a general Value 
 for s) be fubflituted in the lafl of the three Expreffions, 
 and it will become (13^^ -1-9)X99 ■i"?^' Which being 
 divifible by 15, the | Part thereof, or (i3«-t-9)X33"l"2^ 
 (3=429?/ -^323) rauft, consequently, be divihbie by ,5. 
 Whence u is found z=: 3 : Therefore s (zzz 13" 4" 9) = 
 48, and X [z=i 112 -f-9) = 537* So that the three leaft 
 Numbers, anfwering the Conditions of the Problem, 
 are 2685, 3759, and 4833. 
 
 QUESTION CXIX. 
 
 Suppqfing 6jf4-7y-|-8zz=rico ; 'tis required to Jin d all 
 the pojffible Vaiuts of x^ y, and z, in whole Numbers, 
 
 In Queftions of this Kind, where you have three, or 
 
 more, indeterminate Quantities, and but one Equation, 
 
 it will be proper, firft of all, to find the Limits of thofe 
 
 _- . . _, , f. 100 — TV — 8z ^ 
 Quantities. Thus, becaule x z=z -/^ =io-^-z- 
 
 ^"^ ^, it appears that x cannot be greater than 14. 
 
 And, in the fame Manner, it will appear thatjy cannot 
 be greater than 12 ; nor z, greater than 10. 
 
 Now, as a: is a whole Number, by the Queftion, 
 y-\-2z — 4 muft therefore be divifible by 6 : And, as 2z 
 and 4 are even Numbers, it is plain that y muft alfo be 
 an even Number (fince an odd One cannot be divided 
 by an eveh One, without a Remainder). Let y be, 
 therefore, firft expounded by the leaft even Number 
 (2), fo will ^ -{- 2z — 4 become z=z 2z — 2 ; which, being 
 iivifible by 6, it is plain that 2 — i (the H«ilf Thereof) is 
 
 divifible 
 
So 
 
 Algebraical Problems, 
 
 divifible by 3 : and confequently that the feveral Values 
 of 2: (when y z=z 2) are 1, 4, 7, and 10. 
 
 Whence the correfponding Values of a-, by fubftituting 
 above, will appear to be 13,9, 5, and 1. 
 
 Let )' be now taken zz:4; then)'-j-22 — 4 will beizi: 
 22 : And fo, 2, being divihble by 3, the feveral Values 
 of z, in this Cafe, will be 3, 6, 9. But the two firft 
 of Thefe, only, are for our Purpofe, the Laft giving 
 
 By taking y :rz 6, and proceeding in the fame Manner, 
 we (hall get two other Anfwers; wherein 2 will be 2, 
 and 5; and x, 7 and 3. And, by taking yz=:^, two 
 more Anfwers will be found (making 10 in the whole) 
 which are all the Queftion adrrits of; and which, being 
 placed in Order, wil ftai;d as below. 
 
 >■ 
 
 •z 
 
 X 
 
 2 
 
 4 
 6 
 8 
 
 < 4.7.10 
 3.6. 
 
 2 5- 
 1.4. 
 
 13.9.^.1 
 
 8.4. 
 
 7-3- 
 6.2. I 
 
 QUESTION CXX. 
 
 Ifi'/X'\- I9y-j-2i /. 1ZZ400; 'iis propofed to find all 
 the pojJioU Values of x, y and 2, in whole pojitive 
 Numbers, 
 
 When the Coefficients of the indeterminate Quan- 
 tities a, y and z, are nearly equal, as in this Exampjes 
 it will be convenient to 5uh 'iuite for the Sum of thofc 
 Quantities : Thus, let x-\-y-\'':^-=zm; then, by fubtracl- 
 ing 17 times this lalt Equation from the preceding One, 
 we (hall have 2>' -j- 42 =: 400 — 1777? ; and by subftrafi- 
 in^ the triven Equation from 21 times the affumed One 
 X'^y\'Z-zr::.rn,K\\^xt. will remain 4;t-[-«;vzr: 21 ;;z — 400. 
 Therefore, lince^y and 2 can have no Values iefs than 
 Unity, it is plain, from the firft of thefe two Equjitions, 
 ihat40d— 17W cannot be Iefs than 6, and therefore m 
 
z^itk their Solutions. 
 
 400 
 
 or 23: Alfo, becaufe by the 
 
 nor greater than 
 
 fccond of the two laft Equations, 9.1m — 400 cannot be 
 lefs than 6, it is obvious that m cannot be lefs than 
 
 400 -^ ^^ ^ Therefore ig and 23 are the Limits of 
 
 21 ^ 
 
 7n in this Cafe. Thefe being determined, let ^x be 
 tranfpofed in the laft Equation, and the Whole divided 
 
 771 
 
 by 2, and we fliall havejy = 10m — 200 — 2jv -|- ■ • 
 which being a whole Number, by the Queftion, it 11 
 
 771 
 
 evident that — muft- likewife be a whole Number, and 
 2 
 
 confequently tti equal to an even Number ; which, as 
 
 the Limits of w are 19 and 23, can only be 20, or 22; 
 
 Let therefore, 7n be firft taken z=: 20, then y will become 
 — ;iQ — 2x and zfTn — x — yj lo-^x; wher^n x being taken 
 equal to 1, 2, 3, and 4 fucceflively, we fhall have y 
 equal to 8, 6, 4, 2 and z equal to 11, 12, 13, 14 re- 
 peftively; which are four of the Anfwers required. 
 
 Again let 7n be taken zii: 22, then willj>'=z=3i — 2x andz=: 
 X — 9, in which let x be taken equal to 10, 11, 12, 13, 
 14 and 15 fucceflively, and ^^ will come out z= 11, 9, 
 7, 5, 3 and 1, andzz=i, 2, 3, 4, 5, and 6, refpec- 
 tively. Therefore we have the ten following Anfwers 
 in whole Numbers ; which are all the Queftion admits 
 of. 
 
 15 
 6^ 
 
 x-z=i 1 
 
 2 
 
 3 
 
 4 
 
 10 
 
 11 
 
 12 
 
 13 
 
 14 
 
 y= 8 
 
 6 
 
 4 
 
 2 
 
 11 
 
 9 
 
 7 
 
 5 
 
 3 
 
 z = n 
 
 12 
 
 13 
 
 14 
 
 1 
 
 2 
 
 3 
 
 4 
 
 ^ 
 
 QUESTION 
 
8s Algebraical Problems, 
 
 QUESTION CXXL 
 
 To find two Whole Numbers^ zv hereof the Difference of 
 the Squares fhall be yy. 
 
 Let the lefler Number be x, and the Greater at -j- w ; 
 and fuppofe the Number glvai to be reprefcnted by a : 
 So {hd\\[x-\-mY — x' — ^, 
 ihdl IS, 2 7nx'\- mm :=. a \ 
 
 and conlequentlv;t=— ■=: '— . 
 
 2m im 2 
 
 Whence we alfo have x-^-jnz: 1 (=---^- J. 
 
 ' 2m 2 \ 2m f 
 
 But, in the Cafe propofed, a being = yy ^ 
 
 X becomes = > and a'-4-;7Zz=: 1 : 
 
 2m 2 2m 2 
 
 Which being both required in Whole Numbers, it is 
 
 evident, in the firft Place, that m muft be fomc 
 
 77 
 Divifor of 77.; and, fecondly, that raufl be greater 
 
 than ^ ; -^nd confequently m lefs than 9. 
 
 But the Oivifors of 77, be.ow 9, are x and 7 . Which 
 Numbers being \\ rote fuccefiiveiy, in the room of m, 
 the correfponding^ Values of ';t will come out 33, and 2; 
 and -hofe of x-^m, 34 and 9, refpeftively : So that 
 the Qucflion, in the Cafe propofed, admits of two 
 Anfwers, and no more. * 
 
 QUESTION CXXII. 
 
 To find a Whole Number^ to which 12 and 2^ being, fuc- 
 cefjivelyy added, both the Sums fhall befquare Numbers. 
 
 Let z be the Number fought; and affuire x and 
 x-\-m{oT the Roots of the two Squares : 
 
 Then,vill{^:l:JJ=f;^^j. } byth^ducjlion. 
 
 Hence, 
 
with their Solutions. 8g 
 
 Hence, by fubtra6ling the former Equation from the 
 Latter, we get 1 3 zzz (a: + ^Y — x'z=z2.mx -{- mm ; 
 
 and therefore ;c =z= — . Which being a Whole 
 
 27n 2 
 
 Number, m muft be = 1 : Whence ^ z=: 6 ; and z 
 
 (=:^' — 12)= 24. 
 
 QUESTION CXXIII. 
 
 To find three Whole Numbers, Jo that the Sum of the 
 Squares of the Two leajl of them Jhall be equal to the 
 Square of the GreateJL 
 
 It appears, from the Problem preceding the Laft, 
 
 that the Difference of the Squares of ^ — ,and 
 
 is, univerfally, equal to a, 
 
 (a-X-mmY (a — 7n7nY , < ^ , » , 
 
 or -i — ? '- L zzzia ; let a and m be what they 
 
 ^mm . \mm ' ' 
 
 will. Whence it is alfo plain that {a-\-mmy-r:=i{^a — mm^ 
 
 -f- \mma» 
 
 But, fmce it is required to have ^mma^ a fquare Num- 
 ber (as well as the other Two) a muft therefore be a 
 fquare Number ; let it be n\ and then our Equation 
 will become {nn-\-mmyz=z(nn — zraw)*-|-(27w«)*: Wnerewt 
 and n may be expounded by any Whole Numbers, at 
 Pleafure. 
 
 Thus, for Example, fuppofe w=: 1, and nz=: 2 ; then 
 there will come out 5*=3'-|-4'. Again, let m=:2, 
 and «=3, and there arifes (i3)'=(5)* + i^^Y' Laftly, 
 letw:=:2 and «=5, and you will get (2o)*=:(2i)'-j-(2o)'. 
 
 QUESTION CXXIV. 
 
 To finS three Whole Numbers, whofe Squares are in 
 Arithmetical Progrejion, 
 
 Let X, X'\'m, and x-^-n exprefs three fuch Numbers. 
 So (hall {x\-mY^x^z=:{x'\-nY--4^X'\-m)\ by the Nature 
 of the Problem. 
 
 Whence x is found = ■ 
 
 4W — 2» 
 
 G 2 
 
84 Algebraical Problems, . 
 
 h 
 
 Put Am — 2« =^, and «' — 2m^z=.h ; then i? — -^ x -^-in 
 
 a 
 
 — —i . 2ind X ■■+-n:z=z. — . iNow, as the bqua «s 
 
 h b-X-am - b-^an . ... • , t, r 
 
 q[ ^ ^ — ! ^ and -J — are in Arithmetical Proffrel- 
 
 a ' a a 
 
 fion, it is plain that the Squares of their Equimultiples^ 
 b, b'\-am^ and />-{-««, muft be in Arithmetical Progref- 
 fion likewife. From whence, by expounding ?n and n 
 hy different Whole Numbers, fucceffively, as many par- 
 ticular Anfwers as you pleafe, may be exhibited. 
 
 Thus, if 7W iizz 2 and r« = 3 ; then, di being zzr 2, and 
 b^zLi, there will come out i, 5, and 7. But, it w = 3 
 and «=5, we fhall get 7, 13, and ij^ for another 
 Anfwer. 
 
 QUESTION CXXV. 
 
 Suppofing y.'^z=::z^-\-^z-\-^ (where a and h denote given 
 Numbtrs ) ; 'tis required to find the Values of :l and z 
 (ifpojfible) in Whole Numbers. 
 
 Put x-=.'x>-\-m\ then, by Subftitution, 
 
 and confequently 2;=: __ . Which Value, by putting 
 
 , a — n\. a" — ^an-\-n'^ — Ab 
 
 a — 9.W. — wfor m zzn 1 becomesrr: ' i— -— . 
 
 ^ 2 / ^n 
 
 — 2 za^n). Prom which it is evident, that, 
 
 to have the Value of z a Whole Number, n miift be 
 fome Divifor of the given Quantity aa — 4^, and there- 
 fore even or odd, according as a is even or odd. 
 
 Example. Suppofe the given Equation to become x'^z=z 
 z'^'\- 20Z ; in which Cafe, a- being z=: 20, and b z=o, we 
 
 have »=|x(^ 40+«); where the, even, Divifors 
 
 of 400 are 2, 4; 8, lo, 0c. whereof the Second will 
 be found to answer ; the Values of s and x coming out 
 16 and 24, respe£tively. 
 
 Again 
 
 iof( 
 
with their Solutions. 8^ 
 
 Again, fuppofe the given Equation to become ^' z=; 
 
 2'4"*ooz+^ooo : Here we have z=Jx{'~ 20o4-«): 
 
 And the, even, Divifors of 6000, are 2, 4, 6, 8, 10, 
 12, 16, 20, &c. Whereof 4, 12, and 20 fuccced; 
 By the laft of Thefc (which determines the leaft Values) 
 2 comes out :zr: 30, and xzzz 70. 
 
 QUESTION CXXVI. 
 
 Having given x*zz=a'-j-bz-j-cz*, wherein ^i, b and e 
 denote given Whole Numbers ; 'tis required to jind the 
 Values ofx and z (if pojjible) in whole Numbers, 
 
 Put x^zza-^-mz ; then will [a-^-mzYzzza* -\- bz-\' cz' ; 
 
 that is, a'-\-2amz-^n^z^=:a^-]-bz-\-cz\ 
 
 W7U I. ^ — 2aw 
 
 Whence z comes out =: 
 
 mm — c 
 
 Where it is evident, that, in order to have a pofitive 
 
 VaJue, m muft be taken equal to fome Number between 
 
 -i/c, and — .. 
 ^ Qa 
 
 Thus, fuppofing the given Equation to become x'zzz 
 
 64 — i2z-|- 5%', the Value of m, in this Cafe, muft be 
 
 12 
 lefs than y^^, and greater thaM ^. Let it there- 
 fore be expounded by 2 and 1, fucceflively ; 
 
 Txri — 12 — t6m . -c- 1 i24-i6wz , T_. , . 
 
 Whence , or its r-qual — ' { which i». 
 
 mm — 5 < 5 — mm ^ 
 
 here, -the Value of z) will come out 44 and 7 refpeftively; 
 
 and the correfponding Values of x [a-^-mz) arc found to 
 
 be 96 and 15 ; Both which anfwer the Conditions of the 
 
 Problem. 
 
 Gg PART 
 
PART II. 
 
 CONTAINING 
 
 A Variety of Geometrical Problems, 
 with ^^<?/r Solutions : Both by Algebra, 
 and independent of it, from Principles purely 
 Geometrical. 
 
 ALTHOUGH the chief Defign of this Work 
 confifts not in Exhibiting a Number of Rules 
 and Precepts, but in illuftrating Thofe already 
 known, on the Subjeft, by a Set of proper and ufeful 
 Examples ; yet, as we are in this Part to treat of 
 the Inveftigation of Geometrical Problems, without 
 calling in the Afliftance of Algebra (a Thing hitherto 
 very little confidered by Authors, tho* in itfelf very in- 
 terefting and ufeful) it may not be amifs, before we 
 proceed to particular Cafes, to premifc a few General 
 Obfervations on this Head. 
 
 i". In the firft Place, then, it is neceffary, in order 
 to the Conftruftion of Geometrical Problems, that 
 fomething of the Geometric Led fhould be underftood : 
 Thus it will be of Ufe to know that the Place of the 
 Vertex of a Triangle, whereof the Bafe and oppofite 
 Angle are fuppofed to remain conftant, while the other 
 Sides and Angles vary, will always fall in the Circum* 
 ference of a Circle pafliing thro' the Extremities of the 
 Bafe. This appears frorh Euclid B. 3. Prop. 21; and 
 is alfo dcmonftrated in my Elements of Geometry^ B» 3, 
 Prop. 11th. 
 
 It ought moreover to be known, that, the Locus 
 of the Vertex, when the Ratio of the two Sides and the 
 Bafe of the Triangle arc given, or continue invariable, 
 will, aljo^ be the Circumference of a Circle, dividing 
 
 th« 
 
Geometrical Problems. 87 
 
 the Bafc in the given Ritio (For the Demonftration of 
 which, fee Elem. Geometry Prob, xm.p. 220* J. 
 
 If the Sum, or the Difference, of the Squares of the 
 two Sides, together with the Bafe, be fuppofed given, 
 the Locus of the Vertex will, Jii/l, be the Circum- 
 ference of a Circle, in the former Cafe; and a Right- 
 line, in the latter fVid. B. 3. Theorem 20. J 
 
 But, if the Sum, or the Difference, of the Sides, 
 themfelvcs, is given ; then tha Locus will, either, be 
 the Arc of an Eilipfis or an Hyperbola (ai is well known 
 to Thofe who have touch'd upon Conic-Seftions f). But 
 thefe two Lafl are not admitted in the Conflruftion of 
 linear and plane Problems ; of which, only, we purpofe 
 
 to treat The Problems, of the following Col- 
 
 leftion, wherein the Ufe of the Geometric Loci, above 
 fpecified, is particularly exemplified, are the 11, 12, 19, 
 34' 35.48, and^i^ 
 
 2°. When, in the Figure to be conflru£led, the Sum, 
 or the Difference, of two adjacent Sides happens to be 
 given; it will be proper, firft, to form a Triangle, 
 fo that the faid Sum, or Difference, may be one of 
 its Sides ; and, then, to confider, what other Sides, 
 or Angles, will be given, or become known, in Con- 
 fequence Thereof. This Rule is illuftrated in the 1, 
 2, 18, 26, 2f^, and fome other of the fucceeding 
 Problems. 
 
 3"*. It often happens that the Ratio of two, or more. 
 Lines is given, from the Nature of the Figure, or by 
 Hypothefis, though the Lines themfelves are abfolutely 
 unknown : In all fuch Cafes we muff endeavour, by 
 drawing Parallels (or fome other Way)- to obtain other 
 Lines in the fame given Ratio ; fo that, one of them 
 being given, or known from the Nature of the Figure or 
 Problem, the other may alfo become known— The Ufe^f 
 this Rule, which is very extenfive, will particularly ap^ 
 
 * See alfo the Lemma on p. 334 oithe Author sAlgebra, 
 gth Edition. 
 
 f See Prop, xiv. of Hamilton's ingenious Treatife 9n 
 the C$nic Sedions^ p. 99. 
 
 G 4 pear 
 
t% ©EOMETRICAL PROBLEMS, 
 
 pear in the Solutions to the 3, 13, 16, 22, 26, 'X^, V. and 
 56'^ Problems. 
 
 4°, But, if the Lines, whereof the Ratio is given, 
 ftiould happen to lye remote of each other ; then, by 
 the Help of Thofe, we muft endeavour to deter- 
 mine the Ratio of Others, lying nearer together ; , 
 and fo on, till we obtain {if poflible) the Ratio of two 
 Lines, that are both Sides of the fame Triangle ; where- 
 in one Angle and the remaining Side (or fome other, 
 two, Parts) are given — Fpr the better underftanding 
 this Rule, confuk the Solutions to the 49*"* and ^f^ 
 Problems, in particular. 
 
 5°. Laftly, when the Reftangle under two unknown 
 Lines is given, either, a mean Proportional muft be 
 found, or elfe, two other Lines muft be afligned, by 
 forming (imilar Triangles for fomc other Way) fo as to 
 comprehend an equal Re61angle; and fo that, one of 
 them being given by the Nature of the Figure, the 
 Other may alfo become known. This Rule is ex- 
 emplified in the 4, 5, 6, 9, 10, 21, 38, 40, 41, 44, 
 and 53'*. Problems. 
 
 Behdes the above, other Obfervations might be here 
 laid down ; but Thofe already delivered being the moft 
 General that have occur'd to me, I (hall now proceed 
 on in the Refolution of Problems; the proper Bufinefii 
 of this Work. 
 
 PROBLEM 
 
with their Solutions. S9 
 
 PROBLEM I. 
 
 One Leg BC, and the Difference between the other Leg 
 AB and the Hypothenuje AC, of a right-angled Tri- 
 angle ABC, being given ; to find both AB and BG. 
 
 Put BC — ^, AB =a:, and AC=a:+^ ; (b being the 
 
 given Difference) Then, C 
 
 (AC)^ being =(AB)^+ 
 
 (BC)* (Elem. 8. 2. *) we 
 
 have XX -j- '2.bx -\- bb ^^r 
 
 'Kx -[- aa : Whence ^bx 
 
 znzaa — bb 'f and confe- 
 
 , aa — bb aa ^^^ v 
 
 quemly >c = --^=-^ ^_J 
 
 — - . From which AC {x'\-b) is given =1^7-+ - . 
 
 Geometrically, 
 
 If, in AB produced, there be taken BD equa! to 
 the given Difference of AC and AB, and DC be drawn 
 (according to the fecond General Obfervation) it is evi- 
 dent that AD will be equal to AC ,• and the Angle 
 A CD, alfo, equal to the Angle D. 
 
 Therefore, having taken BD as above fpecified, and 
 made BC perpendicular thereto, and of the given 
 Length, and joined D, C ; let CA be fo drawn as to 
 make the Angle DCAizrD ; or, inftead thereof, let a 
 perpendicular EA be erefted on the middle of CD; 
 then the Intcrfeftion of either of thefe Lines with DB, 
 produced, determines the Triangle. 
 
 From this Conftruftion we have the very fame 
 Theorem, for the numerical Solution, as is derived 
 above, from the Algebraical Procefs : For the Triangles 
 ADE and CDB, having each a Right-Angle, and^'D 
 
 * Note: The Q^uotations, in this ^ and the Jucceeding 
 Problems, refer to the Author s Elements of Geometry ; 
 $th Edition, printed for Y , Wingrave. 
 
 common. 
 
9® Geometrical Problems, 
 
 common, arc fimilar. Therefore DB : DC 
 
 DO 
 
 (IDC) : AD (AC)=^=:?^y-^-5^ 
 2Di3 
 
 BC^ 
 
 DE 
 BD 
 
 as before. 
 
 2BD 
 
 2BD 
 
 PROBLEM II. 
 
 The Hypothenufe AC, and the Difference of the two Legs 
 AB andBC, of a right-angled Triangle ABC, being 
 given; to determine the Legs. 
 
 E Put AC = ^, AB = a:, 
 ;C and 'S>C-=zx— b : 
 
 So ftiali XX -\-[x — by — /7V7. 
 
 fEiem. 8. 2.). 
 
 that is, 2xx — ibx-^bh — a/7.. 
 
 Txn 7 ^<2 bb 
 Whence xx — bx — — 
 
 2 2' 
 
 And confequently x zz= 
 
 Geometrically, 
 
 If, in AB there be taken AD equal to the given Dif- 
 ference, and CD be drawn ; then, DB being =: BC, 
 the Angle BDCwiU aifo be =:BCD = Half a Right- 
 Angle. 
 
 Therefore, having laid down AD, and drawn an in- 
 definite Line DC i^ to make the Angle BDE z=: | Right- 
 Angle; upon the Center A, with the given Interval 
 AC, let an Arch be defcribed, inferfcfting DE in C ; 
 from which Point, upon AD produced, let fall the Per- 
 pendicular CB; fo ftiall ABC be the Triangle required. 
 
 The numerical Solution, according to this Con- 
 llru£lion, is very eafy, by the Help of Trigonometry; 
 For, two Sides and one Angle ot the Triangle ADC 
 being given, the other Angles may from thence be 
 found ; and then, all the Angles and one Side (AC) of 
 the propofed Triangle being known, the other Sides 
 AB and BC mayalfobe determined. 
 
 PROBLEM 
 
'\th their Solutions. 
 
 PROBLEM III. 
 
 The Bafe AB and the Perpendicular CD of any Triangif 
 ABC being given ; tojindthe Side EF, cr EH, of the 
 infcribed Square EFGH. 
 
 PutCDz=:a,AB= 
 h, andDI( = EF) = 
 x: Then will Cl = 
 a — X ; and, by Reafon 
 of the fimilar Tri- 
 angles ABC and 
 ECF, we {hall have 
 a\b:'.a — x '. x [z=z 
 EF). Therefore ax 
 - — ' ab — bx; and con- 
 
 fequcntly x 
 
 c 
 
 / 
 
 \ 
 
 \ X 
 
 T'-, 
 
 / 
 
 I 
 
 \ 
 
 i ■ 
 
 / 
 
 
 
 y 
 
 /. 
 
 / 
 
 / 
 
 
 \ 
 
 \ 
 
 \ 
 
 A H D 
 
 d-\-b 
 
 Geometrically, 
 
 The Ratio of EH to EA being given, as CD to CA, 
 EF muft therefore be to-EA in the fame given Ratio : 
 And, if CL be drawn parallel to EF, "Vneeting AF pro- 
 duced in L (agreeable to the 3** General Obfervation) it is 
 Evident, becaufc of the fimilar Triangles, that the Line 
 CL, fo drawn, will be to CA, Jiill, in the fame given 
 Ratio; that is, CL : CA : : CD : CA ; and con fe- 
 quently CL=CD. Whencethc Method of Conftruftion 
 is manifeft. 
 
 PROBLEM IV. 
 
 To determine the Sides of a Re8 angle, EFGH, infcribed 
 in a given Triangle ABC, whofe Areafhall be to That 
 of the Triangle in a given Ratio. 
 
 Put the Perpendicular CDzr::^, the Bafe ABzz;^, and 
 the Altitude EH of the Reftangle z=jx ; and let the 
 given Ratio of ABC to EFGH be that m to «. 
 
 Becaufe 
 
92 
 
 Geometrical Problems, 
 
 Becaufe of the fimilar Triangles ABC and EFC, it 
 ^"'^^C will be, CD (a) : AB 
 
 (b) : : CI (a—x) : EF= 
 
 ( Elem. 12. A, ) 
 
 a "^ 
 
 Therefore EF x EH = 
 
 abx — bx"" . 
 
 ; and conle- 
 
 a 
 
 ab 
 B quently m : n : : — : 
 s 
 
 fby the QueJiionJ, 
 
 Hence ax — xx = — ; and therefore x zzz h 
 
 2m 2 — 
 
 A 
 
 aa naa\ 
 
 4 
 
 zm 
 
 / 
 
 Geometrically, 
 
 The Reftangle HF being to the Triangle ABC in a 
 given Ratio, and the latter of Thefe being aftually 
 given, the Magnitude of the Former is alfo given ; and 
 therefore may be exprefled by a given Re£laogle ABPL, 
 en the Bafe AB ; whofe Altitude KD is to Half That 
 of the Triangle in the aforefaid given Ratio. 
 
 But it appears that the Reftangle DIX^F is to the 
 Re6tangle DIx^C in the given Ratio of EF to IC, or 
 of AB to CD (Elem, i and 12. 4) : and that the Rec- 
 tangle DKxAB is alfo to DKxCD in the fame given 
 Ratio. Therefore, the Antecedents being equal, the 
 Confequents muft likewife be equal, or DIxIC = DK 
 XCD. Whence this Conftruftion. 
 
 Defcribe, upon CD and CK, two Semi-circles ; and, 
 from the Point M wherein the Circumference of the 
 latter cuts AB, let MN be drawn, parallel to DC in- 
 lerfefting the Former in N ; fo (haM MN be the required 
 Altitude of the Reaangle. Since 01X^1 = (IN)' =: 
 (DM)^=DKXCD fElem. xi. 4). as above, 
 
 Thif 
 
u'ith their Solutions. 
 
 93 
 
 This Problem, it is obfervable, becomes impoffible 
 when MN paiTes intirely witl\oiU the leffer Semi-circle, 
 that is. when the given Reftangle is fuppofed greater 
 than half the Triangle. ' The fame thing appears alfo 
 from the Algebraic Solution, in which Cafe the Quantity 
 
 ( — ^ J under the Radical-Sign, becomes negative, 
 
 PROBLEM V. 
 
 To divide a given Right-line AB into fuch Farts AC 
 and BC, that the ReBangU contained under them may 
 be- of a given Magnitude. 
 
 Put ABzn:^, and ACmA:, and let the given Magni- 
 tude, or Content, of the 
 propofed Reftangle be re- 
 presented by the Square 
 V*Ti whofe Side BE, or 
 ED, let be denoted by 
 
 b\ Then will ;<:Xf^ — ^;Z=: 
 hb ', QX XX —axz^z — bb. 
 
 
 "^"^ 
 
 p-F 
 
 ¥. 
 
 
 
 N^ 
 
 1 
 
 ( 
 
 
 
 
 \ 
 
 \ 
 
 \ 
 
 OH C 
 
 B 
 
 Whence x z=:4:\/(i^«? — bb-^ — ^ Y 
 
 Geometrically, 
 
 If upon AB, as a Diameter, a Semi- circle AFB be 
 defcribed, it is evident (by Elem. 11.4.) that a Perpen- 
 dicular FC, drawn from the Point wherein the Circura. 
 ference interfefts DE, will cut AB in the Point requir- 
 ed— It i? plain, from both thefe Solutions, that the given 
 Rectangle muft not be greater than the Square of half 
 the given Line, to be divided. 
 
 PROBLEM 
 
94 
 
 Geometrical Problems, 
 
 PROBLEM VI. 
 
 To a given Line AB, it is required t^ add another Line 
 BC,/^ tJiat the ReSfangle under the whole, compound- 
 ed. Line AC, and the l:'art added^ may be of a given 
 Magnitude, 
 
 Let AC=:fl, BCzzTAT, 
 13 ' 7E 
 
 and 
 
 the Side of the given 
 Square BEDH, ex- 
 prefling the Magni- 
 tude ot the propofed 
 Rcftangle, =^. 
 Then we fhall have 
 [a'\rx) y(^xz=.bb; and 
 confequently x zzn |/ 
 [bb'\-\aa)'^\a. 
 
 Geometrically, 
 
 If, upon AB, a Semi- circle be defcribed, and CF be 
 fuppofed drawn to touch it in D, it is plain, from Elem, 
 Corol foil 3, that(CF)Ms = ACxBC = (BE)Y^J 
 Hypothejisj \ and confequently CFz= BE : Therefore, 
 OF beingr=OB, it follows that OE and OC are like- 
 wife equal. Hence, if to the Middle of ■ AB, we draw 
 EO, and take OC =3»EO, the Thing is done. 
 
 PROBLEM 
 
with their Solutions. 
 
 95 
 
 PROBLEM VII. 
 
 To divide a givin Right-line AB into two fuch Parts, 
 that, the ReBanglc under one oj them AC and another^ 
 given. Line BD, may be equal to the Square of the 
 remaining Fart BC 
 
 Put AB=fl, BD=^, 
 and BCiiz:;c: then will 
 AC zn: a — x. 
 and therefore xxz=!a — x) 
 X ^, by the Qiiflhon. 
 Her>ce xx \- bx z=: ab ; 
 and, confequcntly x zz= 
 ^',ab-^lbb)—\b. A 
 
 B F B 
 
 Geometrically, 
 
 Since fBCVcrr ACX BD fby Hypothecs J it follows, by 
 addina BCx'BDtoEach, that (.BC)^-|^BCxBD--::AC 
 XBD+BCxBD; or that, BCxCD==BDxAB=: 
 (BE)% taking BE a Mean Proportional between BD and 
 AB fby Elem. 14. 5.) But the Reaanglc BCxCD, 
 if a Semi-circle be defcribed upon the Diameter iSD, is 
 known to be equal to the Square of the Tangent CG 
 (Elem, Cor. to 22. 3). Hence (CG)'=CBE/ ; and con. 
 fequcntly CG=;BE : Therefore, FG being alfo = FB, 
 it follows that EC is equal to FE ; whence the Method 
 of Conftru6lion is manifeft. 
 
 PROBLEM 
 
f6 Geometrical Problimi, 
 
 PROBLEM VIII. 
 
 To determine two Lines, whereof the Re&angle Jltall be 
 equal to a given ReBangle ABFE, and the Sum of the 
 Squares equal to a given Square ABCD, 
 
 C Put AB (=BC)z=:a, BF 
 (z=:AE) = /^, and let the 
 two required Lines be de- 
 noted by X and^. 
 
 p Then will xy-rzLah, and 
 XX '\-yyz=:aa^ by the Quef- 
 tion : Whence by adding, 
 and fubtra6ling, the Double 
 of the former of thefe Equa- 
 
 B tions from the Latter, 
 
 we have /^^ + 2^^4-M = (-^+>')^] = ^^ + 2^^. 
 we nave "^^^x— -^xy ■^yyS^^{x—yY\=aa'—zab. 
 
 Therefore /^ +^ = ^!)^;^ +^^f ^ 1 
 
 (ir 1-- 1- f 2xz=Z'\/(aa'X-2ab)'{'\/(aa — 2ab) \ 
 
 From which {^^ ^ J^I^^^Ia^^jZ^l^a-za*)./ 
 
 Geometrically, 
 
 If upon AB a Semi-circle be defcribed, interfc£ling 
 EF ip^> then the Lines joining A, H, and B, H will 
 anfwer the Conditions of the Problem. \ For, the Angle 
 AHB being a Right-one (Elem. 13. 3.) thence is (AH)' 
 4-(BH)"==(AB)^ = ABCD (Elem. 7, 2.) ; and AHx 
 BH r= twice the Triangle ABHJ == ABFE (Elem. 
 Corel, to 2. 2), 
 
 PROBLEM 
 
wiik their Solutions. 
 
 97 
 
 PROBLEM IX. 
 
 To determine two Lines, whereof the ReBangle Jhall he 
 equal to a given Rrilangle ABFE, and the Difftrencc 
 oj their Squares equal to a given Square ABHI). 
 
 Put AB = ^z, BFzr:^; and let x be the greater, and 
 
 ^ the lefler, of the two £, ;, H^ 
 
 Lines required : So D 
 (hall xyzzzicib ; and xx — 
 ^v=:«itf, by the Quejiion. 
 From the former of 
 which Equations we 
 
 have y z=. ; which 
 
 . * A 
 
 Value, fubftituted in the Latter, gives xx 
 
 Hence x^ — a';c'=«*^» ; 
 
 and confequently x r=z y^[ aa-\-a •/'bh -[- \ <»^)] • 
 
 Whence^ will alfo be known. 
 
 Geometrically, 
 
 It is evident, in the firft Place, that the two Lines to 
 be determined will be the Hypothenufe and one Leg of a 
 right-anglerl Triangle (ABG) vvhofe remaining Leg is the 
 given Line AB. And, fince the Re6iangle under thcfe 
 Lines is fuppofed given, another Triangle ACG, fimilar 
 to ABG, muft therefore be alTumed; fo that CG, in the 
 Former, and BG, in the Latter, may be homologous 
 Sides (according to the ^'^ General Ob/ervation ; vid^ 
 p. 88.) 
 
 Hence we have CGx AB=:BGxAG fElem, 19. 
 4 )=:ABXBF fby H.p.) and confeq;;ently CG=:BF ; 
 
 And (o, ACxBC=fCG)% Efem. Cor. 19. 4; be^ 
 ing given =(BF/, the Cdfe under Confideration is now 
 reduced to our 6*'* Problem : Whence we have the fol- 
 lowing Conftru£lion. 
 
 H To 
 
^8 
 
 Geometrical Problems. 
 
 To the Middle of AP, let FO be drawn; and. 
 having taken OC=:OF, let a Semi.circle be defcribed 
 upon AC, cutting BF in G ; fo {hall BG be the lefler, 
 md AG the greater of the two Lines required. 
 
 PROBLEM X. 
 
 The Diameter AB of a Semi-circle being given, to find a 
 Point D in the Perpendicular BC, from whence DA 
 being drawn, the Part thereof, ED, without the Semi^ 
 circle, fhall be of a given Length (HF). 
 
 Put- DE(=BF;=^, AB=^, and AD = a: ; alfo let 
 BE be drawn. Becaufe the Angle AEB is a Right- 
 1 C one, the Triangles ADB 
 
 and ABE are fimilar. 
 And therefore AD ix) : 
 AB [b) : : AB [b) : 
 AE (^ — a). Whence 
 XX — axz=zbh\ and con- 
 fequentiy x z=z. ^{bb -f- 
 
 Geometrically, 
 
 Since DAx^A \% z=l^^^ ( Elem, 19. 4) where the 
 Part DEof DA is iTiven (::i=BF) the Cafe is therefore 
 reduced to our 6*" Problem. 
 
 From whence it will appear, that, if upon BF a 
 Semi-circle be defcribed, and through the Center there- 
 of, AH be drawn, meeting the Periphery in H, art 
 Arch defcribed from the Center A, with the Radius AH, 
 will cut BC in the Point required. 
 
 PROBLEM XL 
 
 Tldjjing the Length of two Chords AB and CD, cutting 
 each other at Right Angles, together with the Di/iance 
 OE of the Point of their Inter feElion from the Centtr ; 
 to determine the Diameter of the Circte, 
 
 Upon the given Chords, from the Center O, let fall 
 
 the 
 
with their Solutions, 
 
 99 
 
 the Perpendiculars OF and OG ; ^nd draw the Radii 
 OA and OD: Alfo put AF (=iAB)z=^, DG 
 ( = |CD)=:^, OE=^ C 
 and AO ( =i= DO ) == a:. 
 
