At the 119° 52′ 15′′, and his alt. then 37° 58′ 51′′. time of the second observation we find his azimuth from the north 168° 20′ 54′′, and alt. 51° 20′ 52′′. The azimuth from the north at the third observation = 73° 5′ 19′′, and alt. 6° 3′ 56′′; from whence we find TA 25.6165, TB = 15.9956, and TC = 188.2156 yards. Now in the triangle AT B, we have given AT, TB, and the ▲ ATB, 71° 46′ 51′′, to find the ABT 71° 49′ 42′′; again, in the triangle TBC we have TB, TC, and the ▲ BTC 95° 15′ 35′′, to find the TBC= 79° 56′ 24"; also the ▲ ABT+ TBC= = 4 ABC = 151° 46′ 6′′. Lastly, in the triangle ATC we have AT, TC and the ▲ ATC= ATBBTC, to find AC = 213.2617. Draw the diameter AD, and join CD; then 3. the angle CAD ▲ ABC ; by simp. geom. 16. - 90° = 61° 46′ 6′′ : hence the diameter A D is found 451 yards nearly, and the content of the field 33 acres. Otherwise by Mr. J. Gawthrop, Leeds. In this question we have given the time from noon, the latitude and the sun's declin. to find his alt. and azimuths at each of those times : 1. Hour 45° 0′, declin. 14° 50′ 45′′, lat. 53° 5', to find.......... 2. Hour 7° 30′, declin. 14° 53 22", lat. 53° 5′, to find.......................... alt. 37° 58′ 51′′ azim. 60° 7′ 45′′. alt. 51° 20′ 52′′ Šazim. 11° 39′ 6′′. 3. Hour 100° 0', declin. 14° alt. 6° 3′ 56′′ 58' 00', lat. 53° 5', to find.......... azim. 106° 54′ 41′′. Now by adding the refraction and sun's semi-diameter 15′ 55′′, we have the apparent alt. of the sun's upper limb as follows: Let T be the place of the tree; then in the triangle ATB we have AT 25.3562, BT 15.8376, and the T71° 46′ 51′′ to find the ▲ TAB= 36° 24′ 2′′; and in the triangle BTC we have BT 13.8376, TC 176.6343, and the T 95° 15′ 35′′ to find BC = 178.6179; also in the triangle ATC, we have AT = 25.3562, TC = 176.6343, and the ▲ T = 167° 2′ 26′′, to find TAC 11° 20′ 32", then TAB TAC= BAC 25° 3′ 30′′. Draw the diam. A D, and join DC; then, DCA is a right 4, 31. Eu. 3, and the CDA 25° 3' 30"; also AC = 178.6179, whence AD = 421.726 yards, and the area is found 28a. 3r. 17p. In this solution the variation of the sun's declination, also refraction, &c. is taken notice of, which makes the result different from the preceding one. C H E P 12. Qu. (42) Answered by Mr. J. Whitley. The figure being described as per question, divide AC in G, so that AG may have to AB the given ratio of PE to DB. Then G will be a given point, and GB G being drawn, meeting ED in F, will be given by position. F AD B By sim. triangles, and the nature of the curve BP H, we have AG :AB:: DF:DB::PE:DB; therefore PE DF; consequently PD FE; and the area BPHA generated by PD, will be equal to the area BGCB generated by FE = the quadrant ACB -A AGB. Again, the ordinate PD will be a maximum when its equal FE is so; which is evidently_the case when a tangent to the ellipse at the point E is parallel to GB; therefore, since GB is given in position, the point E may be found by Prob. 12. Book II. Simson's Conics, and from thence FE = PD, the greatest ordinate, will become known. Ergo Solutum. The same by Rylando. Put A B t, A C = c, B D = x, DP = y, and DB :PE::m: n. By the property of the ellipse, DE = ✔✅(2tx — x2), and, by the question, m:n::x : PE= o and by reduction r =7+ greatest ordinate. tx xx From the equation of the curve we have xy = nxx √(2tx — x2)— for the fluxion of the area, the m fluent of which, or general expression for the area of the curve, is x cir. seg. rad. and versed sine с To find the area geometrically; since the ratio of DB: PE is given, the ratio of DB: DP is given also ; therefore, take DB: DP :: AB: AF, and draw BFG, then the area of the curve will be equal to the triangle A BG. Again, by Messrs. Hine, Maffett, and Winward. Put A Ca, AB = b, m: n the given ratio, DB =2, and DP=y; then m: na by the ellipse, a : b :: √(2ax — x2) : bx 