The fame anfwered by Mr. Allan Murdoch. B D HE mit the GH on AC, and the thing is done. With the given fide AB, and given AD, conftruct the tight angled triangle ABD, from A draw AC perpendicular to AB, meeting BD produced in C, from D demit the DF on AC, and on D as a center with the radius DE DF, defcribe the arc Ee, cutting AB produced in E; join ED; 11 to which from B draw BG, cutting AD in G, from which point de For by fim. triangles DE GB¦¦DA ¦ GA¦¦ DF: GH, but DE DF by conftruction; therefore GB=GH. Other anfwers were-given by all thoje Gentlemen who answered the preceding queflion. 3rd Question answered by Mr. Jofeph Edwards, Jun. Hoxton. Let ABCDE be the path of the waggoner, A the point at which he begins, and E where he ends his journey, according to the queftion; GL the tract in which the team travels. Put 4 1760 yards, the man's rate of hourr; that of the team v; and t to the time of performing his journey. travelling per L The time of travelling round, the team and horfes being the fame, at whatever point in the path the journey may commence, let it be fuppofed to begin at O, when the neareft diftance of the team from O is 2. yards, and end at F; the points (O, F) being in the right line or path GL, call the waggoner P; and at the inftant P fets out in the path OBCEF, fuppofe another tra- BOTA veller Q to commence a journey from in the straight path OG, and with the fame celerity; and when P arrives at K, 2 yards diftant from the end of the waggon, fuppofe Qat G, then GK will be equal BOCK 4 yards; now let Q return in the direction GL, and overtake the given point F, 2 yards before the team, which point will also be overtaken by P in the fame time, because GF = KDEF. G D This being premised, put GF 28 yards a, then because Q approaches the given point G with the relative vel. rv, and afterwards the point Lwith the relative vel. rv, we have a + I a =t whence v Vr Xr−2: =√1239400 = 3520 yards, or 2 miles t per hour the required rate of travelling. The fame answered by Mr. Coultherd. Let x be the number of miles the team travelled per hour, and C and v the points from which they began to EM D move; then when the man had gone 2 yards, or was at B, 8+2x yards before him: and, 104+2x the diftance between 4-X B and M, to which add CB2 yards, the fum will be the number of yards from C to M. 112 4- -X And by a fimilar C B procefs, the distance from M to 0= Other Anfwers were given by Meffrs. Ariftarchus, Adams, Allison, Armstrong, Baron, Bearcroft, Crawford, Dean, Dent, Gale, Hewitt, Harvy, Leach, Langford, Moody, Robinson, Read, Surtees, Thoubren, Tale, and Young. Question 4 anfwered by Mr. T. Coultherd. It is well known, that two chords, having the fame tenfion, and whofe Tengths are to one another as 2 to 3, when made to vibrate, will produce a perfect fifth. And as the folidity of fimilar figures, and confequently their weights, are in the triplicate ratio of their homologous fides; there fore as 313 23 10000: 2962'96 the weight required. Other Anfwers were also given by Meffrs. Ariftarchus, Andrews, Bell, Barns, Gale, Griffith, Hewitt, Robinfon, and Surtees.. Queflion 5 anfwered by Mr.. T. Coultherd.. Let the given line AB-a, tang. A =p, and x=AC; then as rad. atang. ZA DC=px, and CB2+CD2 BD2 D = p2x2+a2. 