5. Qu. (35) answered by Mr. Bruster. In printing this question an error has crept into the enunciation by substituting the word digits for numbers. After rectifying this error the solution may be as follows: Let a' be the given square, and x the less number; then, a' -x will be the greater. But, by hypothesis, a2 — 2x = √/a2 = a, and 2+ a; in each of these equations = + a a2 a2 a2 a 2 a' 2x a2-2x = the less number, and a2±a is the greater. Hence, if the given square be 1, the required numbers are 1 and 3; to the given square 4, the numbers are 3 and 6, &c. By assuming data to render the question possible, answers to this question were transmitted by Messrs. Baines, Dunn, Ford, Hine, Hirst, Maffett, Nesbit, Rylando, Tomlinson, and Winward. 6. Qu. (36) answered by Lucinda and Rylando. Let x, y, and %, denote the numbers required, and let a' their sum; then, by the question x+y 2z by assumption, also x + z − y n2, and y + z − x = a2 = a2 %= 2y 2x m2; hence, x = a2 a2 = m2 + n2 + v2, which assume = (m+s)2 = m2 + n2+v2—s2 2ms+s2; then 2ms =n2 + v2— s2 and m— 2s Take n = 4, v = 6, and s = 2; then m = 12, and a2 — 196 = 142, x = 26, y = 90, and z = 80. Otherwise by Mr. J. Hine and Mr. Maffett, Plymouth. Let x, y, and z, be the three numbers required, then, by the question, x + y + z = a2, x + y − z =c2, x + z -y= d2, and y + z xe'; now by adding the three last equations, we get + y + z = c2 + d2 + c', which must be a square, because its equal a2 is a square; to make c2+d+e2 a square, assume it equal (6 + d)2 = c2 + 2cd + d2, from whence c or d may be where e and d may be any num found, viz. c = 2d bers at pleasure, so that e' be but divisible by 2d. Take e2 64, and d = 2, then c 16, consequently a2 324 is a square, and by subtracting the three last equations severally from the first, we obtain ≈ = d2 and x = a, and the three 2 2 2 numbers are 34, 130, and 160, which answer the conditions of the question. Again by Messrs. Ford, Putsey, and Winward. Put x, y, and z, for the numbers required, and by the question we have x+y-z = a2, x + z—y = b2, and y + z xc2; the sum of these three equations gives x + y + z = a2 + b2 + c2, which is also to be a square, by the question; assume it = (a + n)2 = a2 + 2ann2, and we find a = (b2 + c2 n2) 2n, where b, c, and n may be any numbers at pleasure. Let n = 2, b 8, and c = 10, then a 40, and if from the fourth equation we subtract the three first severally, we ob tain x= b2 + c2 2 a2 + b2 2 =832, y = a2 + c2 2 850, and z = 82, which are three numbers that will an swer, and other answers may be found by taking other even numbers for n, b, and c. It was answered also by Anglicus, Mr. Baines, Mr. Gawthrop, Mr. Hirst, Mr. Nesbit, and Mr. Tomlinson. 7. Qu. (37) answered by Mr. Nesbit, Farnley. Let x2, 23x2, and 49r2 denote the numbers sought, then the sum of their square roots is 13x, which is to be both a square and a cube by the question; put 6 p 13' 13rp, and we have x = where if p= 13, we 3446462296225, find x 13'137858491849, 25x2 and 49x26755066100601, which answer the condi̟tions required. It was answered in exactly the same manner by Messrs. Anglicus, Baines, Ford, Gawthrop, Hine, Hirst, Lucinda, Maffett, Putsey, Rylando, Tomlinson, and Winward. 8. Qu. (38) answered by Rylando, and Mr. A. Hirst. Let the roots of the required squares be denoted by -x and x; then by the question (4x)2 + x = x2 +x+, and x2 x+, are squares, and we have only to make ( — x ) 13 — x2 — — x a square; assume it n'; then ≈ = ————12, and — — x = = 8 n may be taken at pleasure, provided it be less than unity. Ifn, then x = n, then x = and 4 and ; when x; and so on. Otherwise by Mr. Winward. Put rand-x+x for the numbers required, which are obviously squares; then-x+x2+ x = x2+ x + 76 = (x + 4)2 also x2 + (4— x) = x2 4); it only remains then to make 7/8 − 1 x + x2x2- r, a square. Put as for its root, then (4 — ax)2 = 17 −x+= (x a- 1/6 1 1 x = ; where a may be any whole number greater than 2. Take a = 4, then x2 = 1824 and o - 1 x + x2 = 2, two numbers which answer the conditions of the question. And according to one or the other of these methods it was answered by Messrs. Anglicus, Baines, Brooke, Gawthrop, Hine, Maffett, Nesbit, Patsey, and Tomlinson. 