 Then will (OYf =zx'—a' 
 and {OGYzizix' — b'; but ^/ 
 (0F)^+(6G)^ = (0F)^ + 
 (FE)* is r=(OE)^; that is, 
 2^-*— 'tf' — b'^-izzc^ ; and confe- 
 
 uently xz=:y/{ — — - — j 
 
 Whence the Diameter is given r= y/[ 2^* -j- %b^ J^ 2C^] 
 
 Geometrically, 
 
 Since (FE)' + (DG)= is ( = ( OD )' = ( OA)*) 
 =±:(OF)*-f{AF)% it is evident that (FFJ'--(OF)* is 
 given =(AF)'— (DG)*. We are therefore to con- 
 ftruft a right-angled Triangle upon the given Hypothe- 
 nufe OE, whereof the Squares of the two Legs fhalj 
 have the fame Difference as the two given Squares (AF)» 
 and (DGJ\ 
 
 In order, to which, upon OE, let a Serai-circle be 
 dcfcribed ; alfo, from the Centers O and E, with Radii 
 equal to DG and AF, refpeftively, let two Arcs be 
 defcribed, interfering each other in H ; from which 
 Point, upon OE, let fall the Perpendicular HI ; which 
 will interfeft the Semi-circle in (F) the Vertex of the 
 required Triangle : Since it is evident that (FE)^ — (GF)* 
 = (EI)*— (dl)* =fEH)*-^(OH)* = (AF)* — (DG)' 
 {by Conjiru^ion.) 
 
 Therefore, if in EF produced, FA be taken of the 
 given Length, OA (when drawn) will be the Radius 
 of the required Circle. 
 
 H 1 
 
 PROBLEM 
 
loo GlOMETRICAL PROBLEMS, 
 
 PROBLEM XII. 
 
 To draw a Right line to cut two given concentric Cirdts, 
 OAB. OZl^Jothat the Chords, or Parts ojthefaid 
 Line intercepted by thofc Circles, may obtain a criyen 
 Ratio, * 
 
 Put the Radius OA of the greater Circle =5, and 
 the Radius OC of the Leffer^v^ ; and let the given 
 Ratio of AB to CD be That 
 of m io n : Then, denoting 
 ^OE, the Diftance of A B 
 ' from the Center, by x, we 
 haveAE^=:aa — xx, and CE' 
 z=zbb — XX, Therefore, AE 
 being =1 AB, and CE=:| 
 CD, it follows, that, aa — xx : 
 bb — XX : : m* : n*; and con- 
 fequently n* a^ — n*x*z=zm' b* 
 ___, .fmmbb — nnaa\ _ ^, 
 
 — »i*a:*: Whence xz=z\/\ -~J By Means 
 
 V mm — nn / ' 
 
 of which AB may be drawn. 
 
 Geometrically. 
 
 Since the Ratio of AE to GE is given, let OC be 
 produced to F, fo that OF may be to OC in the fame, 
 given Ratio, (agreeable to the 3" General ObfervationJ 
 then. A, F being joined, the Triangle CAF will be 
 fimiiar to the Triangle CEO ; and confequently the 
 Angle CAF a Right one. Hence the following Con- 
 flruftion. 
 
 Having drawn the Radius OC, and in it, produced, 
 taken OF in" proportion thereto, as AB is to CD fas 
 above fpecijied) let a Semi-Circle, upon CF be defcribed, 
 interfering the greater of the two given Circles in A ; 
 from which Point, through C, draw AB, and the thing 
 is done. 
 
 It is manifeft, toth from This, and the Algebraic 
 Solution, that the Ratio of m\.o n (or of AB to CD) 
 
 cannot 
 
with their Solutions. 
 
 toi 
 
 eannot be given lefs than That of O A to OC, without 
 rendering the Problem impoflible. 
 
 PROBLEM XIII. 
 
 To deUrmine the Radii of three equal Circles xA, B, C, 
 dtfcribed in a given Circle HIK to touch each other 
 and likewife the Circumference of the given Circle, 
 
 Let the Centers of the feveral Circles be joined ; and 
 let AO and BO be 
 produced to bife£l 
 BC and AC in D 
 and E: Al fo let the 
 Radius (OI) of the 
 given Circle " be de- 
 noted by a, and 
 That of each of the 
 required Ones by x. 
 Now the Triangles 
 BCE and BOD being 
 fimilar, and CE = i 
 BC, it appears that OD is also = \ OB. But (OB)*— 
 
 [a x^ 
 
 (ODV is = (BD)* ; that is, in Species, (j— a;)' ^ '-' 
 
 4 
 z=zxx. Which, solved, gives x =y^(i2aa) — 3« = 
 
 '^XCVS— 3)- 
 
 Geometrically, 
 
 It is evident that the Right-line IK, joining the Points 
 of Contaft I and K, is the Side of an Equilateral Tri- 
 angle infcribed in the given Circle: And, that, if in 
 OK produced there be taken KL = \ IK, a Line drawn 
 from 1 to L, will be parallel to another Line drawn 
 from B to K ; becaufe the Triangles IKL and BCK 
 (having IKL = BCK, and IK : KL ( : : 2:1):: BC : 
 CK) are equiangular. 
 
 Therefore, in order to the Geometrical Conftru6lion, 
 
 having firft drawn the Radii OH, OI, and OK, to 
 
 H % divide 
 
lOS 
 
 Geometrical Problems, 
 
 ^^^i,;- 
 
 divide the Circumference in three equal Parts, and, 
 taken KL, iil OK produced, equal to f IK; draw LI, 
 and KB, parallel thereto, meeting OI in B ; make OA 
 and OC each = OB ; and upon the Centers A, B, 
 and C, through A, I, K let the three required Circles 
 be dcfcribcd. 
 
 PROBLEM XIV. 
 
 Suppojing AB and ABD to be two Arcs ^f a Circle ^ in the 
 Ratio of 1 to <^\ and that both their Chords AB and 
 AD are given ; 'tis required to find the Radius oj the 
 Circle. 
 
 Let the Arch BD be equally divided in C, and upon 
 AD let fall the Perpendiculars BE and CF ; aifo let AG 
 be perpendicular to the Radius OB : And put AB === a, 
 AD zz= b, and AG (= OB) = x. 
 
 It is evident, becaufe AB, BC and Cl> uxe all equal, 
 thatAE=DF, a^id 'EF — 
 BC = a: Therefore AE 4- 
 DFz=:*— a, ^pd AE zj:;?: 
 
 -!- — '-. It alfo appears tb^^ 
 
 the Triangles ABE and 
 AOG are equiangular, be- 
 caufe the Angle BAD, 
 {landing upon the Arch 
 BD, is equal to the Angle 
 O, at the Center, (land- 
 ing upon AB (=;: \ BDy' : Hence we have, AE» (a) : 
 
 AE [LZI-]: .f,0 (x):OgJ^^^:^M1. But 
 
 \ 2 J 2a 
 
 (AB)»=(OA)*-f (OB)*— t OB+OG; tkat is, in 
 
 Species, a'=x'+x'— i '^ ^ ^ or -^ = ^ax' — bx' 
 
 Therefore x = x/{~r^) = a /(^T^)' 
 
 From 
 
with their Solutioks. 103 
 
 From the fame Equation, of the Radius AO (x) and 
 the Chord fbj of an Arch ABD be fuppofed given, the 
 Chord AB (a J of the Sub-triple of that Arch, may be 
 determined ; But this, by the bye. 
 
 Geometrically. 
 
 The Geometrical ConftrucSlion of the propofed Pro- 
 blem is alfo obvious from the known Value of AE and 
 the Equality of the Angles O and EAB ; and is thus : 
 Draw AD of the given Length, from which take DH 
 ;=: AB ; let the Remainder AH be bifefted by the per- 
 pendicular EI ; to which draw AB fo as to be of the 
 given Length ; and upon the same, as a Bafe, let a 
 Ifofccles Triangle AOB be conftituted, whofe vertica 
 Angle O fliall be = EAB ; then it is evident that either 
 of the equal Sides AO, or BO, will be the Radius of 
 the Circle. 
 
 As to the Trigonometrical Calculation, it is too plain, 
 from the Conftruction, to need any thing further to be 
 faid about it. 
 
 H 4 PROBLEM 
 
104 
 
 Geometrical Problems, 
 
 PROBLEM XV. 
 
 Having the Lengths of two Lines AE and CD drawn 
 Jrom the acute Angles of a right-angled Triangle \BC, 
 to ififectf and terminate in the oppofite Sides ; to de^ 
 ,termine the Triangle. 
 
 Call AE. a\ CD, b ; and 
 
 BD (or AO, x\ Then 
 
 E will (CE)'=^— Af ' ; whence 
 
 {BE)"[=i-(CB)-=^^; 
 ^ 4 
 
 E and therefore fl'[=r:(AEj'i=: 
 
 (BE)- + (BA)-]=*l=ii 
 ^ Hence I5xxz=.^aa — bh ; 
 
 Geometrically, 
 
 If EF be fuppofed parallel to CD, meeting AB in F» 
 the Length thereof, being half Thar of CD, will 
 confequently be given: Whence (ABj* — {^^ i* =(AE)» 
 — (EF)'J is alfo given. 
 
 Therefore the Problem is reduced tp this ; To de- 
 termine two lines AB and BF, in the Ratio of 4 ro 1, 
 fo that the difference of their Squares may he equal 
 to the Difference of the Squares of two given Lines 
 AEand EK. 
 
 Hence, having drawn two indefinite Lines BP and 
 BQ at right- Ang-es to each other, take BG, in rhe 
 Former, equal to £F ; and from the Point G, to BQ, 
 
 draw 
 
with their Solutions. 
 
 105 
 
 dr w GH=AE : So fhall (AB)«— ^BF) r=(AE)*-.(EF)»J 
 =:BH' fEiem. Cor. 8. 2) Therefore, if from any 
 Point M in PB, to BQ there be dtawn MN = 4 MB ; 
 it is evident that a Line, HF, drawn from H parallel to 
 NM, will cut. off BF as required ^This Problem be- 
 comes impofTiblc when either of the two given Lines is 
 greater than the Double of the Other. 
 
 PROBLEM, XVL 
 
 The Length and Pofition of a Right-line DE, drawn 
 parallel to the B ffe of a given right-angled Triangle 
 ABC being known ; 'tu propo/ed to draw another 
 Rig ht-h ne C V from the Vertex of the Triangle, fo that 
 the hart thereof {\G) intercepted by AB and ED, may 
 be equal to (EG) the P^rt of \LY:i ' intercepted by C^L 
 and CF. 
 
 Upon AB let fall the Per- 
 pendicular GH : 
 And put ED=:«, CD = ^, 
 DB=c, and \lQz=x : 
 Then, from the bimilarity 
 of the Triangles CDG ani 
 GHF, it will be, CD [Ij)-. 
 DG [a — x) : : GH (c) : 
 HF — ^X(^-^ J. A 
 
 And therefore 
 
 c^y^a-^xy 
 
 f ^'(HF'+GH^)-=:;t'(GF*). 
 Whence, by Reduction, b'x^ — c^x -\-zc^axz=^c''a^'\'C^b^ ^ 
 
 t?b — cc bo — cc 
 
 and x^'\- 
 
 From which xh found 
 
 bc^[aa-\-hb — cc) — aC* 
 "^ bb — cc 
 
 Gtomtirically* 
 
io6 
 
 Geometrical Problems, 
 
 Qtomttrically , 
 
 Since GF is to GC every-where in the given Ratio 
 of DB to DC, GE, in the required Pofition, muft 
 therefore be to GC, in the fame, given, Ratio. Hence, 
 if, in ED, there be taken EIz=DB, and from the 
 Center I, at the Diftance of' CD, an Arch be defciib- 
 cd, interfetling EC in K ; then a Line CGF, drawn 
 parallel to the Radius KI, will determine both the 
 Length and Pofition of GF : For it is evident that 
 CG : EG : : EK ; El : : CD : DB : : CG : GF ; and 
 
 confequemly that FGz=: GF This Problem appears 
 
 to be impoffible when BD is greater than CE. 
 
 PROBLEM XVIL 
 
 Suppqfing the Arta of a Square BEDF, forrad within 
 a given right-angled Triangle ABC, to be equal to the 
 area of the Triangle ADC, 7nade by drawing Lines 
 from the Extremes of the Hypothenufe to the adjacent 
 Angle Ti of the Square : Uis propofed to determine the 
 Side of the Square, 
 
 Call BC, a\ BA, ^ ; and BE(or BF), a;': Then, 
 CE being =:a — x and AF = ^ — jr, the Area of the Tri- 
 angle CED will 
 
 be=i><il— i 
 
 2 
 
 and That of the 
 Triangle AFD 
 
 But it is cvi- 
 d.nt that BEDF _ ^^ ^ ^ ^ Y. 
 
 4- C5D 4- AFD 4- ADC is = ABG : Which, becaufe 
 ADC tj equal to BEDF, alfo gives «BEDF4- CED 4- 
 
 AFD = ABC. that i,. .x' + ^XC^-f) | ^, X(^-^) 
 
with their Solutions. ip; 
 
 = ±: Hence .v. + 2±^' X ^ =4 ' ""'^ " = 
 Vl^ + ii'^ + i^r^'-'-^- Q.E.I. 
 
 GeomHrically, 
 
 V' UD he produced to meet AC in G, and GH be 
 drawn pcrpe.. iicnlar to AB, it is evident that the Tri- 
 
 an >'\DC ^vill be to the Triangle ABC(^^-^5^j 
 
 a GP • .^'£», or as HF to HB (EUm. 2. aw^ 8. d?/ 
 /f^.' 4, \nd it alfo appears that t '^ x(-'^-^ + ^C) 
 is =: ABx^^C. i berefore it follows that Tr:ang. ADC : 
 HBX(.VB+BC) . . HF • HB • ■ HIV^^H^C . 
 
 AB-i-BC 
 HBx — '^^ — ; ^nd confequently that the Triangle 
 
 ADO = HFx---^-Y^=HFxBK; bytak^^^ BK 
 
 (in AB produced) equal to "■ . *"' — . 
 
 Hence, (BF)^ being ( = ADC) r=HFxBK, the Caje 
 under Conf'deration is reduced to our f^ Problem ; 
 and th'. Geometrical Conftruttion will, therefore, be 
 as foiiOWS' 
 
 Having drawn BG (to bifeB the Angle ABC), and 
 GH perp:pdicu!ar to AB, and aKo taken BK equal to 
 Half die S. m of AB and BC fas above intimated J, let 
 a Seini c^)cle, upon HK, be next defcribed, inter- 
 fering BC in N ; from which Point, to the Middle of 
 BK, V-i NM be drawn ; then make MF:;=MN, and 
 BF will be the fide of the Square; as is manifeft from 
 the Problem above quoted. 
 
 PROBLEM 
 
loS 
 
 Geometrical Problems, 
 
 PROBLEM XVIII. 
 
 Having given the Hypothenufe AC of a right-angled 
 Triangle ABC, and the Difference of two Lines AD 
 and CD, drawn from the Extremes thereof to the 
 Center D of the infcribed Circle ; to determine the re. 
 maining Sides AB and BC, of the Triangle, 
 
 Upon CD, produced, let fall the Perpendicular AH ; 
 C and make AC z= a, AD 
 z=zx, DC = ^, and the 
 given Difference, x — y^ 
 
 It is evident that the 
 Angle ADH is =DAC 
 + DGA=:iBAC + | 
 BCA=z:| a Right Angle ; 
 and therefore AHz=: HD 
 AD X 
 
 CD* + AD'4-2DHxCDz= AC*; that is, in Species, 
 y'-\-x^^xy y/azzria*. Which Equation, by fub- 
 flituti; g X — /; inftead of y its Equal, and denoting ^2 
 by c, becomes x' — 2bx-\-h'-\- x*'^- ex' — chx := a' ; 
 that is (2-|-.')X^' — (2-|-c)x^-^+^*= a* : Whence 
 , aa—bb , taa — bb ,,,., b 
 
 2-\-c ' 
 
 2+C 
 
 Geometrically, 
 
 The Geometrical Conftru6lion of this Problem, as 
 the Angle ADH is given (= | a Right Angle) is ex- 
 ceeding obvious : For, if D£ be fuppofed m DC, fo 
 that AE may exprefs the given Difference of AD and 
 CD, the Angle DEC (fuppofmg CE drawn) will be 
 given z= I ADH. Therefore the Triangle AEC, by 
 Means of the given Angle AEC, and the two given 
 Sides AE and AC, may be conflrufted. And then, by 
 producing AE, and making the Angle ECD = DEC, 
 the Point D will likewife be determined ; and confe- 
 
 quently 
 
with their Solutions. 
 
 109 
 
 qucntly the Radius of the Circle, by letting fall a Per- 
 pendicular DF, upon AC : Whence the Circle itfelf 
 may be defcribed ; and two Lines may be drawn from 
 A and C to touch the fame ; and thereby form the Tri- 
 angle ABCj'flj required. 
 
 PROBLEM XIX. 
 
 Having the Bafe AB, the Perpendicular CD, and the 
 Ratio of the two Sides AC, BC, of a Triangle ABC ; 
 to find the Sides. 
 
 Call AB, a ; CD, ^ ; and AD, x ; and let the given 
 
 Ratio of AC to BC be expounded by That o'l m to n. 
 
 Hence BD=a—x; (AC)' [= (CD)'-f (AD)']z=:^^+JtA: 
 
 and (BC)^ [=(CD)" + (BD)^J — l?b + aa— 2ax -\-xx : 
 
 / And therefore 
 
 ^ ^ ' Q , mm : nn : : hh 
 
 -\' XX '. bh -^ 
 
 aa — 2aX'-\-xx. 
 
 From which, 
 
 by multiply- 
 
 _ ing Extremes 
 
 A ED B O and Means, we 
 
 have m^'b^ + ^*^* — zm'ax -|- m^'x^ = n'b^ -}" ^'^' • 
 
 Whence (ww — nn)\xx — 2mmaxz=z{nn — mm)y^bb — m'^a'; 
 
 zmma ,, m^a^ _^^, . , ^ , 
 
 andA:^: X'^ = — ^^ • Which fol- 
 
 mm — nn 
 
 mma 
 ved, gives x :=: 
 
 mm — nn 
 
 mm-—nn 
 
 ^ UXm.r. — nn/ -^ 
 
 Geometrically* 
 
 The Geometrical Conftru6lion of this Problem is 
 given By Geom. Prob. xiii.*. For, if the Bafe AB be di- 
 vided at E in the given Ratio of AC to BC, and, in 
 AB produced, there be taken, EO, a Fourth Pro 
 
 * See alfo the Lemma on p, 334 of Mr, Simpfons 
 Algebra. J.H.H. 
 
 portional 
 
tio GmuttRtCAL Problems, 
 
 portional to AE — BF, AE, and BE, it is there demon- 
 fir at^d^ that ^wo Lines jiawn from A and B to meet 
 »nv where in the Circumference of a Circle dcfcribed 
 thi'j' E,- from the Center O, will be in the fame given 
 Ratio cf A E to BE. Whence it is evident that the In- 
 terfecHon of the faid Circumference vith a Right-Iine 
 FG, drawn parallel to AB, at the given Diftance DC, 
 will determint the Vertex of the Triangle. 
 
 PROBLEM XX. 
 
 Having the Bafe AB, the Perpendicular CD, and the 
 Difference of the fwo Sides, AC and BC of a Triangle 
 ABC ; tofnd the Sides. 
 
 Making AEziziAB = 5, CD = ^, AC— BC z^^, 
 
 C 
 
 and ED z=2 X, we have AD 
 ^za-^x, BD=:a—x, AC:= 
 ^[J^^[a+xYl a: id BC = 
 i/l^'+{a — xy]; and confe- 
 quently ^\_hh-\-{a--\'xY'] — d — 
 A £ D B l/t^^+(^-^rj-^'hichEqua- 
 
 tion, fquarcd, gives bd-\-{a -\-xy — ^d \/[^^ -|-(<i -J-a:)*] 
 -\-dd :=zhl)-^{a — ^)'; and this, by Reduftion, becomes 
 ^ax-^ddziz^d ^/[^-^^(^-t-^*)]. And this, again fquarcd, 
 produces i6c4'x^-\-Sdddx^d*z=z^ddx{i^^^^^^*aX']-xx); 
 or, i6a*x^'\-d'^z=z^ddy^{aa-{-b5)^4ddxx . Whence x = 
 
 r^dd X (^^ + ^^) — ^n 
 
 ^ L \^aa — ^dd ~"j' 
 
 1 he Geometrical Conftru6lion of this Problem being 
 only a particular Cafe of a more general One, given at 
 large hereafter /'Prc^/'/e^^z 49) I fhall not infort \i here: 
 But obferve, with refpe6t to the Algebraical Solution, 
 that, if d be fuppofed to denote the Sum, inftead of the 
 Difference, of the Sidei, the Value of ;t (or DE) will 
 
 be 
 
zuitk their Solutions. 
 
 Ill 
 
 be given by the very Equation above exhibited ; as it 
 manifeft from the Procefs. 
 
 PROBLEM XXI. 
 
 Having the Bafe h.Ct the Perpendicular CD, and the 
 Re^iangfe of the two Sides AC and BC, of a Triangle 
 ABC; to determine the Triangle. 
 
 By retaining the Notation of the preceding Pro- 
 blem, and making the given G- ^y ^ F^"-^^C G 
 Rcftangle = <:% we have 
 
 -— ;t)M = c'; and theretore 
 
 -— c*\ that IS b^'\-b^Y\;ia'-\- 
 
 Whence a:^4'(2^'^ — 2^2^)X^* ^ 
 -— c-* — ^* — 2rt* h^ — a\ 
 
 Which, folved, gives x z=. ^[aa — vb -h \/ {c* 
 Geometrically, 
 
 The Magnitude of the Refiangle under the two u«- 
 known Lines, AC and BC, being given, two othef 
 Lines muft, therefore, be afligned, containing an equal 
 Reftangle ; whereof One being given, the Other will 
 alfobecomeknown. fVid,0£f^7vatton^,P.SS]. But 
 it is known that the ReBangle under the, given, Per- 
 pendicular CD and the Diam^iter of a Circle circum- 
 fcribing the Triangle, is equal to the Reftangle under 
 the faid, unknown. Sides of the Triangle (^£/f7». 25. 3) : 
 Hence the Diameter of the circurafcribing Circle is 
 given : and from thence the following Conjruftion. 
 
 Find a Third-.Proporticnal to CD and the Side ot the 
 Square, expreffing the Magnitude of the propofed Rec- 
 tangle ; and with the Half Thereof^ from the Point A 
 for B) defcribe an Arch, cutting EF, perpendicular to 
 
 AB, 
 
lis Geometrical Problems, 
 
 AB, in O ; from which Point, as a Center, with the 
 fame Radius, let a Circle ACB be defcribed then a 
 Right-line GFG, drawn parallel to AB, at the given 
 Dittance DC, will interfeft the faid Circle in the Ver- 
 tex of the Triangle. ' 
 
 PROBLEM XXII. 
 
 The Lengths of three Lines AE, BF and CD, drawn 
 from the Angles to the Middle of the oppofiteSides of a 
 . Triangle^ being given ; to find the Sides. 
 
 PutCD = a, BFzzz:^. AE = c, AB=;c,AC=> 
 
 and BC =:z. Since, 
 by a known Pro- 
 perly of Triangles, 
 (AC)^ + (3C/ z=: 
 2 (CD;' -f 2 (AD)' 
 (Elem. 12. 2) we 
 have y^ -)- z* = 2a* 
 -f- I a: , and there- 
 fore y* -)- 2' — 4 ;if *= 2a', Jn the fame manner 
 
 x'-\-y' — {z'z=:2c\ 
 
 From Whence (by taking the former of thefe Equations 
 IVom twice the Sum of the two Latter) there comes out 
 4x*-j-| A:'r=2X('^^*4*2<^' — ^'}- And confequently 
 X = y ^[th •\-2c^ — a') By the fame Argument, y ^r 
 ^ y^(2a* + ^^" — ^'); and z = 4- -/(2a' -j-2^* — c*}. 
 
 Geometrically. 
 
 If CGand CH be drawn parallel to AE and BF/ 
 meeting AB, produced, in G and H ; it is plain, be- 
 caufc CE=BK, and CF=AF, that AG=AB=BH ; 
 and alfo thai CG:=2AE, and CH^naBF. Therefore, 
 the two Sides CG, CH. and the Lme CD, bife6^ing 
 the Bafe of the Triangle GCH being given, the 
 Diagonal CI (=;2CD) of the Parallelogram GCHI (as 
 
 well 
 
miik their SoLUriui>js. 
 
 %v«li as the Sides) will- be givea (EIctju 12. 2). Hence, 
 in order to the conftru6Hon, let a triangle CGI, 
 ^hofe three fides are equal to tlie doubles ot the three 
 given lines. Be conflitmed, and draw GDH to bife6i 
 CI in D; alfo fet off DA and DB, each equal to J of 
 GD ; j in A, C and B>, C, 4i7id the iking is done. 
 
 From this conftru£lion we have the very fame 
 numerical foluiion as from the algebraic procefs : 
 for, fince, 2(GD)^ + 2(CD)* z= (CG)^ + (CH)' 
 (EUm 11. 2.)=4(AEV + 4(BF)% thence is GDcir: 
 y'' t(AEy-|-2(BFy — (CD)^,, and confequently AB 
 (=^GD)=| /[2(AE)^ + 2(BF)^— (CD)^j. By the 
 conftruttion it alio appears that no one of the three given 
 lines rauil be greatei: than the Siirn of the other two. 
 
 PROBLEM XXIII. 
 
 All the Sides of a Triangle ABC being given ; to find the 
 Perpendicular CD, the Segments of the Bafe AD and 
 BD, together with the Area oj the Triangle, 
 
 Put AC=«, AB=^, BC = c, and AD = ;^, then 
 B©z3:^ — x\ andc* — 
 
 [b -^xY (= CD^)=«^ /^ ^^^^C 
 
 — x^ ; that is, t' — 
 b^ -|- ikx — x^ ■zna' — 
 x". Whence i.bx -z^ 
 a* -V-b^ — c% and x ziz. 
 aaX-bb — cc 
 
 2/^ 
 
 E E 
 
 Now(CD}^=fAC)'— {AD)'=(AC+AD)x(AC— AD) 
 , , aaA-bb — cc\ ^ aa-^bb — cc^ 
 
 — \<'-\ — —I — ^^''- -^ 
 
 )y.{<^ 
 
 ih 
 
 •)= 
 
 Aa-\-2ab-\-bb — cc — aa'^2ab — bb^cc [b -\' oY — c* 
 
 Tb ^ Tb — Tb 
 
 c-^{b-^aY 
 
 Hence 
 
114 Geometrical Problems, 
 
 Hence CD=^[[(*+<.)'_c']x[^' -(*-«)=]]: and 
 the area (^^^^)=J/[:(A+<«)'-c']x[^'-(*-'')-]]. 
 
 Geometrically , 
 
 From the center A, with the radius AC, let a femi- 
 circle ECF be Hefcribed, cutting AB produced in E 
 and F ; fo thar BF may be the Sum, and BE the dif- 
 ference ot the Sides AC and AB : alfo let EC and FC 
 be drawn. 
 
 Then will (FC)^=FExFD (=2AFxFD); and EC 
 =:?^EvE3 (=2AExED), by Elem. CoroL to 20. 4. 
 
 Alio (BC;^= (BF)*-4-(FC-'(2AFxFD)— -aFBxFD 
 :i=:(BF;*-2ABx^I^ f Elem. Cor. 2.^(79. e.j: and likcwife 
 (BC)'=fBF:)^-HE'C)*(2AExEE))-2EBxEDi=:BE»-f 
 2ABxEE)- 
 
 Hence it appears that aAB-f FD is =(BF)*— (BC)», 
 and 2AB X ED = (BC)*— 'BE)*: and confequently. 
 that [(BFHBC)'JXr ■BC)HBE)*J=4(ABj*XFDxED 
 :=:4(AB)'X(i^C)V^J>' ^^<'w»- Cor. to 19. 4.). Therefore 
 
 ^^|^=il/[L(BF;HBCnx[(BC)SBE/]] = 
 
 the area of the triangle, as before. 
 
 From whence it appears^ that the area of any triangle 
 will be determined by finding the Difference between tkt 
 fquare of Any one of its fides and the Squares of the fum^ 
 and difference, of the other two \ and then taking I of 
 the fquare root oj the produB arifing by the multiplication 
 of the f aid differences into each other. 
 
 PROBLEM 
 
with their SolutK)NS. 
 
 J'i 
 
 PROBLEM XXIV. 
 
 Having all the Sides of a Triangls ABC/ to find ihc 
 Radius oj Its injcnbcd CircU DEF. 
 
 From the center O, to the angular points and th« 
 points of contaft let lines be drawn; and, upon BO 
 produced, let fall a perpendicular AG. 
 
 It is plain, in the firft 
 place (hecaufe OD := OE 
 =OF) 'hat AD = AF, 
 BD=BE, andCF=CE: 
 therefore, by addition, 
 BD+CF(=BE+CE) 
 = BC : take each of 
 thefe equal quantities 
 from AB-f-AC, and there 
 
 will remain AD-}- AF=: 
 
 AB + AC— BC; from -^ ^ ^ 
 
 whence (AD being zzzAF) we get AD (or AF) =: 
 
 ^ . By fubtrafting of which from AB, 
 
 and from AC, we alfo have BD — ^t — 
 
 2 
 
 ,^_ AC+RC— AB 
 andCF= i— . 
 
 2 
 
 Moreover, it is evident that the triangles AOG and 
 COF are fimilar : for the fum of all the angles at the 
 center, DOE -|- DOF -f- FOE being :^^ ^ right angles, 
 the fum of their halves, BOD -|- DO A + COF, muft 
 be = 2 right angles = BOD -f DO A -f AOG ; and 
 confequently COF=:AOG. 
 
 Now let the Values of AB, BD, and CF (found 
 above) be denoted by <2, ^, and f, refpeftivelv ; and put 
 OD (OE=OF) z=x : then, it will be BO {^{H+xx)): 
 
 (x) : : AB fa-^b ): AG=z ^rrrr^; and BO 
 
 OD 
 
 I 2 
 
 y{bb+xxy 
 
 BD 
 
ti6 Geometrical Problems, 
 
 BD : : AB : BG = •4^^fP-'^* Therefore OG(BG- 
 
 BO ) = ~t^,-- ^(^^4.,,)=^:zi:i. But, 
 
 AG : OG : : CF : OF; or ax^bx : c^— ;tjt : : c:x; 
 whence ax' -f- ^;v» =: abc — cx^ ; and confequcntly x zzr 
 
 
 Geometrically. 
 
 Let DO and AG be produced to meet each other in 
 H. Then, by reafon of the fimilar triangles, it will 
 be, AB : HO ( : : AG : OG) : : CF : FO ; and 
 therefore by alternation and compdfition, AB -I- CF : 
 CF : : H04fO(HD) : FO (OD; : : HDxOD : (OD)». 
 'But HDxOD is =ADxBD fEUm, 24. 3) : there- 
 fore we have AB+CF : CF : : ADxBD : (OD)*=3 
 ADxBDxCF , - , - 
 
 AD^-BD+CF' '^^ ^'"^ ^'"^' ^* ^^^°^"- 
 
 From this conclufion, the rule in common praftice, 
 for finding the area of a triangle, having the three 
 fides given, is eafily deduced ; for it it evident that 
 the area of the triangle ABC is equal to the radius 
 (OD) drawn into the half fum of the fides (AD+BD 
 4-BF^; that is=v^' (AD4-BD-|-CF)X ADxBDxCF]. 
 Where AD, BD, and CF, a the differences between 
 the half fum and each particular fide. 
 
 PROBLEM 
 
iviih thtir Solutions. 
 
 117 
 
 PROBLEM XXV. 
 
 The Radius of a. Circle and the Tangents of two Arcs 
 thereof being given, to determine the Tangent of the 
 Su?n ofthofe Arcs, 
 
 Let AB and BC be the propofed arcs, whereof die 
 ■given tangents are AD and CE; and 
 4et the former of thefe be continued 
 out to meet the radius OC, pro- 
 duced, in F ; fo fhall AF be the tan- 
 gent of AC, the fum of the faid arcs. 
 
 Now, calling AG, r; AD, m\ 
 CE, «; AF, x\ and FO, y, and 
 making DO perpendicular to FO ; we 
 have (by fimilar triangles) O F (y) : AO 
 
 (r) : : DF r^— 7«; : DG = ^^^II^' 
 
 Alfo OF ^^'^ : AF ,^;^; : : DF (x-^m)\ FG: 
 
 XX 
 
 mx 
 
 from which laft we have OG (==; OF— FG) =>' — 
 XX — mx yy — xx-\-mx rr^mx 
 
 y 
 rr-\-mx 
 
 '= — = (becaufe yy — xxz=:zrr). 
 
 CEfnJ^^nd confequentlyr'jf — r'm=r::r^n-\-mnx. Whence 
 rrxjm+n) ^ (AO)-x(AD+CE) 
 
 — rr^mm ' ^ — (AO)' — ADxCE* 
 
 y 
 
 But OG ( 
 
 I 3 
 
 Othertui/e. 
 
ii8 
 
 Geometrical Problems, 
 
 Otherwije, 
 
 Let AD and AE be the tangents of the two arcs AB 
 and AC. and BF that of 
 their Turn BC ; alfo let EG be 
 drawn perpendicular to OD, 
 interfefting the radius OA 
 in H. Then, by reafon of 
 the fimilar triangles, it will 
 be AG : AD : : AE : AH. 
 And, OH (AO — AH) : DE 
 ( : : OG : GE ) : : OB 
 
 (AO)^X(AD+AE) , - , , 
 
 rPT, — T7^ rTT:^^-rr7: — rr^: rr^J bccauie, by the 
 
 AO'— AOx^^^ AG' — AD X ^^ 
 
 firft proportion AOxAHmADXAE. Which conclu- 
 
 fion is the very fame with that above. 
 
 If the tangents of the Arcs AC and AB ffig. i.) 
 ^vere to be given, in order to find the tangent of their 
 
 difference BC ; then by the proportion ^ : 
 
 (AOVxDE 
 
 rx — rm 
 
 rrxi^ 
 
 : r : 72 (above derived) we (hould have n 
 
 ■m) 
 
 -\-KX 
 
 ; or CE = 
 
 (AO)^X(AF— A D) 
 (AO)'-fAFxAD 
 
 PROBLEM XXVL 
 
 The Rai'o of the Sines DE, FG of two Arcs AD, AF, 
 of a given Circle, together with that of their Tangents 
 AB, AC being given; to find both the Sines and the 
 
 Tangents. 
 
 . Put the radius AO :z= a ; and let the given ratio of 
 AB to AC be that of m to n ; moreover let DE be to 
 
 FG 
 
with their Solutions. 119 
 
 FG as /> to ^ ; and call AB, x; and AC, y: then 
 (by fimilar triangks) (OB)^ (a^+Ar^) : (AB)* (;c ) : : 
 
 ^ / \ / \ aa-\~xx 
 
 and(OC)'(fl'+y): (AC)^(>') 
 
 Therefore, ^;; /^^ Queftion.p^ : 
 f I ! r— : --x — ' ^^^ 
 
 .X 
 
 / 
 
 D 
 
 M/ 
 
 <£ 
 
 
 />^ 
 
 aa-\'XX ' aa -J- vy' O G. "E ^^ 
 
 confequently pY X(«'+'*')=^'-*'X(«' +>')• But, by' 
 
 the queftion, we alfo have, mi n: ',x : 7 = — ; 
 which value, fubftituted in th« preceding equation, 
 
 gives —- — X(^ +'*)=f'* X{'=*M )• whence 
 
 mm \ Tfiin ) 
 
 «*f 'X(^"*+^') = ^*X(^''»'+"*^') ; and a: = - X y' 
 l nnpp-mmqq \ ^^^^ ^^^^^ ^^^ ^^ ^^^ ^^ ^^^ ^^^^ 
 \ qq^pp J 
 given. 
 
 Geometrically, 
 
 Since the ratio of AB to AC is given, as /« to 72 ; 
 GK (fuppofing K to be the inierfettion of FG and 
 OB) will be to GF, in the fame given ratio : and, if 
 (agreeable to the 3** general obfcrvarion) KH be drawn 
 parallel to AO, meeting OF in H, OH will be to OF 
 JHU in the fame given "ratio. 
 
 Again, if Dl be drawn parallel to AO, mee'ing OF 
 in I; then OK : OH f: : OD (OF) : OI : : FG : DE) 
 : : q : p ; whence OK is alfo given ; and (mm thence 
 the following conilruftion. 
 