2 ax + x2; alfo CB X BD2 XCD3 p3 x3 Xa -x X p2x2 + a2 B C Other anfwers were alfo given by Mefrs. Ariftarchus, Adams, Allison, Armstrong, Baron, Crawford, Crowie, Dent, Dean, Gale, Grhih, Marvy, Hill, Kirby, Langford, Life, Murdock, Orpheus, Oitter, Pearfon, Robinson, Reed, Surtees, Simfon, Thoubren, Tompkins, Young, and the Propojer. Question 6 anfwered by Mr. T. Coultherd. Let the furface of the water np, by the immerfion of the cone, be raised to m o, and put rs=x; then, as E FGH: GvE F—rs:qr=30 — 3 x, and the folidity of the fruftrum E Frs=1000 G H m x 3 x 1.2889 inches, the height required. Other anfwers were alfo given by Ariftarchus, Allifon, Adams, Baron, Crawford, Dent, Furby, Gale, Griffith, Hewitt, Langford, Life, Moody, Orpheus, Robinfon, Reed, Surtees, Thoubren, Yates, and Young. Question 7 answered by Mr. Surtees, Minor, Alftone, Cum berland. Let x2, nx2, n2x2, be the three numbers; then per question n2 + n +1X x2 =, or n2+n+1=; put ne- -1 its fide, then `n2 e2 — 2 ǹ e +1=n2+n+1, hence n= 2e+1 , and the three taken any number at pleasure, and e greater than one. The fame anfwered by James Gale, Efq. Edgeware Road, London. Suppofe the numbers to be x2, x, and 1; then x2 + x + 1 muft be a fquare number, the root of which put x-n: then x2+x+1= x2-2 nxn2, or x + 1 2n xn2, and by tranfpofitión x + n2 1 If n be taken 2,, then x will be The fame answered by Mr. James Thoubren, Student in Mr. Rutherford's School, Lancaster. Let 1, x, and x2 reprefent the required numbers, then 1+x+x2 muft be a square number, the fide of the fquare of which put 2 2m+1 2 x, then m2 ·2 mx+x2 or 1+x=m2. 2m x, whence x= m2 2 m 71' 2m+1 8 64 7' 49 ly, and by taking m3, the numbers will be 1, 2 The fame anfwered by Mr. T. Coultherd. refpective Let x2 be the firft number, n their ratio, and each of the other numbers will be represented by nx2, and n2x2; then per queft.; n2 + n + 1 Xx2, or n2 + n + 1, must be a square number, affume n + a for its -- root, and n2+n+1 = n2. 2 a na2 and n= aT Xx2, and 2a+ a may be taken any numbers at pleasure, provided a be greater than one. Other anfwers were also given by Meffrs. Ariftarchus, Adams, Baron, Dent, Edwards, Griffith, Hewitt, Johnfon, Langford, Moody, Orpheus, Robinson, Reed, Simpson, and the Propofer. 8th Question answered by Mr. J. Edwards, Jun. Hoxton. Let BD be the given befecting line, upon which defcribe a fegment of a circle, to contain the given angle; and on BD conftitute the rectangle BDLK = to the given area; join B & C, the point of interfection of the circular feg. with KL; and in CB produced, take AB BC; join DA. and DC, and ADC is the A required. For AB BC by confir. and by the 38 and 41, Euc. 1, the A ADC DBKL = given area by conftruction. When the circle cuts KL in two points, MC, as in the prefent cafe, the queftion admits of two folutions; and when only touches it, the area is a maximum, and when it neither cuts nor touches it, the prob. is impoffible. The fame anfwered by Mr. John Surtees, Minor, Alftone, Cum berland. Conftr. Let M be equal to the fide of a fquare, expreffing the given area of the triangle. On CD, the given bifecting line, defcribe the fegment of a circle capable of containing the given angle; and at C erect Cr DC, and equal to a third proportional to. DC and M; throughr || to DC, draw dB Arc, cutting the circle in A; join AC and AD, and in AD produced, take DB = AD; join CB, and ABC will be the triangle re- B quired. For the BCD = ▲ DCA (Euc. 6. 1.) = ABCD + DCADCX Cr=M2 equal to and lines; and the angle CAB and DC are bifecting line by conftruction. D DC X Cr, whence the the given area by conftr. to the given angle and Other anfwers were alfo given by Mers. Ariftarchus, Adams, Baron, Coultherd, Dent, Gale, Griffith, Hewitt, Johnfon, Murdock, Orpheus, Robinson, Thoubren, and the Propofer. 9th Question anfwered by Mr. Jofeph Edwards, Jun. the Propofer. PQ2, draw BC, making the KBC the given /Q; take KA KB, and from the point C, where BC cuts the indefinite FO, draw. CA, and ABC will be the triangle required. |