For let D E be the sine, and CE the persed sine of the arc DC, and since ab = 2 bc, by sim. triangles, AB= 2BC, but by the property of tangents to the circle, BC=BD, therefore, BDSDA, and by Simp. on Max. et Min. the rectangle DEX CE is a maxi. mum,** Inference. The arc DC=120°. For drawing the radius DO, since AB = 2BC, ABC is evidently half an equilateral triangle whose side is AB;, there. fore, the żB = 60°, the supplement of which is the angle DOC = arc DC = 120° : Otherwise by Messrs. Baines, Brooke, Cattrall, Cummins, Gawthorp, Hirst, Nesbit, Putsey, Rylando, and Webster. Let radius=1, and put the versed sine = r; then, the sine =^(2x x), and xv (2 x 2) is to be a max.; make its fluxion = 0, and by reduction we hare *=* = versed sine of 120°, the arc required. 10. Qu. (65) Answered by Mr. Váyheeg, the Proposer. Analysis. Suppose A the thing done, and 2 that PAČis the circle required; through P and 0, the centre of the given circle, draw the right line CPEOF; then since AP, PD, PB is given, AP, PE, PF is given also. For PD, PB=PE, PF, and because the point P and the circle EDBF are given in position, PE, PF is given; therefore, A Pis equal to a given line, but one extremity P of it is given, consequently, the other extremity A must be in the circumference of a circle whose radius is A P. Consequently AP, PE, PP is equal to AP, PD, PB, the given solid. : : Again, by Mr. Whitley. The point Pand the circle EDBF being given, the rectangle PD, PB is given; and since the solid AP, JA BLAD PD, PB is given, À P is given; therefore, a circle described about the centre P, with the given radius AP, will be that required. !! * Otherwise by Rylando. in .. Let P be the given n ' . point, and QBD the X given circle'; draw the IP tangent PQ, and produce it to S, so that QP. PS may be equal, to the given solid; then with radius SP-PT=PA, and centre P, describe the circle STA, and the thing is done. For the given • solid is QP. PS = QP2 x PT = PA. PD. PB, by a known property of the circle. Ś In the same elegant manner it was answered also by Messrs. Brooke, Butterworth, Gawthorp, Hine, Maffett, Nesbit, Putsey, Tomlinson, and Winward. 11. Qu. (66) Answered by Messrs. Eyres and Rylando. Let DR, CS, be the radii of : PA the given circles, and P a point in the required locus; from P draw the tangents PD, PC, and join PR, PS, RS; then, DK it is evident that when PD? + DR2 is constant, PD? + DR: ☺ (PR) + PC + CS? (PS), ;'. that is P R + PS’ is constant; and because R and S are constant the locus of Pwill be a circle. Cor. 1. If the difference DP CP: be constant, the locus of P will be a right line perp. 10 RS. : Cor. 2. By Mr. Whitley. When R and S coincide, RS vanishes, so that, in this case, the locus of P is a given circle, concentric with the given circles; hence question the 18th (72) is only a particular case of the above, This question was answered also by Messrs. Brooke, Butterworth, Gawthorp, Hine, Maffett, Nesbit, Putsey, Vayheeg, Whitley, and Winward.:: i s 261251. -12. Qu. (67) Answered by the Proposer. * Because AE. DC is, by the question, to be a max. and EDP, DAP, are right angles, AE = A D . AP; therefore. A D » APX DC. must be a max. or AD2 x DC a B max, which is known to be the case when AD=2DC. Hence we have the following .. · Construction Divide Ač in D, such, that AD: 2 DC, join PD, and draw D E perp. to PD, intersecting BA produced in E, join EC, and the triangle EDC will be a max. as required. Otherwise by Mr. Rylando, Let A BC be the given triangle, P the given point, in the side A B, and draw the rest of the lines as in the figure above. Put AE=X, PA =m, and AC=; then, AD= mx, the area of the triangle CAE= pr, and the area of the triangle DAE=1XNmx; therefore, by the question 1px-1C7mx is a max. and its fluxion = 0, whence x is found = ÅD=p, and the area of the triangle = le-2p. To construct the 27 m problem; make AD={AC, join PD, and bisect it in m, draw mn perp. to PD, meeting PE in. n, with centre n, and radius Pn, describe a circle, which will cut PA produced in E, the point required. .. Answers were also given by Messrs. Gawthorp, Hine, Maffett, Nesbit, Putsey, Whitley, and Winward. 