 In any radius OF of the given circle, take OH to 
 
 OF in the given ratio of ?» to w ; and upon HF let a 
 
 femi'Circle be defcribed: take alfo a fourth pro- 
 
 I 4 portional 
 
120 
 
 Geometrical Problcms, 
 
 .portional top, q, and OH; with which, as a radiw, 
 from the center X), defcribe an arch, cutting the 
 -femi-circle in K ; and, having drawn HK, make OA 
 parallel thereto. 
 
 PROBLEM XXVII. 
 
 The Sid^. of tkc Square, and the Radius vf the CircU^ 
 (ivfcribedin a Right an ^ItdTricngie ABC, being given\ 
 to determine the Triangle. 
 
 Draw the diagonal BD, of the fquare ; and from the 
 rcjiiei-jO of the givep circle, to the points of contact, 
 Q draw the radii OO and OP, 
 and upon the hypothenufe AC 
 let fall the perpendicular BQ : 
 calling the fide of the fquare, 
 a; the radius of the circle, b, 
 and A Q, x ; then becaufe 
 of the parallel lines, we fhall 
 iiave FG (a^b) :,BF (a) : : (OD 
 
 :BD::)OPW:BQz=^ 
 whence DQ = \/(BD' — BQ') 
 
 . aabb \ . . - 
 
 {a— by 
 
 Let it, for brevity- fake, be denoted by c ; and let BQ 
 
 (-^)= d: then we fhall have, AQ(a:) : BQ(df) : : BQ 
 
 dd 
 {d):CQ=:~; and »Uofby EU?n. x8. 4.) AD {x+c) : 
 
 CD (^—c):: AB : BC : : AO/'.^;: BQ /^i^; whence, 
 by multiplying extremes and means, we get dx-^-cd^iz 
 
 dd — ex; and, from thence, ;cz= 
 
 ^X(d-c) 
 
 d + c 
 whereof every thing elfe is readily found. 
 
 By mean* 
 
 Geometrically. 
 
with their Solutions. 
 
 jt*i 
 
 Geometrically, 
 
 The geometrical conftruftion, and the trigonome- 
 trical folution of this problem are very eafy. For, 
 iince the pofition of the point D with refpeft to the 
 circle, is given ; a right-line DPA drawn from thence 
 (by EUm, 21. 5.) to touch the circle, will determine 
 the triangle: and then, from the given lines OP 
 and OD, the angle D may be found : by means 
 whereof and ABD (= \ a right angle) together 
 with the given fide BD, all the reft will becxmie 
 known. 
 
 PROBLEM XXVIII. 
 
 71? determine the Sides of a regular Pentagon andDecagoUt 
 infcribed in a given Circle, 
 
 Let AB, BC, CD, ^c. be fides of the decagon, 
 and AC a fide of the pentagon : and let AD be 
 dr^wn interfering the radius OB in P. It is evident 
 in the firft place, that 31 ^ n li 
 
 D 
 
 the Angles BAP and 
 OAP, ftanding on 
 the equal arches BD 
 and BF, are equal to 
 one another, and alfo 
 equal, each of them, 
 to the Angle AOP, 
 infifting on the arch 
 ABr£/<fW. 10.3.) And 
 fecondly, that the 
 triangle' BAP (as 
 well as APO) is an 
 ifofccles one, because the perpendicular A?iC. makes 
 equal angles BAG. DAC with the two fides AB, AP 
 of the triangle. Hence it appears very plain that aU 
 the three lines AB, AP, and OP are equal among 
 themfelves ; and likewife that AO (OB) : AB (OP) : : 
 OP (AB) : '2»Y(by Elem. 18. 4.) feeing the angle BAG 
 
 is 
 
saft Geometrical Problems, 
 
 is bifcfted by AP. Moreover, by letting fall tbc two 
 perpen iiculars PQ and CM, tipon AO -^nd ABM, the 
 triangles BCM and A Q (.s BC is = /B~ AP, and 
 the angle MBC = MAD z=z PAQ; .vill appear to be 
 equal in all relp; dts ; and 'o. BM hein^t—AQ) = 
 iAO, we have (AC;' {z==z['^C)' + < ABY -{-'-cA x AB, 
 fE/em. 11 2.)=:(BC;*4-(AB)4-AOx "B, Bui, by 
 the abo-e prcpor en (AB;' is z=z fiO X ^P : tliere- 
 fore, by writing AOy^^' ^^ '^^ roun of (AB)*, we 
 get{ACj=(BC)'4- AOxAB+AOx3?=(BGj'+ 
 (AO)*: whence (BC being f^ift found) AC win alfo 
 become known. — If AO be now denoted by a, and AB 
 by x^ then from the equality of (AB)' and AOxBP, 
 you will bave;t*z=:^x(^ — "^i » ^^^^ which x w>i, be 
 
 found =y^f^^- \; and from thence AC =; 
 
 As to the geometrical conftrn£lion, it likewifc very 
 eafily follows from above. Fcj fince the fide of the 
 decagon appears to be equal to utQ greater part of the 
 radius divided according to extreme and mean pro- 
 portion ; and the fquare of the fide of the pen- 
 tagon exceeds that of the decagon by the fquare ot 
 the radius; the following folution (given by Dr. Barrow^ 
 as an improvement upon Euclid's) is manifell:. 
 
 Draw the radius OR, at right-angles to the diameter 
 AF ; and, having bife^ed the radius AO in Q, let 
 OS be taken, in AJ^, equal to the diflance QK : fo 
 ftiall OF : OS : : OS : FS fElem. 19. ^.) and confe- 
 quently OSmAB the fide of the decagon. 
 
 Andbecaufe(RS)'(rupporing RS drawn) is z=: (OS)' 
 -^OR)\ it is plain alfo that RS will be equal to the 
 fide AC of the pentagon. 
 
 PROBLEM 
 
zouk their Solution's. 
 
 12^ 
 
 PROBLEM XXIX. 
 
 Having the Hypotkenufe AC of a right-angled Triangle 
 ABC. and alfo the Radius of the infcribed Circle 
 DEFG / to find the two Legs AB and BC. 
 
 From the center D of the given circle, io the points 
 ofcontaa, let DE, DF, and DG be drawn ; alfp 
 draw aD and CD : JNJ 
 
 and put DE ( = 
 DG = DF) =z=a,, 
 AC = /^, AB=;c, 
 and BC=;^. 
 
 It is evident, that 
 CE muft be=:CG 
 r=^ — a ; becaufe 
 the right - angled 
 triangles CDE and 
 
 CDG, having DE=:DG, and CD common, are equal 
 in all refpc6ls. In the very fame manner is A F — AT? 
 
 eA 
 
 I 
 
 G \ N. 
 
 
 \j 
 
 
 
 F 
 
 B H i 
 
 Therefore^ — a'\'X — a-z=ib ( rzzAC) ; from which 
 equation we have x-^-y =:^-|- 2^z. But, from the pro- 
 perty of right-angled triangles, we alfo have xx -^yy — 
 Ifb, And, if from the double of this, the fquare of 
 the former equation be fubtra6led, there will remain 
 XX — 2xy-\-yyz=2bb — /^ab — ^aa. 
 
 From whence, by cxtrafting the fquare root, on 
 both fides, we get x — yz=z\/{bb — ^ab — ^aa). Which laft 
 equation, added to, and fubtra£led from, the firft, 
 gives 2xz=:'iii-\-b'\'\/{bb — ^ab — /{aa), and 2y:zs^2a-^b — 
 j^ ( bb — ^ab — j^aa ) . 
 
 Geometrically, 
 
 Seeing the difference between each ]eg of the 
 triangle and the adjacent fegment of the hypothenufe, 
 is equal to the radius of the circle, it is plain that the 
 .fum of the two legs (AB -f-BC) will exceed the fum 
 
 of 
 
Ml 
 
 Geometrical PfioRLfiMS, 
 
 of the two fcgments (or fhe whole hypothenufc) by 
 twice he radius; and will therefore be given r=f AC 
 4-2BF. Whence, if Bl be fiipnofcd equal to BC, Al 
 will likewife be givenz=: AC-[-2BF, and the angle 
 BlC=:halfa right-one. Hence the following con- 
 flruftion. 
 
 Draw an indefinite line, in which take AH equal 
 to the given hypothenufc, and HI equal to the diameter 
 of the given circle ; alfo make the angle A IN equal 
 to half a right angle ; and from the center A, with 
 the interval AH, let an arch be defcribed, meeting 
 IN in C ; from which point upon AP let fall the 
 perpendicular CB ; fo fliall AB and BC be the two 
 legs required. 
 
 PROBLEM XXX. 
 
 The Perimeter ^ and the Area, of a right-angled Triangle 
 ABC being given ; to determine the Triangle, 
 
 Put the given perimeter (AB + BC-f- AC) r=/', the 
 N \C area ( f AB X BC ) 
 
 = a^; and let half 
 the fum of the legs 
 AB and BC be de- 
 noted by X, and half 
 their difference by 
 ^ : then, AB be- 
 ing =:a4-;', BC = 
 X — jy, and ACzrr 
 A F B H <1T p~zx, we (ball have 
 
 C^+ylXt^— >') ( = ^^^XBC) =: a^'; and [x +JVJ^+ 
 ^x — •j/)V,=AC*)=(/? — tx)^', that is, by redu£iion, xx — 
 »=2«* ; and zxx-^-zyyzizpp — j^px-\-/^xx. 
 
 Now, by the addition of the latter of thefc 
 equations to the double o\, the former, there arifes 
 j^x^pp — J^px-\'i{xx -f- 4^' • whence x comts out =^ 
 
 CutH-— 1 A-U— . From which value that of y =z 
 
 ^ [x*'^9id^), wiU lilfecwife be given. 
 
 Geometrically. 
 
mth their Sx>LUTiQiirs, Atf 
 
 Geometrically, 
 
 If from the center D, of a circle infcribed in the 
 trbnfijle. lines be fuppofed drawn to the angular 
 points, thi-* propofed triangle will, by that means, be 
 divided into three others, ADB, BDC, an ADC; 
 whofe bafcs arc the three fides of the firft triangle, 
 and their perpendiculars all radii of the said circle. 
 From which it is evident that the triangle ABC is 
 equal to a rectangle under half the fum of its three 
 fides (which we will here exprefs by the given line 
 AO) and the radius DF of the infcribed circle; and 
 confequently that the radius DF will be given by 
 taking a third proportional to AQ and the fide (a) 
 of the fquare expreffing the given area. Whence, 
 making QH and QI, each, equal to DF, fo found; 
 it vyill appear from the preceding problem that AH 
 will be ZZ2 the hypothenufe, and AI z=z to the fum of 
 the two legs, of the propofed triangle : which quan- 
 tities being both given, the method of conftru£lion i* 
 laaimittifromtke lajl problem* 
 
 PROBLEM 
 
126 
 
 Geometrical Problems, 
 
 PROBLEM XXXI. 
 
 The Sum^ or Difference of the two Legs AC and BC of 
 a right*angled Triangle ABC being jriven, together 
 with the Sum, or Difference of the Hypothenufe AB 
 and a Perpendicular CD falling thereon from the 
 Right-angle ; to find all the Sides of the Triangle, 
 
 Put AC 4-BC= J, AC — BC =r d, AB -fCD==^. 
 
 AB— CD=9r; 
 then will AC 
 
 2 
 d 
 
 ^-—'^ but (AB)'=(AC)*+(BC)^ ; and AB X CD (= 
 fi area ABC) =:ACxBC : which, in fpecics, give 
 
 and/?'— ^'=ri' — d\ 
 
 By adding, and fubtracling the double of the laft of 
 thefe equations to and from the former, we have thefe 
 two other equations. 
 
 '^^^' 3p'+2-pq—f=:is\ 
 and, — />'-i-2^^+3^'==:4<^'' 
 from which, when any two of the quantities, J, d^ 
 pt q, are given, the other two will, easily, be deter- 
 mined. 
 
 Thus, let J and p be given ; then, from the former 
 equation, we have q^ — zpqzrz'^p^ — 41' ; whence q' — 2pq 
 +/''=4^'— 4-^% and/:— ^=2v/(/>*— s*): therefore CD 
 
 {=J^)=^[f-^s% and AB=:^-v/(^W). 
 
 If //and q be given; we fhall have CDrr4/(y'— df=); 
 becaufc;?' — q^z=:s — d\ or p'-^y -zizq^ — d' (p. above.) 
 
 But, 
 
with their Solutions. %%y 
 
 But, if s and (j be given, then will 3/^' + 2^f =S 
 4J'-j-^'; which, folved, gives /?=:Jy^(3Jj-[-^7) — tf* 
 
 Laltly, if ^ and ;? be given, we ih^H have g^^-f-^/'f 
 ::z:4^*4^^ ^'^^ confequently f=:jV^(3^^4-/'/') — ^P* 
 
 Geometrically, 
 
 The very fame properties whereby the algebraital 
 folution is above brc. jghf out, lead us aifo to geome- 
 trical conflrucrions ol the feveral cafes of the pro- 
 blem under confideraiion. Bur it will be fufficient, 
 here, to exhibit that of the cafe, wherein the fura of 
 the tegs (AEt and the d<ifference of the hypothenufc 
 and perpendicular (AF) arc fuppofed given, being the 
 n^oft difficult. 
 
 Thus, becaufe (AC)'+(BC)^=rAB)% 
 and 2ArxBC (z=2AHx^Dj=2ABxBF ; 
 we {hall, bv adding iheleeqiMl quantities, 
 have (ACy+(BCy+-ACxBC (oi (AE)», fElem,6. 2.) 
 = rAB/42 .3xBF=:(ABj^+2AB x(Ai^— AF]=: 
 3(AB)^— 2ABxA^'- 
 
 Therefore, if an arch, from the center A, with the 
 radius AE, be defcribed ; and, fr.m its inrerfefclioa 
 (Hj with FH perpendicular to AE, another arch be 
 alfo defcribed, with the radius 2AE, cutting AE pro- 
 duced in N ; then the hypothenufe AB of the required 
 triangle, will be | of the line AN thus determined. 
 
 For (HN)'f4(AE)*) being = • AN)'+ fAH)^(fAE)») 
 — 2ANxAF/^£/e/?j. 10. 2.}"and(herefore3fAE)— (AN)* 
 — 2AN X AF ; it is plain, if AB be tbken — ^AIST, 
 that 3(AE)'=,sABX3AB— 6ABx AF : and conlequent- 
 ly that (AE)*=3AB'— 2AB XAF, the veryfamezts 
 ubove. 
 
 From the value of AB, thus given, what yet re- 
 mains to be done, will be effected with great facility. 
 For, if in FH there be taken FK=FB, and KC be 
 
 drawn 
 
«98 
 
 GiOM-iTRiCAL Problems, 
 
 ^I9^v(n parallel to AN, intferfefting a femi-circle, de- 
 fcribed upon AE, in the point C, that point will, it is 
 evident, be the vertex of the triangle. 
 
 LEMxMA. 
 
 tja Line be drawn from the Vertex to any Point in the 
 Bafe of a Triangle^ the Sum oj the two Solids under 
 the Squares of the two Sides and the alternate Seg. 
 nients of the Bafe, will he equal to the Solid under 
 the whole Bafe and its two Segments, together with 
 the Solid under the fame Bafe and the Square of the 
 dividing Line, 
 
 Let ACD be any propofed triangle, and BD the 
 
 dividing line ; then, I 
 fay, that (AD)* X BC+ 
 
 (cd;^xab=acxas 
 
 XBC+ACx(BDr 
 For, if AB and CB 
 be bifefted in M and N, 
 and a perpendicular DE 
 be let fall upon AC, it 
 is known, 
 
 that 
 
 M B EN C _^^ ^, 
 
 /(ADV— (BDV=ABX2ME\ Elem. 9. cf i. 
 
 \(CD)MBD)'=BCX2NE/ Cor. 2. 
 Whence it follows, 
 
 that (ADrxBC — (BD)*X BC= AB xBC x^ME; 
 and (CD)*XAB — (BD^X ^"^^ = AB xBC X 2NE. 
 
 Let thefe equal quantities be added together, and the 
 fums will alfo be equal : 
 
 that is, (AD)'X BC-f (CD^X AB — ACx(BDj' (= 
 ABxBCX'-iMN) z=:AQx^CXAC ; and confequently 
 (AD^XBC +(CD)' xAB =ACxAB xBC + ACx 
 (BDV. Q. E. D, 
 
 CoROL. I. Hence, if ABr=BC, then will (AD)* 
 +(CD)*=2(AB)^+2(BD)^ ^^^^^^ 
 
with their Solutions. li^ 
 
 CoROL. 2. But, if AD=DC, then we (hall have 
 (AD)^XBC+(CD)^X AB = (AD)* X BC + (ADj'x 
 AB =(AD)^xAC : Whence (AD}*xAC = ABxBC 
 XAC+(BD)'XAC ; and confequcntly(AD)'=ABx 
 BC + (BD)\ 
 
 CoRoL. 3. Laftly, if the Angle ADB z= the Angle 
 CDB (or AD : CD : : AB : BC. Elem. 18. 4,) then 
 ADxBC being =CDxAB, it follows that(AD)'xBC 
 = AD X CD X AB ; and that (CD)* X AB = AD X 
 CDxBC. Hence (AD)*XBC + (CD)*XAB = AD 
 XCDx(AB4.BC)=:ADxCDxAC=ABxBCxAC 
 •UfiQYxP^C (p. above)', and confequently ADxCD 
 E=:ABxBC4-(BD)'. 
 
 PROBLEM XXXII. 
 
 From three given Points^ A, B, C, in the fame Right- 
 line, to dramas many Lines, to meet in a fourth Point 
 D, fo as to obtain a givin Ratio among themfelves'. 
 
 Call AB, a ; BC, ^ ; AC , c ; and AD, x ; and let AD, 
 BD, and CD be, D . _____ 
 
 in Proportion to / \\ 'rr 
 
 one another, z.% p, x \ \ Z» 
 
 q, and r, refpec- 
 tively. Then, BD 
 
 being = ^ , and 
 
 CD= ^, we Jj; 
 
 (hall, by the preceding Lemma^hd^vt x"^ X^-t-*"— X^ 
 
 PP 
 
 abc+cx—t 
 
 PP 
 Whence (^/?'-|-ar'— c^*)X^'=«^f/''; 
 
 And 
 
13^ Geometrical Problems, 
 
 And confcquently x z=z V\/\ i ^ ■ V 
 
 \bpp-\-arr — cqq/ 
 
 After the fame manner the Problem may be refolved, 
 
 when, inftead of the Ratio, the Sums, the Differences, 
 
 or the Re6langles of the three Lines, are given. 
 
 Geometrically* 
 
 If from any Point E, in BD, two Lines EF and EG 
 
 be fuppofed drawn, fo as to form the Angles BEF and 
 
 BEG, refpeaivelj, equal to BAD and BCD, it is 
 
 evident, from the Similarity of the Triangles BEF, 
 
 BDA, and BEG, BDC, 
 
 , ^ jfBEiBF: :BA:BD 
 mat <^^^. BQ. . BQ . 3^^ 
 
 And, confcquently, that BF : BG : : BC : BA. There- 
 fore, if BF be taken=BC, BG will be = AB. 
 Moreover, from the abovementioned, fimilar, Ti:ianglcs, 
 we have /BD : AD (: : ^ : ^) : : BF (BC) : FE 
 we have <^^^ : CD (: : ^ : ;) : : BG (AB) : GE. 
 Whence FE and GE are given; and from thence the 
 following Conffru£iion. 
 
 Take BFiz=BC and BGz= AB ; alfo take a Fourth 
 Proportional to q, /?, and BC, and another to q, r, 
 and AB ; And, with Thefe as Radii, from the Centers 
 P and G, let two Arcs be defcribed; and, to their In-^ 
 lerfeftion E, draw BE and FE ; alfo draw AD, making 
 the Angle BAD zm BEF, fo (hall its Interfeaion with 
 BE, produced, be the Point of Concourfe required. 
 
 The Trigonometrical Calculation, from the Con- 
 ilruaion, is very fhort and eafy : For, all the Sides of 
 the Triangle FGE being given, the Angle F may be 
 found ; then, in the Triangle BFE, two Sides and the 
 included Angle being known, every Thing elfe is readily 
 
 determined. It may be obferved that there is another 
 
 Conftru6hon of this Problem ; by Means of the Imer- 
 feftion of two Circles, fo defcribed that Lines drawn 
 from the given Points to meet in the Peripheries There- 
 of, may obtain the given Ratios but the Method given 
 gbove I Jook upon as preferable, — As to the Limitations, 
 
 it 
 
xuitk their Solutions, 
 
 *3i 
 
 it is plain the Problem becomes impoflible when the two 
 Circles, defcribed from C and F, do not meet each other; 
 that is, when q is given either, lefs than the Difference, 
 
 t r. r P^ . ra 
 
 jox greater than the Sum of— and ^. 
 
 PROBLEM XXXIII. 
 
 The two Sides AD, CD, of a Triangle being given in 
 Lengthy together with the Length of a Line DB divid- 
 ing the Bafe AC in a given Ratio ; to determine the 
 Bafe^ or Linefo divided. 
 
 Call AD, a ; BD, b ; CD, c ; and AB, x ; and let 
 the given Ratio of AB to 
 BC be that of « to n. Hence 
 
 BC = A and AC(=j; 
 
 m 
 
 * mj 
 
 nx\ {m'\-n)y^x 
 
 And 
 
 m 
 
 therefore, by the lemma, 
 nx , \mJ^n)y^x 
 
 m 
 
 m 
 
 m m 
 
 Whence, by Redu6iion, mna^'^-m^c^ — my(m^n)yb'^-^ 
 (m^n)xnx\ 
 
 fmnaa-^mmcc ^ mhh \ 
 
 And xz=.\/\—i — ; — ^ , J. 
 
 ^ \ [m-^n)xn n / 
 
 If DB be fuppofedto bifeft the Bafe ; then m and n 
 
 being equal, x becomes =:y( — ~ bb). 
 
 But if DB be fuppofed to bife6l the vertical Angle, 
 we Ihall have /»;«(:: AD : CD) :: a:c (Elem. i8. 4). 
 
 K2 
 
 Whence, 
 
132 Geometrical Problems, 
 
 Whence, by writing a and c indead of 771 and 71, 6\xx 
 
 iLquation becomes x zzn v [ -, — ~ ■ — J — - 
 
 Geometrically, 
 
 In any Right-line EG, at pieafure, let there be taken 
 EF and FG in the given Ratio of AB to BC ; and, 
 from the F*oints E, F and G, let three Lines» be drawn 
 to meet in a Point D (by the lajl Problem) To as to ob- 
 tain the given Ratio of AD, BD, and CD, refpe6live- 
 ly : Then, if DA be taken of the given Length, and 
 ABC be drawn parallel to EFG, it is manifeft that 
 ADC will be the Triangle required. 
 
 PROBLEM XXXIV. 
 
 Supfofing two Sides AC, BC, of Triangle ABC, to be 
 given in Length ; and that two Lln^j BE, AD. drawn 
 from the oppofite Angles, to cut ^ff givtr. Segments AE, 
 BD, are equal to each other; 'tis pro pofed to deter?mne 
 the oth*r Side AB of the Triangle. 
 
 Put CE ==:.?, AE=/^, CD = c,BD = fl^, CA=:/, 
 CB=^, ABz=;;c, and AD 
 (or BE) rm z. Then, by the 
 Lemma, we have the two fol- 
 lowing Equations. 
 
 g'Xb-^x'Xci=zfab^fX z\ 
 
 f'X^-t-x'x^^gcd-j-gx 2*. 
 
 From the former of which, 
 
 multiply'd by g, let the latter. 
 
 multiply'd by /, be fubtra£l- 
 
 A. B cd» ^"^ there will arife bg^ — 
 dp4- ag-x^-^ cfy z=i ahfg — cdfg. 
 i r A^hL-cd)Xfg\df^^hg^^ 
 Therefore ^-/(^ ^-^^ ^j. 
 
 Geometrically* 
 
mtk their Solutions. 
 
 U 
 
 Geometrically. 
 
 If DF be drawn parallel to BE, the Ratio Therao 
 to BE (or AD) will be given, as CD to CB : and CF 
 will likewife he to CE in the fame given Ratio, and 
 therefore will be given in Length. Hence it is evident 
 that the Pofition of the Point D with refpeft to the Side 
 AC (firft laid down) will be determined by the Inter- 
 feftipn of two Circles ; One defcribed from the Center 
 C, with the given Interval CD ; and the Other, by 
 Alg. p. 334. fo that Lines drawn from F and A, to 
 meet any-where in the Periphery Thereof, may obtain 
 the faid given Ratio of CD to CB. Through whick 
 Point, fo determined, the other Side CB of the Tri- 
 angle muft be drawn ; and then, AB being joined, the 
 Thing is done. 
 
 PROBLEM XXV. 
 
 All the Sides of a Trape%ium ABCD, ttbout which a 
 Circle may be defcribed ^ being given in Levgth ; to de^ 
 ttrmine the Diameter of the Circle* 
 
 Let the Diagonal AC be drawn, and upon BC and 
 AD let fall the perpendi- 
 culars AE and CF. And 
 put ABr=fl, BC=/^, CD 
 z=zc, AD = fi^, and BE 
 
 Now the external Angle 
 CDF being equal to the 
 internal, oppofite. Angle 
 B (Elem, 34 3.) the 
 Triangles ABE and CDF 
 are fimilar : And there- 
 fore AB (a) : BE (xj 
 
 ' : CD (c) : DF==-. But (AB)^+(BC)' 
 
 K3 
 
134 
 
 = (ACV 
 
 Geometrical Problems, 
 
 (ADf +(CD)^ + 2ADXDF; that Ts, iff 
 Species, aa-^- bb — ^bx =ct-J-^^-J 
 
 From which Equation x is found=- 
 
 aa-\-bb — cc — dd 
 
 zb -\'2cd 
 
 Whence AE=y'(^^ — x;c) and AC= ^/laa'\-bb — zbx) 
 will alfo be given : And then it will be AE : AC : : AB : 
 the Diameter fought (Eleml 25. 3.) 
 
 Gtomctrically, 
 
 Whence <[bC : AD 
 
 If the two Diagonals of the Trapezium be drawn, 
 interfering each other in E, the Triangles ABE and 
 DEC as well as CEB, and DEA, will be equiangular 
 (Elem. 12. 3. 
 
 - '- — BE; CE 
 BE:AE. 
 J^ And, fmce the 
 
 Ratios of CE and 
 AE to BE are 
 thus given, it 
 follows, that, if 
 any Line FG be 
 drawn parallel to 
 AC, the Parts 
 thereof GH and 
 FH, intercepted 
 by BC, BE, and 
 BA ( produced ) 
 
 will be to BH in the fame given Ratios, 
 ^ , , JAB :CD::BH: GH 
 Or that <[bC: AD::BH:FHf 
 
 and confequently, that, if BH be taken =:A B, GH 
 will be given z=. CD, and FH= 
 the following Conflruflion. 
 
 BC 
 
 Whence 
 
 Makr 
 
with their Solutions. 
 
 i3J 
 
 Make FH a Fourth-Proportional to BC, AB, and AD 
 (by EUm, 13. 5} ; and, in the fame Line produced, 
 take HGzzuCD : Then fby Alg. lemma on p, 334.) let a 
 Circle IK be defcribed, fo that two Lines drawn from F 
 and G, to meet any-where in the Periphery Thereof, may 
 obtain the given Ratio of AB to CB. And from H, 
 as a Center, with the Radius AB, let another Arch be 
 defcribed, interfering the Former in B ; draw BF and 
 BG, in which fet off AB and BC of the given Lengths; 
 then, thro' the three Points, A, B, and C, let a Circle 
 be defcribed, and the Thing is done. 
 
 PROBLEM XXXVL ' 
 
 All the Sides and^ the Area of a Trapezium ABCD bein^ 
 given ; to determine the Trapezium, 
 
 Suppofe the Diagonal AC to be drawn; fuppofc alf^ 
 CE and CF to be per- ~~ 
 
 pendicular to AD and 
 AB : And make AD 
 =.a, DC = ^, BC = c, 
 ABzn^, the given Area 
 ABCD=r% DE — ;t, 
 and BFzu^y. 
 
 Then will aa^bb-\- 
 2fl^z= (AC)* = i:c+^^ 
 -\-2dy fElem. 10. 2.) and 
 
 a\/{bb—xx)-\-d^/[cc-^ A B 
 
 ^jy)=2ADC+2ABC=27^ (by the OueJiionJ, 
 
 Put r'-}-^' — a^ — b'z=z2f\ and, by the lirft Equation, 
 you will have ax — dy — -f. 
 
 Moreover, by fquaring the two laft Equations, and 
 then adding them together, you will have 
 a^b^-\'C d'-\-2ad^(bb — xx) x y^i^c — yy) — ^adxy^rz /^r* 
 -f/'. Which, by dividing by 2ad, and makincr g __, 
 
 ■-yy)=g + xy. 
 
 This, fquared, gives b^c^ — by — c^x^^=z:g''\-^gxy i 
 
 K4 Which, 
 
tus 
 
 Geometrical Problems, 
 
 Which, by writing "^ in the Room of, its Equal, 
 
 X, will became t'c'-^by-^ ^I2l[^t/i ' _ ^i J^ 2gy X 
 
 2c'df 
 
 aa 
 
 -^--L^: Whence, putting ^V—~-^i^ g^zzzh, 
 
 +f=i,a„d3'+^?^ 
 
 a^ 
 
 -{-/y* ; and therefore^ =:v/r--—l — j^ 
 
 2apr , , 
 
 — ^ =/, we get hzzz ky 
 h . kk\ k_ 
 
 2/" 
 
 Geometrically, 
 Bccaufe (AD)'+(DC)^ + 2ADxDE(=r( AC)* )= 
 
 p TV (AB)^ + (BC)^ + 2ABx 
 
 1] BF (Elem. lo. 2.) it is 
 plain that sAD X DE ^ 
 » 2AB X BF is given = 
 (AB)'+(BCV~-(AD)' 
 
 -{CD)^ 
 
 Find (by Elem. 2 and 
 
 3. 6.) a Line L whofe 
 
 fquare fhall be=:fAB)* 
 
 +(BC)MAD)^-(CD)- 
 
 ^ fo fhall 2ADX DE^ 
 
 ao 
 
 ABxBF 
 AD 
 
 2ABxBF=:I*, and confequently DE 
 
 aAD* 
 
 But .^ , fuppofing PQ perpendicular to AB, 
 
 and BP a Fourth Proportional to AD, AB, and BC, 
 appears to be s=BQ: And ABxCF (fince AD : AB 
 : : BC : BP : : CF : PQ) will alfo be equal to AD x 
 PQ (Ekm, 10. 4.) 
 
 Hence 
 
zuith their Solutions. 137 
 
 Hence we have DE~BQ= a ri » ^"^ ADx^G 
 
 + ABx CF (= 2r')=:ADx EC + AD X PQ : 
 Whence, by taking a Third Proportional fTJ to 2 AD, 
 and Z, and Another fVJ to | AD and r, there refuhs 
 DE— BQ=r; and FC4-PQ=^. 
 
 Therefore the Problem is reduced to this ; to jind two 
 Angles CDE, PBQ,y^ that the Sum of their Sines CE, 
 PO, and the Difference of their CofiUes DE, BQ (an^ 
 fwering to given^ but unequal^ Radii, DC, BP) may be 
 both given iluantities. 
 
 And the Conftruftion thereof (which is exceeding 
 evident) will be a« follows. 
 
 Draw two Lines cutting each other at Right-Angles 
 in M, in which take MPnr ^, the given Sum of the 
 Sines: andMN = r, the given Difference of the Co- 
 fines : Then, from the Centers P and N, with the 
 
 given Radii PB^z=: ?^~ j and DC, let two Arcg 
 
 be defer ibed, interfe6ling each other in B : through 
 which Point draw OA parallel to MN; and join B, 
 P, and B, N ; fo fhall PBO and NBO (=CDE) be the 
 two Angles required. Which being known, the Tra- 
 pezium itfelf is very eafily conftrufted. 
 
 This Problem, it may be obferved, becomes impof- 
 fTble when the two Arcs, defcribed from the Centers P 
 and N, do not meet, but fall fhort of each other : that 
 is, when the given Area is greater than That of a Tra- 
 pezium of the fame given Sides, infcribed in a Circle, 
 determined by the preceding Problem. 
 
 But, belides the above, there is another Limit, for 
 the leaft Value of the Area (except in one particular 
 Cafe) and the Problem will be impoffible when one of 
 the two Circles falls wholly within the other, as well 
 as when it falls wholly without it : But this Laft de- 
 pends upon the particular Order of joining the given 
 Lines ; whereas the Cafe is otherwife with refpe6t to 
 the firft, or greateft Limit. 
 
 PROBLEM 
 
138 
 
 Geometrical Problea«, 
 
 PROBLEM XXXVII. 
 
 To divide a given Angle BAG into two Parts BAI and 
 CAI, fa that their Sines BE and CF may obtain a 
 given Ratio ; fuppofe that oj m to n. 
 
 Put the Chord BC of the given Angle, =1:^, and the 
 
 Part thereof BD inter- 
 cepted by AI, = X : 
 Then, becaufe of the 
 fimilar Triangles BDE 
 and CDF, It will be, 
 BD fx) : CD (a^xj : : 
 BE : CF : i m : n, by 
 Hypothejis, 
 
 ma 
 
 Therefore nx-zzzmYXO- — x] ; and confcquentlv^r= ■ 
 
 ' ^ ' m-f-n 
 
 From whence, and the given Angle ABD, the Angle 
 BAD will be found. 
 
 The Geometrical Conftruftion (like the Algebraical 
 Operation) may be performed by dividing the Subtenfe 
 BC in the given Ratio of m to n : But the following 
 Method is preferable. 
 
 In AC take AG z=zfn; and, in B A produced, take 
 AHurz/z; then a Line AE drawn parallel to That join- 
 ing the Points G and H, will divide the Angle as re- 
 uired. For, by Trigonometry, AG f?n) : AH fnj : t 
 ine AHG (BAE) : Sine AGH fCAEJ : From this 
 Conftruftion the numerical Solution is exceeding eafy ; 
 both the Sides AG and AH, and the included Angle 
 HAG being given. 
 
 t 
 
 PROBLEM 
 
with their Solutions. 
 
 139 
 
 PROBLEM XXXVIIL 
 
 To divide a given Angle B AC into twofuch PartsBAG, 
 CAG, that the ReBangle under their Sines^ BE and 
 CF (to a given Radius J may be of a given Magnitude. 
 
 Let AH be perpendicular 
 to the Chord BC ; alfo let 
 BH (=CH)z=^, AH = 
 b, BE X CF = c\ and HD 
 :=:x ; luppofing AG to in- 
 terfea BC in D. 
 
 Becaufe of the fimilar 
 Triangles DAH, DBE, and 
 DCF. 
 
 S^'^[y[f^}+xx)) : AH [b] : : BD (^+;c) : BE 
 
 we have -^^ aD(/(^^+;c4) - AH {b) : : CD(^-:;.)': CF. 
 Whence, by compounding the two Proportions 
 
 l/"~'T~XX 
 
 From which Equation x is found == b \/(^xK^ — I " 
 
 By means of which and the foregoing Proportions both 
 BE and CF will become known. 
 
 Geometrically, 
 
140 
 
 Geometrical Problems, 
 
 Geometrically* 
 
 If CI be fuppofed perpendicular to AB, and FK to 
 CI; tjhen, the oppofite Angles CNF and ANI be- 
 in^ equal, their Complepients 
 NCF and NAT will likewife 
 be equal : And therefore, the 
 Triangles CFK and ABE be- 
 ing equiangular, we have AB 
 : BE : : CF : FK, or ABx 
 FK = BE XCF=:c* (accord^ 
 ing to Obfervation 5.) whence 
 L B ^^ is given. Therefore, if, 
 from the Extremity of the 
 Chord CFM, there be drawn MN perpendicular to 
 CI, the Length thereof, being twice That of FK, 
 will alfo be given ; and, from thence, the following 
 Conllruftion. 
 
 Take IL a third Proportional to \ AB and the Side 
 fc) oi the given Square, exprefling the Magnitude of 
 the propofed Reftangle : Draw LM perpendicular to 
 AB, meeting the Arch BC in M ; then a Line AG 
 drawn to bife6t MC will divide the Angle B AC. as re- 
 (juired. 
 
 The Numerical Solution, from hence, is very toncife 
 and eafy : For, having found the Value of IL (by di- 
 viding the Meafure of the given Reftangle by Half the 
 Radius) let the Co-fine AI of the whole, given, Angle 
 be added thereto ; then the Sum AL will be the Co-fine 
 of (BM) the Difference of the two, required. Parts. — 
 This Problem becomes impoffible when IL is given 
 greater than IB ; that is, when the Re6langle propofed 
 is greater than Half the Reftangle under AB and BI. 
 
