III. REBUS, by Mr. Geo. Stevenfon, Master of the Boarding-School, Well Boldon, Durham. Suppofe Aurora's offspring bright, (Which in verdant meads you'll often find) Ladies no doubt but in your mind, What then before your eyes you fee. IV. REBUS, by Mr. William Boyer, of Leyland, mar Preston. To the beauteous dame for whom in days of yore, So many Trojans fell in ftreaming gore; Two-thirds of it, and then you've at command V. REBUS, by Mr. S. Oxley. To two-thirds of a bird of which poets have fung, See the Charades in the Supplement. I. QUERY, by W. H. Hall, Efq. Barrifier at Law. Why is found more diftinct at night than in the day-time, though our faculties and the doctrine of founds do not change? II, QUERY, by Mr. James Williams, of Colyton School. ** Whence or how arifes the found commonly called the finging of the tea-kettle? III. QUERY, by Mr. Henry Lee, of Bingham. Would it contribute to the happinefs of the married ftate, were divorces more eafily to be obtained? IV. QUERY, by Mr. J. Jackfon, of Hutton-Rudby School. It is no less remarkable than true, that in the splitting of marble blocks, freeftone, and even flinty rocks, cavities have been found, containing one or more live toads in them, which on being exposed to the air, have foon after died. Can any account be given, how thefe have been generated, or how fuftained, in fuch a fituation? V. QUERY, V. QUERY, by Mr. W. Bearcroft, of Nawton. What is the reason that commonly the air is colder about the time of fun-rife and fun-fet, than it is either before or after? VI. QUERY, by Mr. William Clark, of Wiftow, near Selby, Yorkshire. What fyftem of philofophy gives us the moft convincing and demonftrative proof of the immortality of man? The Editor congratulates bis numerous readers on baving this year, at the frequent folicitation of many of bis learned contributors, made room for a much greater number than ufual of their very ingenious compofitions, which it bas always bitherto been with beart-felt reluctance that he was obliged to fupprefs for want of room. This defirable end be bas attained by the publitation of a Supplement to the Ladies' Diary this year, price 6 pence, (to be bad jeparately of the fame perfors who fell the Diary itself), containing an alpbabetical lift of all the Enigmas that have been printed fince the commencement of the Diary in the year 1704, with a great variety of the original compofitions of our correspondents. And this Supplement being intended as an experiment this year, it is hoped that all true friends of the Diary will promote the fale and knowledge of it as much as in their pozver, that the Editor may judge bosv far it is agreeable to bis readers to continue fo great an improvement of the Diary in future gears. There will be eight prizes, to be determined by lot as ufual, viz. two of 8 Diaries and Supplements for the Solution of the Prize Enigma, two of 8 Di, and Sup. for the general Solution of the Enigmas; two of 6 Di. and Sup. for the Solution of the Rebufes, Queries, &c. alfo one of 10 and one of 8 Diaries for the Solution of the Prize Question.—All our correspondents Letters must be fent before the ft of May, and the separate Solutions of the Prize Enigma and Prize Question before Candlemas Day, and all franked or poft paid, or they will not be meceived, many having being rejected last year on that account. They are also ftill requested to make their compofitions as brief as rb⋅y can; and muft obferve to fend Solutions with every thing new that they propofe. In answer to feveral who enquire for the early Diaries, it is obferved that the republication of the whole till the year 1773 inclusive, in 5 vols. by Dr. Hutton, may be bad at Robinson's or Baldwin's, in Paternoster Row, viz. be Poetry in 2 vols. and the Mathematics in 3 vols. also bis Mathematical Mifcellany in 1 vol. ANSWERS ANSWERS to the MATHEMATICAL QUESTIONS, I QUESTION 863 answered by Mr Tho Woolston, ROM the fecond equation take twice the firft, and we have 13 FROM the fecond equatioes the wit equation take the ad, and we have y7-2: fubftitute thefe values of y and z in the 3d eq. and we have x-x8: whence we find x3; and confeq. y 1, and 1216; therefore the word is CAP. Let this ornament, ye fair, The fame by Mr Geo Roope, of Tring Academy. By taking the ift equation, and the double of it, from the ad, wa..... get y=332x, and x = -13; there fubftituted in the 3d equ. give 158x-6x2992. Hence ≈ 16, 3, and y = 1; and the ornament is a CAP. The fame by Mr Ja Scholefield, Schoolmaster at Brumley From the 2d equation take double the 1ft, and xx+13; take this from the firft, and y7-2x; place thefe values of x and y in the 3d equation, which then becomes 6x2. -2x48; hence x3, y=1, z=16, and the answer a CAP, an ornament becoming all modeft women. Various other ingenious answers were given by Mers Jas Adams, Al lenfis, Amicus, Rd Ball, Rob Barwick, Mrs Eliz Baufor, Wm Bearcroft, Gen Befwick, John Birib, John Boden, R Bretberick, John Brownrigg, Wm P Burman, John Burrow, WC, John Cansfield, Jobn Cavill, Geo Clayton, S Clement, Tho Cock, John Cooper, Jos Caving, Rob Crofier, John