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[[left margin]] 113) [[\left margin]]
   Again, by the said Cor. [[numerator]] tx-1 [[\numerator]] / [[denominator]] x+t [[\denominator]] = [[numerator]] y [[\numerator]] / [[denominator]] z [[\denominator]] (by putting y = tx-1, and z = x+t) = Cotang. of ½ < PEa: Whence, as before, [[numerator]] dy [[\numerator]] / [[denominator]] cz [[\denominator]]= Tang. of [[numerator]] aPE ~ PaE [[\numerator]] / [[denominator]] 2 [[\denominator]] , [[numerator]] δy [[\numerator]] / [[denominator]] 2z [[\denominator]] = Tang. of [[numerator]] aPE + PaE [[\numerator]] / [[denominator]] 2 [[\denominator]], and 1 - [[numerator]] dy [[\numerator]] / [[denominator]] cz [[\denominator]] x [[multiply]] [[numerator]] δy [[\numerator]] / [[denominator]] 2z [[\denominator]] ÷ [[numerator]] dy [[\numerator]] / [[denominator]] cz [[\denominator]] + [[numerator]] δy [[\numerator]] / [[denominator]] 2z [[\denominator]] =  [[numerator]] mz [[superscript]] 2 [[\superscript]] [[\numerator]] / [[denominator]] y [[superscript]] 2 [[\superscript]] [[\denominator]] (putting m = [[numerator]] c2 [[\numerator]] / [[denominator]] dδ[[\denominator]]) = Cotang. of aPE=nPE [[underline]] per [[\underline]] Quest. and ∵ [therefore] = [numerator]] a∝-bβx [[superscript]] 2 [[\superscript]] [[\numerator]] / [[denominator]] ∝bx-aβx [[\denominator]]. Hence by Reduction, & putting n = m∝b + maβ, nxz[[superscript]] 2 [[\superscript]] = a∝y [[superscript]] 2 [[\superscript]]- bβx [[superscript]] 2 [[\superscript]] y [[superscript]] 2 [[\superscript]]: by restitution and dividing by bβt [[superscript]] 2 [[\superscript]], I get x [[superscript]] 4 [[\superscript]]+ [[numerator]] n [[\numerator]] / [[denominator]] bβt [[superscript]] 2 [[\superscript]] - 2/t x [multiply] x [[superscript]] 3 [[\superscript]]  [[numerator]] 2nt + bβ - a∝t[[superscript]] 2 [[\superscript]] [[\numerator]] / [[denominator]] bβt [[superscript]] 2 [[\superscript]] [[\denominator]] x [[superscript]] 2 [[\superscript]] + [[numerator]] nt [[superscript]] 2 [[\superscript]]  + 2a∝t [[\numerator]] / [[denominator]] bβt [[superscript]] 2 [[\superscript]] [[\denominator]] x  = a∝.  Solved x= 
     = Tang. ° ' "; whence, seeing N is in the first quadrant from A, As 50": 1 Year :: 360°-2x:    years, to which add 1760 gives [[strikethrough]] ??   for which [[\strikethrough]] for the year required.

        Scholium I.
If in these figures, E be the North Pole, P the Zenith of a given place A and N two known stars, at a[[insert]] n [[\insert]]y given or supposed time; then the Angle NEn=AEa [[insert]] will [[\insert]] be the time from that given or supposed in which these stars shall both appear upon the same vertical circle Pna; the time found will be before or after that given or supposed according as

[[left margin]][ V. the proportions on p.A7 and A9. 




By two of these Obser. [[superscript]] s [[\superscript]] of 4 known *s [[assume meant "stars"]], the latitude of the place; time; &c may be found. [[\left margin]]

    But the most natural & easay time for that given or supposed is when one of the stars is upon the meridian; wherefore, in figs. 41. & 42. let P be the North Pole, Z the zenith, BbF, MaO the path of two known stars, and Zba an azimuth [[strikethrough]] circle [[\strikethrough]] or vertical circle. [[strikethrough]] Suppose the upper star at P. [[\strikethrough]] The upper Star is supposed upon the meridian at B in fig. 41. when the other is at A; and the lower upon the meridian at A, in fig 42. when the other [[strikethrough]] B [[\strikethrough]] is at B. Wherefore to find the declination of the plane passing through these two Stars when they both come upon the same vertical circle Za, at b and a; bP, aP are the co-declinations, [[strikethrough]] ? [[\strikethrough]] aPb=APB, the [[strikethrough]] ? [[\strikethrough]] difference of their Right Ascen. from b let fall the perpendicular bC upon aP; then (as in the solution to the above quest at p.124) [[strikethrough]] ? [[\strikethrough]] Rad.:Tang. bP::cos.bPA:tangPC; and Pa-PC=Ca; then again S.Ca:S.PC::tang bPa: tang.ZaP; and SZP(=co-lat. of the place):S.ZaP::S.Pa:S.aZP, the declination of the plane from the North; whether East or Westward is determined by the Rule on p.53. ___ Should the time of this observation be required; proceed as in the solution just referred to, for here are the same  [[underline]] data [[\underline]] and [[underline]] quosita [[\underline]], as in that question. ____ Fig. 41 is when both are upon a Soth [[sic: South]] Azimuth Za, and 42. when upon a North Azimuth Za; but should one have a North [[strikethrough]] aspect [[\strikethrough]] and the other a South [[strikethrough]] one [[\strikethrough]] aspect, & both at the same time upon the [[insert]] same [[\insert]] vertical plane; as the one at a and the other at β, then add 180°. to the lesser, or subtract it from the greater. Rt. Ascens. their places will be reduced to the same aspect, [[strikethrough]] and [[\strikethrough]] upon the same Azimuth circle [[insertion]] as at ∝ & β[[\insertion]]; and the difference between this sum or difference and the Right Ascens. of the other star is the angle ∝Pβ, with which, & their co-decl. ∝P, βP proceed as before.

          Scholium II.
     In the last Scholium there is no necessity to have the R.A. and declin. [[superscript]] n [[\superscript]] of the Stars, provided their Longitudes and Latitudes are known; for if P be the pole of the ecliptic, then aP, ∝P, bP, βP, & bPa, βP∝ will be the co-latitudes, and difference of their Longitudes instead of the co-declin. [[superscript]] s [[\superscript]] & difference of [[strikethrough]] ? [[\strikethrough]] R.A. also ZP will then be equal to the colatitude of the place [[strikethrough]] p [[\strikethrough]] [[underline]] plus [[\underline]] 23°29'.

Transcription Notes:
symbols in single brackets: [divide symbol], [squared], [therefore] e.g. x's in this may be x's or sigma symbols -- can't tell without better understanding of the math that's being done. Uppercase As and Bs may mean uppercase alphas and betas -- some of the uppercase As look like delta symbols too -- figure names look like A1 or [delta]1, kind of back and forth; not sure which is correct, except his other uppercase As are very carefully drawn, making those fig. names sure look a lot more like deltas to me.