 PROBLEM 
 
with their Solutions, 
 
 i4» 
 
 PROBLEM XXXIX. 
 
 Jo divide a given Angle MCN into twofuch Parts MCD, 
 NCD that their Tangents AD, BD may obtain a ginen 
 Ratio ; fuppofe That of m ton. 
 
 the Radius CD=a, the Tangent of the given 
 
 C 
 
 Put 
 Angle MCN = ^, and 
 That (AD) of the 
 Part MCD = a:. Then 
 it will appear, from 
 Problem 25, that the 
 Tangent BD of the 
 remaining^ Part NCD 
 is truly exprefled by 
 
 aa-\-bx 
 
 Hence we have 
 
 m 
 
 there- 
 
 aa-\-hx 
 fore nx X {aa-^-bx) zzz md^ X (^ —•^)- ^^°"™ which x '\i 
 
 found =« ^ [^+f!^±^'')]_if+'"'X^' 
 
 Q.nb 
 
 Geometrically. 
 
 w 
 
 Th^ Ratio of AD to BD being given, as m to «, 
 tlie Ratio of" tlicir^i^n^ AB to their Difference AE (fup- 
 pofmg DE = DB) will hi! given, as w -f" " to tw— w. 
 And if NG be drawn parallel to BA. meeting CE 
 and CA in R and G, the whole Line NG will be to 
 the Part GR in the fame Ratio of m-^-n to m-~n : And 
 it is evident, that, if two Perpendicul irs NP and RQ 
 be let fall from the Points N and R upon AC, they will 
 likewife be in that Ratio. Whence the following Con- 
 fl;ru6lion. 
 
 In NP, perpendicular to MC, take PH a Fourth 
 Proportional to »t4-n, m— ??, and PN ; draw HR pa. 
 
 rallel 
 
14* 
 
 Geometrical Problems, 
 
 rallel to CM interfering the Arch MN in R, and draw 
 CD to bifeft NR, and the Thing is done. 
 
 For the Trigonometrical Calculation, it will be 
 m-^-n : m—n : : NP : RQ ; 
 
 that is, as m-\-n is to m — w, fo is the Sine of the whole, 
 given, Angle to the Sine of the Difference of its two, 
 required, Parts, whence the Parts therafelves will be 
 known. 
 
 PROBLEM XL. 
 
 To divide a given Angle BAC into two Farts BAP and 
 CAV^fo that the ReBangle under their Tangents BD 
 and CE (to a given Radius AB) may be of a given 
 Magnitude. 
 
 If the Radius AB (or AC) be denoted hy a; the 
 
 Y^ Tangent BK of the given 
 
 Angle BAC, by b; and 
 
 the Tangent BD of the 
 
 Part BAP, by x; then 
 
 the Tangent CE of the 
 
 remaining Part CAP will 
 
 a^yj[h-x) 
 
 be reprefented by t-t — 
 
 ^ ^ aa-^-bx 
 
 (See the Note to Prob, 25). 
 
 And we fhall, therefore, 
 
 have^:^:#<f(=BD 
 aa'\-bx 
 
 .,LX{aa^bx) 
 
 X C E)= c\ Whence ^a: — ^^ =" 
 
 {la 
 >^xxz=.cci whence x = id±^ {ldd---cc). 
 
 : Which by making dz=:b 
 
 aa 
 
 ' — , becomes dx 
 aa 
 
 Qeometricilly. 
 
ft/zM M«V Solutions. 14I 
 
 Geometrically, 
 
 If a Line DF be drawn to make the Angle BDF 
 equal to CAE, the Triangles BDF and CAE will be 
 fimilar: and we fhall, thcreFore, have ACv BF — HP 
 xCEz=zf* (according to Ohftrvation g, p. 8g) whence 
 BF, and confequently AF, will be given. Moreover the 
 Sum of the Angles BAD and BDF being equal to the 
 given Ang''^ BAG (by Hypothejis) and the fame Sum 
 -j-ADF ^iqrai to a Righf-Angle (by Elem. Cor. j^to 
 10. 1.) it is evident that ADF is equal to the Difference 
 between the faid given Angle BAG and a Right-One. 
 
 Therefore, having taken BF a Third Proportional to 
 AB and the Side (c) of the given Square, make the 
 Angle AFO = the Angle FAG ; and from (O) the Tn- 
 terfeftion of AG and FO, let a Circle be defcribed 
 thro' A and F, interfering the Tangent BK in D, 
 and D ; from either of which Points draw AD, and 
 the Thing is done, 
 
 ■ In order to the Trigonometrical Calculation, \Qt the 
 Piameter GH be drawn to bifeft the Arch AF in H, 
 and let DM be perpendicular thereto: Then, having 
 found AF, it will be, as AN (| AF) : DM (NB) : : 
 Sine AH (Co-fme NAO) : Sine CD (=Co-fme iDD) 
 =Co-fme of the Difference of the two required Angles 
 BAD and DAI ; whence, as their Sum is given, the 
 Angles themfelves will be known. 
 
 This Problem becomes impoflible, when the Circle 
 OAFG neither cuts, nor touches, the Line BK ; 
 that is, when the given Reftangle is greater than the 
 Square of the Tangent of Half the propofed Angle, 
 and the Angle itfelf is acute ; or, when the faid Rec- 
 tangle is lefs than the Square of half the Tangent of 
 the propofed Angle, an4 the Angle itfelf is obtufe. 
 
 PROBLEM 
 
Hi 
 
 Geometrical Problems, 
 
 PROBLEM XLL 
 
 To draw a Line T>Y.parallel toagiven Line Al.foas to 
 interfed two other Lines AB, hC, given by Pofition^ 
 and thereby ferm a Triangle ADE of a given Mag- 
 nitude. 
 
 Let the given Area of the Triangle be denoted by 
 ^ a*; and let the Sines of 
 
 the given Angles DAE, 
 ADE, and AED (to 
 the Radius r) be ex- 
 prefTcd by m^ », and 
 pi refpeftively : Then, 
 making EP perpendi- 
 cular to AD, and caJl- 
 LB ing AD, x ; we have 
 (by plane Trigonometry) 
 As p : m : : X 
 
 — ; And r 
 P 
 
 DE = 
 :DE 
 
 Hence — X -=:fl* 
 rp 2 
 
 And therefore ,= ^f2Zff) ^a^f ^'P). 
 ^ \ mn J ^ \mn^ 
 
 Geometrically, 
 
 Let AF, perpendicular to AB, be the Side of a Square 
 equal to the Triangle ADE ; then, if AL be taken z=z 
 sAF, and FK be drawn parallel to AB, &c, it is evi- 
 dent that the Triangle ALK, being =(AF)% will alfo 
 be equal to the Triangle ADE. Moreover, by making 
 KM parallel to AI, it will be (ADj' (AM)' : : ADE 
 (ALK) -': AMK (Elem. 24. 4) : : AL : AM (Elem. 1. 4.) 
 ^ : ALx AM : (AM)'. And confequently (AD)' = AL 
 
 XAM 
 
with tknr Solutions. 
 
 Hi 
 
 X AM. Therefore, AD being a Mean Proportional 
 between AL and AM, upon the former of Thefe let 
 a Semi-Circle x^NL be defcribed, interfering MN. 
 perpendicular to AB, inN; make ADzzzthe Chord 
 AN, and draw DE parallel to AI, and the Thing i« 
 done. 
 
 PROBLEM XLII. 
 
 Thro' a given Point P, between two Lines AB and KO 
 given by Pofition, to draw another Line DE, /^ that 
 the Triangle h"^^, formed thereby^ may be oj a given 
 Magnitude, 
 
 Let PG and PF be parallel to AC and AB, and PQ 
 and ER perpendicular 
 to AB : Alfo let AG 
 and PQ (which are 
 given by the Pofition 
 of P) be denoted by 
 a and by refpeftively, 
 Then, calling AD, 
 X, and denoting the 
 given Area by t% we 
 
 ftiall have, DG [x^a): PQ(/^): : AC W : ER 
 
 x-a 
 And therefore -f- x - (zi=:ER X i AD;z= c'. 
 
 Whence xx — =z 7-; and comequently a: =z 
 
 b ' '^ 
 
 Geometrically, 
 
 If upon AF, a Parallelogram AFHI be conftituted, to 
 contain the given Area, it will appear, by taking away 
 AFPMI from each of the equal Quantities AH and 
 ADE, that the Remainders PHM and PFE + IDM 
 will likewifc be equal : And fo, thefe three Triangles 
 
 L being 
 
J 46 
 
 Geometrical Problems, 
 
 being all fimilar to one another, it follows that (PH)' will 
 be=:(PF)^+(ID)^ Whence this Conftruaion. 
 
 From the Center F, with the Interval PH, let an 
 Arch be defcribed, cutting PQ perpendicular to AB, 
 inN; make ID izzPN, and draw DPE, and the Thing 
 is done. For it is evident that (?H)% (FN)%is =i (PF)^4- 
 
 (D1)%(PN)\ Hence it alio appears that the Problem 
 
 will be impoflible when PH is lefs than PF ; or 
 when the Triangle propofed to be conllru6led is lefs 
 than twice the Parallelogram AFPG. 
 
 PROBLEM XLin. 
 
 Thro* a given Point P, between two Lines AB, AC 
 given by Po/ition, to draw another Line DC, fo that 
 the Sum of the Parts, AD -f- AE, cut off thereby, 
 from the two former , may be a given Qjudntity, 
 
 Let PF (parallel to AB) be denoted by a, and PG 
 
 (parallel to AC) by 
 b ; alfo let AD + 
 AE ::=z c, and AD 
 rzz X : Then, by 
 Reafon o\ the pa- 
 rallel Lines, we 
 fliall have DG {x— 
 a) : PG {b) : : AD 
 
 {.) : AE= ^" 
 
 X — a 
 
 And therefore-— -4. x (z=AE -f- AD) = c. 
 
 From which, by Reduftion, xx — [c-\-a-b)y^X'=-=.^gLC ; 
 
 and confequently ;t=: 
 
 c^a—b 
 
 i^((i±±±-_.) 
 
 Geometrically, 
 
with their Solutions. 147 
 
 Geometrically 
 
 The Triangles DGH and PFE being fimilar, the 
 Reftangle under DG and FE will therefore be equal to 
 the given Re6langle under AF and AGz=(AR)'; by 
 taking AM m AF, and making AR a Mean Pro- 
 portional between AM and AG. Moreover, it MM 
 be taken = the given Sum of AD and AE, it is mani- 
 feft that the Sum of DG and FE (whofe Reftangle is 
 above given) will, alfo, be given r::^: GN. 
 
 Therefore, having defcribed a Semi-Circle upon GN, 
 let RS be drawn parallel to AB, interfefting the Peri- 
 phery thereof in S ; from which Point, upon AB, let 
 fall the Perpendicular SD, and thro' P draw DPE, and 
 the Thing is done. For DGxDN = (DS)^ = (AR)'= 
 AMxAGz^AFxAGz^DGxEF: Whence EF=: 
 DN ; and confequentlv AD+AE (==AD-f AF+DN) 
 = MxN. 
 
 It is plain, this Problem becomes impofTible, when RS 
 falls below the Semi-Circle GN ; or, when the Sum of 
 AD and AE is fuppofed to exceed That of AG and 
 AF by lefs than the Double of a Mean Proportional be- 
 tween AG and AF. 
 
 Much after the fame Manner the Problem will be 
 rcfolved, when the Difference, the Ratio, or the Rec 
 tangle of AD and AE is given. 
 
 L f PROBLEM 
 
X43 
 
 Geometrical Problems, 
 
 PROBLEM XLIV. 
 
 Through a given Point, P, betwixt two Right-Lines 
 AD, AE, given by Pojition, fo to draw a Right. 
 lineEFC that the ReSiangie(BPxC?) under the Paris 
 TherecJ, intercepted by that Point and thofe Lines ^ may 
 be of a given Magnitude, 
 
 Let PF and PH be parallel lo AC and AB, and PG 
 
 perpendicular to AB : 
 Then, calling PF, a ; 
 PH, b', FG, c; and 
 BF, x\ we have (BP1> 
 '=Laa-\-xx — 2CX (Eltm. 
 
 9, 2). 
 
 And, by fimilar Tri- 
 angles, BF [x] : BP : : 
 
 PH (b): CP=ixBP 
 
 -n j\ Confequently B P X 
 
 A F G 
 
 b • b 
 
 CP= - V BP* =- X {aa+xx-^2cx). Hence, if the 
 
 X X 
 
 given Value of BP x CP be denoted by d\ then will 
 -r- =«fl-f-^-^ — 2^-^' • Which Equation, folved, gives, 
 
 dd , , ffdd y 1 
 
 Geometrically, 
 
 The Re6langle under two unknown Lines being 
 given, another Line mufl therefore be found, or affum- 
 ed, under which and fome given Line in the Figure, 
 an equal Rc6langle may be contained. fVid, Obfer, 5. 
 p. 88.) 
 
 As 
 
with their Solutions. 149 
 
 As, in the prefent Cafe, the Line AP is given, both 
 in Length and Pofition, let the Reftangle under it, and 
 a Part, PQ, of the fame Line produced, be therefore 
 afrumed^riBPx CP; then the Confequencc will be, 
 
 that, befides obtaining PQ = — r-^, the Triangles PQG 
 
 -A.P "^ 
 
 and PBA (fuppofing QC drawn) will alfobe fimilar (bc- 
 caufe AP : BP : : CP : PQ) And ^ fo, the Angle PCQ 
 being zrz the given Angle PAB, it is evident that a Seg- 
 ment of a Circle dcfcribed upon FQ fi?y Elem. 22. ^.) 
 capable of containing the faid given Angle, will interfeft 
 AE in the Point (or Points) required. 
 
 This Problem will, it is manifeft, be impofTible, when 
 the Circle defcribed as above, falls fhort of AE ; or, 
 
 according to the Algebraic Solution, whenf ^-(-c) -a' 
 
 is negative ; that is, when the propofed Rcftangle is lefs 
 than 2b X (<2— 4 or, its Equal 2PH X (PF—FGj 
 
 L 3 EROBLEM 
 
i,5o Geometrical Problems, 
 
 PROBLEM XLV. 
 
 Thi Area [a^] of a right-angkd TriangU ABC, wko/e 
 Sides are in Arithmetical ProgreJJion^ being given , t9 
 ddermine the Triangle. 
 
 Put the greater Leg ABitza:, and the common Dif- 
 ference =:z >' ; fo ftiall BC = x — y, and AC :=z x -f- )' 
 And therefore 
 ax^yYz=:z[x-yY-\-x^\ ^ .^^^ Qucfllon. 
 
 From the former of which 
 Equations we have xx -{- 
 ^xy-\- yy z^z'i- xx — 2.xy-\-yy ; 
 -and confequently /^y z=z x ; 
 Whence, by fubflituting for 
 X in the fecond Equation, 
 ^ wc get sj^'XS)' =^* • Eiom 
 
 which y is given z=z\/ — ; and x (r=4;') =4 V^ y- 
 Therefore BC = 3 y^ £ AB =1=4 y/^'. and AC 
 
 = 5/ r 
 
 6* 
 
 Geometrically, 
 
 It is well known that (AC-f-BC)xfAC— BC)=[AC/- 
 (BC)^ is=:(ABV (Elem. Cord, toy. 2). And. by the 
 Queftion, AC -f BC is 3= 2AB (becaufe AB is an 
 Arithmetical Mean between AC and BC). Therefore 
 2ABx(AC~BC)z=:(AB)' ; and confequently AC-BC 
 rr: J AB : Take thefe equal Quantities from the equal 
 Quantities AC -4- BC and 2AB, and the Remainders, 
 2&C and 1 ^ AB, will be equal; and confequently 
 2AB = 4BC. But.l ABxBC(=|BC X BC) = 
 (BOy (= a')i and therefore ^(BC)'=:i- {BD]\ 
 
 Hence 
 
with their Solutions. 
 
 ^5* 
 
 Hence, if BH be taken to BD [a] in the Proportion 
 of 310 2, a Mt-an Proportional BC between DB and 
 HB, will be the lefTer Leg of the Triangle required; 
 whofe greater Leg BA, being in Proportion thereto, 
 as 4 to 3, is alfo given from hence. 
 
 PROBLEM XLVL 
 
 The Area (a') of a right-angled Triangle ABC, whofe 
 Sides are in Gcoinetrical ProportioUy being given ; to 
 determine the Tnans-ie, 
 
 iMake AC z= a: ; then BC =z= — , and AB = 
 
 X 
 
 ^ 
 
 ' / 
 
 C 
 
 Sy 
 
 /A// 
 
 / 
 E 
 
 \ 
 
 v\ 
 
 u 
 
 W 
 
 E 
 
 1 
 
 ? 
 
 G 
 
 ^ \ * XX f 
 
 2a^ 
 Therefore, — : x : : x : 
 
 X 
 
 Hence ;v' =z ^/{ x^ + '^— ) ; ;e* = — X ■ - : 
 
 X ^ \ ^ XX ) XX ^ XX ' 
 
 x^ — j^a^x^ = 16a* ; x'* — 2a^=:2 a^'i/20 ; and x = ^X 
 
 ^{'. + ^20). 
 
 Gecmetrically^ 
 
 Since, by Hypothefis, AB : AC : : AC : BC, there- 
 fore is (AB)^: (AC)':: (AC)- (BC)Y^/^^.C^r.i /,p 11.4) 
 
 But AC', fuppofing CD perpendicular to AB, 
 is equal to AB X ^^ * ^"^ i^^Y ^4"^^ ^^ AB x BD 
 fElem, Cor. to 19, 4). 
 
 Therefore (AB)' :ABx AD: :ABxAD :ABxBD; 
 or, AB : AD : : AD : BD. 
 
 Whence the following Conftruflion. 
 Draw any Line EG, at pleafure ; which divide at F, 
 fby Elem, 19, 5.) according to Extreme-and-Mean Pro- 
 portion (fo that EG : EF : : EF : GF) ; ered the Per« 
 
 L 4 pendicular 
 
15^ 
 
 Geometrical Problems, 
 
 pendicular FC, and upon EG let a Semi-circle be de- 
 fcribed interfering it in C ; join E, C, and G, C ; 
 and let AB (by Proh. 41.) be drawn, parallel to EG, 
 to cut off the given Area ABC [z=z a"), and the Thing 
 is done. For it is manifell that AB is divided, by CDF, 
 in the f?.me Proportion with EG ; or that, AB : AD : : 
 AD : BD, as above, 
 
 PROBLEM XLVII. 
 
 Suppojing one Side AC, and the oppojite AngU ABC 
 oj a Triangle to be given ^ together with the Side DE, 
 cr DF, of the injcribed Rhombus EF; to Jindjrom 
 thence the other two Sides of the Triangle. 
 
 Conceive a Circle ABCI to be defcribed about the 
 Triangle, and the Diagonal of the Rhombus pro- 
 meet the Circumference thereof in I : Alfo 
 let AI and CI be drawn ; 
 which will be equal to each 
 other (Elem, Cor. 12. o^.J as 
 being the Subtenffes o\ the 
 equal Angles IBA and IBC, 
 form'd by the Diagonal and 
 Sides of the Rhombus : And, 
 fmce the Arcs Al and CI, 
 as well as the Chords, arc 
 equal, the Angles IAD and 
 IBA, infifting upon Them, rnuil likewife be equal ; 
 and confequently the Triangle IAD fimilar to the Tri- 
 angle IBA. 
 
 But the vertical Angle AIC of the Ifofceles Tri- 
 angle ACI (as well as the Bafe AC) is given, being :rz: 2 
 Right-angles — ABC ( Elcm. Cor, 17. 3.) whence lA will 
 be given (by plane Trigonometry), 
 
 Put therefore, lA z= «, BC ==: b, and IB =x ; then, 
 iTom the Similarity of the Triangles above fpecified, we 
 ihall have, ^ -^ f ID ) : a (lA): : ^ (IA]:;v(IB ) 
 vnencejtAC — bx-zizjaa ; and confequently a:z= '^{aa-\-\bb) 
 -^\b. From which, and the known Values of lA and 
 the Angle ABI, the Value of AB, ^c, will alfo be 
 known. Geometrically, 
 
with their Solutions. 153 
 
 Geometrically. 
 
 The Geometwcal Conftru6lIon hereof, as the Rec- 
 tangle BIX^I (by the foregoing Proportion) is given =1: 
 (AI j% is obvious from Problem 6 ; and is thus : 
 
 Having, upon the given Side AC, defcribed a Seg- 
 ment of a Circle ABC capable of contain'^g the given" 
 Angle ABC (EUm. 22. 5.) ; and in the other Segment, 
 AlC, conliituted the Ifofceles Triangle AlC, make IG 
 perpendicular to lA, and equal to Half (BD) the given 
 Diagonal of the Rhombus; alfo, in IG produced, take 
 GH equal to the Diftance GA ; and, through H, from 
 the Center I, let an Arch be defcribed interfering the 
 Circle ABC in B; then draw .BA and BC, and the 
 Thing is done. 
 
 PROBLEM XLVIII. 
 
 from tzoo given Points A and B, to drazo two Lines 
 AC and BC, to meet in a Right-line DE given 
 by Pojition^ and form an Angle ACB of a given 
 Magnitude, 
 
 Make AM and BN perpendicular to DE ; and let 
 AF be fuppofed 
 parallel to BC : 
 Then, calling 
 AM, a ; BN, 
 b ; MN, c ; and 
 MC, x; il will 
 be as^ : c — x 
 (NC) : :a :MF 
 
 — 6 ' 
 
 But MF and MC are Tangents of the Angles MAF 
 and MAC; which Angles together (becaufeof the pa- 
 rallel Lines AF and BC) make an Angle CAF equal 
 to the given Angle ACB : Therefore if the Tangent 
 of the faid given AngJe, to the . Radius AM, be de- 
 noted 
 
J54 Geometrical Problems, 
 
 AM' — MC X MF — ^* 
 
 This.inSpc. (;c+'SlI^)^a^ " , \ 
 
 cies, gives ^ ^ _^ \^-^--^>y+'^OX^ _j. 
 
 <2Ca: — /iA'AT a^ — ex -j-xx *"" * 
 
 a» 
 
 if 
 
 Which, by Reduclion, becomes ^izif IXff . fff _ 
 vVjv — cx-{- ab. Whence, making c -|-- -^ — // 
 
 and ^ X ( ^) =/» ^^ havey"=A;A: — ^^ ; and con- 
 
 fequently ;c =r|i4-/(/-f| dd). 
 
 When the Line MN is parallel to That joining the 
 given Points A, B ; then, BN (b) becoming zn: hM(a) 
 
 we have, in this Cafe, dz=ic^ and/ -=: a^ ; and 
 
 therefore xzzz-^ "t V^( fl:^+|t'<:). Which may 
 
 ferve as a Theorem for finding the Segments of the 
 Bafe of a Triangle (and confcquently the Triangle itfelfj 
 when the whole Bafe, the Perpendicular, and the vertical 
 Angle are given. 
 
 As to the Geometrical Conftrnilion of the General 
 Problem, it is extrcmeiv obvious ; fmce a Segment of 
 a Circle defcribed upon AB (by EUm, 22. 5.) capable of 
 containing the given Angle, will interfefct DE in the 
 Point (or Points) re(juired. Whence it alfo appears 
 that the Problem will be impoflible when the Circle falU 
 fhort of the Line FE ; and, confequently, that the 
 Angle ACB will be the greaieft poffible when the Circle 
 touches the faid Line ; or, wlien DC is a Mean Propor- 
 tional between DA and DB. 
 
 PROBLEM 
 
wif-h their Solutions. 
 
 ^55 
 
 PROBLEM XLIX. 
 
 To find a Point C in a RighfAine DE, given by Pofition / 
 Jo that two Lines CA, CB being drawn Jrom thence 
 to two given Points A and B, the One BC Jhali ex- 
 ceed the Other AC by a given Difference d. 
 
 Let AR and BS be perpendicular to DE ; and let' 
 thefe Perpendi- 
 culars, together 
 with RS (which 
 are all known 
 from the Pofiti- 
 on of DE) be 
 denoted by a^ 
 by and r, refpec- 
 tively : Then, DA B 
 
 putting ACz=^, we have RC ( =z= /[(AC)'— (AR)*] 
 z=ii/{xx -^aa); and SC ( = /[(BC)^ — (BS)*] = 
 'y/{xx -\~2xd -^ dd — bb)z=c — ^(xx — aa). 
 
 From which Equation, by fquaring both Sides thereof, 
 we get xx-^2dx-^dd — bbzz^cc — 2Ci/{xx — aa)-\-xx — aa. 
 
 This contrafted becomes 2c^(xx — aa)zzzbb — aa — dd 
 
 •\-cc — 2dx ; or, ^/[xx — aa]z=jn — nx ; by dividing by 2r, 
 
 bb — aa — dd-X-bc . d 
 
 and putting •> = — = m^ and — z=.n, 
 
 2C C 
 
 Therefore, by fquaring again, xx — aazzzzmm — 2mnx 
 '■\-nnxx\ whence (i — nn)y^xx '\' zmnx-^iaa-^-mm^ or 
 
 2mnx aa-\~mm 
 
 XX ~T~ ' ' -" •—— —————, 
 
 ' 1 — nn 1 — nn 
 
 / aa-^-mm 
 And contequentiy x =: y^l 
 
 
 mn 
 
 •nn 
 
 Geometrically^ 
 
1^6 
 
 Geometrical Problems, 
 
 Geometrically, 
 
 If AB be drawn, and bifefted in F, and CM be fup- 
 pofed perpendicular thereto, it is well known that FM 
 
 « -S- ^ ^fC^r. ii.tOQ. 2. : Therc- 
 
 2AB ^ 
 
 fore, fnice BC — ACnziPQ, and, confequently, BC 
 
 + ACr=PQ + 2AC, It follows that FM = 
 
 (P0 4-2AC)xPQ __. (PQr , ACxPQ 
 
 2AB — 2.iB "^ Ad * 
 
 p / TJ ^ E ^^^^» ^^"^* 
 
 hQ ^^ both AB and 
 
 ^^ PQ are given. 
 
 the Value of 
 
 a Third Pro- 
 portional to 
 2ABandPQ) 
 is alfo given. 
 Let this, 
 therefore, be 
 ex pre fled by 
 
 FG ; fo that FM may be = FG+ ^?^^^/S : Then 
 
 a'\CvPO 
 
 It is plain that GMizz 1> '" < ; or that, GM is to AC 
 
 Aij 
 
 in the given Ratio of PQ to AB. 
 
 But the Ratio of GM to HC (fuppofmg GH perpen- 
 
 dicular to AB) is alfo given by the Pofition of the Line 
 
 DE ; whence the Ratio of HC to AC (where Both belong 
 
 to the fameTriangle AHC) will be exhibited : For, taking 
 
 GIrziPQ, and making IK perpendicular to AB, it will 
 
 HC GM,_GM 
 
 quently HC : AC : : HK : AB 
 
 But 
 
 be 
 
 pendicular to AB, it w 
 
with fhe'r Solutions. 1^7 
 
 But, if KL be fuppofed parallel to AC, meeting HA 
 produced (if need he) in L, it is manifeft that HC : AC 
 : : HK : KL. Therefore KLzzzAB ; whence the fol- 
 lowing Conflruftion. 
 
 Having taken FG = ^^^', GI=PQ, and ereaed 
 
 the Perpendiculars GH and IK, and alfo drawn \IK 
 fas above fpecijiedjy from the Center K, with the Inter- 
 val AB, let an x^rch be defcribed, cutting HA produced 
 in L ; and, having drawn LK, make AC parallel 
 thereto ; which will cut the given Line DE in the Point 
 required. 
 
 The Trigonometrical Calculation, from this Con- 
 ftruftion, may be as follows : 
 
 Having computed FG^zn J apd fubtrafled it 
 
 from FD and FA, the Remainders GD and Gx^ will 
 be given; and it will then be as GT> : GA : : Tang, 
 DHG (or Co-tang. D) : Tang, AHG ; whence the 
 Angle LHK is likewife known. And, fince Sin, DHG 
 (or Co -fin. D) : Rad. :: IG (PO) : HK, the Value 
 of HK (as well as ThoTe of KL and the Angle HKL) 
 will be known : From which the other two Angles of 
 the Triangle HKL being found, the Angle DAC ( — 
 ECA— D=HKL— D) will alfo be obtained. 
 
 After the very fame Manner the Problem may be re- 
 folved, when the Sum, inlfead of the Difference, of 
 the Lines AC and BC is given. 
 
 As to the Reflri8;ions of the laft Problem, it is evi- 
 dent that the given Diiferenco muft never exceed the 
 • Diftance AB : But when the Line DE palfeth between 
 the Points A and B, the Limit will be ftill lefs ; but 
 is eafily determined, in any Cafe, from the given Po- 
 fition of DE. It is a little remarkable, that the above 
 Solution fails in that particular Cafe, only, wherein the 
 General Problem becomes moft fimple ; that is, when 
 DE is perpendicular to AB. But here the Operation 
 will, alfo, become more fimpte and expeditious : For 
 
 the 
 
1^8 
 
 Geometrical Problems, 
 
 the Pofuion of MC (with which DE is fuppofed i& 
 coincide) being aftually given, the Lehgth ot AC = 
 
 — -^ fp, above J will alfo be known. And, if 
 
 ■with this, as a Radius, an Arch be defcribed, from the 
 Center A, it will interfeft the Perpendicular MC (or 
 DC) in the Point required. 
 
 PROBLEM L. 
 
 From two given Points A, C, in the Diameter EF of a 
 given Semi-circle^ to draw two Lines to meet in tht 
 Circumference, fo that One of them CD may exceed the 
 Other AD by a given Difference^ not greater than the 
 Dijiance of the two given Points. 
 
 The Radius BD being fuppofed drawn; put ABzira, 
 
 BC = ^ AC=z=c. 
 BD=:r, AD=x, 
 and CDz=A:-f-//; 
 d being the given 
 Difference. 
 
 Then, by the 
 Lemma at p. 128, 
 we fhall have x^^b 
 '■\-[x-\'dY y^a:=:abc 
 -\- cr^ \ or, bx^-^- 
 ax* -\- 2adx -}- ai* 
 z=iabC'\-cr*. 
 add 
 
 Whence **-f 
 
 i.adx 
 
 ab-^-r'^ 
 
 And confcqucntly jr == ^/^ab-^-r^ j— — : 
 
 Geometrically, 
 
 is alfo deducible 
 For it is evidenl 
 
 The Geometrical Conflruftion 
 from the Lemma above fpecified. 
 
 RC 
 from thence, that (AD)*X ^+(CD)'is=AC xBC + 
 
with their Solutions. 159 
 
 ^^y^^^^y ^ ACxBC+ACx BG (by taking BG= 
 AB 
 
 ^S") = AC X (BC+BG) = AC xC G. 
 
 Therefore, having taken BG a Third Proportional to 
 AB and BD ; and GL a Mean Proportional hetween 
 AC and CG ; deaw BK perpendicular to GF, meeting 
 the Circumference of a Semi-circle, defcribed upon 
 AC, in K ; and, having drawn AKM, and takea 
 AH equal to the given Difference of AD and CD, 
 upon H as a Center, with the Radius LC, Let an Arch 
 be defcribed, interfering AK in M ; from which Pomt 
 upon GF let fall a perpendicular MN : then, if from 
 A, with the Radius AN, another Arch be defcribed, it 
 will interfe£l the Arch of the given Semi-circle in the 
 Point, D, required. 
 
 For, AB : BC : : (AB)» : ABxBC (={BK)>), Elem. 
 19. 4 : : (AN)', (AD)S : (NM)^=(AD)'x41- ^"^ 
 
 (NM)S ^(ADj=x-|^)> +(HN)«=(HMr=ACxCG 
 
 RC 
 
 (by ConJir.J = (AD)'X J^+ (CD)Y/'. above) , 
 
 Therefore CD = HN — AN + AH = AD - j- AH ; 
 and confequently CD — AD = AH= the given Dif- 
 ference, by Conjirudion. 
 
 The Method of Calculation, from this Conftruf>ion, 
 
 isfufficiently eafy : For having computed BG r= ^^) 
 
 and HM (=/[CAx(CB+BG)] and alfo the Angle 
 BAK (which is had by the Proportion AB : BC : : Squ. 
 Rad, : Squ. Tang. BAK) ; you will then have, in the 
 Triangle HAM, two Sides and one Angle ; whence 
 every thing elfe is readily determined. 
 
 PROBLEM 
 
i6o 
 
 Geometrical Problems, 
 
 PROBLEM LI. 
 
 From the Middle F, and the two Extremes K and B, oj 
 a given Right-line AB, to draw three Lines to meet in 
 a Point C, in a Right-iine DE giren by Pofitionyfo 
 as to be in G&omttrical Proportion^fo that AC : FC 
 : : FC : BC. 
 
 Becaufe (BC)'+(AC)'z=2(FC)'+2(BF)' (Elem, ii. 
 2.)- And BCX ACz=:(FC)^ (by Hyp, and Elem, lo. 4.) it 
 is evident that (Be)*—2BC xAC +(AC)^=2(BF)^ or 
 (BC — AC)' r=:2(BF)'=(BG)% by taking BG as the 
 Diagonal of the Square whofe Side is BF. 
 
 -E Hence BC— AC 
 is given zz= BG : 
 And fo the Cafe 
 in Queftion is re- 
 duced to Problem 
 49 ; to which I 
 fhall therefore re- 
 fer for the remain- 
 ing Part of the 
 Solution. It ap- 
 pears from hence that a Point in the Circumference of 
 a given Circle, whofe Center is in the Line AB, may 
 be fo determined, by the lajt Problem, tliat three Lines 
 drawn from thence to the three given Points A, F, 
 B, fhall be in Geometrical Proportion. 
 
 PROBLEM 
 
with their Solutions. 
 
 161 
 
 PROBLEM LII. 
 
 From two given Points A, B, within a given Circle^ to 
 draw two Lines AC, BC, to 7neet in the Periphery 
 Thereof, fo that the Sum of their Squares. may be a 
 given Quantity. 
 
 Through the given Points let EF be drawn, meeting 
 the Circumference of the ^ 
 
 Circle in E and F; parallel 
 to which, draw the Dia- 
 meter PQ ; and let the 
 Chord CG be drawn to 
 cut EF and PQ, at Right- 
 angles, in D and H. 
 
 Put EF=a, EA = 3, 
 EB = c, DH=^, ED = ;t, 
 and DC =7 ; and let the given Quantity, (AC*-f- 
 (BC)% be denoted bye': Then will DF=:i7—;f, DA 
 z=ib — x, DB=:<: — AT, DG=jy + 2fl^. 
 But EDx 'F = CDxE)G (Elem, 21. 3). 
 And (DA)'+(DC)^+(DB)^+(DC)*=(AC)'-f-(BC)^ 
 fElem. 8. 2). Which, in Species, 
 
 give /^x(^--.^)=7X(:v-f2^) 
 
 or 
 
 (*-^r+>-+(^-^)"+/ 
 
 {ax — XX =.yy -\-2dy 
 2xx-\-2yy — 2bx — 2cx :zz 
 
 ee — bb — cc. 
 Whence, by aclding the double of the former Equation 
 
 to the Latter, we get 2ax — 2bx — 2cxz=zee — bb — cc-\-j^dy; 
 . . , bb-X-cc-^ee a^b — c „ , 
 
 and confequently^= — !— r J : — X ^ =/ + 
 
 bh-^-cc — ee „ . a — b — c 
 
 gx, by makmg — y ■ =:/,>and 
 
 \d 
 
 2d 
 
 M 
 
 Now, 
 
tSK 
 
 Geometrical Problems, 
 
 Now, the Value of y thus found being fubftitufed 
 in the firft Equation, there arifes a;c — xx-z=iff-\-^fgx 
 
 or. i^'^gg)X^x-^{2dg-\-2fg ^a)xx = -^J'- id/. 
 From vvhichj by making w =; i "hggt n z=: a- 2dg - 2^, 
 
 and p =/x(/+ 3^> the Value of jf is found z^— ± 
 ^ V Amm m J 
 
 /^7t%?n 
 
 Geometrically. 
 