9. Qu. (39). We have not received two solutions to this question in which the resulting numbers are exactly the same; -the difference arises from our correspondents not all making use of the same table of the sun's declination, and also from some of them allowing for refraction, &c. and others not. P Solution by Mr. A. Hirst, of Marsden. 3/2 109 Ꮓ H In the oblique angled spherical triangle PZS we have PZ=36°45′ the co-lat. ZS= 65° 32′ the co-alt. (allowing 2' for refraction) and PS S 83°40′ the co-declination, to find the P 56° 46′56′′; hence the time of observation was 3h. 47m. 8s. P. M. Again, his azimuth or the PZS 114° 0′ 12′′; therefore the declination of the wall is 24° 0′ 12′′ easterly. Draw PH perp. to SZ produced; then in the right angled spherical trian. we have ZP, PZH, and the right angle, to find HP, the height of the style 33°7′ 58′′; and H Z the distance of the substyle from the meridian = 16° 53′ 49′′, and the angle HP Z the plane's difference of longitude 29° 3′ 47′′. For the distance of the hour lines; rad. : sin. PH= 33° 7′ 58′′:: tang. HP 11: tang. H 117° 47′ 47′′ = the distance of the hour line eleven from the substyle. In the same manner all the rest may be found, and have been actually calculated by all who answered it; we, however, deem it unnecessary to insert them, as they would take up a great deal of room, and are on the whole useless. Very nearly as above, the question was answered by Mr. Baines, I. B. of Lincoln, Mr. Gawthrop, Mr. Hine, Mr. Maffett, Mr. Nesbit, Rylando, Mr. Tomlinson, and Mr. Winward. D E 10. Qu. (40) answered by Mr. Whitley, Rotheram. Let ADB be the given semi-circle, A the extremity of the arc from whence the motion commences, D and C any two contemporary positions of GF C B the bodies, DC the line connecting them, and E their common centre of gravity, which, since the bodies are equal, will be in the middle of DC, or DE EC. Draw DG, EF perp. to AB, and put the rad. AO= r, arc AD=z, A F = x, and GD = y; then, by the question, the arc ADB: AB :: arc Á D≈ z : A C :: 1.57081 :: 1:11.5708 = m (suppose) then AC mz, hence FCFGmz — -x, and AG = 2x− mz=AO-G O — r — √ √ ( r2 —y); therefore x={ • m2 + { r - {(y), and, taking the fluxions, we žyj ; but from the nature of get x = 1 mx + ( area AFE rmy j + y2 j 1% mr2 + 8 4 √(r2 — y3) -), the correct fluent of which is +~) √/ (r2 −y2) which when y = r becomes 1.5708 (1.57 8 + 1.5708) 4 x 1.5708 r2 x .3555, the double of which is r2 x .711 in this case 1777 square yards the area of the space AEBA, which taken from 3927, the area of the semi-circle ADB, leaves 2149 yards for the area of AEBDA. Note. The length of the curve is found in page 251, No. 13, Math. Companion: and scarcely different from the above were the solutions sent by Messrs. Hine, Maffett, Rylando, Tomlinson, and Winward. A very neat solution was also sent by Mr. Gawthrop, which we are sorry we have not room for. Other answers were sent by Messrs. Anglicus, Brooke, and Nesbit B 11. Qu. (41) Answered by Mr. Nesbit, Farnley. In the annexed figure let T denote the tree, A, B, and C, A the points in the circumference of the field defined by the extremities of the shadows, and let TN be the meridian line. By well known rules in spherics we find the sun's azimuth from the north at the time of the first observation = D C |