Cullyer, Jas Cunliffe, John Dalton, Rd Dening, G Dixon, Edw Emes, Rev L Evans, Evoc Egroeg Semaj, M Fleck, Tho Gell, Griffith, Henry Holme, Jonathan Hornby, J Hunt, Jno JackJon, Wm King. Tho Mafon, Jas Metcalfe, Wm Mudge, Jos Nendick, Geo Robarts, Chr Robinson, lex Rowe, Ifaac Saul, Joe Sherwin, Tim Simpson, no Smith, Ges Stevenfon, no Surtees, Hen Taylor, Fno Howard, Matt Terry, Mifs Janetta Todd, Jno Unwin, Walton, Rd Waugh, Wm White, Abel Whiteboufe, Jas Williams, Jas Wood, B Worship, Jos Youle, and Jas Young. II QUESTION 864 answered by Mr R Bretherick, of Those who would fee this problem refolved in a general manner for all polygons, may, I prefume, have their curiofity abundantly fatisfed by perusing the Scholium at pa. 8 of Dr Hutton's elegant Treatife on Menfuration. But for the fake of those who are not in poffeffion of that book, I have copied one of the methods there laid down, C Suppofe A B C to be one of the triangles which conftitute any regular polygon: Then, "as radius I tang. L CAP AP PCX AP = t, fuppofing AB=1; then t(APX PC)= the AACB, and nt the polygon; where n is the number of fides. So that, by finding the tangent of the CAP, by the table of tangents, and multiplying it by the number of fides, of the product will be the A multiplier required. Hence we obtain 9365640 the multiplier for the undecagon, and 11.196152 for the duodecagon. P The fame by Mr John Dalton, Teacher of the Mathematics, Kendal. B 1 In the example for finding the multiplier for a dodecagon, Hawney feems to have fallen into feveral mistakes. After finding the perpendicular on one of the fides, he multiplies it by 5 or, half the base or fide of the decagon, which would give the area of one triangle, or of the whole; but by misplacing the decimal point, he in effect multiplies the said area by 10, and concludes he has found the whole area of the dodecagon, when he has only 12 or 5 of it. In like manner he has mistaken of the area of the endecagon for the whole area. In one or other of thefe ways is the answer also given by Mesrs Jas Adams, Allenfis, Amicus, no Afpland, Rd Ball, Rob Barwick, Mrs Elix Baufor, Wm Bearcroft, no Birch, no Boden, Jno Brownriggs Fno Burrow, WC, no Cansfield, no Cavil, Tho Cock, John Cooper, Fos Cowing, Fno Cullyer, Jas Cunliffe, Rd Dening, G Dixon, Rev L Evans, M Fleck, Griffith, Henry Holme, Jonathan Hornby, Fno Howard, no Jackson, Wm King, no Lowry, Jas Metcalfe, Wm Mudge, Geo Robarts, Chr Robinson, Gee Roope, Alex Rowe, Isaac Saul, JasScholefield, Tim Simplon, Fno Smith, Geo Stevenson, no Surtees, Henry, Taylor, Matt Terry, Mifs Janetta Todd, Walton, Rd Waugh, Wm White, Jas Williams, The Woolfton, Jos Youle, and Jas Young. III QUESTION 865 anfwered by Mr M Fleck. Since the triangle is right-angled, the hypothenule 2b must be the diameter of the circumfcribing circle, which is given; then if 2 s be the fum, and 2d the difference of the legs', the radius of the infcribed circle will bes-b. Then fince the given diftance g is the hypothenuse, and d and s- b the legs of another right-angled triangle, we have 3b2-g2 g2 = d2+s— · b2 = 3 b2 — 2 sb; hence s= and d 2 b 2b2s2. Which being now known, then s+d the greater teg, and sd the lefs. The fame anfwered by Mr S Clement, Schoolmaster, Arundel. 2. and H E 442; therefore 2 ay+b2 a2, and y = 2ab2 becomes known, A G Mr Fno Burrow, of Bolton Field, conftructs the Prob. thus: With the given radius A D defcribe the femicircle A CB; and with the fame center and given diftance of the centers draw the other femicircle GEF; draw A EH to touch GEF, and make the angle HAC HAB; laftly join BC, and ABC B is the triangle required; as is evident. G D Ingenious folutions were also given by Mers Adams, Amicus, Afpland, Allenfis, Ball, Birch, Boden, Bretherick, Cansfield, Cavill, Clayton, Cock, Cowing, Cullyer, Cunliffe, Dalton, Evans, Griffith, Hornby, Howard, Fackfon, King, Mudge, Robarts, Roope, Roqve, Saul, Schilefield, Smith, Stevenson, Surtees, Taylor, Terry, T Todd, Waugh, White, Williams, Woolfton, Youle, and Young. IV QUESTION 866 answered by Amicus. The lever A E D B and weight w will reft in any pofition, provided their common center of gravity be fupported; and becaufe E it can then only be fupported by the prop, when the lower fide A B of the lever is ho- I rizontal; if r be the prop, and PC perp. in F w K B. to A B meet the axis FH of the lever in c, draw w A E parallel to CP cutting the axis in 1, and C L parallel to A B L ; then I may be confidered as the place of the weight, c the common center of gravity of the weight and fruftum, and o the center of gravity of the fruftum. By the queft. AE = 6 = 3, BD = 6, AP LC=3, FHAK = 30 = a, BK = 2, and AK: BK :: 1:05 2n = 1, or 2 B Kna = 3, AB 3609, AK AB CL: CI=3003747e, IL 21F CL, FI==075, and FC3078747g. Now in order that c may be the common center of gravity of the weight and fruftum, it is manifeftly neceffary that ew FO- - gx folidity or weight of the fruftumro-gx {n2 a3 + b2a+ba2 = b2 a2 + 7a3b+ { n2 a4 — } n2 a3 g —b2ag -nba2g b2 a × 17a — 28g, or w=3174495 cubic inches of oak 91 8538 lb, Averdupois. 12 20 |