-- c=Verfed Sine of the L PZO 137:—=1.7313 b=Nat. Sine of ZP=37 Degrees = 6018 the Co-Sine of the same .1986 Nat. Sine of ZO ; √1-x2 = =its Co-Sine. Then by Page the 70th of Ronayne's Algebra the Verfed Sine of PO is=1ms + √22-22x2. And the Comx n2 —n2x2. Sine of the Difference of the Sides ZO, ZP by Page the 369th of the fame Author is bx + √ d2 — d2x2. Confequently the Versed Sine of their Difference is equal 1— bx - √ d2 ~ d2x2. Laftly by Theorem at Page the 120th, Volume the 2d, of Ozanam's Curfus, b: 11mx+ √ 122 — n2x2 - 1 + bx+√d2-d2x2:c and multiplying Extreams and Means we have bcx=bx¬mx+ √n2—n2x2 +22x2. Let p = bcm-b; Then is px = √ n2 — n2x2 + vd2-d2x2. And p2x2=n2—n2x2+d2 ・d2x2+ 2 √d2x2- {d2 n2x2 + d2 n2x*• Again put q -122 —d2 s = +ď2. p2 + n2 + d2. Then s x2-q= 2 √ d2n2 — 2 d2n2x2+d'n'x; and by Involution s2x2 s2x*—2qsx2 +q2 = 4d2 n2 — 8d2 n2x2+4d2 122x4 By Tranfpofition 444} And by Subftitution -99. b; and extracting the Root x = d 7116 the Natural Sine of 45: the Co-altitude ZO. So the Altitude was 44: 38; and d the Sun's Declination 14: 22 North; which was on the 18th Day of April, and the Time was 2 o'Clock in the Afternoon. QUESTION the 9th, answered by R. D. ET2=CB 24. Inches; b=AB30.; d=7854; xED; (as in Fig. 9.) then by the Property of the LE Parabola 3 за 3a2 = to the Solidity of the Conic-Fruftum CEDPSB. The Fluxion of this, viz. 4dbax 12dbx,x 16dbx3 3 a заа and by Multiplication, Tranfpofition, &c. 4x3-3x2=aaa. Therefore x will be found 10.9298 Inches, and the leffer Diameter = 21.8596, its Height equal to 23.7781 Inches, and the Content 103.242 Wine Gallons. Which was required to be found. QUESTION the ioth anfwered by R. W.E-ft-b. DRAW the Lines (as in the Scheme Fig. 10.) and let HF GFDC=d; AC=a; AP=x; PM⇒y; Then by the Nature of the Queftion it will be, as, AC (a): DC (d) :: AP (x) : HL = dx Wherefore LI a quently the Locus fought is an Elliptis, whofe conjugate Semi-Axes are AC and CD, for by the known Property of the Ellipfis it will be QUESTION the 11th, answered by I. T. IT is evident that a great Circle drawn thro'K perpendicular to vpm will bifect pm in (Vid. Fig. 11.) becaufe by the Qu. the Triangle pKm is equicrural, Then in the oblique Sphe rical d rical Triangle Pmp are given Pm=57: 0, and Pp=36:00 d to find pm, 35:6 the each other and the An given me d 22:46.an d and the Lmn41: 7 to find me d 922 = 14:45 and the L m n o = 51:11 Pizx, From the Side Pm fubtract mn, remains Pn d 34:14. In the Right Angled Triangle Pnx are given Pn and the Pix to find Px26:00 and nx = 23: 6 In the Right-Angled Triangle PKx are given PK = d 108: co and Px 26: 00 to find Kx Kx fubtra&+nx remains K, d 110: 7. From d In the Right angled TriangleK 4 m, givenK ¢=72 : 10 d and pm 17:33 to find Km Kp the Diftance of the d Island from each of the Ports = 73: 7 Laftly, In the Oblique Triangle KpP are given KP= .d 108: 0 and Pp=36: oo & Kp = 737 to find the An gle KPp=14:54 The Difference of Longitude between the Ifland and the Port p. and adding to it the Angle pPm gives the Difference of Longitude betwixt the Ifland and d · the Fort m.= 54 : 54. 2. E. I. QUESTION the 12th, anfwered by J. T. I Et Mm. Ss (in Fig. 12.) reprefent the Tranfverfe and Conjugate Axes of the Elliptis which we feek, Qb, AR the two Conjugate Diam'. given ; and the Angle contained ACb 63: Degrees, it is demonftrated by the Writers on Co nicks, nicks, that if a Tangent be drawn to any Point of an Ellipfis, as A and the Diameter AR be drawn, and any other 2 Conjugate Diameters be produced 'till they meet the faid Tangent as in D and E That DAX AE to the Square of the Semidiameter CQ. the Conjugate to CA, Hence it is plain that if A O be erected perpendicular to DE and CQ That DOE will be a right angled Triangle, and defcribing a Circle upon DE as a Diameter, it will pafs thro' the Point O, and likewife thro' the Centre C, because the ZDCE. is alfo a right Angle. Draw now the Lines as in the Scheme, viz. CB perpendicular to DE, AP perpendi cular to CE, and Af perpendicular to CD, then because CA is known to be 27.5 Feet, and the LCA BLACb= 68 Degrees, CB will alfo be known 25.5 which call b, and BA 10.3=c; Put AK=x; AO=CQ=41.=a; Then b2+c2 +2cx +x2=(CK2=KO2=) a2 + x2· a2b22 Ergo x = 2C 44.9; and the Radius of the Circle=60.8; TOKE add KB, and fubtracting it from the fame gives DB=5.6 and BE=116. and CB being before known we may find CD=26.1=p; and CE118,8=q; But the fimilar Triangles DCE, DfA give this Proportion, viz. As DE: DC DA: Df 3.4 and As DE: CE :: DA:fA CP 15.5 From CD fubtract Df remains Cf=AP=22.7 which call ; CP, m; the Semi-tranfverse Diameter; the Semi-Conjugate c; Then it's evident, that because the Subtangent PE is= m monftrated by Ward, Page the 371. and ƒÐ as is de Confequently t√mq= 42.9 and e√p=24.4 Now for the Geometrical Conftruction. Draw the Diameters AR, Qb, making the given Angle ACb, produce CA and on it take AV to a 3d Pro portional portional to CA and Cb: And having drawn thro' A the right Line AE parallel to Qb, from the Point n the Middle of CV Erect the Perpendicular 2K meeting AE in K; upon the Centre K thro' C and V defcribe a Circle cutting AE in the Points E and D. Through which Points and the Centre C draw EC, DC producing them indefinitely. Then draw AP, Af parallel to CD; CE, and having taken Cm a mean Proportional between CP and CE, and CS a mean Proportional between Cf and CD; make CM Cm and 'Cs CS I fay then that Mm and Ss will be the Transverse and Conjugate Axes of the Ellipfis QUESTION the 13th, anfwered by J. N. PU } Fig. 13. Then is Gw√x2-c2, GS = b-x, Hn=d—y ; HwVy2-c2. с As: 1 C: = LDG w. = LAGS. Ergo As Gw: WD: GS: SA. ie. √x2-c2: cb-x: wD:: ASSB, Ergo AB= like Procefs we fhall find BC= 2cb-2cx and by a 2cd-2cy As Dw: DGAS: AG. That is cx √ y2 — c2 2 bx-x? Now multiply each Part of the Equation by its respective Denominator, and reducing we have bx -- 2 Now involve both Parts of the Equation, and multiply crofs. |