ANSWERS to the last Arithmetical QUESTIONS. QUESTION the ift, answered by W. W. b= 2000. First Man's Share Put b Then C= d = 86 362. Third Merchant's a2. e3. Second's Shareb-a-g3. d+e=21+e. = e2 + 42e+441. Ergo In the firft Step for a substitute its Value. €3 ·e=c+e2 +420 + 441. - 8 F 91 e3 + e2 + 43eb-c441 (=1197.) And by converging Series we fhall find e 9. hence e3 729. The third Merchant's Gain ; a2 = 900. The firft Merchant's, and the Second's Gain 371. Crowns. QUESTION the 2d, answered by Step for y and e we fubftitute their refpective Values a bove found: Then a=g. That is QUESTION the 3d, answered by W. W. PUT e and 응+ { b = 10. C= 20. = +c. Or ae=acze ca2-264 By the first Step we have zeba2 + 2ba. 2e ca2 20a; Hence ba2 + 2ba ⇒ ca2· -2ca; ba + 2b = ca — 26. ca - ba= 2c + 2b; a= 2c+26 c-b 6. The Number of Men, and e=240. The Number of Pounds. QUESTION the 4th, anfwer'd by I. N. LEt d=143; 4=AC (in Fig. 3.) a+d=BC; Then is AB=√2a2+2da+d2 by the 47. 1. Euc. but the Triangles A CB; ADC; BDC; are all fimilar, Hence Vza2+2da+da:a: aa √2a2+2da+d2 2a2+2da+d2: a+d:: a+d: à2+2da+d2 = AD. And BD ́za2+zda+d2 : √ za2+åda+d2 plying Extreams and Means together we have 3 a a =2a24da+2d2. Or a2-4da=2d. and by Extraction of the Root a√6d2 +2d=636.27 Yards; AC. And BC= 779.277; AB = 1006.04; The greater Segment BD 603.62 S Yards. The Triangles Area = 247918.02 Square Yards, equal to 51 Acres, and 35.63 Perches. The Geometrical Construction of this Question may be thus effected. In Fig. 3. make BH 3. and HG=2.; and upon BG as a Diameter, defcribe the Semi-Circle GFB. Upon H erect the Perpendicular HF and draw GF, and = and 3F and produce it towards C, alfo produce BG towards A: Then make FI FG, and draw GI; again fet off the Given Difference of the Legs from B to E and draw EA parallel to IG until it interfect BG in A: From A draw AC parallel to GF, until it interfect BF continued as in C; Then will ABC be the Triangle required. From hence it is eafy to deduce a Trigonometrical Solution. QUESTION the 5th, anfwered by R. G. SUPPOSE the Thing done (as in Fig. 4) and the Sides of the Triangle ABC being given, its Area will be 裙 found 2498.3, and the Angle BAC 78: 52 therefore the Area of the Triangle AHG must be 1498.98; and EF being drawn parallel to BA, ED will be found 9.98 which makes EA 19.98; Put this a. and FD = 50; AG — x ; c = 1498 98 then is HO= and by Gimilar Triangles, as a + :p × Hence px= 20 x . px2-2cx=2ca, andx=√2ca+ c2 76.01 AG; and AH=40.26; HG= 78.4 QUESTION the 6th, anfwer'd by I. N. Raw the Lines (as in Fig. 5.) and put AC the Hypothenufe;s=DE Dp the Side of the infcribed Square; a BS the Perpendicular let fall from B; e Cw the Difference of the Segments of the Bafe. Then Bta-s; and by fimilar Triangles, Asa-s:sa:h, hence ba bssa ; and a= bs. =Gs; and As; Now because the Triangle ABC is 2 right-angled we have ASX CS BS. that is bb-ee aa 4 reduced and the Root extracted we have e= This alfo may be folv'd by a Quadratic Equation when the Sides of the infcribed Square are parallel to the Legs of the primary Triangle. Thus, (In Fig. 6.) Produce the Side of the Square BC to G, and from E erect EG perpendicular to EF, meeting BG in G; and let fall EH perpendicular to BG. Then call AD AB, a ; FE,b; CG, x; CE,y; And BG is a+x; FC = EG=by For the Triangles CBF, CEG, CHE, EHG are fimilar, alfo the Triangles CBF, EHG are equal Becaufe BC=EH. Then by fimilar Triangles, as a(CB) : b—y (CF :: y (CE) ; * (CG). Ergo ax=by—y2. But GC2= GE + CE2. that is x2-b2 2by2y^- ̧ Or 2 ax. therefore b2x2 = 24x; and x = yax. √b2+a2 = a; HA QUESTION the 7th, anfwered by R. F. Aving fuppofed the Problem refolved draw CD paral lel to PM, and put a AB; b= AD; c = DC; d = DB; x = A P; 2 = z = PE; y PM; PF; PB4-x; Then the fimilar Triangles APE, ADC. And BDC, BPF. will give thefe Proportions. As AP (x) PE (2) :: AD (b): DC (c). Ergo bz=cx. As BD (d) DC (c) BP (ax): PF (u). and duacex. But by the Conditions of the Problem we have PE (≈): PM ( y ) :: PM (y): PF (u). hence zuy Now CX by the first Proportion = and by the Second u = b ; In the third Proportion for z and a substitute an Ellipfis, of which AB is the Tranfverfe Diameter, and and Oits Centre: For by the common Property of the Ellipfis ADX DB: DC2: AU: KO, confequently AP × PB: PM: AD × DB: DC. that is in Algebraic Terms ax-x2:y? :: bd: c. and by multiplying Extreams and Means we have' ac2x-c2x2 bd fame Equation as that before found. =yy. The very 2. E. D. QUESTION the 8th, answered by I. T. FIRST, For the Geometrical Conftruction fet off (in Fig 8.) the Latitude and Azimuth as ufual, then becaufe the Sum of the Declination and Altitude is 59. Degrees the Sum of their Complements is 121 Degrees; Lay of the faid 21:00 from Z tox, and draw the Meridian PxS, and whatthrough the Centre and the Point x draw fm, ever fx measures upon the half Tangents, lay of the fame upon the Tangents from the Centre to m Then fetting one Foot of the Compaffes in m, extend the other to E, and where it cuts the Azimuth Circle ZxN as at O will be the Point through which the Meridian POS muft pafs, which will compleat the Triangle. TRIGONOMETRICAL CALCULATION. In the Triangle NS are given NS37: 00, and 1 d Nx 59:00, and the Angle SNx=43: 0. To find Sx. and the Angle SxN. Then in the Triangle POx are given Px to the Complement of Sx to 180 Degrees, and the Angle P XOLS N. Now becaufe Zx is-Complements of the Altitude and Declination, and PO+ZO= alfo to the Complements of the Altitude and Declination, Take away Zo common to both, and there remains x PO; and confequently the Angle xPOLPxO. But the two Sides being equal, a Perpendicular let fall from upon Px as Ov will bifect the Side Px; Hence Pu and the Angle at P being given, PO may be readily found, which is the Complement of the Declination, and from thence all the rest, D ALGE |