13. Qu. (68) Answered by Mr. Nesbit, Farnley Academy. · Put the area, length and weight of one beam respectively A, L, and W, that of the other a, l, and w, and let S, s, G,g, be as in the question; then, by art. 225, SX A XGSXaxg. but Marrat's Mechanics, TX W T xw when the lengths are equal or L'=l, the weights are as the areas, or W:w:: A :a, therefore, the KAR DEPARTMENT. above equation, under these circumstances, becomes sxG=Sxg, that is, S;s::G:g. 2. E. D. And in nearly the same manner it was answered by Messrs. Gawthorp, Harvey, Rylando, and Whitley, 14. Qu. (69) Answered by Mr. G. Harvey, jun. Plymouth. Analysis. Suppose the thing H done, and that ABC is the tri. N angle required; because NC, the diff. of the seg, of the base, and NCF, the diff. of the angles at the base are given, HF, the diam. of the circumscribing cirele is given; and because LK-MC is given, ML drawn parallel to HF is given in position, and the centre of the inscribed circle is in this line, by the question ; but the centre of the in. scribed circle is known also to be in the line CF, it is therefore at S, their point of intersection. Now, 'by prop. 35. Mod. Geom. Student. CS X SF =HF X SL and CS, SF, HF are all given; therefore LS is given; hence, if a circle be described from centre S with radius SL thus found, and CA, CB, be drawn from C, touching it, and cutting the circle ACB in A and B, and A, B, be joined, ABC will be the triangle required; as is evident from the analysis. ,, The same, by Messrs. Rylando and Whitley. :: ANALYSIS. Suppose the thing done, and that ABC is the triangle required; then, since the diff. of the angles at the base, and the diff. of the seg. of the base are given, the triangle CDF is given; draw CH perp. to CF, meeting FD produced in H, then HD, and the diam, HF are given. " But KL is given, and from thence EL; also by prop.:11. Mod. Geom. Student. EL = EK X EI; therefore, El becomes known, and of course IK; but : HD X DE=EK X IK; therefore, D E becomes known, and the method of construction is evident.. :: Solutions were given also by Messrs. Butterworthin Eyres, Hine, Webster, and Winward., ; 15. Qu. (70) Answered by Mr. Nesbit, Farnley Academy. In the annexed figure let HO denote the horizon, EQ.. the equator, 69 the tropic of Cancer, N the north and S . EX the south pole, F the town of Farnley, and D the point in which the sun's upper limb h would appear to a person elevated to A, on June 21st, at midnight. Now, from the co-latitude = 36° 13', substract the declination = 23° 28', and we obtain 12° 45', the depression of the sun's centre below the horizon at midnight, from which take 15' 47" + 35 = 48' 47", the semi-diameter and refraction, and add 9" for parallax, and we get 11° 56' 22"* = LOC 69 = LACD. Then, in the right-angled triangle A DC, right-angled at D; as the nat. co-sine of 11° 56' 22* .: CD= 3985 :: rad. : AC = 4073.1141 nearly"; from which take 3985, and there remains AF = 88,1141 miles, the height to which a person must be elevated, to see the sun's upper limb at the given time. : Put CF=r=21040800 feet, the radius of the earth; AC=a=21506042.448 feet, the distance fallen from; 8 = 1614 feet, half the velocity or force at F; D = the velocity required, and t= the time of falling from A to F; then (see Dr. Hutton's Course, vol. ii. p. 335; Marrat's Mechanics, art. 665; or Vince's Flux. art. 81) v=V4grx =5411 feet per second.. rond is Again, with the radius AG=GC, describe the semicircle ABC; draw B F perpendicular to AC, and join BG; then, we have BG=10753021.224, and GĚ= 10287778.776, to find the angle BGF = 16° 54 564, and side BF — 3128751.6. We also find the watc ABC= 33781691.4773; and, as 180°: 33781691.4773 :: 10° 54' 56" : 3174644.883 the arc A B; then, per a AB +BF a sidem, ut supra, + = 4 * CF 17.3" = 853", the whole time of falling to the surface, it! |