 That the Locus of the Vertex of a Triangle, whereof 
 
 the Bafe and S^um of the 
 Squares of its e Sides 
 are given, is the Cir- 
 cumference of a Circle, 
 defcribed from, the Mid^ 
 die, of the Safe as a 
 Center, is, evident, yr(7w» 
 El^m. 11. B ; bccaufe the 
 lyine drawn from the 
 Xertex to the Middle of 
 the B^fe, is an invariable 
 Quantity^ 
 
 Therefore, having- bire£^ed AB with the Perpendi- 
 cular IMN, take MN and MR equal, each, to Half the 
 Side (e) of the given Square ; draw NR, apdj from A 
 to the Perpendicular MI, draw AK equal to NR ; and 
 upon the Center M; at the Diftance MK, let an Arch 
 be defcribed; which will meet the Circumference of the 
 given Circle in the Point C, required. For AC, 
 BC, and MC being drawn, it is ewdent that (AC)"+ 
 (BC)'=2(AM)* + 2(CMf=2(AM)'+2(MK)'=2(AK)' 
 =2(RN)^=4(RM)*=r:c\ In this Problem it is rc- 
 quifite that MC =y^[itf' — (AM)'] fhould be greater 
 than the Difference, and lefs than the Sum, of the 
 
 Radius 
 
with their Solutions. 
 
 163 
 
 Radius of th6 given Circle and the Diftance of the Point 
 M from its Center. 
 
 PROBLEM LIII. 
 
 To draw a Lint PQ thro' a given Point A, Jo that the 
 Redangle of two Perpendiculars^ falling thereon from 
 two other given Points ^ B and C, may be of a given 
 Magnitude, 
 
 Let BA and CO be produced to meet in E ; and 
 upon BE let fall the Per- 
 pendicular CF; which, 
 and the Segment AF, 
 are both given, by the 
 Pofition of the propofed 
 Points. 
 
 Put, therefore, CF 
 =a, AF=r^, AB=f, 
 and EF = a: : Then 
 yfi\\C^z=:y/{aa'\-xx)\ 
 
 And it will be (byfimilar Triangles)^ 
 
 asCE:EF::AB;BP= '''' 
 
 ^[aa-\-xx) 
 
 xyXh^x) ^ 
 
 bx -\' XX 
 
 ^{aa -|- xx) 
 
 *ndCE;EF::AE:EQ=- ^, . .. 
 
 Whence CQ(=CE— EQ)=/(fltf4.;c;f; 
 
 aa — hx 
 ^{aa -j- XX) 
 
 And,confequently,BPxCQ=— ^^^^ = g' ; fup- 
 
 ofing g" to denote the given Magnitude of the propofed 
 
 poiing g 
 Reftang 
 
 aac 
 
 From this Equation, by making </=:7-t — » ^^^ f =ss 
 M 2 
 
164 Geometrical Problems, 
 
 T~-'^ the Value of a: is found z=.\d±>/{\(id-^fg). 
 Whence every Thing elfe is readily determined. 
 
 Geometrically, 
 
 If QM be fuppofed parallel to AB, the Triangle 
 CQM will, it is plain, be fimilar to ABP ; and confe- 
 quently ABxQM=: BPxCQ (EUm. 10.4.) =^': 
 Whence OM is given ; and, from thence, the following 
 Conftru6iion. 
 
 Find a Third-Proportional to AB and the Side (g) of 
 the given Square; and, in BF produced, take FD equal 
 to the Double thereof ; draw DG parallel to FC, inter- 
 fering the Circumference of a Circle, defcribed from the 
 Center A with ihe Radius AC, in the Point G; and, 
 having drawn CG, draw PQ perpendicular thereto. 
 
 For the Trigonometrical Calculation, it will be AF: 
 AD (AF+FD) : : Co.f. HGC : Co-f. HG, the Dif- 
 ference between the Angles BAP (HAQ) and CAQ : 
 From which, as their Sum is given, the Angles themfelves 
 will be known. — This P'-oblem become5 impoflibie when 
 FD is greater than FH ; that is, when the propofed 
 Reftangic is greater than Half the Re6}angle under AB 
 and FH. 
 
 PROBLEM 
 
with their Solutions. 
 
 16^ 
 
 PROBLEM LIV. 
 
 Two Lines AB, AC, drawn from the fame Point A, 
 being given both in Poftion and Length; to draw 
 another Line PQ thro' that Point, fo that two Per. 
 pendiculars BP, CQ, falling thereon from the Ex- 
 tremes of the two given Lines, may form two Triangles 
 ABP, ACQ, equal to each other. 
 
 Let BA and CQ be produced to meet in E; and 
 
 upon BE let fall the 
 Perpendicular CF: And 
 put CF = a. AF=^ 
 AB=c, and EF = ;*r. 
 The Areas of fimilar 
 Triangles being, in Pro- 
 portion, as the Squares 
 of their homologous 
 Sides ; and both the 
 Triangles AEQ, ABP 
 being fimilar to ECF; 
 
 wc 
 
 CFxEF(ifl;c) : : (AB)'(^' 
 
 have (CE)' 
 Area ABP = 
 
 [aa-^-xx) 
 \ac*x 
 
 ax 
 
 aaJ^xx 
 : [x-^by [=:(AEj^] : Area AEQ: 
 
 This, taken from the Area AEC (= 
 
 And, aa^xx 
 
 aa-^xx 
 |<,X(«+*) leave, theAreaACQ=ii><if^±*l><(ffr:ii). 
 
 Which being equal to ABP, by Hypothefis. we there- 
 fore have [x-\-b)y^[aa — •bx):=zc^x : From whence, makiug 
 
 dz=: 
 
 cc— aa 
 
 f ^, X is found = ^/[aa'\'\dd)^/\d. 
 
 M3 
 
 Geometrically. 
 
i66 
 
 Geometrical Problems, 
 
 Geometrically* 
 
 If hb be fuppofed perpendicular^ and equ^al to AB, 
 and bp perpendicular to PQ, the Triangle hbp, being 
 fimilar to ABP, will alfo be equal to it ; and confe- 
 qufntly equal to ACQ : And, it thefe equal Triangles 
 be, fucccflively, taken from the Trapezium AQC^, the 
 Remainders/?/^ CQ, and KCb will likewife be equal. 
 
 i 
 
 Hence it appears that the Parallelogram under Qb and 
 DE, whofe Angle is CDE (fuppofing DE parallel io^ 
 and an Arithmetical Mean between, CQ and bp) is 
 equal to the given Triangle AC^; and confequently 
 the Altitude Thereof equal to half Th«rt of the Tri- 
 atjgle. Whence it is evident that the Point E muft fall, 
 fbmewhcre, in a ^Line FI drawn thro' the Middle of 
 AD (of AG) parallel xobQ (EUm. 2. 2). But the Point 
 E, fince the Angle AED is a Right-one, will likewifie 
 fall in the Circumference of a Semi-circle defcril-)ed upon 
 the Diameter AD (EUm, 13. 3). And therefore FE, 
 being a Radius, mull be equal to AF ; and confequently 
 DG = Afi ; fuppoiing DC produced to ropet AQ 
 in G. 
 
 Therefore, in order to the Geometrical ConftruiEliori, 
 having made A^ perpendicular, and equal to, AB, and 
 drawn AD to the middle of C^ (as above indmaled) let 
 DG, in DC produced, be taken equal to AD ; and 
 from G, through A, draw GP, and the Thing is done. 
 
 It often happens that the Demonftration of a Geome- 
 trical Gonftruftion, to be the moft neat and elegant, 
 proceeds upon Principles very different from Thofe 
 whereby we firft arrived at fuch Conftruftion. The 
 
 Cafe 
 
with ikiir SoLUtioS^. i'^j 
 
 Cafe above is an Inftance of it : Where from the fimilaJ" 
 Triangles, it is manifcft that GQ f A/>) : QC : : Gp 
 (AQ) :ptf; and therefore i AQxQC =i A^X£^ = 
 i BPxAP. g. E' i). 
 
 Note. If AB and AC be two Semi- conjugate Diame- 
 ters of an Ellipfis, then the Line PAO, determined as 
 above, will be the Pofition of the Greater Axis : And, 
 if, upon the Diameter C^, a Circle be defcribed inter- 
 feaing AD in H and K ; then AH, and AK will be 
 equal in Length to the greater, and lefler, Semi-Axis, 
 refpeftively. 
 
 From whence the moft ufeful Properties of the 
 Conjugate Diameters of an Ellipfis may be very eafily 
 deduced. Such, as that, the Sum of the Squares of any 
 two Conjugate Diameters, is equal to the Stim of the 
 Squares of the two Axes : And that, any Parallelogram 
 defcribed about the Conjugate Diameters of an Ellipfis 
 is equal to the Reftangle under the two Axes j and fo 
 forth. But thefe are Matters not altogether proper to' be 
 infilled on in this Place. 
 
 However it will not be impropei' to obferVe, that 
 the laft Problem is always pbffible, except in thofe two 
 Cafes, wherein the given LitifcS are perpendicular, an4 
 ibrm one continued Line. 
 
 M 4 PROBLEM 
 
x6H 
 
 Geometrical Problems, 
 
 PROBLEM LV. 
 
 I/, at the Extremes, M and m, of two given Right-lines 
 
 AM and Am, making a given Angle MAm, two Perm 
 
 pendiculars MC and mC he ereded-; 'tis propofed to 
 
 find the Dijiance of their Interfedion, C.from A the 
 
 given angular Point. 
 
 Since the Angles AMC and AwC a^e both Right 
 .Ones, the Circumference of a Circle, defcribed upon ihe 
 
 Diameter AC, will pafs 
 thro' M and m. Therefore, 
 if Mm be drav;n, and a 
 Perpendicular mE be let 
 fall upon AM, the Tri- 
 angles AwC and ;?2EM will 
 appear to be equiangular ; 
 becaufc the Angles ACm 
 and EM»J, infilling on the 
 fame Arch Am, are equal ; 
 and AwC is equal to wEM, being both Right-angles. 
 
 If now the Ratio of wE to Am (which is given be- 
 caufc the Angle EA/w is given) be denoted by That of s 
 to r; and the Ratio of A£ to Aw» by that of c to r ; 
 
 we fhall have AE =«^X Aw ; and therefore (Mm)' ( = 
 
 (AM)» + [kmY — 2AM X AE) = (AM)* -f (Aw)* — 
 
 — X AMx Aw : Whence, by Reafonof the fimilar 
 
 Triangles above fpecified, it will be, (wE)*: [mAf ( :: J* 
 
 :r«) : : (Mw)': (AC)'=^ x[(AM)'+(Aw)*]~^'x 
 
 AMx Cw ; whence AC is given. 
 
 As to the Geometrical Conltruflion, it is indicated 
 by the Conditions of the Problem, without any fort of 
 Argumentation. 
 
 PROBLEM 
 
with their Solutions. 
 
 109 
 
 PROBLEM LVI. 
 
 To find a Point C, fro7u whence three Right-lines drawn 
 to Jo many given Points A, B^ and iL^ Jhall obtain 
 the Ratio oj three given Quantities a^ b, and c, re- 
 Jpeiiively, 
 
 The given Pointg being joined ; make AB 3=/^ AE 
 = ^, and AC =: ;v : Then ^y, 
 
 BCbeing=:^"l, and EC 
 
 ex 
 
 (by Hypothefis) we 
 
 alfo have AM ( = i AB 
 AC--BCV ^ 
 
 ^ 2AB ;— 2 ^ -^ 
 
 (aa—bb) yx 
 
 i.aaj 
 AC^— EC 
 
 B 
 
 (Elem. 9. 2; ; and Km ( = ^ AE 4- 
 
 >\ _ [aa — ee\ V ;t' 
 
 -]:= -^ + ^aa " ' ^»PP<>fi"gCMand 
 
 2AE / 2 ■ 2aa^ 
 
 Cm to be perpendicular to AB and AE, rcfpeftively. 
 
 aaf 
 
 Hen 
 
 ce, putting 
 
 aa 
 
 f [aa—hh] X y^ 
 2 "*" laaf 
 
 -bb * aa — t 
 CM) -A- """^ 
 
 2A' 
 
 z: ky and 2: — 
 we get XX — 
 
 Q.h%—fh; and therefore Aw (=^-4-^] = ^ -f 
 
 Now, fince (by the lajl Problem) [hM)* ^ (KmY — 
 tic ss 
 
 -XAMxA»2=-— X (AC)' (where j and c denote 
 
 the Sine and Co-fine of the given Angle MAw, to the 
 Radius r) we Ihall, from hence, by fubftituting the 
 
 above 
 
X70 Geometrical Problems, 
 
 above Values of AM, hm, and AC, obtain the follow, 
 ing Equation, 
 
 From the Refolution of which the value of z, and 
 from thence the Pofition of the Point D, will be de- 
 termined. 
 
 Geometrically, 
 
 I! in AB, there be taken AF = a, and FH be drawn, 
 
 to make the Angle AFH 
 
 ==ACB and meet AC 
 
 in H ; then the Ratio of 
 
 FH to AF (by Reafon 
 
 of the fimilar Triangles 
 
 AFH, ACB) being the 
 
 fame with That ofBC 
 
 to AC, it is evident that 
 
 FH is given =:^. 
 
 Moreover, if HG be drawn, making the Angle AHG 
 
 3=AEC, it will likewife appear, that both HG and 
 
 AG are given: 
 
 For, fmce AG : AH : : AC : AE, 
 and AF : AH : : AC : AB ; 
 it follows that AG : AF : : AB : AE : whence AG is 
 given : And then it will be a : e ( : : AC : EC) : ; 
 AG : HG. Whence HG is given; and, from thence, 
 the following 
 
 ConJlruBion, 
 
 Take AT=:a, and make the Angf* AFG =:zAEB; 
 alfo take AI=e, and draw IK parallel to FG ; more- 
 over from the Centers F and G, with the Intervals ^ 
 and AK, defcribe two Arcs, and from the Point H of 
 their Interfeflion draw HF and HA, then a Line, BC, 
 drawn to make the Angle ABC =s AHF, will cut AH, 
 produced, in the Point required. 
 
 As to the Trigonometrical Solution, it is too obvious, 
 /rom the Conflru6lion, to need an Explanation. But 
 
 it 
 
with their Solutions. 
 
 171 
 
 il will be proper to take Notice that the Problem itfelf 
 becomes impoffible, when the two Arcs, defcribed from 
 the Centers F and G, neither cut, nor touch, each 
 other ; that is, when the Diftance FG is, either, greater 
 than the Sum, or J^fs than the Difference, of b and 
 AK. 
 
 PROBLEM LVII. 
 
 Three Points, A, B and E, being given^ to determine a 
 fsurih Point C,/o that Lines (AC, BC, EC,) drawn 
 Jrom thence to the three former ^may have given Dif- 
 ferences, (Provided the Difference gf no two of 
 •the faid Lines be given greater than the Diftance 
 
 of the two given Points from whence they arc 
 
 drawn.) 
 
 Suppofmg the given Points to be joined, put AB zi-a 
 
 BC = ^ 4" Z'* ^"^ 
 EC z=. X -f-^( where 
 ^ and g reprefent the 
 given Differences). 
 Then if upon AB 
 and AE, the Per- 
 pendiculars CM and 
 
 Cm be let fall, it 
 
 will be (by a known A VI B 
 
 Property of Triangle,) AB [a) : BC + AC (2x4-;^) 
 
 : : BC — AC fpj : BM -r- AM = ?£L±Z^: 
 
 a 
 
 Whence AM =: i d- 2£i±iZ: And, by the very 
 
 fame Argument, Amz=:ib -JdlS^, Which two 
 
 Values, by putting f a-JL-=^J^ and | i^ — ?| = ^ (for 
 
 the fake of Brevity) will become/-^- , and^-^. 
 
 Moreover 
 
i/a Geometrical Problems, 
 
 Moreover, if the Sine, and the Co-fine of the given 
 Angle MAm, to the Radius r, be denoted by s and e, 
 refpeftively, it will appear, frem the Problem preceding 
 
 the Laft, that (AM)* + (A;n)' -^x AM xAm =z— X 
 (AC)- or. (in Species) (/--^^)'-f(^-|i)«-'i 
 
 X (/- ^)x{s — ^t)~''T" ^^'""^^ Equation may 
 be.eaucea.o(^^ + i|-i-_;i)x.'- 
 
 (•? +^ -f -4-)x- = -/-=.-^^: 
 
 Whence, by fubftituting for the Coefficients of the 
 Powers of x, &c, the Value of a-, will be found, and 
 from thence the Pofition of the Point C. 
 
 Geomttiically, 
 
 If AB be bifefted in F. and FG be taken a Third 
 Proportional to » AB and PQ (BC — AC), it is evident, 
 
 from PrebUm 49, that GM= A^^^Q . 
 
 And, for the very, fame Reafons, if AE be bifefted 
 in/, and^ be taken a Third Proportional to 2AE and 
 
 RS(EC— AC), wefhallalfohave^;7z=:: J^SlX^, 
 
 Whence it appears that GM is to gm^ in the given 
 
 . PQ RS .^^ ABxRS 
 Ratio of ^ to ^,orofPQ to ^ .or 
 
 laftly, of GI to ^2; by t^ing GI=:PQ, and giz=z 
 ABxRS 
 
 AK • 
 
 If 
 
with their Solutions. 
 
 ^73 
 
 If therefore GH and ^H, and alfo IK and zK, be 
 drawn, interfering in H and K, it is manifeft that 
 the required Point C muft fall, fomevvhere, in the 
 Right-line dc pafTmg thro' H and K ; (ince, in this 
 Cafe (and no other) it will be GM : Gl : (:: HC : HK) 
 ''"gm'.gi; or, alternately, GM : ^m : Ol : ,fri. By 
 
 Reafoning in the very fame Manner, from the Triangles 
 ACB and BEC, the Pofition of another Right-line^ 
 wherein the Point C falls, may alfo be determined ; 
 whofe Interfe6lioH with de will confequently be the 
 Point required. 
 
 But, inafmuch as the Cafe, from the pofition o{ de 
 thus given, is now reduced to our 49*" Problem, the 
 remaining Part of the Solutionis like wife given, /r<?;w 
 thence^ by a different Method : According to which, 
 and what is above demonftrated, we have the followinf^ 
 
 ConJiruBion. 
 
 Having taken fg a Third Proportional to 2AE and 
 RS ; and gi a Fourth Proportional to* AE, AB, and 
 RS ; and having alfo taken FG a Third Proportional to 
 2AB and PQ, and fet off GI= PQ, and drawn the 
 Right-line de through the Interfeftions of the Perpen- 
 diculars 
 
^7'i 
 
 Geometrical Problems, 
 
 diculars GH, ^H ; IK, zK, as ahoittjptcijitd ; let an 
 Arch of a Circle, from the Center K, at the Diftancc 
 of AB be next defcribed, and through A let HN be 
 drawn, meeting it in L ; then a Right-line AC drawn 
 parallel to That joining the Points L and K, will cut 
 de in the Point required ; as is evident from the above- 
 mentioned Problem. By means of which the Trigono- 
 metrical Solution will alfo be brought out. 
 
 In this Conftrnftion, the Solution of the Problem 
 for defcribing a Circle to touch three given Circles, is 
 determined. 
 
 LEMMA. 
 
 If from the Vertex of any plane Triangle ABC, a Right- 
 line CD be drawn^ to divide the Bafe AB in the Ratio 
 of two given Numbers n and m (Jo that AD : SD : ; 
 n;m)/ then the Sum of the Multi-pies of the Squares 
 of the two Sides of the Triangle, whofe Fa&ors are the 
 f aid given Numbers ^ taken alternately, will be equal t9 
 the Sum of the ReSangle of the two Parts of the Bafe 
 and the Square of the dividing Line, repeated as often 
 as there are Units in the two propofed Numbers 
 (that is, m times (AC)* 4- n times (BC)' = m-4-n times 
 [ADxBD+(DC)']. 
 
 be bifefted in M and N, and 
 upon AB let fall the Per- 
 pendicular CE ; then will 
 (AC)^— (DC)^=AD X 2ME 
 (Elem, 9. 2). Whence it 
 is evident that m times 
 (AC)* —m times (DC)' = 
 w times AD X^ME. 
 
 And, by the very fame 
 Argument it appears that 
 A M E n ^s[ B 
 
 n times (BC)' — « times (DC)*is=:»timesBDx 2NE. 
 
 Therefore, 
 
with their Solutions. 17^ 
 
 Therefore, by adding tbefe equal Quantities together, 
 m times (AC)' + ^ times (BC)* — m -\'n times (DC)'= 
 m times AD X 2ME -|- n times BD X 2NE. But m : 
 ni: BD : AD fhy Hyp J : : BDxaNE : AD X 2NE 
 (Elem. 1. 4); and therefore « times BD X 2NE = ;© 
 times AD X 2NE. 
 
 Lefc the latter of ihefe equal Quantities be wrote 
 above, in the Room of the Former; fo will m times 
 (AC)"-}- w times (BC)' — w+w times (DC)^ zzztti times 
 AD X *ME + ^ ti"^cs ^^ X 2NEsiz: m times 
 (ADx2ME+ADx2NE) = /w times i\DxAB=?w+» 
 times AD X ^^ (becaufe, by Conftru£lion m : m-\-n : : 
 BD : AB) ; and confequently w times (AC)' -jk ?i times 
 (BC)'=7»4-w time8(DC)'4- ?»+ wtimesAD x BD 
 
 S. E. D. 
 
 It is plain from hence that, if m and n be fuppofed 
 to denote two given Lines, inflead of Numbers, the 
 Sum of the Solids ;;2 X (AC)' and n X (BC)'will be equal 
 to the Sum of the Solids {m'^n) X (DC)' and {m-\-n) y 
 ADxBD- 
 
 PROBLEM 
 
1/6 
 
 Geometrical Problems, 
 
 PROBLEM LVIII. 
 
 From any Numher of given Pohits, A, B, C, D, &c. 
 ^0 draw as many Lines AP, BP, CP, Sec. to meet in 
 a Right- line MN given by Po/ition, fo that the Sum 
 of all their Squares may be a given Quantity. 
 
 fr«m all the given Points, let 
 fall Perpendiculars, AQ, 
 BR, CS. <3c. which Per- 
 pendiculars, as well as the 
 Diftances between Them, 
 are all given from the Po- 
 fition of MN : Put, there- 
 fore, AO =J, BR =zb, 
 CS = c, DTz=^,EV = €, 
 QR=r, QS=:i.QT=^, 
 &c. Alfo put QP=a:, and 
 the given Sum of all the 
 propofed Squares =i^\ 
 
 Then RP being zn x— r,SV :=:x — s, &c, 
 
 {ATy=zaa 
 
 Wc {hall have, 
 by Elem. 8. 2. 
 
 yz=zaa -^xx 1 
 
 )' =ibb - - XX — 2rx ^ rr I __ z, 
 Y-=zcc --XX — ^sx ^ss r"~ 
 ') z=dd-\-xx — 2tx + // J 
 
 5C. &C. 
 
 (BP)' = 
 (CP! 
 
 Put aa-^bb'\-cc^ dd &c. z= ot\ r -fj -^t &c. =z /3, 
 and rr-^ss -f- tt &c. = 7' ; i^nd let the Number of the 
 given points be denoted by n : 
 Then our Equation will be reduced to 
 M*4.nx'-— 2^x ^y' = k': 
 
 found =^±W^^^^^^*+^) 
 n ^ \ n ^ nnJ 
 
 From whence .» is 
 
 Geometrically, 
 
 If PQ be drawn to bifeft the Dillance AB, then 
 
 wilUAP)*+(BP)^=2AQxBQ4-i(QP)%^>'M^^^w^^. 
 
 6 And, 
 
with their Solutions. 177 
 
 And, if QC be drawn, and QR be taken equal 
 to J Thereof; then will 2(QP)*4- (CP)^ = 3QRXCR 
 gfRP)', by the Same, Whence, by adding thefe equal 
 Quantities together, and taking 2(QP)' (common] away, 
 we have (AP)^ + (BP)^ +(CPy z= 2 AQ X BQ + SQ^ 
 XCR+3(RP^ B 
 
 Again, by draw- ^ yQ\ C 
 
 ing RD, and y^^y-^^K''' \ 
 
 taking R S = ' y^ *'•••. V- X \ 
 
 I Thereoi we ^^^"^^-.^ ^\ '••X \ 
 
 have 3(RPJ^ + ^"'"■^</^-/Xv \ 
 
 DS + 4(St')', h j \^ . 
 
 tne Lemma. .Tip T ^ 
 
 And therefore ■^ / V 
 
 fAP)' + (BP)-+ S y\ 
 
 (CP)'+(OPJ-= ^i/ D 
 
 PN 
 
 2AB x.BQ-h3QRxCRH:4RSxDS+4(SP), by 
 
 the Addition of equal Quantities. 
 
 In the fame manner, if SE be drawn, and ST be 
 taken =j Thereof, we fhall have 4(SP)^ + (EP)'rz:? 
 ^STxET+5(TP)\ Whence, again, by the Addition 
 of equal Quantities, (AP)^+(BP)'4-fCP)'+ (DP^^-f- 
 CEPj^^aAO X BQ + 3QR X CR + 4RS X 03+ 
 5STxET + 5(TP)'. 
 
 Hence it is evident (without proceeding further) that, 
 let the Number of the given Points be what it will, 
 the Square of the last of the Lines QP, RP, SP, TP, 
 &c, drawn as above, will always be a^zW/z Qi-antity ; 
 bccaufe the Sum of all the ReHangles 2 AG X BQ, 
 3QRXCR, 4RSXDS, &c, (or That of theiT Equals 
 ABxQB, QCxRC, RDxSD, %.) is given, by th« 
 Pofition of the Points A, B, C, D, '^c» 
 
 N 
 
fy^ Geometrical Problems. 
 
 In order, therefore, to the Geometrical Conftruftion, 
 let a Rectangle be conftru£l:ed fby Eiem. 8. 6.) equal to 
 the Excefs of the given Sum of the Squares above the 
 Sum of the faid Reftangles: Then find a Mean Pro- 
 portional between the Length of the Re6langle, fo de- 
 termined, and that Part of its Breadth defined by the 
 Number of the given Points; with which, as a Radius, 
 from the laft of the Points Q, R, S, T, &c, let the 
 Circumference of a Circle be defcribed ; which will in- 
 lerfeft the Line MN, in the Point P, required. 
 
 If, inftead of the Right-line MN, the Circumference 
 of a Circle (or any other Curve-line) given -by Pofi- 
 tion, be propounded (wherein the Lines, drawn from 
 the given Points, are required to meet), the Method of 
 Conftruftion will, it is manifeft, be exa61:ly the fame. 
 
 And it may not be amiss to obferve farther, that if 
 
 «X(AP)' + ^X(BP)- + cx(CP)' +</x{DP)". ej?f- 
 (where a, S, c, d, &c, reprefent known Quantities, 
 either Numbers or Lines) be fuppofed given, the Locus 
 of the Point P will, JHllt be the Circumfersnce of a 
 Circle, determined after the fame manner, from the 
 premifed Lemma. 
 
 For. i^v (AQ:BQ::i:a 
 
 ,.'< ^y J OR :RC : ic -.a + d 
 
 making m^gs :DS::^ 
 Cafe y.j. 
 
 this \^ai^ I c^r" . T^'TT 
 
 and then proceeding as above, it will appear thata>^(AP)' 
 + ^X(BP)'+cX(CP)^+'^c. is equal to (fl4-^)xAQ 
 
 SD4-(a4-^+c+./+0XSTxTE+ (a+^+.+^+.)X 
 (TPj*. Whence it is c\'ident that TP is a give* 
 Quantity. 
 
 PART 
 
PART lit 
 
 WHEREIN 
 
 The THEORY of GUNNERY, or the 
 Motion of Projectiles, is confidered. 
 
 IT is, ufLially, taken for granted, by Thofe who 
 treat of the Motion of Projeftiles, that the Force 
 of Gravity near the Earth's Surface is every-wherc 
 the fame, and afts in parallel Direftions; and that the 
 EfFeft of the Air's Refiftance upon very heavy Bodies, 
 fuch as Bombs and Cannon Balls, is too fraall to be taken 
 into Confideration. 
 
 That the Error arifing from the Suppofition of Gra- 
 vity afting uniformly, and in parallel Lines, muft be 
 exceeding fmall, is very obvious ; becaufe, even, the 
 greateft Diftance of a Proje6iile above the Surface of 
 the Earth, is inconfiderable in Comparifon of its Di- 
 ftance from the Center, to which the Gravitation tends : 
 But then, on the other hand, it is very certain, that 
 the Refiftance of the Air, to very fwift Motions, is 
 much greater than it has been commonly reprefented. 
 
 Neverthelefs, if the Amplitude of the Proje6lioD, 
 anfwering to one given Elevation, be firfl; found 
 by Experiment (which our Method fuppofes to be 
 done), the Amplitudes in all other Cafes, where the 
 Elevations and Velocities do not very much differ 
 from the Firft, may be determined, to a fufficient 
 Degree of Exaftnefs, from the foregoing Hypothefes : 
 Becaufe, in all fuch Cafes, the Effe6ls of the Re- 
 fiftance will be nearly as the Amplitudes thtmfclves ; 
 ai}d, were They aaurately fo, the Proportions of the 
 
 N 2 Amplitude*, 
 
iZo The TriEORY ^ GuNNtRY, or 
 
 Amplitudes, at different Elevations, would then be tke 
 very fame as in vacuo. 
 
 For this Reafon, and to avoid having Recourfe to 
 Principles and Calculations no- ways adequate to the Ex- 
 perience and Underftanding of Beginners, for whofc 
 Ufe this little Traft is chiefly intended, I fliall, in what 
 follows, conform to the Method of other Writers, fo 
 far, as to take no Notice of the Air' s Refiflance ; but 
 confider the Motions as performed in vacuo, 
 
 Nbw, in order to form a clear Idea of the Subjeft 
 here propofed, the Path of every Proje£lile is to be con- 
 fidered as depending on two different Forces ; that is to 
 fay, on the impellent Force, whereby the Motion is firfl 
 began (and would be continued, in a Right-line), and on 
 the Force of Gravity, by which the Projeftile, during 
 the whole Time of its Flight, is continually urged down- 
 wards, and made to deviate more and more from its firft 
 Direftion. 
 
 As whatever relates to the Track and Flight of a Ball 
 (neglefting the Refiflance of the Air) is to be determined 
 from the Aftion of thcfe two Forces, it will be proper, 
 before we proceed to confider their joint Effefl, to pre- 
 mife fomething concerning the Nature of the Motioa 
 produced by Each, when fuppofed to aft alone, inde- 
 pendent of the Other ; to which End the two firfl, of the 
 f«ur following Lemmas^ arc prcmifed. 
 
 LEMMA 
 
thi Motion df Projegtiles. iZx 
 
 LEMMA I. 
 
 Every Body, after tht imprefs*d Force, whereby it is 
 put in motian, ceajes to aB, continues to move uni- 
 formly in a Right-line ; unlefs it be interrupted by 
 fome other Force or Impediment, 
 
 This is a Law of Nature, and has its DemonflratioB 
 from Experience and Matier-of-Fa6l, 
 
 Corollary. 
 
 It follows from hence that a Ball, after leaving the 
 Mouth of the Piece, would continue to move along the 
 Line of its firft Direftion, and defcribe Spaces therein 
 proportional to the Times of their Defcription, were it 
 not for the Aftion of Gravity ; whereby the Dirc6lioii 
 ii changed, and the Motion interrupted. 
 
 LEMMA IL 
 
 The Motion, or Velocity, acquired by a Ball, in freely 
 defending from Rejl, by the Force of an uniform 
 Gravity, is as the Time of the Dejcent : and the 
 Space fallen thro\ as the Square of that Time. 
 
 The firft part of the Lemma is extremely obvious: 
 For, fmce every Motion is proportional to the Force 
 whereby it is generated, That generated by the Force 
 of in uniform Gravity muft be as the Time of the 
 Defcent ; tecaufe the whole Effort of fuch a Force 
 is proportional to the Time of its Action ; that is, as 
 the Time of the Defcent. 
 
 N3 T« 
 
l|2 
 
 The Theory cf Gunnery, #r 
 
 q'- 
 
 i 
 
 g 
 
 To dcmonftrate that the diftances defccnded arc pro- 
 A. portional to the Squares ot the Times, 
 *7 let the Time of falling thro* any pro. 
 ^^ pofed diftance AB be reprefented by the 
 Right-line PQ ; which conceive to be 
 ^ divided into an indefinite Number of 
 very fmall, equal, Particles, reprefented, 
 each, by the Symbol m ; and let the Di- 
 ftance defcended in the Firft of them be 
 he ; in the Second cd ; in the Third 
 dt ; and fo on 
 
 Then the Velocity acquired being 
 always at the Time from the Begin- 
 ning of the Defcent, it will '' at the 
 Middle of the firft of the faid Particles 
 '^'B be reprefented by \m ; at the Middle 
 of the Second, by i f t^:"; at the Middle of the 
 Third, by 2 ^ ;«, Be, Which Values conftitute the 
 
 Series 2. 35, 5™, 7^ 9^ 0,. 
 
 2 2 2 2 2 
 
 But, fince the Velocity, at the Middle of any One of 
 the faid Particles of Time, is an exa6l Mean between 
 the Velocities at the two Extremes thereof, the corre- 
 fponiiing Particle of the Diftance AB may be therefore 
 confidered as defcribed with that mean Velocity : And 
 fo, the Spaces Ac^ cd^ de^ ef^ &c, being, refpe6lively, 
 
 edual to the abovcmentioned Quantities — , 4- H- 
 ^ 2 2 2 * 
 
 77S ■ » 
 
 '— , £3c, it follows, by the continual Addition of 
 s 
 
 Thefe, that the Spaces Ac, Kd, Ag, Af, &e, 
 fallen thro* from the Beginning, will be exprefled by 
 
 m Am gm i6m 2gm -, ^,,, . , 
 
 -- . t^ , ^ , , -^-, &c. Which, ife evi- 
 
 a * 2 t 2 2 
 
 dently, to one another in proportion, as i, 4, 9, 16^ 25, 
 
 &c, that is, as the Squares of the Times. Q. £, D. 
 
 COROL. 
 
thi Motion c/ Projectiles. i8| 
 
 Corollary. 
 
 Seeing the Velocity acquired in any Number fnj 
 of the aforefaid, equal, Parti<;les of Time (meafured 
 by the Space that would be defcribed in one fingle 
 Particle) is reprefented by n times m, or n w, it will 
 therefore be, as i Particle of Time, is to n fuch 
 Particles, fo is nm, the faid Diftance anfwering to 
 the former Time, to the Diftance, n^m, correfponding 
 to the Latter, with the fame Celerity, acquired U 
 the End of the faid n Particles. Whence it appears 
 
 that the Space (Jound above) thro' which the Ball 
 
 falls, in any given Time riy is jufl; the Half of That 
 \n*m) which might be uniformly defcribed with the laft, 
 oj greateft, Celerity, in the fame Time. 
 
 Scholium. 
 
 It is found, by Experiment, that any heavy Body, 
 near the Earth's Surface (where the Force of Gravity 
 may be confider'd as uniform) defccnds about i6 Feet, 
 from Reft, in the firft Second of Time, ^ 
 
 Therefore, as the Diftances fallen thro' are proved 
 above to be, in Proportion, as the Squares of the 
 Times ; it follows that, as the Square of i Second, i« 
 to the Square of any given Number of Seconds, fo is 
 16 Feet, to the Number of Feet a heavy Body will 
 freely defcend in the faid given Number of Seconds, 
 Whence the Number of Feet defcended in any givea 
 Time will be found, by multiplying the Square of the 
 Number of Seconds by 16. 
 
 Thus the Diftance defcended in «, 3, 4, 5, &c. 
 Seconds, will appear to be 64, 144, 256, 400 F, ^c, 
 refpeftively. 
 
 Moreover, from hence, the Time of the Defcent 
 thro' any given Diftance will be obtained, by dividing 
 the faid Diftance, in Feet, by 16, and extrafting the 
 fquare Root of the Quotient; or, which comes to the 
 
 N 4 fame 
 
^Si The Theory of Gunnkry, «r 
 
 fame Thing, by extrafting the fquare Root of the 
 whole Diftancc, and then taking | of that Root for 
 the Number of Seconds required. Thus if the Di- 
 stance be fuppofcd 2640 Feet ; then, by cither of th« 
 two Ways, the time of the Defcentwill come out la* 
 84 Seconds, or 12": 50'". 
 
 It appears alfo (from the Corol.J that the Velocity per 
 Second (in Feet,J at the End of the Fall, will be deter- 
 mined, by multiplying the Number of Seconds in the 
 Fail by 32 : Thus it is found that a Ball, at the End 
 of 10 Seconds, has acquired a Velocity of 320 Feet 
 
 per Second. After the fame Manner, by having any 
 
 two of the four following Quantities, viz, the Force^ 
 the Timt^ the Velocity, and Dijiance, the other Two 
 inay be determined : But this not being abfoli^tely 
 neccffa^y in what follows (though equally ufeful in 
 other Difquifitions) I (hall put down the feveral Rules, 
 or Equations below, in a Note *, to be taken or 
 omitted, at Pleafure. 
 