2) nx EP = b a and, ·√(2 ax—x2) the fluxion of the area, or xy = √√ (2ax— x2) a the correct fluent of which, when ab, is where A is the quadrant of a circle whose rad. is a. b Also the fluxion of✔ (2ax — x3) — =0, by reduction we determine the greatest ordinate as above. Mr. Gawthrop's solution to this question is very neat, and purely geometrical. Mr. Nesbit's name was omitted among the answers to the 11 Qu. No. IV. and Mr. Putsey's to Qu. 10. We wish to observe to our correspondents in general, that as we have only a few days in which to collect and arrange all the materials for the Enquirer, it is next to impossible but that some trifling errors and omissions must occur; we are as careful, and take as much pains as we can to avoid both, but we are human, and it will be sufficient, we hope, to observe that the errors and omissions are few, and the latter were never intentionally made. From the 20th to the end of the month is all the time we are allowed. Rylando is requested to send for any book he may think proper, the price of which does not exceed half a guinea. NEW MATHEMATICAL QUESTIONS TO BE ANSWERED IN NO. VII. 1. Qu. (56) By Mr. Dunn, Broughton. FIND three square numbers in arithmetical progression whose common difference is a cube number. 2. Qu. (57) By Mr. J. Baines, Horbury Bridge. What number is that whose square root, cube root, and biquadrate root are in arithmetical progression ? 3. Qu. (58) By Mr. Putsey, Pickering Academy. The radius of a circle inscribed in a right angled plane triangle is 2 feet ;--it is required to determine the sides of the triangle, such, that a heavy body may occupy one second of time in descending down the hypothenuse, when the plane of the triangle is perpendicular to the horizon? 4. Qu. (59) By Mr. R. Maffett, Plymouth. Find two square numbers such that their sum, the root of the greater number added to the less, and the root of the less number added to the greater, may be all squares. 5 Qu. (60) By the same. A person applied to a surveyor to measure him a field in the form of a right angled triangle; but the base and perpendicular being surrounded with water, he could only obtain the length of the hypothenuse = 100 chains, and that of a path-way bisecting the angle at the base, and terminating in the perpendicular 80 chains now Mr. Cross-staff contends that from these data the area cannot be had, and he has bet the gentleman the price of a new field-book, that none of the Theodolite gentry who write to the Enquirer, can from thence find the content. Pray, gentlemen, show him that, for once, he is wrong. 6. Qu. (61) By Mr. J. Baines, Horbury Bridge. On what day of the year, in the lat. of 53° 45′ N. will the shadow of the point of a gnomon describe an hyperbola, whose transverse diam. is to the conjugate as 9 to 11 ? 7. Qu. (62) By Rylando. To describe a parallelogram, which shall have its area and perimeter respectively equal to the area and perimeter of a given triangle. 8. Qu. (63) By Mr. Putsey. At the moment a heavy body was let fall perpendicularly from the top of an inclined plane, another body was projected directly up it from the bottom, with a velocity equal to that which the same body would have acquired by descending freely down the length of the plane;-to determine their nearest approach; the length of the plane being 50, and its height 40 feet? 9. Qu. (64) By Mr. Jesse Winward, 2nd R. L. Militia. What are of a circle is that, the product of whose sine into its versed-sine is the greatest possible? 10. Qu. (65) By Mr. Roger Vagheeg. PORISM. Having a circle given in magnitude and position, another circle may be found, given also in magnitude and position, so that if through a given point P, any line be drawn cutting the circles in Ă, D, and B, the solid under A P, P D, P B, may be always of the same given magnitude. |