Complt Lat. of the intended Port. Alfo there is given LA 200 Leagues 10°, AB = 120 6o, and BC= 80 Leagues 4°, which laft being taken from NC leaves NB = 41° 25. Therefore in each of the Triangles ALN and ABN, there is given the three Sides to find the Angles ANL 15°34′ and ANB 8° 57'. Hence the Latitude is 45° 58', and Longitude 42° 12' Weft from the Lizard. The fame answer'd by Mr. Geo. Godhelp. Let N represent the North Pole, L the Lizard (in Lat. 50) L'A the firft Courfe, A P the fecond, BC the third, and P the Port. Firft, in the Right angled Triangle PNC there is given the Sides PC (250 Leagues) 12° 30', and NC 45° 25', and the Right Angle at P, to find the Side PN=44° 01'43" the Comp. of the Lat. of the Port, and the Angle PNC the Port's Diff. of Long. from C17° 41′ 29", and CN-CB= NB=41° 25'. Now in the Oblique angled Spheric Triangle BNA there are given all the Sides to find the Angle at N 8° 57'. And the Diff. of Long. made by Sailing on the Parallel LA will be found 15° 33′ 26". Hence the Port's Diff. of Long, from the Lizard is found 42° 11′ 55′′ W. and Lat. 45° 58′ 57′′ North. Solutions to this Queftion were also received from Mr. Harland Widd, Mr. Wm Pirfon, Mr. J. Holden, Mr. Abr. Lord, Mr. John Nichols, Mr. Stephen Hartley, Mr. Tho. Garrard, Mr. James Hogard, Mr. Alexander Rowe, Mr. Wm Kingstone, Mr. John Williams, and San&tus Harmon. (5) Question 115. anfwered by Mr. Jof. Orchard. Let VC =20=c, p= Parameter= bb C 31.36; CB 48,9845 CH==EF; Then will HBc-x, and per Property of the Curve † × c—x = HE'; the Sub 2 2x -2x 2 2 A I t per fim. Triangles LHE and LCO, √xx:: 2 X :P -- 2x x √ px x OC the Radius of the in 2 feribed Sphere. Again, (per 47 E. 1.) VLH)2 + HÐ2= LEP2+4pc 4px; and per fim. Triangles LHE and OE=OC the Radius as above; Thefe Values made equal Radius is 13.3243. 2. E. F.. The fame anfwer'd by Tabularius, the Propofer. 62 C Let VC=31.3=b, BC=20=c, 49a, HB BT=x, HE=y, CH FE=c-x, CT-TE=c+x by fim. Triangles HE: TE:: Ft: Et :: FE: OE; viz. c+x::c-x: У =OF..CO=y quation of the Curve ax = y2.. ax or 3x2-2cx+axc. Now putting x and by the E · 26x + 2x2 = c2—x2 a-2c 3 = ±2m according as a may be greater or lefs than 2c, we get x = 3 I + m2 + m=11,644 -1.510,144 in the pre fent Cafe, whence y 22,2947, and thence the Radius CO OE13.326 nearly. The fame answered by Mr. Rich. Gibbons. = Let us with Ward, Shircliffe, and other Gaugers, make this Solid to be generated by the Parabola round its whole Ordinate (which to me feems to be the Propofer's Meaning) then we have CB 20 c, and CV 31.3 b, to find r=Radius of its infcribed Semicircle = COOE; Put 2x Subtangent HB, then by common Properties, Asc: b2 the Square of HE, hence the Squ.of TE is= +4x3Square of CT by common Properties of the Cir cle; and therefore c2+2 cx+x2; hence x = 10,1464, x62 and e-x=9,8536, and 497,0163 its fquare Root x2+c2-2cx+x2 22 is22.2939 HEz. Hence by 47 E. 1. Let BEV reprefent a Semi-Parabola, which by revolving about its Semi-Ordinate CV = 31,13=6, will generate a Semi-Parabolic Spindle (the Radius of whofe Base will be CB 20 c) at the fame Time that the Semicircle CE will generate an inferibed Sphere. Let ET be the Tangent to the Circle and Parabola at the Point of Contact; then, by the Property of a Circle ET=TC and ET2 TC, and by the Property of a Parabola BH BT.. calling BH (x) and HE (y), y2+4x2=EH)2+HT)\2=c+x2 = ET2; but c¿2::x:y2== which fubftituted for y gives 4x + C 2 = 2 62. 