 LEMMA 
 
 * Let the Space freely defc ended by a Ball, in thefirfi 
 Secand of Time (which is as the accelerating Force) be de- 
 tiotedby f ; alfo let t denote the Number of Seconds wherein 
 any Dijiance, d, isAefcended; and let v he the Velocity^ 
 per Secondy(it the F'nd of the Defcent : Then will 
 2d 
 
 V = 2ft =: 2 y^(fd ) =z= — 
 
 (d \ V 2d 
 
 . ^ vv tv 
 
 "^ 4f 2 
 d V vv 
 tr 2t 4d* 
 All which Equations are very eefth deducedfrom thetw» 
 original Ones, dz=z hu and vz=:zn, already demon/irat-, 
 (d^ the Former in the Propqfition itfelf; and the Latter, 
 in (he Corollary to it ; by which it appears that the 
 
 Meafure 
 
ihc Motion of Projectiles. 
 
 1% 
 
 HMQ, perpendicular 
 
 LEMMA III. 
 
 A Ball, projeEled in the Dire&ion of a Right-line AC 
 making an Angle with the Horizon AB, is, by Means 
 of Its Gravity, defected continually, more and more, 
 from its frjl Direction ; but the Celerity with which 
 it approaches any Perpendicular [HC) to the Horizon, is 
 neither increafed nor decreafedby the Action of Gravity. 
 
 For, let a line 
 Horizon, be con- 
 ceived to move pa- 
 rallel to itfelf, to- 
 wards BC, along 
 with the Ball : 
 Then, as the Gra- 
 vity always a£ls in 
 this Line, it can 
 have no Effect in 
 making it, either, 
 move fafter or 
 flower towards BC ; 
 but is wholly em 
 
 H 
 
 ploy'd in drawing down the Proje£lile along the fame, 
 from its firft Direftion AC ; and thereby caufing it to 
 defcribe a Curve-line AMB. 
 
 COROL. 
 
 Meafure of the Velocity at the End of the fir Jl Second is 
 2f ; whence the Velocity (v) at the End oft Seconds mufi 
 confequently be exprtffedby sf X t or 2ft. 
 
 Having proceeded thus far, Ifhall here take the Oppcr» 
 tunity to point out, and reBify an inadvertent Expreffion, 
 at p, 230 oj my Book of Fluxions*, relating to this Sub- 
 je^, — It is there [aid, byway of remark, that, what- 
 ever Ratio the Times have with Rcfpeft to the Diftances 
 defcended, &c, the fame alfo will the Velocities have. 
 
 Edition of \jp,o. 
 
 being 
 
i86 n^ Theory ^Gunnery, #r 
 
 Corollary. 
 
 Hence it appears that the Proje6lile, at the End of 
 a given Time, is in the very fame vertical Line HG, 
 ds it would be in, if Gravity was not to aft; and that 
 ^ the horizontal Diftance AH, as welj as the Hypothenufe 
 
 AG, is proportional to the Time wherein the Projeftile 
 aftually moves through the Arch AM, correfponding 
 to the faid Diftance. 
 
 being proportional to the Times. Which Ohftrvation^ 
 it is clear from the Words thtmfelves, ought to be re- 
 Jirained to thofe Cafes^ zuhere the Velocities are pro- 
 portional to the Times ; that is, where the accelerating 
 Forces ar^ equal. Therefore, in/lead of the Jaid Ohfer- 
 vation, as it nowjlands, read. What is above demon- 
 ftrated with refpeft to the Times, holds alfo in the Vc» 
 locitiei, when the accelerating Forces are equal. 
 
the Motion of Projectiles, 
 
 187 
 
 LEMMA IV. 
 
 The Dejlections arijingfrom the ABion of Gravity, cr the 
 vertical Dijiances, MG, BC, intercepted by the Path 
 of the Ball and the Line of Diredion, are in propor- 
 tion, to one another^ as the Sauares of the correj pond- 
 ing Psrts AG, and ACj of trie f aid Line of Diredion. 
 
 Conceive a Line GML to be carried along, parallel 
 to BC, fo that its extreme Point G may trace out the 
 Line of Dire£lion 
 AC, in the very 
 fame manner as 
 the Projeftile itfelf 
 would defcribe it, 
 were it not to be 
 deflefted therefrom 
 fey the Aftion of 
 Gravity : Then, 
 fmce, by the pre- 
 ceding Lemma, the 
 Projeftile is always 
 in the Line GML, 
 and the Force of 
 Gravity is, wholly, 
 employ'd in urging 
 it downwards, along 
 that Line, the Ef- 
 feft produced by 
 the faid Force, or 
 the Diftance MG of 
 the Ball from the Extreme Point G, at the End of any- 
 given Time, will confequently be the very fame, as if 
 the Line GML (inftead of being carry'd uniformly to- 
 wards CB) was to have continued in its firft Pofition 
 AR ; and the Ball fufFeired to defcend from Reft along 
 that Line ; the Force employ'd being the fame in both 
 Cafes. But it is proved, in Lemma 2, that the Spaces AP 
 and AR that would be defcribcd in defccnding, freely, 
 
 N 6 fron* 
 
M 
 
 the Theory ^Gunnery, or 
 
 from Reft, arc as the Squares of the Times ; Therefore 
 their Equals GM and CR are likewife as the Squares of 
 the fame Times ; that is, of the Times wherein AM 
 and AMB are defcribed : Which are to each other as 
 (AG) to (AC)^ fl^y CoroL to Lem. 3^ 
 
 COROL, I. 
 
 If B be taken as the Point where the Projeftile im- 
 pinges on the Horizon AB, and in BC (perpendicular 
 to AB) there be taken CI=GH; then a Right-line, 
 drawn from I to A, will cut the vertical Line HG in the 
 very Point (M) thro* which the Center of the Proje6UIe 
 paffeth. 
 
 For (AC)' : (AG)' : : BC : MG, as above. 
 and (AC)' : (AG)' : : (BC)' : (HG)', byftm. Triangles. 
 Theref. BC : MG : : (BC : (HG)', by Equality of Ratios : 
 and confequently MG X ^C = (HG)% 
 
 This, turned into an Analogy, gives BC : HG : : 
 HG : MG ; and fo likewife is BC to CI (becaufe of the 
 parallel Lines). Whence it is evident that HG and 
 CI are equal to eaeh other, in all Pofitions of HG : 
 By Means whereof, as many Points in the Curve, a& 
 you please, may be determined. 
 
 COROU 
 
M.e Motion q/^ Projectiles. 189 
 
 COROL. II. 
 
 If GNS be drawn, parallel to AB, interfcaing 
 BC in N, and meeting the Circumference of a Semi- 
 circle, defcribed upon BC, in S ; it will further ap- 
 pear, that the Height (HM) oF the Ball is every- 
 where a Third- proportional to BC and NS. For, 
 fince BC : HG : : HG : MG, it follows, by Divifion. 
 
 that, BC : CN : : HG (BN) : HM = ^ ^^ • = 
 
 ^-~-f Ekm, 19. 4.). Hence it is evident that the Projec- 
 tile will be at its greateft Height, when it has performed, 
 juft, the Half of its Flight; fmce it is well known that 
 
 (NS)»\ . 
 NS (and confequently^-^p- j is the greateft poflible 
 
 when it coincides with the Radius QR. 
 
 It appears moreover that the Heights hm, HM, of 
 the Ball, in any two vertical Lines, equally diftant from 
 That pafling thro' the higheft Point E of the Trajeftory, 
 are equal ; becaufe the corrcfponding Ordinates ns and 
 NS arc equal, as being equally diftant from the Center 
 Q of the Semi-circle. — Laftly, it is apparent vhat the 
 
 greateft Altitude DE is = |BC ; becaufe^--— i-, when 
 
 NS coincides with QR, becomes == ^ = 1 BC. 
 
 Which may be otherwife made out, by confiderlng, 
 that, as AF is but the Half of AC, its Square will be 
 only I of the Square of BC ; and therefore FE only 
 I of BC (by the Lcm.) And fo, DF being = 4 BC (by 
 Jim. Triang.J DE, as well as FE, muft be =1 BC. 
 
 PROPOSITION 
 
igo 
 
 The Theory ^Gunnery, *r 
 
 PROPOSITION I. 
 
 The Amplitude, or horizontal liange of a Piece, with d 
 given Charge of Powder, at an Elevation of j^^ Be- 
 grees.being known, from Experiment ; to determine 
 the Elevation fo as to hit an Obje&, at a given Di. 
 fiance, on the plane of the Horizon ; the Quantity of 
 Powder remaining the fame. 
 
 Let PQ be the given Amplitude, at an Elevatioi? 
 (QPR)ot45 Degrees; let AB be the given Diftance 
 ^R of the Objta, and BAD the 
 required Elevation : In AB, 
 produced, take AO zz:i PQ, with 
 which a? a Radius, from the 
 Center O, let a Scmi-circlr 
 AME be defcribed, and let 
 AD, produced, meet the Peri- 
 phery thereof in H, join E, H, 
 and make 1^% and HI perpen. 
 V. dicular to AE. 
 
 Since the Charge of Powder, or the Velocity a( 
 both the Elevations QPR and BAD, is fuppofed to be? 
 the fame, the Times of Flight, during which the Di- 
 ftancei PR and AD would be uniformly delcribed with 
 that Velocity, will therefore be to each other, direftly 
 a$ the iaid Difooces ; and confequently (PR)' : (AD)': : 
 
. the Motion of Projectiles- 191 
 
 RQ : DB : foy Principles already explained, vid, Lem, 
 Qand^). 
 
 But fPR)^ z= (PQ)^ + (RQ)^ = 2(RQ)^ (becaufe the 
 Angle P (=45°) z=z Z R). Hence the above Proportion 
 becomes 2( RO)^ : (ADj^ : : RQ : DB ; from which we 
 have (AD}'=:5RQ X DB=AExDB, i?y ConJlruElion. 
 
 Moreover, from the fimilar Triangles ABD and 
 AHE, we have AE : EH : : AD : DB : Whence 
 EH xAD =AExDB=(AD)^ fp, above) and confe- 
 quently EH = AD : And fo, the Triangles EHI and 
 ADB being equiangular, it is plain that HI is alfo equal 
 to AB. Whence follows this eafy 
 
 ConJlruBion, 
 
 With an Interval equal to the given Amplitude, at 
 the elevation of 45", let a Semi-circle AME be de- 
 fcribed ; make OM perpendicular to the Diameter 
 thereof, in which take ON equal to the given Diftance 
 AB, and through N, parallel to AE, draw HN^, inter- 
 fering the Circumference in H and h : then either of 
 the Direftions AH, or A^, will anfwer the Conditions of 
 the Problem. g. £. /. 
 
 COROL. I. 
 
 If OH be drawn, the Angle EOH will be =: 2EAH 
 (Elem, 10. 3.) and it will be, as OH (PQ) : HI (AB) : : 
 Radius : Sine EOH ; that is, in Words, as the given 
 Amplitude, at 45" Elevation, is to any other propofed 
 Amplitude, fo is the Radius^ to the Sine of twice the 
 Elevation correfponding to the Latter. From whence 
 it is evident, that the horizontal Amplitudes, at different 
 Elevations, are to one another as the Sines of the Doubles 
 of the faid Elevations ; and that, the Amplitude of the 
 Projeftion at an Elevation of 45 Degrees (when HI coin- 
 cides with MO) is the greateft poffible. 
 
 C@ROU 
 
192 The Theory of Gunnery, #r 
 
 COROL. 11. 
 
 Since it is found above that AE X DB (=(AD)*) 
 =:(EH)% it follows, that (AE)': (EH)^: : (AE)*: AEX 
 r)B : : AE : DB : : I AO (I AE) : C V [\ DB. vid, CoVoL 
 '2 to Lem. 4,) that is, as the Square of the Radius is to 
 the Square of the Sine of the Angle of Elevation, fo is 
 Half the greateft Horizontal Amplitude, to the grcateft 
 Altitude of the Projeftile. 
 
 Hence it appears, that the Diftance which the Ball 
 would afcend, if projefted in a vertical Dircftion (ufually 
 called the Impetus) is juft one Half of the greateft Am- 
 plitude; fince, in this Cafe, the Sine of the Elevation 
 becomes equal to the Radius, 
 
 Therefore, as a Body (in vacuoJdXctnAs and defcends 
 with the fame Celerity ; and feeing the Diftance AG, 
 cxprefling the perpendicular Afcent, is as the Square of 
 the Celerity at A, fp. Lem, 2, J; it follows that the 
 greateft horizontal Amplitude AO, being =: 2 AG, ii 
 alfo as the Square of the fame Celerity. 
 
 From whence, and Corol. 1, it is manifeft, that the 
 Amplitudes, when both the Elevations, and the Ve- 
 locities, differ, will be to each other in a Ratio com- 
 pounded of the Ratio of the Sines of the Double Elc- 
 rations, and the Duplicate Ratio of the Vciocitjci. 
 
 PROPOSITION 
 
the Motion of Projectiles. 
 
 19s 
 
 PROPOSITION Ih 
 
 The greatefl horizontal Amplitude, and the Angle of Ele- 
 vation being given ; to find at what Di/lance the Piece 
 ought to be planted, to hit an Object, who fe Height 
 above, or DepreJJion below^ the Level of the Piece is 
 alfo given. 
 
 Let BC be the perpendicular Height, or deprcflTion 
 of the propofcd Obje6}, and AB the required Diftance ; 
 let BC, produced, meet the Line of Direftion AD in 
 D, and let P be the Place where the Path of the Pro- 
 jeftile (produced) meets the Level of the Piece: Make 
 PQ perpendicular to AP, and CN parallel to AD. 
 
 B N 
 
 By the laft Propofition it will be, as Radius : Sin. 2 A 
 : : the greatefl horizontal Amplitude, to the Diftance 
 AP ; which, therefore, is known. 
 
 Moreover, it appears from the fourth Proportion, in 
 Corol. 1, toLem. 4, that PQ : BD : : BD : CD. 
 „ , /PQ : BD : : AP : AB\ y r i ^ ■ i 
 ^"^ \BD : CD : : AB : AN/ hM'^^^ Triangles. 
 Therefore, by Equality of Ratios, 
 AP: AB:: AB: AN; 
 
 O Whence 
 
194 ^^^ Theory of Gunnery, or 
 
 Whence AP : PB : AB : BN (hy DivifionJ 
 and confequently APxBNrzzABxPB. 
 
 Let AP be now bifefted, in O ; then AByBP (or 
 its Equal APxBN)beIngz:iz(AO)^ ^.[Ohyr EUm.S. 2.) 
 we therefore have {OB)^z=(AO)^4: AP-fBN = AO X 
 (A0i2BN) ; whence AB is likewife known, g. £. /. 
 
 Corollary. 
 
 If the Amplitude AP, the Elevation PAQ, and the 
 Diftance AB of any perpendicular BCD fronj the 
 Place of Projeftion A, be suppofed .given ; then the 
 Height, or Depreflion of the Ball in that Perpendicular, 
 may, from hence, be found. 
 
 For it is proved that AP : BP : : AB : BN ; from 
 which BN is given : Bur, as the Radius^ is to the 
 Tangent oi ^}:IC (or BAD), fo is BN to BC. 
 
 PROPOSITION III. 
 
 Having the greatejl horizontal Amplitude, or Range of 
 -a Piece, with the Df lance and the Height (or De- 
 prefjionj of the Objed, to find the Angle of Elevation^ 
 
 Let BC (in Fig. 1 and 2) be the perpendicular 
 Height, or Depreflion of the Objeft, AB the given 
 T M _ H horizontal Diftance, 
 
 P ^^"XZ^^-^^^fi^^^^^ and AH the required 
 
 Direftion : Alfo let 
 N // ir\ / \ ^E PQ (Fig. 3.) be the 
 
 greateft horizontal 
 Amplitude, anfwer- 
 ing to 45** of Ele- 
 jS^ BO ~ 'V vation, (vid. Carol. 
 
 1. Prop. 1.), Draw AC, in which produced (if need be) 
 take AG=PQ; make MOO perpendicular to AG, 
 meeting AB produced (if neceflary) in O ; and from 
 the Center O, with the Interval OA, let the Circum- 
 ference of a Circle be defcribed, imerfe£ling AG pro- 
 duced. 
 
the Motion of Projectiles. 
 
 195 
 
 duced in E ; join E, H, and let HI, AN, and QR, b« 
 
 perpendicular to AE, 
 
 AO, and PQ re- 
 
 fpeftively, and let 
 
 BC be produced to 
 
 meet AH in D. 
 
 It appears, from 
 what has been al- 
 ready delivered, that 
 (AD)': (PRj': : DC 
 : RQ. Whence, 
 
 (PR)^ being z=2(PQf 
 
 =2(AG)'= i r AE)% and RQ= PQ= \ AE (by Con- 
 Jiruhronj we have (AD)* : i (AE)' '~ 
 
 : : DC : I AE; and confequently 
 (AD)^=:AEx DC, 
 
 But, the Triangles ADC and 
 AEH' being equiangular (becaufe 
 zADC=zDAN=2:E.^"dDAC 
 common) we likewife have AD : P 
 DC : : AE : EH : and therefore AExDC=ADxEH 
 = (AD)' fp, above) whence EH = AD ; and fo, the 
 Triangles ADB and EHI being equiangular, we like- 
 wife have HI =: AB ; and from thence this eafy 
 
 ConJiruElion, 
 
 Having defcribed the Circle AHEF, as above direft- 
 cd, and drawn MG perpendicular to AE, take therein 
 G 71 equal to the given horizontal Diftance AB, and 
 thro' w, parallel to AE, draw HA, cutting the Circum- 
 ference of the Circle in H and h ; then either of the 
 Direftions AH, or AA, will anfwer the Conditions of 
 the Problem. 
 
 From this Conftruftion we have the following Cal- 
 culation, i-'/z. As AB : BC ; : AG TPQ) : GO ; which, 
 added to, or fubtrafted from G« (AB),£ives On, 
 
 O 2 Then, 
 
196 
 
 The Theory of Gunnery, or 
 
 Then as AG (PQ) \ On : : Co-fint of OAG ; C^?- 
 fine HO«:=z: (HAA) the Difference of the two required 
 Elevations : From which the Elevations themfelves will 
 be known : And from whence, if BC be fuppofed to 
 vanifti, the Solution to Prof), 1. will be obtained ; being 
 only a particular Cafe of That above. g. E, L 
 
 COROL. I. 
 
 Hence the Time of the FUg;ht may likewife be deter- 
 mined: For, S. AHE (ACD) : S, HAE : : AE : 
 
 (l~XH^ XAE=) HEz= AD fp,above)^^\i\c\i being 
 
 proportional to the Time of the Flight fvid. Lem. 3,) 
 it follows that the faid Time will, always, be as 
 
 o TTAF 
 
 becaufe AE is fuppofed conftant. There- 
 fore, if the Time 
 of the perpendicular 
 Afcent, or Defcent, 
 thro' the Impetus 
 A, be fonud, and 
 denoted by J", it 
 is evident that the 
 Time fought will be 
 truly reprefented by 
 5. HAE _ ^ 
 
 57ACi3 X ^^' ^^^^ 
 is the Co-Jine of the 
 Objeft's Elevation 
 above, or DeprefTion 
 below the Lcvel of 
 the Piece, will be to 
 F the Sine of the Ele- 
 vation of the Piece 
 above the Objeft, 
 I? as twice the Time 
 of the perpendicular Afcent, or Defcent, is to the true 
 Time of the Flight. 
 
 COROL. 
 
i^t Motion 0/ Projectiles. 197 
 
 COROL. II. 
 
 If the Elevation of the Piece, together with tba 
 DiHance^ and the Height '(or Depreflion) of the Ob- 
 je6l be given, the Impetus may, from hence, be 
 alfo found. 
 
 For fir ft, it will be AB : BC : : Radius : Tang. 
 BAC; whence (as BAD is given) EAH will Hkewife 
 be known. 
 
 Then, S, EAH (CAD) : S. AHE (ACD) : : HE 
 (AD) : AE. Alfo S. ADC : Rad{u3 : : AB : AD. 
 
 Therefore by compounding thefe proportions, we 
 have S. CAD x S. ADC : Rad. X 5. ACD : : AB : 
 AE ; which is 4 times the required Impetus fvid. 
 Corol. 3. to Prop, i).' 
 
 CoROL. III. 
 
 Moreover, if the Elevation, and the Impetus be 
 given, the Amplitude of the Proje6tion on an afcend- 
 ing, or defccnding Plane ACE, whofe Inclination is 
 given, may from hence be derived. 
 
 For, 5. AHE (ACD) : S, EAH (CAD) : : AE : 
 EH (AD). And 5. ACD : S. ADC : : AD : AC. 
 
 Bv the Compofition of which Proportions we have 
 n 5. ACD : S. CAD X ^- ADC : : AE: AC ; whence 
 AC is given. 
 
 COROL. IV. 
 
 Hence, alfo, may the Ratid of the Amplitudes, on 
 the fame Plane, at different Elevations, be deduced : 
 For the firfl; and third .Terms, of the lafl: Proportion, 
 continuing invariable^ the Ratio of the 2'* and 4*'', will 
 likewife.be invariable ; that is^ the ReBanglc under the 
 Sine of the Elevation above the Plane, and the Qq^ 
 Jine of the Elevation above the Horizon, in any one 
 Cafe, will be to the like Reftangle, in any Other, as 
 the Amplitude in the former^Cafe, to the Amplitude in 
 the Latter. 
 
 O 3 CoROL. 
 
tgS 
 
 The Theory of Gunnery, or 
 
 COROL. V. 
 
 But, if the Elevation be fuppofed conftant, and the 
 Plane's Inclination to vary ; then, fince, by the above 
 
 D • A^- ' rn 5. CAD x<^- ADC 
 
 ' Proportion, AC is umverfally = — " Tnr^ — X 
 
 LI otnc r^KjXJ 
 
 AE (where AE and the Angle ADC are fuppofed con- 
 ftant) it follows that the Amplitude will be, barely, as 
 
 n y n A r^ ' *^^^ ^^» inverfely^ as the Square of the Co- 
 
 fine of the Inclination of the Plane, apply'd to the Sine 
 of the Elevation above the Plane. — If both the Incli- 
 nations, and the Elevations, differ, it will appear, from 
 the fame Equation, that, the Amplitude will be, uni^ 
 vtrfally^ as the Redangle of the Sine of the Elevation 
 above the Plane, and the Co-fine of the Elevation above 
 the Horizon, apply 'd to the Square of the Co fine of 
 the Plane's Inclination. 
 
 CoROL. VI. 
 
 Since it is proved that HI is, always, equal to AB, 
 it is evident that, when the Former coincices with 
 MG» and thereby becomes a Maximum, the Latter will 
 alfo be a Maximum : In which Circumftance AC will 
 Jikewife be a Maximum; and the Point D will then 
 coincide with M and H fas in the annexed Figures J be- 
 caufe AD and EH are always equal to each other. 
 
 dmh 
 
 Therefore, fince, in this 
 Cafe, the Angle HAE 
 is (= E) = NAH, it 
 appears that the Ampli- 
 tude, 6n any inclin'd 
 Plane, will be the great- 
 eft pofTible when the 
 ABO ^ Line of Dircaion AH 
 
 hiff.Sts the Angle EAN, included between the Plane 
 and the Zenith. 
 
 CoROL. 
 
the Motion of Projectiles. 
 
 »99 
 
 COROL. VII. 
 
 Hence the greateft Amplitude, on any inclin'd 
 Plane, with a given Impetus, may be determined : 
 For the right-angled 
 Triangles AOG and 
 HOB, having AO = 
 HO, and the Angle N 
 O, common, are equal 
 
 in all refpefts : There- 
 fore it will be, as the 
 Tangent of AHG (or 
 BAH, the Piece's Ele- 
 vation) is to the Tang, 
 of CHG (or CAB, the Plane's Inclination) fo is AG 
 (twice the given Impetus) to the Difference CG be- 
 tween the faid Double Impetus and the Amplitude 
 fought. 
 
 CoROL. VIII. 
 
 Hence, alfo, if the greateft Amplitude, on an in- 
 clin'd Plane be given, the Impetus may be found : For, 
 it will be as the Radius^ is to the Sine of the Plane's 
 Inclination BAC, fo is the given Amplitude AC to the 
 Difference (BC or CG) betwixt it and twice the Im- 
 petus. Vid, CoroL 2. to Prop, 1, 
 
 COROL. IX. 
 
 But, if, inftead of the Plane's Inclination, the per- 
 pendicular Height, or Depreffion of the Obje£l above, 
 or below the Level of the Piece, be given ; then, AC 
 being i=AG4:BC, and (AB)'[=(AC)'— (BC)'] = 
 (x'\G)'^l2AGx BC, the greateft Diftance AB, at 
 which the Ball can poffibJv_ hit the Objeft, will there- 
 fore bc= -/[AGx(AG4.2BC)]. From which, as all 
 the Sides of the Triangle aBC are given, the Angle 
 
 O 4 BAC 
 
20O Thi Theory of Gunnery, &c, 
 
 BAC will Ukewife be kaown ; and from thence, the 
 Elevation, BAH, by CoroL 6. 
 
 From the latter of the two Cafes here confider'd 
 (where the Objeft is fuppofed below the Level of the 
 Piece, as in Fig. 2.) the greateft Amplitude of a Ball, 
 projefted from a given Height above the Plane of the 
 Horizon, is given : hc\r\g:=.\/\^hO X 1^^+ 2^^)] f^^ 
 above) in which BC denotes the given Height. 
 
 CoROL. X. 
 
 But, if the horizontal Diftance AB is given, and it 
 
 be required to determine the greateft Height the Bali 
 
 can poffibly reach in the Perpendicular BCD; we 
 
 fhall then have, HG (AB) : AG : : Radius : Tang, of 
 
 the Elevation (BAH) or AHG) ; and, as jR^^zmj ; Tavg, 
 
 BAC(=2BAH C/D90): : AB : BC ;_vvhich therefore 
 
 is known. But, becaufe AC=AG+BC, and (BC)» 
 
 ---(AC)'--(AB)\ the Value of BC will, ^/^, be truly 
 
 _ , - AG-c/D AB - 
 cxpreffed by — ^^^ 
 
 COROL. XI. 
 
 Laftly, if there be given the perpendicular Height^ 
 or Depreflion, of the Objeft, and its horizontal Di- 
 ftance, in order to determine the Elevation, and the 
 haft Impetus, to hit the Objeft : Then it will be as 
 AB : BC : : Radius : Tang, BAC ; whence the Ele- 
 vation BAH is alfo known, by CoroL 6 : And, as* 
 Radius : Tang. AHG (BAH) : : HG (AB) : AG ; the 
 Half of which is the Impetus,^^ Prop. 1. Coroi, 2. 
 
 Here 
 
Here follow the praSlical Solutions of the fe- 
 
 vera! Cafes depending on the foregoing 
 
 Theory. 
 
 I, Of Proje^ions made on the Plane of the 
 Plorizon. 
 
 PROBLEM I. 
 
 The greatejl Amplitude of a Piece being known (from Ex- 
 periment) to find the Amplitude^ at any ptopo fed Ele- 
 vation, 
 
 Solution. 
 
 As the Radius, is to the Sine of double the pro- 
 pofed Elevation, lb is the given, to the required, Am- 
 plitude (by Prop. 1. Cor. i.) 
 
 Ex. Let the gr. Amp. be 8coo Feet, and the givea 
 Elev. 30° : 16. 
 
 Then, as Radius lo' ocoo 
 
 to Sine 60° : 32' 9- ^3q8 
 
 fo is 8000 '8*9030 
 
 Amp. req. 6963 F. 3-8428 ^ 
 
 PROBLEM II. 
 
 The Impetus, or the greatejl Amplitude (which is the DouhU 
 thereoj) being known, to find the Elevation, tojrikean 
 ObjeSl at a given Dijiance, 
 
 Solution. 
 
 As the greated Amplitude, is to the given Diftance ; 
 fo is the Radius, to the Sine of the Double Elevation (by 
 Prop. i.CoroL 1). ^ 
 
 Ex, Let the gr. Amp. be 7500 F. and the given 
 Diftance 5620 F. 
 
 Thfea 
 
202 The Theory ^Gunnery, ot 
 
 Then, as 7500— c^* 8750 
 
 is to 5620 3" 7497 
 
 fo is Radius lo* 0000 
 
 to the fine of 48° : 32, or 131® : 28' — 9. 8747 
 Therefore 24° : 16' is the lower, and 65** : 44' the 
 higher Elevation, required. 
 
 PROBLEM III. 
 
 The Angle of Elevation, and the Dijiance of an OhjeSI, 
 being given, tojind the Impetus^ fo as to finke the 
 ObjeB, 
 
 Solution. 
 
 As the Sine of twice the Elevation, is to the Radius, 
 fo is the Diftance of the Obje6t to twice the Impetus 
 (by Prop. 1). 
 
 Ex. Let the Elcv. be 32° : 12' and the given Dill. 
 6500. F. 
 
 Then as Sine 64^* : 24 g* 95,51 
 
 is to Radius lo* 0000 
 
 fo is — 6500 3- 8129 
 
 to 7208 3' 8578 
 
 Whence 3604 is the Impetus required. 
 
 PROBLEM IV. 
 
 The Amplitude^ at any one known Elevation being given, 
 to find the Amplitude at any other known Elevation, 
 
 Solution. 
 
 As the Sine of Double the firfl Elevation, is to the 
 Sine of Double the Second, fo is the Amplitude at the 
 Former, to that at the Latter, (by Prop. \). 
 
 Ex. Let the firft Elev'. be 25° : \2\ the 2^ 36* : 
 15', and the given Amp. 52^0, F. 
 
 Then 
 
the Motion of Projectiles. ao| 
 
 Then, as Sine ^o"" : 24' — Co. Ar. o' 1132 
 
 to Sine— 72" : 30' 9* 9794 
 
 fo is 52^0 F. ■ 3* 7201 
 
 10 the Amp. req. 6407 F. il3- 8127 
 
 PROBLEM V. 
 
 The Amplitude of the Projedion^ with a given Quantity 
 of Powdery being known ; to find the Requifite of 
 Powder^ fo as tofirike an Objeii at a given Dijiance\ 
 the Elevatien remaining the fame. 
 
 Solution. 
 
 As the given Amplitude, is to the propofcd Diftance^ 
 fo is the given Weight, or Quantity of Powder, to 
 the Quantity fought, nearly *. 
 
 Ex. Snppofe the Requifite of Pcwder to throw a 
 Shot 4000 Feet, at 45° Elevation, to be 16 lb. What 
 Quantity is neceflfary, to flrike an Obje6l at the Di- 
 flance ot 5000 Feet. 
 
 Here it will be, as 400 : 5000 : : 16 : 20 lb. the 
 Quantity fought. 
 
 * In this Solution, the Velocity communicated to the 
 Ball, isfuppofed to be in the Subduplicate Ratio of the 
 ^antity of Powder : which is notfhiBly true, efpecially 
 in large Charges ; fince a confiderable Part of the 
 Powder^ infuch Qafes, is blown out, unfired : There are, 
 hefides, other Reafons to be affigned, why, the Velocity 
 cannot be exa£ily in the Proportion above fpecified. 
 
 PROBLEM 
 
204 The Theory of Gunnery, or 
 
 PROBLEM VI. 
 
 The Dtjlanci of the ObjeB, and the Elevation of thi 
 Piece ^ keing given ; to determine the Time of the Flight 
 
 Solution. 
 
 As the Radius is to the Tangent of the Elevation, fo is 
 the given Diftance of the" Objc61:, in Feet, to the Square 
 of 4 times the Number of Seconds^ required (vid. p, 184. 
 and 187). 
 
 Ex. Let the Elevation be 32°, and the Diftance of, 
 the Objeft 5280 Feet. 
 
 Then, Radius lo* 0000 
 
 Tang. 32" ~ 9' 7958 
 
 Dift. 5280 ,^- 7226 
 
 . , 57' 44 — : : Y 7592 
 
 I of which, or 14% 36, is the Time required. 
 
 But, when the Elevation is 45 Degrees Twhich is 
 commonly the Cafe, in Throwing Bombs, &c. ) then 
 I of the fquare Root of the Difiance, in Feet, will give 
 the Number of Seconds taken up in the Flight. The 
 Knowing of which will be of Ufe in adjulling the Fufe. 
 
 II. Gf Projedlions, when the Ohjedt fired at, 
 is above, or below, the Level oj the Piece. 
 
 PROBLEM Vn. 
 
 The horizontal DiJIance, and the Angle of Elevation, or 
 DepreJJion of an O^jeB. being given, with the Ele- 
 vation 0} the Piece ; tofnd the Impetus, Jo as to hit 
 the ObjeB. 
 
 Solution. 
 
 As the Reftangle of the Sine of the Elevation of the 
 Piece above the Obje6>,. and the Co-fine of its Ele- 
 vation 
 
the Motion of Projectiles. 205 
 
 vatlon above the Horizon, is to the Reftangrle under 
 the Radius and the Co-fine of the Obje£l*.«' Elevation, 
 or Depredion ; To is | of the given horizontal Diftance 
 of the Obje6f, to the Impetus required [by Prop, 3. 
 Corol. 2). 
 
 Ex. Let the horizontal Diftanee of the Objeft be 
 5600 Feet, and its Elevation 8° : 1,5' ; and let the Ele- 
 vation of the Piece be 32° : 30' : Then, 
 
 24° : 15', Co-ar. of its Sine o* 3865"^ "^ 
 
 32° : 30', Co-ar. of its Co. f. o* 0740 | ^ 
 
 Radius ■ : : — lo* 0000 y "* 
 
 8° : 15', its Co-fine 9' 9955 1 ^ 
 
 1400 F. ■ : 3" H^O 2 
 
 4000 P. the Imp. req^ 2(3. 6021 
 
 PROBLEM VIII. 
 
 The horizontal Diftance, and the Angle of Elevation^ or 
 Deprejjion of an ObjeB, being given, together with 
 the Impetus ; to find the Elevation of the Piece^ te hit 
 theOhjeB. 
 
 Solution. 
 
 As the Radius is to the Tangent of the Objeft's Ele- 
 vation or Depreflion, fo is twice the Impetus to a 
 fourth Number; which add to, or fubtraft from, the 
 given horizontal Diftance, according as the Obje6l is 
 elevated, or depreffed : Then fay. 
 
 As twice the Imp-etus is to the Sum, or Remainder, 
 fo is the Co-fine of the given Elevation, or Deprefiion, 
 to the Co-fine of an Angle ; which added to, and fub- 
 trafled from, the Angle included between the Objcft 
 and the Zenith or (vertical Point), gives the Double of 
 the Complements ol two different Elevations, whereby 
 the Ball may hit the Objeft (See Prop. 3). 
 
 Ex. Let the horizontal Diftance of the Objeft be 
 5600 Feet, and its Elevation 8° ; 15' ; and let the given 
 Impetus be 4000 F. 
 
 Then 
 
2o6 The Theory of Gunnery, or 
 
 Then, as Radius - — = — lO'oooo 
 
 is to Tang. 8° : 15' • ■ — 9* 1613 
 
 fo is 80C0 '■ 3*9031 
 
 to — 1160 — 1|3' 0644 
 
 This added to 5600 gives 6760 : 
 
 Therefore, as Bgoo 3- qo<^i 
 
 is to 6760 3-8299 
 
 fo is Co-f. 8° : 15' — q- 9955 
 
 to the Co-f. 33" : 15' 9. 9223 
 
 Which, added to, and fuhtrafted from 81" : 45', gives 
 115°: 00', and 48° : 30', respeflively : the Halves of 
 which are 57° : 30' and 24° : 15' ; whofe Comple- 
 ment 532°: 30', and 65° ; 43', are the two Elevations 
 required. 
 
 PROBLEM IX. 
 
 The Impetus, and the Angle of Elevation, being given ; t6 
 find at what Dijlance the Piece ought io be planted, 
 to hit an OhjeB, whuje Dijlanse above^ or below, ihe 
 Level oj the Piece ^ is alfo given. 
 
 Solution. 
 
 As the Radius, is to the Sine of twice the given Ele* 
 vation, fo is the Impetus, to half the horizontal Am- 
 plitude, at that Elevation (byProp. 1). 
 
 And, as the Radius, is to the Co-tangent of the Ele- 
 vation, fo is twice the perpendicular Height or De- 
 predion, of the Obje6l, to a Fourth-Proportional; which 
 take from, or add to, half the horiionial Amplitude, 
 according as the Obje6l is elevated, or depreffed ; then 
 find a Mean Proportion between the Half- Amplitude, 
 and the Sum, or Remainder; which, added to the faid 
 Half-Amplitude, gives the Diftance fought / by Prop, 2), 
 Ex. Let the Impetus be 3000 Feet, the Elevation 
 40*, and the Hcighth of the Objeft 200 Feet. 
 
 Then, as Radius lo- 0006 
 
 to Sine— 80° 9- 9933 
 
 fo is — 3000 $■ 4771 
 
 to 29^4 :M3-4 04 
 
\ 
 
 (he Motion of Projectiles. ' * ^07 
 
 And, as Radius — lo* 0000 
 
 istoCo-t. 40° 10 0762 
 
 fo is 400 2 6021 
 
 to 477 i;2. 6783 
 
 Now, the Difference between the tv^o Niumbers 
 above found is 2477; whofe Log. being added to that 
 of 2954, and the Sum divided by 2, the QMOUv-nt will 
 be the Log. of 2705, the required Mean Proportional : 
 Whence the Diftance fought comes out 5659 Feet *. 
 
 III. Of Proje^ions on Planes, inclind to that 
 of the Horizon, 
 
 PROBLEM X. 
 
 Tht Inclination of the Plane^ and the Elevation and 
 Impetus of the Piece, being known; tojind the Am- 
 plitude oj the Proje&ion. 
 
 Solution. 
 
 As the Square of the Co-fine 9f the Plane's Incli- 
 nation to the Horizon, is to the Re6langle of the Sine 
 of the Elevation above the Plane and the* Co- fine of 
 the Elevation above the Horizon, fo is 4 times the 
 Impetus, to the Amplitude of the Proje6lion (^y Prop, 
 3. Cor. 3j. 
 
 Ex. 1. Let the Impetus be 4000 Feet, the; Elevation 
 32° : 30', and the Afcent of the Plane 8* : 15'. Then 
 the Elevation above the Plane will be 24° : 15' ; and 
 the Operation as follows. 
 
 * The Operation. 
 
 Log. 2477= 3* 393. 926 
 Log. 2954 = 4- 470, 41 1 
 
 2 )6- 864, 337 
 
 Log. 2705 3' 432, 169 
 
 + ^954 
 Sumz=:56^g =: Di/i. required. 
 
2o8 The Theory cf Gunnery, or 
 
 8* : 15', Co-Arith. of its Co-f. — o -0045^ 'S 
 
 The /ami repeated o* OO45 [ ^ 
 
 24°: 15'— its Sine 9* 6135 > ^ 
 
 32° ; 30' — its Co-f. 9* 9260 ^ 
 
 16000 F. — its Log. 4- 2041 J^ o 
 
 5658 feet the Amp. req". ^Is' 75^^ 
 
 Ex. 2, Let the Elevation and Impetus be the fame 
 as in the laft Example ; but let the Plane in this Cafe 
 have a Defcent of 8° . 15' (inftead of an equal Afcent). 
 
 Then the Operation will ftand thus, 
 
 8* : 25', Co-Ar; of its Co-f. o* 0045"^ '^ 
 
 Thi fame repeated o* 0045 | Z§ 
 
 40° : 45', its Sine 9* 8147 \ ^ 
 
 32" : 20', itsCo.fine 9* 9268 j -^ 
 
 16000 F. its Log. 4* 2 -41 J 2 
 
 9007 F. Amp. req''. ^ l3'954^ 
 
 PROBLEM XL 
 
 The Inclination of the Plane^ the Elevation of the Piece, 
 and the Amplitude of the P.rojeSion, being given ; 
 to fnd the Impetus. . 
 
 Solution. 
 
 As the Reftangle of the Sine of the Elevation above 
 the Plane and the Co-fine of the Elevation above the 
 Horizon, is to the Square of the Co-fine of the Plane's 
 Inclination, fo is the given Amplitude, to 4 times the 
 required Impetus (by Prop. <^.Cor, 3). 
 
 Ex. Suppofe the Plane to have an Afcent of 8* : 15',: 
 and that the Amplitude thereon, at an Elevation of 
 32° : 30', is 5658 Feet. Then, the Elevation above 
 the Plane being 24° : 15', we (hall have, 
 
 24* 
 
the Motion £/ Projectiles. 209 
 
 24° : 15' Co-Ar. of its Sine o- 3865% 'g 
 
 32^ : 30' Co-Ar. of its Co-f. o* 0740 :§ 
 
 8°.: 15' its Co.f. 9- 9955 > I 
 
 The fame 9* 9955 | -^ 
 
 5658 F. its Log. 3- 7526 J 3 
 
 16000 F. -2I4' 2041 ' 
 
 I of which, or 4000 Feet, is the Impetus, req*. 
 
 PROBLEM XIL 
 
 The Inclination of the Plane, the Impetus, and the Am* 
 plitude being given, to determine the Elevation* 
 
 Solution. 
 
 As the Radius is to the Co-fine of the Plane's Incli- 
 nation, fo is the given Diftance, on the Plane, to the 
 horizontal Diftance correfponding : From which, the 
 Impetus, and the Plane's (or Obje£l*s) Elevation or De- 
 preflion, the Elevation of the Piece may be found, by 
 Prob, 8. 
 
 Ex. Let the Plane have an Afcent of 8" : t^', and let 
 the given Amplitude thereon be ^658 Feet, fuppofing 
 the Impetus to be 4000 Feet. 
 
 Then, as the Radius *— 10; 0000 
 
 istothe Co-fmeof 8°: 15'- 9- 9955 
 
 fo is 5658 Feet 3- 7526 
 
 to — 5600 Feet 113- 7481 
 
 the horizontal Diftance of the Place where the Ball im-- 
 pinges : As for the reft of the Operation, it is exaftly 
 the fame as in the Example to ProJf, 8 ; for which 
 Reafon it will be needlcfs to repeat it here. 
 
 PROBLEM 
 
£io r>^<; Theory r/ Gunnery, er 
 
 PROBLEM XIII. 
 
 Having the AngU of EUvatien, or Depre/Jion of the Ob- 
 jeEly together with the Elevation^ and Impetus of the 
 Piece : to determine the Time of the Flight, 
 
 Solution. 
 
 As the Co-fine of the Elevation, or Depreflion of the 
 Objeft, is to the Sine of the Elevation of the Piece 
 above the Objeft, fo is Half the fquare Root of the Num- 
 ber of Feet in the Impetus given, to the requir*d Nunaber 
 of Seconds in the Flight. 
 
 Ex. Suppofe the Elevation of the Obje6l to be 8" ; 
 30' ; That of the Piece 45°; and the Impetus 3600 Feet. 
 
 Then, the fquare Root of 3600 being 60, we have 
 As the Co-fine 8° : 30 — Co-Ar. o* 0048 
 
 is to the fine 36° : 30 ' 9* 7744 
 
 fo is 30 1- 4771 
 
 to 18-04 Seconds req". i|i- 2563 
 
 IV. of tfie Maxima and Minima, in the 
 Motion of ProjeSliles, 
 
 PROBLEM XIV. 
 
 The greatefi horizontal Amplitude being given^ to fnd 
 the greatejl Amplitude on a Plane iokofe Inclination to 
 the Horizon is alfo given 
 
 Solution. 
 
 Take lialf the Angle included between the Plane and 
 the Zenith ; the Complement of which is the required 
 Elevation (by Prop. 3. Cor, 6). Then fay, as the Tan- 
 gent of the faid Elevation, is to the Tangent of the Plane's 
 Inclination, fo is the given Amplitude, to the Difference 
 between it and the Amplitvide fought (by Prop, 3. 
 
 Corol, 7). 
 
 E 
 
M<; Motion ^ Projectiles* ftii 
 
 Ex. Let the great«j(l horizontal Range be 8000 Feetj 
 and the Inclination of the Plane 12" : 30', defcending. 
 
 Here the Angle included between the Plane and the 
 Zenith, or vertical Point, being 102° : 30', the Half 
 thereof will be 51° : 15', and the Elevation 38° ; 4^'. 
 
 Therf. as Tang. 38° : 45' 9' 9045 
 
 is to Tang. — 12°. 30' 9* 3457 
 
 fo is 8000 F. 3* 9031 
 
 to 2210F. 3* 8443 
 
 Which, added to 8000 (becaufe the Plane defcends) 
 gives 102 lO Feet, for the true Anfwer, in this Cafe. 
 
 PROBLEM XV. 
 
 Thegreatejl Amplitude^ on an inclined Plane^ being given^ 
 tojind the greateji Amplitude, on the Plane of the Horizon. 
 
 Solution. 
 
 As the Radius, is to the Sine of the Plane's Inclination, 
 fo is the given Amplitude, to the DiflFerence between it 
 and the required Amplitude fby Prop. 3, CoroL 8). 
 
 Ex. Let the Inclination of the Plane be 12° : 30', 
 defcending, and the given Amplitude 10210 Feet. 
 
 Then, as Radius lo* 0000 
 
 is to S. 12° : 30 9* 33c53 
 
 fo is 10210 4* 0090 
 
 t6 ^2210 -3' 3443 
 
 Which taken from 10a 10, gives 8000 Feet, far the ixut 
 Diftancc fought. 
 
 P % PROBLEM 
 
fii2 The Theory ^Gunnery, tr 
 
 PROBLEM XVI. 
 
 The Impetus^ and the perpendicular Height^ or Deprejftottf 
 of an ObjeQ being given, to find the greateji horizontal 
 Di fiance at which the Bail can, pojjibly, hit the Ob- 
 ject and aljo the Elevation anjwering thereto. 
 
 Solution. 
 
 Take the DifFerence of the two given Quantities, it 
 the Objeft be elevated ; but otherwife, their Sum ; 
 then find a Mean Proportional between the Impetus and 
 faid Difference, or Sum ; the Double of which will be 
 the Diftance fought (by Prop. 3. Cor. 9). 
 
 For the Elevation, it will be, as the Diftance thus 
 found is to the Height, or Depreflion of the Objeft, fo 
 is the Radius to the Tangent of an Angle ; which added 
 to, or fubtrafted from, 90 Degrees, refpeftively, gives 
 the Double of the required Elevation (by Prop, 3. 
 Cor. 9). 
 
 Ex. Let the Impetus be 4000 Feet, and the Depref- 
 fion of the Objeft, below the Level of the Piece, 2210 
 Feet. 
 
 Here we are to take a Mean Proportional between 
 4000 and 62 10 ; which is =: y/ (4000 X 62 10) ==4984 ; 
 whofe Double 9968 is the required Diltance. 
 
 Moreover, we have, as 9968 8* 9986 
 
 is to— 2210 3* 3443 
 
 fo is Radius lo- 0000 
 
 to Tang. 12** : 30 9* 34<57 
 
 whence the Elevation appears to be 38^* : 45'. 
 
 From this Problem, the greateft Amplitude of a 
 Ball, projefted from a given Height above the Level 
 of the Horizon, is given. 
 
 PROBLEM 
 
(he Motion of Projectiles. aig 
 
 PROBLEM XVII. 
 
 Having the hoj-izontal Diftance of the ObjeB^ together 
 with its perpendicular Height, or Deprepon, above or 
 below the Level of the Piece ; to determine the leafi Im- 
 petus^pofjible^ whereby the Ball may reach the O'bjeB ; 
 and alfo the Elevation correfponding. 
 
 Solution. 
 
 As the horizontal Diftance of the Obje6l is to its per- 
 pendicular Height or Depreflion, fo is the Radius to the 
 Tangent of an Angle ; which, added to, or fubtrafted 
 from 90 Degrees, gives the Double of the Elevation 
 (by Prop. 3. Carol, 6). 
 
 And, as the Radius is to the Tangent of the Ele- 
 vation, fo is the given horizontal Diftance, to twice the 
 Impetus required fby Prop. 3. CoroL ii). 
 
 Ex. Let the horizontal Diftance be 9968 Feet, and 
 the Diftance of the Objeft below the Level of the 
 Piece fl2io Feet, 
 
 Then, as--'996p 3* 9986 
 
 isto 22io 3' 3443 
 
 fo is Rad. 10. 0000 
 
 to Tang. 12** : 30' 9- 3457 
 
 Therefore the Elevation is 38° : 45'; and it will be 
 
 As Radius -lo' oooo 
 
 is to Tang. 38" : 45' 9- 9045 
 
 fo is — = 9968 ■ 3* 9986 
 
 to 8000 3*9031 
 
 The Half of which, or 4000 Feet, is th« Impetus fought. 
 
 PROBLEM 
 
114 The Thiory of Gunnery, (3c. 
 
 PROBLEM XVIII. 
 
 To find the great eji Height a Ball can ^ P^llP'^^yi ^^^ch in 
 a Perpendicular to the H.orizon ; the Impetus^ and the 
 Diftance of the Piece from thefaid Perpendicular, 
 being given* 
 
 Solution. 
 
 Find a Third Proportional to the Impetus and halC 
 the given Diftance, which fubtraft from the Impetus, 
 and the Remainder will be the Anfwer fby Prop. 3. 
 CoroL 10). 
 
 Thus, if the Impetus be 4000 F. and the given Di- 
 ftance 3000 Feet, then it will be, as 4000 : 3000 : : 
 3000 : 2250; which taken from 4000, leaves 1750 
 Feet, for the greateft Height the Ball can poflibly reach 
 in the propofed Perpendicular, with that Impetus, 
 
 PART 
 
PART IV. 
 
 EXHIBITING 
 
 A new, and very comprehenjive Method for 
 extradling the Roots of Equations in Num-- 
 bers ; by increqfing the Dimenfions of the 
 unknown ^antity. 
 
 IN this Method (as In the common Method by con- 
 verging Series's) the Value fought is, firft of all, to 
 
 be nearly eftimated (either from the Equation itfelf, 
 or from the Nature of the Problem whence it is derived) ; 
 and fome unknown Quantity (as z) muft be affumed to 
 cxprefs the Difference between that Value (which we 
 will denote by r) and the true Value required : And 
 then by fubflituting r ±-2;inflead of its Equal, in the 
 given Equation, a new Equation will emerge, aflfefled 
 only with z and known Quantities. 
 
 The Equation being thus prepared for a Solution, 
 multiply it into two, or more fucceeding Terms of the 
 Series i -f- Az -f Bz^ + Cz^ 4- Dz^&c. (according to 
 the Degree of Exaftnefs Neceflary) and let the Coef- 
 ficients of the homologous Terms of the new Equation 
 hence, arifing, wherein the Square, and the next, higher 
 Powers of z are concerned, be made equal to Nothing: 
 By which Means the Value of A, &c. will be deter- 
 mined, and as many Terms of the Equation deftroyed, 
 at the fame time, as there are Terms in the affumed 
 Multiplicator minus one. And the Terms involving the 
 fuperior Powers, which yet remain, being of very imall 
 Value, may alfo be reje6>ed: Whence the Equation 
 will be. reduced to a Simple One : From which the 
 Value of 2 will be found. 
 
 P 4 Let 
 
2l6 The RESOLtTTION of EQUATIONS. 
 
 Ex. Let the Equation propounded be ^('=2:50. 
 
 Here, a; being fomething greater than 7, put 7 -j- « 
 srrj;; then the given Equation will become 49-4- 142 
 
 or, — 1 +14^ 4" ^' = ^' This, multiply 'd 
 by 1 4. Az+Bz', 
 
 gives< * — Az 4-14^2*+ Az» ♦ V=:o. 
 
 \ * * __ Bz' 4-MB2^+Sz* J 
 Whence, by equating the Coefficients of the like Terms, 
 we have 1 -f- j 4 A — B = o, and A -^ 14B =: o : 
 
 From which Eq uations A is found rr: ; and B 
 
 197 
 
 97 
 And, by fubftituting thefe Values above, 
 
 wc have — 1 A-fiA A ^^ X ^ 4 = o. 
 
 • V ^ » i^-j^ ^^ 197 
 
 Which, by reje£ling the exceeding fmall Quantity 
 
 
 9' 
 Hence — 197 4" 2772Z: 
 
 becomes — 1 +/'i4 + ' , -■] X ^ = 
 
 J q7 
 and z zzn =,0710678, nearly: Which Value is 
 
 2772 ^ ' ^ 
 
 true to the lafl Decimal Place : And, if more Terms of 
 the Series 1 -f- Az-J- Bz' -^Cz^ &c. had been taken, the 
 Conclufion would have been ftill exa6ier in Proportion. 
 
 Ex. 2. Suppofe the given Equation (when prepared for 
 a Solution) to be — 2 4" <5^ — z' =s c. 
 
 Here, if four Terms of the general Series 1 -j- Az -}- 
 Bz' -J- Cz' £eft-. be taken, and multiply'd by — 2 -f- 52 
 — 2^ our Equation will be cbang'd into the following 
 One, 
 
 viz. 
 
{ 
 
 The Resolution of Equations. 217 
 
 '— 2 — 2AZ — 2BZ'— .2Cs3 * ) 
 
 * -■ * * z^^hz^&c.S 
 
 Where, by comparing the homologous Terms, we havfe 
 iB=5A, 2C=5B— 1, and 5C=:A : 
 
 Hence 25B — 5 (== loC) = 2 A ; but, by the firft 
 ©f these Equations, B = "i- ; therefore — ^ — — 5 = 
 
 2 
 10 
 
 *A, and confequently A = 
 
 ^ ^ 121 
 
 Let this Value be now wrote inftead of A ; by which 
 means our laft Equation for the Value of z (neglefting 
 
 the Terms — - Bz*— Cz') is reduced to -— « + (5 -~ ) 
 
 Xz=o. 
 
 242 
 Whence z comes out =zz— ^z=o. 414 nearly. 
 
 Let there be now given the general Equation, 
 
 Then, multiplying by (1 + Az) the two firil Terms 
 of the Series, only, 
 
 we^get I IZ. pAz+ aAz^ &c. f =^' 
 
 Here by making b -{* aA =0, A is found =: ; 
 
 and our Equation becomes — /> -f" f ^ + •^)X2=p: 
 
 p ap 
 
 Whence z is given = , ,^ : — — j-r- . 
 
 ° a -\- op aa ■\-bp , 
 
 a 
 
 The Value of 2, here determined, taking in two 
 
 Terms of the given Series, az-|-^z*-[-cz^ 6?r. I call an 
 
 Approximation of the fecond Degree as the common 
 
 Method of Converging Series's, which takes in the firft 
 
 6 Term 
 
new 
 Equation 
 
 ai8 Ti^^? Resolution ^Equations. 
 
 Term only, may, by the fame Rule, be cail'd -an Ap- 
 proximation of the Firft Degree.) 
 
 But to obtain an Approximation of the Third De- 
 gree, or fuch an One as (hail include three Terms of the 
 ©riginal Series, let the given Equation, 
 — p •\- az-^hz^'-^-cz^ &c, z=:o, be now multiply'd by 
 three Terms of the afTumcd Series i -j- Az -j- Bz' (^c : 
 M^hence there ctrifes this 
 
 — P -\- ^^ -f" ^2;*-|- cz^ &c.'\ 
 
 * * —^px^J^^az^&c, } 
 
 Where, -by equating the homologous Terms, we have 
 b-^- ha — B/?=z:o, and <: -(- A/^ -f- Ba =0. 
 
 Let the former oF ihefe Equations be multiply'd by a, 
 and the Latter by p ; and then add the two Produfls 
 together ; fo ftiall 
 ab-\' ha* 4"/'^ + hbp z=:0 ; and confequently A n= — 
 
 — ~r-r-. Whence z (■=. — r Vs Hkewife ffiven. 
 
 aa-^ bp \ a — phj ^ 
 
 To have an Approximation of the Fourth Degree, 
 four Terms of the* Series i+Az-j-B^^ &c, mult be 
 taken, for a Multiplicator : By which means the Equa- 
 tion given will be 
 
 {— /> 4. az •\- bz'^ cz' + dz^ &c. "J 
 * — hpz -hhaz'+ hhz' -A-hrz* &c, I 
 * * -B/;z*4- Baz'-i-mz^&c f""* 
 * * * - Cpz'^Caz'^c. J 
 
 Here we have 
 
 p P 
 
 Ba ^ hb 'c 
 
 o := Cfl + B^ + Ac + d^, 
 From whence, exterminating C, there arifes 
 A^ . c . B^ . Ac , J 
 
The Resolution ^Equations. 219 
 
 or(-+7)XB+(-+-)xA+- + -=o. 
 
 Which, by fubftituting the Value of B, 
 
 becomes (^+^)X(~+^)+(j-h-)xA+^ 
 
 + ^=:o;thatis/i4-i^ + i>lxA+'^+tU^ 
 ^ a \PP p CL} pp^ ap^ p 
 
 Multiply the Whole by<2;7-; then will [a^'\-2abp-\'Cpp) 
 
 X A. -^ a"^ 4" ^'z' + ^^p "T ^/P'=o •■ 
 
 WhenceAisfound = - -_^±ifi±W)XH:^. 
 
 And z=i= — ^-T-, ^J 2w the preceding Cafe, 
 
 By the fame Method, an Approximation of any 
 higher Degree, to take in as many Terms of the pro- 
 pofed Series as you pleafe, may be derived. 
 
 For, it is evident, from above, that in all Cafes 
 whatvel", 
 
 Bwillbe=— + i, 
 p p 
 
 P P V 
 Ca . B^ A<: d 
 
 ~T p '^ p 7' 
 
 _ Da .a . Be . Ad ^ e 
 
 &c. Be. ^ 
 
 where the laft Value is, always, equal to Nothing. 
 
 Therefore 
 
520 The Resolution oJ Equations. 
 
 Therefore, 
 by making 
 
 and 
 
 ^ P 
 
 ~ p 
 
 ~ p 
 
 I "" P 
 
 r b 
 
 r— ^1+f 
 ~ P 
 
 s 
 
 -br- 
 
 — p 
 at'\'b5-\-cr~^dq-\-f 
 
 — -p 
 
 the Value of A will be determined, by dividing the laft 
 of thefe quantities q, r, j, t, l£c. by the corresponding 
 Quantity of the upper Series Q, R, S, ^c, and changing 
 the Sign of the Quotient. 
 
 Whence we get 
 
 •r, 
 
 — ? — , ^c, for fo many 
 
 1 
 a . r 
 
 
 different Values of z ; whereof each is more exa£l 
 thanth« preceding One. 
 
 Thefe 
 
The Resolution of Equations. 221 
 
 Thefe Equations are cafily derived from Thofe above, 
 •xpreffing the Relation of the Quantities A, B, C^ 
 
 For, by writing Q and q, inftead of their Equals, in 
 the firft of thofe Equations/'fi = — — }--) it becomes 
 
 » 
 Which Value of B, wrote in the 2** Equation, 
 
 ^ B« , A^ , c . 
 C= h-. gives 
 
 C=:^2A4_^+^+£=:RA+r; by fubftitut- 
 
 P \ P P P 
 ing R and r in the Room of their Equals. 
 
 Moreover the third Equation, by fubftituting for C 
 
 and B. becomes D =^^+!:2 + *-SA+ ^+^ 
 
 V ^ P P V P 
 
 I ^_SA4-J. 
 
 After the very same manner E = TA -f- ^ E = 
 VA-J- V, &c. And by putting all thefe feveral Values, 
 
 fucceflively, equal to Nothing, we have — i--, — — 
 — — , ^c» for fo many different Approximations of 
 
 o 
 
 the Value A. Which being fubftituted in the general 
 
 Equation z :=: __ . -» the very Expreflions before 
 
 given, for the Value of z, are obtained. 
 
 It will be proper, now, to fhew the Ufe of the fe- 
 veral Approximations, or Theorems, derived above, 
 by a few Examples. 
 
 In 
 
22a The Resolution ef Equations. 
 
 In the firft Place, then, let the Equation given be 
 
 lO. 
 
 This Equation, by making 2,-}- z =z:^,will be trans- 
 formed to — 2 -}- 1 2z -j- 62^ -|- z^ = o : Which being 
 compared with the general Equation 
 
 — p '\' az -\- bz"" -\- cz^ "4" ^^'^ ^^' =^ ^» 
 we hav^e p:z=. 2, a z=z 12, ^z=r«6, t r= ', d tzirO, &c. 
 
 Therefore by the firft of the three Approximations, 
 
 at ;?. 217) — A /^= - ) is found = ^. 
 
 Whence 2 \z=l —~f~ ) comes out z= - == o. 1538, 
 nearly. 
 
 But, according to the fecond Approximation, or 
 Theorem, the Value of — A^=-^-^^ V will be =- 
 
 ^ — -^, — SZ- : And confequently that o\ z f — 
 
 144 -J- 12 78 n y ^ 
 
 a ^ Ap ^ 505 
 
 Laftly, the Value of — A, according to the third 
 Approxunation being = -_^-p-^^— p-— =:: 
 
 144 X 6 + 96 __ 36x6 + 24__48 jjj^ 
 
 144X12 + 144X2 + 4 .36X14+1 101* 
 correfponding Value of z will therctore come out z=z 
 
 •^ = o. 154434 : Which Number is true in all its 
 
 ^54 
 Places. 
 
 The very fame Conclufions will likewife be brought 
 out, from the general Solution. 
 
 For, 
 
The Resolution of Equations. 223 
 
 For, p being here = 2, a ziz 12, b z=: 6, c = 1, 
 
 dz;:zo.^c. (as before)^ 
 
 wehaveQ/'=:-- ) = 6, 
 
 g (a R + /^Q+r n 505 
 ^ » y 2 ' 
 
 Alfo,,(i) = 3, 
 
 2 
 
 \ V )' - 
 
 
 Therefore ^=:=1, I_.37L £— H2— i?.. and con- 
 
 For a fecond Example, fuppofe loz— z^ .«_, ^^ qj. 
 — 2 -J- iqz — z^ r:;:: o. 
 
 In this Cafe ;7z=2, a=:io, ^=0, C:=. — 1, dz=zo, &c. 
 And therefore, for an Approximation of the fourth 
 
 Degree, we have — A( - , ; -r—. — ^ ^ ■ ' — -^)zr= 
 
 ^o * V a^ -Y Q.abp ■\- cpp J — 
 
 II_£2 3::^ ZZA ; and confequently z (= • ] 
 
 1000 — 4 249 V a — pA) 
 
 — ^49 . — Q 2008045; which Value is true to the 
 
 1240 
 laft Figure. 
 
 For 
 
«24 The Resolution of Equations. 
 
 For a third Example let there be given the Equation 
 a' -j- 2fix^ ~l" 432^ =^ 2272. 
 Here the Value of x appears to be about 4 ; let there- 
 fore 4 -|- z be wrote for x ; whence our Equation is 
 reduced to 
 
 96 + 768Z + 482^ + 2' = o. 
 Which being compared with 
 — ;; 4" az -\- bz^ -J^ c%^ '\- dz^ Wc. := o. 
 we have, in this Cafe, 
 )P = — 96, J = 768, ^ = 48, c = 1, flf s=: o, ^c. 
 
 Or, f==_8,*=-i.£=-JL,^,. 
 
 P V ? 96 
 
 Hence Q(^) = -8 
 
 yp pi 
 
 
 
 V ;» ' ;»/ ^ 96 yo 
 
 (o-r,bq.d\ 380 
 
 M-1-r; = 4-4.=-^'^ 
 
 \ p ' p * pi 12 
 
 Therefore X= J- I 2^3 J^ i _ 
 
 inerciore ^ ^^. ^ _^^^^^^«. ^^^. ^ _ 
 
 380 X 8 _ 3Q^° . 
 483S5 ""4S385' 
 
 From whence z = ^ — =: — o. 1 2598, nearly ; 
 127 
 
 or, z=— ^-g-_= — o. 1259894, m<7r<r nearly, 
 or, z = -^-gy^ = — o. 1259894802,/?// nearer. 
 
 Laftly, 
 
ri^<r Resolution? ^ Equations. ss^ 
 
 Laftly, let there be given the Equation 
 
 1 1_J 1- f &c.z=z^ 
 
 a 2-3-4 A'S'i's:^' 2-3'4-5-67-8- a 
 
 Then, making z = x\ &c. we have 
 
 — l+Z r. &C. z=zO, 
 
 12 12X30 12X30X56 
 
 Here p = 1 , a z=z 1 , i> = — ' — » ^ =: — , fi^n=: • 
 ^ ' ' 12 12x30 
 
 : &c. 
 
 12X30X56 
 
 And, by fubftituting thefe Values in the third General 
 Approximation fvid, p. 219), we have — A 
 
 \ aaa ■\- ^abp -j- cpp ) 
 
 _i + _L-4._l I 
 
 21 12x30 12x12 12x30x56 
 
 I2"ri2X30 
 
 30X56— 56-~i40+^__ M^5 
 
 10X30X56 + 56 16856' 
 
 Therefore % (= ~Z.-~)=i.^— 1.09661 ; 
 V a—pK' 15371 ^ 
 
 and A:(=y^z)=i'047i9. 
 
 After the fame manner the Roots of other Equations 
 may be approximated : But I (hall here Ihew, how the 
 general Theorems themfelves may be rendered more 
 commodious, for certain particular Cafes, by Means of 
 a proper Transformation. 
 
 It is known, if two Quantities be, refpeftively, in- 
 
 creafed, or decreafed by two other, fmall, Quantities, 
 
 ' near^ in the fame Ratio with the two Firft, that the Sums, 
 
 or Differences will ftill be in the fame Ratio with the two 
 
 > firft Quantities, very near. 
 
 K Q Wherefore 
 
ai6 Thi Resolution of EquAXioKs. 
 
 Wherefore, feeing the Numerator of the Fraftion 
 — 'V^ i exprefling the 2''. General Value of — A. (vid> 
 p. 218) is in proportion to the Denominator, nearly, as 
 b to a, or cp to —^ ; let cp be therefore taken from 
 
 the Numerator, and -~- from the Denominator ; agree- 
 able to the above Obferyation : By which Means the 
 
 FraQion itfelf will be transformed to rr— r t=: 
 
 aa-f-pp — acp 
 
 ___=__; fuppofingr=-.-^. 
 
 And, in the very fame manner the FraQion 
 
 ffi+(fiiiJlXZ+^.expreffingthethirdVaIueof 
 aaa -J- 2app -J- cpp ^ ° 
 
 — A, 16 changed to -^ — / . ^^^ z — =3 
 
 aaa -f- 2at?p -|- cpp — aapp 
 
 aah'\'{ac-\'bb)wp ., . ad\ 
 
 a^ZX%sr ^ (byputang.= c--). 
 
 But this laft Value is ftill capable of a further Re- 
 daftion : For, the Ratio of the Numerator to the De- 
 nominator being That of — ^ to — ^ -f- j/?% nearly f ai 
 
 appears from the preceding Cafe) let, therefore, the 
 former of tliefe Quantities be fubtracled from the Nume- 
 rator, and the latter from the Denominator : Whence 
 the Fra6lion itfelf becomes 
 
 r a rat 
 
The Resolution of Equations. 227 
 
 ==xrT-7r-r rr^^^ ; by putting wz=l — 
 
 r ^ cb — ad ^ 
 
 V a bb — ac) 
 
 Now, to exemplify the Ufe of the Theorems thus 
 transformed, Jet the Equation 96 + 768Z + 482*4- 2» 
 *=: o (given at p. 224) be here refumed : 
 
 Then, in this Cafe ;? being = — 96, a =1:768, ^ — 
 
 48i <^=i, <s^ = o, ^c, we have rf=: %\:=z ^^ s 
 
 a bJ 2,ji^ 
 
 bJ ' a ar ) %^' 
 
 Whence, according to the firft Approximation, 
 
 -A(=: . ^ \^ 48 , _J^ . 
 a-^rp) 768 — 4 191 
 -And, according to the Second, 
 — Ar=: ^^M^\f>^')y^P \ 768x484> (t, H)y-~96 
 aa-\-[b'\-aw)y^ p) 768 X 7684.(484.24) X -96 
 
 768 — * 
 
 ~ 768xi6 — 72x2 ^^"^ <^ividmg every Term by 48) 
 = 7^3 
 1 2 1 44 
 
 Henee the Value of z (= ) come* out zss — 
 
 ~ - A ^ 
 P 
 
 }9^ — r — o, 1259894, nearly: or equal to p^ 
 
 1516 ^ 96389 
 
 — o. 1259894802, more nearly. 
 
 Thefe Conclufions agree with Thofe before giren, by 
 the former Method. But the laft general Approximations, 
 containing the feweft Dimenfions of the Quantity p, will 
 commonly be found to have the Advantage, in point of 
 Expedition, when the Value of that Quantity confifts of 
 fcTcral Decimal Places, and alfo when the Coefficients 
 
 Q2 
 
228 The Resolution of Equations. 
 
 a, by c, d, of the Powers of the unknown Quantity 2, arc 
 related to each other according to fome known Law. 
 
 Of this Kind are the Coefficients of fuch Series's ai 
 arife in extrafting the Roots of pure Powers ; and in 
 thefe Cafes the general Theorems, or Equation them- 
 felves, are capable of being rendered ftlll more com- 
 modious. 
 
 Let there be afTumed the Equation x^'zzizk (which in- 
 cludes all the Cafes of pure, or fimple Powers, accord- 
 ing to the Value of the Exponent w). 
 
 Then, by affuming r nearly equal to a:, and making 
 rX(i+2)=^» wefhallhave f°X(i+'^)" = ^ J 
 
 and therefore — — -f" ( 1 -}- z)" = o ; that is 
 
 k , , , n n-^i , , n n — 1 «— 2 
 r"' ' '1^2 12^3 
 
 &c. =0 ; or laftly — P + z^ z' -J X 
 
 z^ -[-&€. =z=o ; by dividing the Whole by n, and 
 
 putting p:=r:, 
 
 nr" 
 
 Here (by a Comparifon with the general Equation 
 
 — p-\'az-\-hz'&c. =0) we have<2z=:i» ^3= ■■, czzn 
 
 n — 1 n — 2 , n — t n — 2w — 3 
 2^3 2 '^ 3 4 
 
 Theref. r ( r )=="- ;7^:=3— ^> 
 
 \a b J 2 3 6 ' 
 
 ad\ n-2 fn-i «-.S\ n-i «-fi 
 
 / ad\ n-2 fn-\ «-Q\ n-i « + i 
 /^ £\ n-^ 
 
 n—i n — 2 1 
 
 Whence, 
 
The RisoLUTioN of Equations. 229 
 
 Whence, for an Approximation of the third Degree, 
 
 But, for an Approximation of the fourth De<rree, 
 « A / ^^+ [c^rbw)y:p \ _ ixfa:;j_)-f-(y z-i)xr2«-i)xfi/> ! 
 V"~'^^+(/5'+^a;)X// 1+1. «^ 
 
 ind z 
 
 ■. + "-^xP+^^^?i^'x.. 
 
 Hence it is manifeft that the Root ^, of the pro- 
 pofed Equation x"" z=ik, is equal to 
 
 Or equal to, r^ ^^Xd+i"/') 
 
 .+^*x.V-i^^i^x/ 
 
 77Z<7r^ nearly. 
 
 But both thefc Exprcffions, in Cafes where ;? is a 
 proper Fraftion, will be better adapted to praftice by 
 
 71 /"" 1 1 
 
 making -r— „ = (vzzz-y and fubftituting - for p 
 its Equal: Whence (after proper Reduftion) 
 -y 2 — ' » nearly ; 
 
 »z^rd nearly. 
 
 To fhew now the Ufe and great Exaflnefs of thefc laft 
 Approximations, by an Example, let it be propofed to 
 extraft the fquare Rpot of 441. Here the general Equa- 
 tion Jf" =4, becoming .^' =: 441, we have n z= 2- 
 
 ^ = 441, and vlzzz , , ; fuppofmg r to be 
 
 affumed =: 20. 
 
 Q 3 Therefore 
 
sgo Thf Resolution of Equations. 
 
 Therefore by the firft Approximation, 
 
 , 41 1641 , 67281 ^^ , 
 
 nearly. 
 
 67280 
 
 And, by the Second, x z=i2o*\' —' 00. =: 21 — 
 
 rr-cT- * more nearly, 
 
 2758481 ^ 
 
 Again, let there be given the Equation x^ z=r 500 : 
 Then,a{ruming r = 8, wc have v( ^^^^ \z= ? ^^^* 
 = — 128. 
 
 Hence, by the firft Approximation, 
 
 16 ^^ — 768 — 10 3032 ' 
 
 93700527, nearly; 
 
 And, by the Second, ;t = 8 ^ ^ 253 
 
 128x251 + 1 
 8 — -' ll^ = 7- 9370052599/ "^ore nearly. 
 
 All the different Approximations hitherto delivered 
 were, originally, derived by multiplying the given F qua- 
 lion into a certain Number of Terms of t,.e aiiutn^d 
 Series 1 + A2r4" ^^' + ^^^ ^^"- But Lherc are other 
 lylethods (though, perhaps, nono fo general) by which 
 the fecond, third, &c. Dimcndons of the unknown 
 Quantity may, in like fort, be deftroyed (without aflfum- 
 ing any Series) and from thence the Value of thar Quantity 
 approximated, to whdt Degree of Exa6tners you pleafc. 
 
 Let there, for Inftance, be given the Equation, 
 
 2z -j-2^ = 1, or z'zzz I — 22 : 
 Then, byfquaring both Sides thereof, there arife* 
 
 And if inilead of the laft Term 42', its equal, 
 1 --22, be fubftituted, you will have «* = 5 — 12a : 
 
 Thii, 
 
The Resolution of Equations. 231 
 
 This, fquared, gives ^s' = 25 — laoz-j- 1442' z= 169 
 — 4082, by fubftituting for z% as before. 
 