2 +x, and confequently 3x2 + C 20 2c2¬b2±√ba—+b2c2 +16c2 6c =10.146276. whence)=√62*\22,293727, and CD=HE HE 13-3245135 is the Radius required. This Queftion was truly folv'd by Harland Widd, Mr. Cha. Smith, Mr. Geo. Godhelp, Mr. Abr. Lord, Rimfide, Mr. Steph. Hartley, Mr. Wm Spencer, Mr. John Watchorne, Mr. Tho. Garrard, Mr. Wm Enefer, and many others, as may be feen in the Catalogue. (6) Question 116 folv'd by Mr. Wm Kingston. Conftruction. Draw the indefinite right Line AP (fee Fig.following) take two Radius's in the Ratio of the Line bisecting the Bafe, and That bisecting the Vertical Angle, and draw the two Arches op and mn. Draw RT at Right Angles to AP, and Af and Ag thro' the two Points of Interfection. In RT take Lb the given Diff. of the Segments of the Bafe, and thro 1. draw DE parallel to AP till it meet AG produced in E; draw AD at Right Angles to AE; bifect DE in S, and defcribe the Circle ADE; continue Af till it meet DE in F, fects the Bafe, becaufe DE paffes thro' the Centre of the Circle at Right Angles to BC, and by fimilar Triangles as Aggh: AG: GH; and as Af: fb:: AF: FH, but as Ag and Af are in the given Ratio by Construction, and BC parallel to RT; whence AG and AF are in the fame Ratio, and FH to LH. This Question was very methodically folved by Mr. Wm Pirson, Mr. J. Holden, Mr. Cha. Smith, Mr. Abr. Lord, Mr. Wm Spencer, Mr. John Watchorne, Mr. Tho. Garrard, Mr. Rich. Gibbons, Mr. John Williams, and Mr. Paul Sharp. A (7) Question 117 anfwered by Mr. Tho. Garrard. P Difference between CP PO, and (per 47 E. 1.) 2 A y 2 √x2 - 2 dx + a2 +2 + the whole Perimeter, which (per Du) is to be a Minimum. Therefore by making a flow, we have x2+2dx+d2+y« xx x 2 √x2+2ax+x2+;2 + √x2+zdx+d2+; 2 = 0, which Equation being divided by x, brought out of Surds and properly reduced gives xo. Confequently CPOD and AC AC=BD, and the Expreffion for the Perimeter will become ولو + is a Minimum. In H and fquar ing both Sides the Equation &c. 6—a ayya add, Solv❜d y= 12,8 very nearly. Hence AC=BD=13,1072, AB 9,678888, and CD≈ 15,321112. The fame anfwer'd by Mr. Tho, Mofs. Let ABCD be a Trapezoid with unequal Sides, whofe Area and Difference of the parallel Sides are given; which in order to demonstrate Geometrically, that 2. the Perimeter thereof (AB+BD+DC+CA) may become lefs, and yet the given Difference AI of the parallel Sides, and likewife the given Area (ABCD) remain the fame. Let fall the Perpend. CK, draw CI parallel to DB, bifect AI in L, moreover draw LF parallel to CK meeting DC produced in F; join F1 and draw BH parallel thereto. Now 'tis plain that LB= AB+CD; and because of the parallel Lines IF and BH, IC and BD, AB and FD; theTriangles FCI and HBD are fimilar and equal in all Refpects; and therefore FC=HD, and confequently FH CD; whence it appears that the two parallel Sides (and confequently their Difference) their nearest Distance to each other, and the Area may remain the fame, and yet the two unequal Sides may vary: Therefore 'tis plain that CI (DB) +AC, the Sum of the two Sides joining the Extremities of the parallel Sides of the Trapezoid; or which is the fame, the Sum of the Sides AC, CI, of the Triangle ACI will be the leaft poffible (and confequently the whole Perimeter of the Trapezoid) when they are equal to each other. Moreover it is further manifeft that the required Trapezoid, whose Area and the Difference of the parallei Sides are here given, can be no otherwise varied than either to have four unequal Sides, or else those two Sides joining the two parallel Sides to be equal to each other; but, in the firft Cafe it is proved above C that |