 Here, rejefling z^ (on Account of its Smallnefs in 
 Comparifon of the other Terms) we have 4082 = 169, 
 
 and therefore z z=. — ^ = 0,414213, nearly ; which 
 
 is true to the laft Figure, inclufive. But, if you would 
 have the Anfwer flill nearer the Truth, let the above 
 Equation 2^ z=z i69 — 4082 be, either, multiply'd, 
 again, by itfelf, or into fome one of the preceding 
 Ones. Thus, if it be multiplied into 2* = 5 — 122;, 
 you will have 
 
 z" = 845 — 40682 4" 4^9^^' = 5741 — 138602 : 
 
 Where, 2" being reje£led, % is found = - ^L - '■ 
 
 = 0. 41421356, £^r. 
 
 Again, if there be given the Equation 2^ =: 32 — p ; 
 then, by fquaring both Sides thereof, we have 
 2^^=92' — Qpz-^p^: 
 
 And therefore zTz=z^^^-\- 6pz* — p^z =: — 6/7z'-|- 
 (pp -\- 27) X 2 — 9/7 J by writing ijz — gp inftead of 
 its Equal 92^ 
 
 Now, to the triple of this laft Equation, let the 2*. 
 Equation, multiply'd by 2/?, be added : 
 
 By which Means z* will be exterminated, and you 
 then will have 32' -|- 2/72^ =: (8i - ^pp) y^z- zjp-^'zp^. 
 Whence (rejefting 32'-[- 2pz^) the Value of 2 is found 
 
 J^7-^PP)XP ^ nearly. 
 
 {9—pp)x9 : , 
 
 Varioui other Expedients might be ufed, to extermi- 
 nate the 2**, 3^ 6?c. Powers of the unknown Quantity ; 
 but what is already delivered may fufficc. 
 
 PART 
 
PART V. 
 
 GIV tNG 
 
 Some Accomit of the Nature o/"Fluxions ; 
 together with the Invejligation of the fun- 
 damental Rules, 
 
 1. TN the Do6lrine of Fluxions all Kinds of Magnl- 
 JL tudes are confidered as generated by the con- 
 tinual Motion of fome of their Bounds or Ex- 
 tremes ; as a Line by the Motion of a Point ; a Surface 
 by the Motion of a Line ; and a Solid by the Motion 
 of a Surface. So likewife Time may be confider'd 
 as reprefented by a Line, increafing uniformly by the 
 Motion of a Point : And, as Ouantiiies of all Kinds 
 whatever are capable of Increafe and Decreafe, They 
 may be reprefented, in like manner, by Lines, Surfaces, 
 or Solids, conceived to be generated by Motion. 
 
 2. Every Quantity thus generated is caWd a Fluent^ 
 or Flowing Quantity : And the Magnitude by which any 
 Flowing Quantity would be uniformly increafed^ in a given 
 Time, with the generating Celerity at any propofed Po. 
 
 fition, or Injtant * (was it from thence to continue in- 
 variable) is the Fluxion of the /aid ^antity at that Po- 
 fition, or Injiant. 
 
 3. Thus, let a Point m be conceived to move from 
 A, and thereby generate a Right line km, with a 
 Motion any how regulated ; and fuppofe the Celerity 
 
 771 thereof, at any 
 
 ■^ lion R, to be 
 
 fuch, as would, (was it to continue invariable from that 
 
 Pofition) 
 
 * Some Authors define the generating Celerity itfelffand 
 not the Magnitude it would produce ) to be the Fluxion : 
 tut ufe that Magnitude as thejMeafure ofthefaid Cele^ 
 rity or Fluxion : Which is, in effe^l, coming to the fame 
 Thing. 6 
 
Somi Account of Fluxions. 23^ 
 
 Pofuion) be fufficient to defcribe, or paf$ uniformly over 
 the Diftance Rr, in the given Time allowed for the 
 Fluxion : Then will the faid Diftance Kr truly exprcfs 
 the required Fluxion of the Flowing Line Aw, in that 
 Pofition. 
 
 4. It appears from hence, that, when the generating 
 Motion is uniform, the Fluxion, and the Increment 
 aBually defcribed in the given Time, are one and the 
 fame thing: But, if the Velocity cominually in- 
 creafes, or decreafes, the Fluxion muft then be either 
 lefs, or greater than the faid Increment, or the Space 
 a&ually defcribed : Since an Increafe of the Velocity 
 muft necefTarily caufe an Increafe in the Diftance gone 
 over, and vice verfd, 
 
 5. It appears moreover, from the above Definition, 
 that Quantities, which flow, or increafe together, fo as 
 to continue, always^ in a conftant Ratio, have their 
 Fluxions, likewife, in the fame, conftant Ratio. 
 
 Although^ in forming a juji and JiftinB Conception of 
 the Nature^ and Quantity of a Fluxion^ the Conf deration 
 of Time is, abfoluteiy, neceffary (on which, even, our Ideas 
 of Velocity depend), yet in the Bufinefs and Application of 
 Fluxions, it 2S not always requi/ite,thatfome vulgar for 
 common J Meafure of Time (as a Second, Minute, Hour^ 
 &c.)fhould be propounded for the ProduBion of Fluxions 
 of the Quantities under Conjideration. A Line generated 
 by the uniform Motion of a Point, it is obferved above, 
 may be taken as a proper Reprefentative, or Meafure of 
 Time : And that Interval of Time (be it what it will) 
 wherein the Line,fo generated, is augmented by any 
 Length, or Fluxion, affigned, may be taken as the Time un- 
 derflood in the Definition, allowedfor the ProduBion of 
 the Fluxions of all other Quantities that have any Rela- 
 tion to, or Dependance upon, the faid uniformly generated 
 Line, And thefe Fluxions themf elves, by means of the 
 faid Relation and the given Length, or Fluxion, may be 
 alfo truly exhibited, independant of any particular, known 
 Meafure of Time ; as zuilli hereafter, be fully made to 
 appear. 
 
 To 
 
»34 Some Account ef Fluxions, with the 
 
 To illuftrate This by a particular Example, fuppofc 
 two Lines, hm and Cn, to be fogeneraicd, by the uni- 
 form Motion of two Points m and n, that the Latter of 
 them fhall be always equal to the Double of the former : 
 Then, taking R, S, and r, j, are cotemporary Pofitions 
 of the faid generating Points, CS will be the Double 
 of AR, and Cs the Double of Ar, by fuppofition ; 
 ivhencc Sj, the Fluxion of CS, muft of Confequencc 
 be the Double of Kr, the Fluxion of AR. 
 
 m 
 
 -V ! 
 
 r 
 
 s ^ 
 
 It is equally plain, on the other hand, that if the 
 Ratio of the Fluents Am, Cn is variable, that of the 
 Fluxions muft alfo vary.— -Thus, if, while the Point 
 m continues to move uniformly on, at the Rate of 
 one Inch (Foot, Yard, &c.) in a Second of Time, 
 the Motion of the other Point n be fo regulated that 
 the Number of Inches (Feet, Yards, &c.) in the 
 flowing Line Cn generated thereby, may be always equal 
 to the Square of the Number of Thofe in Am defcrib- 
 cd by the former Point m ; then, in this Cafe, it is 
 manifeft, that the Ratio of the Fluents Am, Cn is a 
 variable One ; and that the Ratio of the Fluxions varies 
 alfo ; fedng the Diftances i, 4, 9, 16, 25, &c. de- 
 fcribed in 1, a, 3. 4, 5, ^c* Seconds of Time, by 
 the Point n, increafe much fafter in proportion than 
 i» 2, 3, 4, 5, ^c, the correfponding Diftances gone 
 over by the other Point m, moving uniformly. 
 
 6. Hitherto Regard has been had to the Fluxions of 
 
 Lines; But the Fluxions of Superficies and Solids are 
 
 confidered in the fame manner, and ar« compre- 
 
 6 hended 
 
hivejligation of the principal Rules, 235 
 
 bended with equal Facility. Let a given Right-line mn 
 
 S s G 
 
 Br 
 
 /L 
 
 A^ 
 
 =m 
 
 F 
 
 be conceived to move parallel io itfelf, with an equable 
 Motion, from the Pofition AB, and thereby generate 
 the flowing Re^^angle AwwB ; let alfo the Diftance 
 Rr be taken (as above) to exprefs the Fluxion of the 
 Bafe A;72, and let the Reftangle Rr^S be completed : 
 Then, this Reftangle being the Space that is uniformly 
 defcribed by the generating Line mn^ in the Time that 
 Km is in like manner increafed by Rr, it will therefore 
 be the true Fluxion of the flowing Reftangle A«, by 
 
 the Definition. Art, 2, 
 7. The Generation, 
 angular, or curvilineal, 
 
 m^ 
 
 A 
 
 Pi 
 
 and the Fluxion of any tri- 
 Space ASR, are conceived in 
 like manner ; by fup- 
 pofing a Right-line tiiN 
 to be carry'd along, con- 
 tinually parallel to itfelf, 
 fo that the intercepted 
 Part Thereof mn (which 
 is itfelf a variable Quan- 
 tity) may pafs over, and 
 thereby generate the 
 Space ARS propound- 
 ed. And the Fluxion 
 
 r 
 
 of the Space thus generated, if Rr be taken as the 
 Fluxion of the Bafe (or AbfcifTa) Am, will be truly ex- 
 preifed by the Reftahgie (Rj) under Rr and RS ; as we 
 ihall have Occafion to (hew more at large hereafter. 
 
 8. From what has been thus far delivered, it will not 
 be difficult to form a juft Idea of the Fluxion oi a Solid : 
 But it is time we now come to fhew the manner of de- 
 termining the Fluxions of Algebraic Quantities ; by 
 means whereof all Others, of what Kind foever, are 
 explicable : In order to which it will be requifue, firfto! 
 all, to premife the following Obfervations. 
 
 1, That, 
 
t^6 Some Account of Fluxions, zviih the 
 
 1, Thatf the final Letters u, w^ a:, y^ z of th^ 
 Alphabet are ufually put Jar variable Q^uantities ; and the 
 initial Letters a, b, c, d, &c. Jar invariable Ones : 
 Thus, the variable Bafe hm of the flowing Reftanglc 
 AwzkB (in Art. 6.) may be reprefented by x, and the 
 invariable h\\\i\xdi^ mn, by a, 
 
 2, That^ the Fluxion oj a ^antity reprefented by a 
 fingle Letter is commonly expreffed by the fame Letter with 
 
 a Dot, or Full-point, over it : Thus the Fluxion of x 
 is denoted by x ; and the Fluxion of y by y. 
 
 3, That, the Fluxions of all Quantities (having any 
 Relation to each other) are always to be taken, as con- 
 temp or aneow^, or fuck as may be generated together, with 
 their refpcBive Celerities, in one and the fame Time, 
 
 PROPOSITION I. 
 
 ^. The Fluxion of any variable Algebraic Quantity 
 being given, 'tis pr op of ed to find the Fluxion ef the 
 Square of that Quantity, 
 
 Let two Points m and n be fuppofed to move, at the 
 fame Time, from two other, fix'd, Points A and C, 
 along the Right-lines AC and CD, in fuch a manner, 
 that the Meafure of the Diffance Cn, defcribed by the 
 Latter, may be, always, equal to the Square of the co- 
 temporary Diftance Am, defcribed by the former Point 
 »i,moving with an equable Celerity. 
 
 ' ' ^ "^ 
 
 C s e S B 
 
 I — — I — I — J ■« 
 
 n 
 
 Moreover, let R and S be any two contemporary 
 Pofitions of the faid Points ; and, fuppofing the de- 
 fcribed Diftances AR and CS to be denoted by x and^, 
 let the Lines jf and y be taken to reprefent the Spaces that 
 would be uniformly paffed over, in the fame given Time, 
 with the Celerities of the faid Points at R and S : So 
 
 fhall 
 
p 
 
 Invejiigaticn of the principal Rules, £37 
 
 fliall thofe Lines exprefs the Fluxions of the variable 
 Quantities Aw and C/2, when the generating Points wi 
 and n arrive at the forefaid cotemporary Pofitiong R and 
 S [by the Definition, Art. 2). 
 
 Furihermore, if r and s be confidered as any other 
 correfponding Places of the propofed Points, and the 
 Interval rR be denoted by v ; then, AR being z=. x, 
 andAr, z= x — v,we fhall have CS (rrzy) = ;t*, and 
 Cj ZZ3 (x — v)% by Hy-pothejis \ and confequentiy Sj ( — 
 CS — Cj) =ir ixv — vv. 
 
 From which it appears, that, while the former Point 
 m moves, uniformly, over the Diftance v, the other 
 Point n pafleth over a Space expreffed by 2xv — w. 
 
 But this laft Diftance, fmce the Velocity of the gene- 
 rating Point n increafes continually (See Art, 5.) is jefs 
 than the Space that would be uniformly defcribed, in the 
 fame Time, with the Velocity at S ; and greater than: 
 That which would be defcribed with the Velocity at s ; 
 and, therefore, is equal to, and may be taken to ex- 
 prefs, the Space which might be uniformly gone over 
 by the Celerity at fome intermediate Point e, between 
 / and S, in the fame Time. 
 
 Therefore, feeing the Diftance [jixv — vv) that might 
 be defcribed with the Celerity at the faid intermediate 
 Point f , is to the Diftance [v) defcribed by m, in the 
 fame Time, as 9.x — v to Unity, it is evident that the 
 faid (mean) Celerity at e, rnuft be to the Celerity of w, 
 in the fame Ratio ; and confequentiy, that, in the Time 
 the Point m would move over the given Diftance x, the 
 other Point «, with its Velocity at e, would defcribe 
 the Diftance 2x'x — vx : Since the Spaces defcribed in 
 equal Times, by uniform Motions, are known to be 
 ai the Velocities of the faid Motions. 
 
 This 
 
 The above Method of Invejiigation hath been reprefented 
 as bearing a near Affinity to the Methods of Prime and 
 Ultimate Ratios : Againf which fo many Objections have 
 beenjiartedy by the celebrated Author of the Anah/I^ as to 
 embarrafs andjiagger a great Number of P erf on s ; who 
 
238 Some Account of Fluxions, with the 
 
 This being determined, let r be now fuppofed to co- 
 incide with R, and s with S, by means of the Arrival 
 of the generating Points R and S ; then ^, being al- 
 ways between s and S, will likcwife coincide with S ; 
 
 and 
 
 art not well apprized how far the f aid ObjeBions are 
 
 jujiifahle, nor wherein their main Force conjijls. It is 
 not my De/ign to take Part in a Difpute^ on which enough 
 hath been already faid by Others : Though, that the Method 
 itjelfis perfeclly Scientific, I believe no One, that under - 
 
 /lands it, zuill deny; but whether the great Inventor has 
 therein exprejfedhimf elf with all the Caution and Accuracy 
 he was capable of, is another Quejiion. Had he call* d That 
 a hiMiTi't^G'KATio which he names an\}\i\mzit-Ont, the 
 ingenious Author above mention' d might not, perhaps, have 
 
 ybKWfl^/?^^;wyi?r his Ghofts of departed Quantities. How- 
 ever, be this as it will, thofe ObjeBions have nothing at all 
 to do with our Method of Invefigation given above. The 
 prejjing (and only) Difficulty, in the Bujinefs oj ultimate 
 Ratios, conjijls, it is known, in confidering the Values of 
 the Quantities compared, in their ultimate State, wherein 
 their Ratio is Juppojed to be taken : For, if we look upon 
 them as real magnitude, it is objeSed, that their ratio will 
 notjlridly agree with the ultimate ratio ajjigned : andif 
 on the other hand^ they be taken as mere nothings, we then 
 lofe the very idea of proportion. But our invejligation, as 
 is already objerved, is not embarrajfed with any Juchdif 
 
 Jiculty for, though the dijlance S s indeed, deer cafes con- 
 tinually, and even vanifhes when the generating point n 
 arrives at S ; yet the velocity of that point, which is the 
 quantity in que/Hon, neither vanfhes, nor afjumes a new 
 law ; butjlill continues to increaj'ein the fame manner as 
 before. — It may, pojjibly, be obje£fed, that, as the meafure 
 of the faid velocity is, originally, derived by means cf the 
 dijlance nS, we cannot retain a clear idea of it, when that 
 dijlance is vanijhed out of the equation. But with equal 
 reafon, it might be urged, that, we can have no jujl con- 
 ception of the dimenfons and true proportion of a build- 
 ing, after the Jcaffolding by means of whichit was raifed, 
 is taken away. 
 
InveJHgation of tht principal Rules. ^39 
 
 and the forefaid Diftancc 2xx — vx , that might be uni. 
 formly defcribed with the Velocity at e (now at S) will 
 become, barely, 2xx ; ^fi\\ich (by Art. ?.) is equal to (j^) 
 the Fluxion of Cn or x\ From whence it appears, 
 that the Fluxion of the Square of any variable, or 
 flowing Quantity, is found by multiplying twice the Root, 
 or Quantity itfelf, into its Fluxion. 
 
 The /ami otherwife. 
 
 10. Let the Diftancc Ar be denoted by u^ and let 
 other Things remain as before: Then, CS being = 
 
 H- 
 
 OC 
 
 A 
 
 ^ 
 
 -*-' 
 
 R 
 
 7n 
 
 B 
 
 ^ 
 
 
 iv 
 
 XX, and Cji;= uu, by Suppqfitien^ the Diftance jS, de- 
 fcribed in the fame Time with rR (z= a: — u) will there- 
 fore be truly expreffed by^jv — uu, or its Equal (x-\-u) 
 X [x" — w)* Which Diftance, as the Velocity of n con- 
 tinually increafes, rauft be greater than That which 
 would be uniformly defcribed, with the Celerity at J, 
 in the fame Time. Whence it is evident, that the 
 velocity at s is to the, uniform, velocity of m, in a lefs 
 ratio than that of {x-\-u) X (x — «) to ^ — «, or of [x-^-u) 
 to unityy or laflly, of (^ -j- ^} X -^ ^o •* (becaufe the 
 velocities of uniform motions are as the fpaces def- 
 cribed in equal times,) Therefore, fmce the uni- 
 form velocity of m (meafured by the Diftance moved 
 over in a given time) is defined by i, it is plain that 
 the meafure of the velocity of the other point at 
 s (or the Diftance that might be uniformly defcribed 
 in the fame given time) will be lefs than (^ + «) X •*• 
 Which quantity being, itfelf, lefs than 2x y^ x (becaufe 
 u is lefs than x) the celerity at s muft confequently be 
 lefs than ^xx^ take the point s where you will, on this 
 fide of S. . 
 
 Let 
 
f40 Some Account of Fluxions, with the 
 
 Let now r' and / be any other cotemporary pofitioni 
 of the two points, on the other fide of R and S, and let 
 hr be denoted by «; : fo fhall tjie diftance S/, def- 
 cribed in the fame time with R/' fzu — xj, be truly ex- 
 prefTed by ww — xx, or its equal {w-\'x) X (^ — x) (by 
 hypothtjis,) Which diftance, as the velocity of the def- 
 cribing point n continually increafes, muft, evidently, 
 be lefs than that which would uniformly arife from the 
 celerity at /, in the fame time. Whence it is alfo 
 plain that the faid velocity at /, is to the uniform 
 velocity of m^ in a greater ratio than that of (z^ 4"*"^) X 
 iw — X) to {w — x), or q{{w-\-x) X ^ to ^. But thcquan- 
 tity [w-\-x ) X ^ is, itfelf, greater than 2xy^ x, becaufew 
 is greater than x ; and fo the meafure of the faid ve- 
 locity at J"' muft confequently be greater than 2xx.- 
 
 Therefore, fmce the velocity increafes continually, 
 from C to D ; and feeing the meafure thereof, be- 
 fore the arrival of the generating point at S, is every 
 where lefs, and afterwards every where greater, than 
 %xx ; it is manifefl, that, at S, it can be neither leffcr 
 nor greater, but muft have, or pafs through, the very 
 value, or degree, expreffed by <ixx. ' Q. E. I. 
 
 If the line Am (x) be fuppofed to be generated with 
 an accelerated, or a retarded motion, inftead of an uni- 
 form one, it will readily appear, from the firft of the fore- 
 going methods, that the required fluxion of a:', fuppofmg* 
 to denote the meafureof the velocity at R, vii\\,Jiill, be ex- 
 pounded by 2XX. 
 
 For the Spaces rKfvJ and jS (2xv — vv) defcribed 
 in the fame time, being to each other, in the Ratio of 
 X to txx—vx, the mean celerities of the generating 
 motions, at certain intermediate points between the 
 extreme ones r, R, and j, S, muft be likewife, in that 
 ratio : which ratio, when, v becomes = o, and the points 
 coincide, will become that of ^ to 2xx. 
 
 PROPOSITION II. 
 
 11. The Fluxions X and y^ of two Jlowing Quantities, x and 
 y, being given; it is propofed to determine the Fluxion 
 ojthe ReBangUy or ProduB, xy, oj the faid Quantities, 
 
 Let z be-, always, equal to the fum of the two propofed 
 
 quantitiei 
 
Invejiigation of the principal Rules, 2j^i 
 
 quantities ;v and ^; then, the fluxions of equal quantities 
 being alfo equal, 2; muft ber—i-f-/* Moreover, fince 
 Z \s z=zx "^y, we fhall have zz z=:zxx -{- ixy-^yy ; and 
 therefore xyz=.\zz^—\xx — \ yy. But the Fluxion of 
 \zz — \xx — \yy (and confequently that of its equal xy) 
 appears, from the preceding propofition, to \iQ.zz-xx-yy\ 
 which, by writing A" -^-y, and i-j-J^, in the room of their 
 equals 2; and z, will become [(^-j-y) X (>^-f"j') — ^^ — y)'\ 
 •zzz yx -\. xy, the required fluxion of xy. Hence it is ap- 
 parent that the fluxion of the produ6l, or reftangle, of any 
 two flowing quantities, is exprefTed by the fum of the pro- 
 dufts arifing from the multiplication of each quantity into 
 the fluxion of the other. 
 
 12. From the fluxion of a reftangle, above determined, 
 the fluxion of a fraftion, ^, is very eafily deduced. 
 
 ^ y 
 
 For, by putting xz=i^ (the propofed fraftion) and mul- 
 tiplying both fides of the equation byjy we have xy z=zz; 
 and therefore xy ^ yx z=: z, as above, (the fluxions of 
 equal quantities being, neceflarily, equal). From this 
 equation, by tranfpofing xy, and dividing the whole by y, 
 
 ss xy , *. , . . z . . 
 
 we get ;c r= ^ *. this, by writing »— in the room 
 
 z zy y^ — zy 
 of its equal x, becomes x ==. — . ^—z=z — : which 
 
 value is therefore the true fluxion of x, or, its equal, 
 — , the fraftion propofed. 
 
 y 
 
 13. Moreover, from the fluxion of a re£langle, the 
 fluxion of the continual produft of three, four, five, or 
 any other number, of flowing quantities, may be deter- 
 mined. 
 
 Thus, let the fluxion oiyzu, where thenumber of faftors 
 is 3, be firfl; required : then, by putting xz=.zu, our given 
 expreflion will be reduced ioyx\ and its fluxion will be 
 yx -J- xy {by Prop. 2.) But, x being = zv, and there- 
 fore X 'z=zzv'{''i^^ (h the fame )^ if thefe values be fub- 
 
 R ftituted 
 
«42 Some Account »J Fluxions, with the 
 
 ftituted in yx -^-xy^ it will become y X [zV'\-vz) -J-s^jfzr: 
 yzv -^yi-v '\-yzv, the true fluxion of yzUy required. 
 
 Again, it the fluxion of y%vwt where the number of 
 faftors is four, was to be demanded ; then, by making x 
 z=.zvw, the quantity propofed will be reduced io yx : and 
 its fluxion will therefore beexprefledbyjy;c-4""^j'J which, 
 becaufe x\%:=izvw, ^ndixzzzvw -^zvw-\'zvw (as ap- 
 pears from abovdj will be likewife exprefled by^ X f*^^'' 
 -^zvu) J^zvw) -\-zvwy, or yzvw -\- y%vw y%vw-\-yzvw. 
 
 In the fame manner the fluxion o\ yzvws^ will appear 
 to be yzvw's-\- yzvzus -■\- yzvws -^yzvzus -\-yzvws \ and 
 fo of others. 
 
 14. From the fluxions thus determined, the fluxion of 
 any power of a variable quantity (whofe exponent is a 
 whole pofitive number) is very readily obtained ; nothing 
 more being herein required, tlian to confider all the faftors 
 as equal among themfelves. Thus, the fluxion o{ yvw 
 being found yvio -^ yvw -^yuWy it is plain, if both v and 
 w be fuppofed equal to^, that the fluent, or quantity pro- 
 pounded will become^^, and its fluxion yyy-^yyy-^yyy 
 
 Thus, alfo, becaufe the fluxion of yzvw is yzvw-^yzvzff 
 '\-yzvw -j- yzvWf it appears that the fluxion of ^^ will be 
 truly exprefled by 4y^y' And, from the given fluxion of 
 yzvws, that, of^* will in like manner appear to be sy*y* 
 From whence the law of continuation is manifefl: ;'tne 
 fluxion of y being univcrfally expounded by py^ — ^jy, 
 
 15., This lafl: general conclufion, which isof very great 
 importance in the bufmefs of fluxions, being the refult of 
 feveral dedu6lions, whereby its truth and evidence may, 
 perhaps, lofe a part of their force, a more direft inveftiga- 
 tion of the Jame may not here be amifs ; though the former 
 method will, doubtlefs, appear the mofl: eafy and proper for 
 beginners, to whom, the manner of working by general 
 indices, is not plain and familiar. 
 
 Conceive two points m and n to move, at the fame time, 
 from two other points A and C, along the right-lines AB 
 and CD ; and let every thing be fuppofed as in Prop. 1 ; 
 only, let the meafure of the dittance defcribed by the point 
 n be, always, equal to the p power, (inftead of the fquare) 
 
 of 
 
InvejUgation of the principal Rule's. 243 
 
 •f that defcribed by the other point m^ moving uniformly. 
 
 V — 1 I ~ 1 
 
 A r fi r B 
 
 • « ^ — ^ » 
 
 y D 
 
 Then, fmce by hypothefis, the value of CS is here zzr 
 A'P, and that of C.y — u^, (fee the fecondfolution to the fore, 
 f aid prop.) it is evident that the diftance jS, defcribed in 
 the fame time with rR [z=zx — mJ, will be truly defined 
 hyx^ — w^ or its equal [x — v) X (^""'4"^" — ^u-\-x^ — ^m* 
 
 Whence, fuppofing x to denote the meafure of the uni- 
 form velocity of m, it will appear, by reafoning as in the 
 faid propofition, that the meafure of the velocity of w, at 
 any place j-, on this fide of S, mufl be lefs than k x [x^ — ' 
 
 -\-x^ ^u-\-x^ ^u^ \-u^ ^), and confequently 
 
 lefs than x x px^ ^ ; feeing each of the (p) terms of the 
 faid feries (the firft only excepted) is lefs than x^ ^ u being 
 lefs than;>r. 
 
 Again, by confidering r' and / as two other, contempo- 
 rary pofitions, beyond R and S, and reafoning in the fame 
 manner, the meafure of the velocity at s' will appear to be 
 greater, now, than the abovefaid quantity ;tX/^-*'' ^• 
 
 Therefore, as the velocity increafes continually, and 
 feeing the value thereof, before the point arrives at S,is 
 every where lefs; and afterwards, every where greater, 
 than xy^px^ ^ [or px^'^^x) it is evident, that, acS,itmufl 
 be neither leffer nor greater, but cxaftly equal to px^ — ^x\ 
 which quantity is therefore the true fluxion of x^iand 
 agrees exaftly with tha*: determined above. 
 
 16. In bringing out the foregoing fluxion of x*y the 
 value of p, in both methods, was taken as a whole number ; 
 nevcrthelefs the conclufion itfelf holds equally true, when 
 
 p is a fraftion, as in the notation of roots. To make this 
 
 3 
 appear, let the quantity ;v* ( = y' x^) be propounded, in 
 
 order 
 
 R s 
 
»44 '^^''^^ Account of Fluxions, with the 
 
 order to determine the fluxion thereof : which, by writing 
 
 I for/7, in the general fluxion px^—'x, comes out Ix ^ x. 
 
 Now to prove that this is the true fluxion, put^zziA: |, 
 the quantity given; and then, by Iquaring both fides, you 
 will have jy'=Af^ ; which, in fluxions, gives izfy — -^x^x (as 
 has been already fliewn.) This, by fubftitutmg for jy, be- 
 comes 2xly='^x'-x , Whence, by divifion, y^rf x^ x, 
 the very fama as before. 
 
 But, to demonftrate the fame thing in a general manner, 
 m - ^^ 
 
 fuppofe the fluxion of ^ " to be required [m and n being 
 
 m 
 any whole numbers, whatever) put y — .y " ; and then by 
 raifing both fides to the power n^ you will have y" — .y° ; 
 
 which, in fluxions, gives ny'' — ^y^=z?nx x ; and con- 
 
 m r 
 
 fequemly.;. =-X— =T-=-X ^--^=~X 
 
 ^ x_ 7n^^ yx X 
 
 y 
 
 m 
 
 ^^ — * m m 
 
 = 71 X ;«'' • 
 
 ^m X, 
 
 Which value, of the fluxion of ^ 7 is evidently the very 
 fame, as that arifingby expounding/? in the general fluxion 
 
 fpx^ — *A;^by — ; which was to beproved. 
 
 17. Now, from what has been thus far delivered, the 
 following pra61ical rules, for determining the fluxions of 
 algebraic quantities, are obtained. 
 
 RULE I. 
 
 To find the Fluxion of any given Power of a flowing 
 Quantity. 
 
 Multiply the fluxion of the root by the exponent of the 
 given power ^ and theprodud into that power of the fame 
 root which arifes by fubtraHitig unity from the given 
 txponent. 
 
 The 
 
Invejligation of the principal Rules, 245 
 
 The reafon of this rule is feen above ; the rule itfelf 
 being nothing more than j?xx^~^~^ (the fluxion of x^) ex- 
 prefled in words, 
 
 RULE ir. 
 
 To find the Fluxion of the Produft of feveral variable 
 Quantities, multiplied together. 
 
 Multiply the Jluxion of each^ by the vrodu£f of the reji 
 of the quantities ; Jo fhall the Sum of all the produ&s 
 thus arifng be the true fluxion required. 
 
 The reafon of which is, likewife, evident from what has 
 been already delivered. See Art. 13. 
 
 RULE IIL 
 
 To find the Fluxion of a Fraftion, arifing from the Divi- 
 fion of one variable Quantity by another. 
 
 Fro7n the fluxion of the numerator, drawn into the 
 denominator, fubtraSl tKe fluxion of the denominator, 
 drawn into the numerator ; and divide the remainder by 
 the fqu are of the denominator, 
 
 y% — %y . z 
 
 This appears from i, the fluxion of—, detcr- 
 
 yy y 
 
 mined in Art. 12. 
 
 Though I might here, with propriety enough, put an end 
 to this part, as my protefled defign therein extends no far- 
 ther than givmg the young beginner fome account of the 
 nature, and firit principles, of fluxions, together with the 
 invefligation of the fundamental rules exhibited above ; 
 neverthelefs, as different ways of bringing out the fame 
 truths have often a very good effe61, and feeing fluxions of 
 all quantities whatever(wheiher powers, fractions, £]j?c.Jare 
 deducible from the fluxion of a reftangle, Ifliall, therefore, 
 fubjoin a different method, whereby the faid fluxion of a 
 reftangle f given by Prop, 2.) may be invelligated : in 
 order to which it will be requifite, firft of all, to premife 
 the following " 
 
 LEMMA. 
 
446 Some Account of Fluxions, with ihz 
 
 LEMMA. 
 
 18. Tht Fluxion of a curvilineal Space hl^.^, generated 
 bytheOrdinate RS for the intercepted Part of a Right- 
 tine RT (moving par alUl to ^tfelf is equal to the Kec- 
 tangle (Rs) under the /aid Ordinate, and the Fluxion 
 (Rr) of the Abfciffa KK, 
 
 For, let a Right-line 7««, of the fame length with RSin 
 the propofed pofition PQ, be conceived to move from 
 thence, parallel to itfelf, jr 
 with the fame celerity 
 that the generating line 
 itfelf has in that pofition : 
 by which means the rec- 
 tangle PriQ will be uni- 
 formly generated, with 
 the very celerity by 
 which it begins to be ge- 
 nerated, or, by which the 
 fpace ARS is encreafed 
 in the propofed pofition 
 
 PQ; fmce both the A R 77Z 7^ 
 
 length, and velocity of mn, are the fame as thofe of RS in 
 the faid pofition. Hence the reftangle fo generated muft 
 confequently, be the true fluxion of the fpace ARS, by the 
 definition. 
 
 But the fame thing may be otherwife made to appear, 
 froiti a different, and more logical method of arguing ; by 
 proving that the required fluxion can neither be greater, 
 nor lels, than the faid reftangle. 
 
 Thus if the line mn (while it moves uniformly on to- 
 wards rs) be fuppofed to increafein length, the area P^wwQ, 
 
 generated 
 
Invejiigation of the principal Rules. 247 
 
 generated thereby, will evidently be greater than that 
 
 which would uniformly 
 arife in the fame time, 
 with the given length at 
 the firlt pofition PQ; 
 fince the new parts, pro- 
 duced each fucceeding 
 moment (as the genera- 
 ting line continues to 
 lengthen) aregreaterand 
 greater. 
 
 And, in the fame 
 
 manner, if the line mn, 
 
 A R generating the fluxion, 
 
 be fuppofed to decreafe, in length, from the given pofition 
 PQ, it is equally plain that the generated fpace FmnQ will 
 be iefs than the cotemporary fpace that would, uniformly, 
 arife with the given length at the firft pofition PQ. 
 
 Therefore, feeing the fluxion (or the fpace that would 
 uniformly arife from the generating celerity at the propo- 
 fed pofition) is Iefs than any fpace that can bedefcribed, in 
 the given time, when the line «?« increafes, and greater than 
 any fpace that can be defcribed, when the faid line de^ 
 ereafes; it muft confequently be equal to that fpace which 
 will arife, when the length of the faid line, from the given 
 pofition, is fuppofed neither to increafe nor decreafe ; that 
 is. when the generated fpace PmnQ is a, re£langle, as in 
 the preceding figure. 
 
 PROPOSITION III. 
 
 19. To deiermi?ie the Fluxion of the FroduEl or ReBanqU, 
 ^ of two variable (Quantities (x andy,) 
 
 Lat two right-lines DE and FG, perpendicular to each 
 
 other 
 
248 
 
 Some Account of Fluxions, &c. 
 
 to 
 contin- 
 
 move from two other 
 
 lines 
 
 other, be conceived 
 BA and BC, 
 ually parallel to them- ^^ 
 felves, and thereby ge- f 
 nerate the variable rec- ' 
 tangle DF : let the path ^'^ 
 of their interfeftion, or 
 the place of th' angle H, 
 be the line BHR, divi- 
 ding the generated rec- 
 tangle DF in two parts 
 BDH and BHF: more- 
 over let Tid fxj and F/ 
 
 fy) be the fluxions of ^ -X' J ' ^i: cl C 
 
 the fides BD fxJ and BF fyj ; and fuppofe dm and/« to 
 be drawn parallel ,and equal, to DH and FH, refpeftively. 
 Then, fmce, by the preceding lemmas ihe fluxion of the 
 fpace or area BDH, is truly exprefled by the reftangle Vim 
 (=A:_y), and that of the fpace or area BFH, by the rec- 
 tangle Fw (rrryjf), it follows, becaufe equal quantities have 
 equal fluxions, that the fluxion of the propofed reftangle 
 ^jvfzziBDH -^ BFH) is truly exprefled by xy -\-yy, the 
 very expreflTion before determined. See Art, 1 1 . 
 
 It may, perhaps, be expefted, that I fliould now give 
 fome inftances of the ufe and application of the theorv 
 hifherto explained, in the refolution of problems : but 
 having infifted very largely on this head in my Do&rine 
 and Application oj Fluxions, I fliall take the liberty to re- 
 commend that work, to the perufal of fuch as are defirous 
 of farther information in the matter, Thofe, for whofe 
 ufe the above account is in a more particular manner de- 
 figned, may have the opportunity of being inftrufted in the 
 praftice, by proper examples, without the trouble q[ turn- 
 ing to other books. 
 
 F I N I S. 
 
 friiitcd by T. C. HANSARD, t'eterborough-court, rl««l Street, LONIXNT. 
 
 V 
 

 mummMm 
 
 3K^*?a7ii^ 
 
 ^^" ER ^TY OF r\LIT^OFN 
 
 i 
 
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