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[[right corner]] (124 [[/right corner]]

[[left margin, written in red ink]][[strikethrough]] within the red crotchets amounts to nothing.[[/strikethrough]]
 [[/left margin, written in red ink]]

[[strikethrough]]
ad/bmc x - a/btc x; wherein y[[superscript]]2[[/superscript]] + x [[superscript]]2[[/superscript]] = 1(Rad.); a/c = tang. PE = T; b/d=[[scribbled over]] tang. co[[/scribbled over]] t. nEa=τ; ∵(Tτ/m x + T/m y - T/bt x = 1; or rather τ/m x + 1/m y - 1/bt x = 1/T, or, by dividing by T/m, better thus) y + τ - 1/bt m * x = 1/T m. ____) ∵ y + d/b - m/bt * x = cm/a; where 1/m is the Cot. nE, b/d = tang. nEa, [[strikethrough]][[?]][[/strikethrough]] and c/a = cot. PE; [[insertion]] from [[/insertion]] which [[strikethrough]][[?]][[/strikethrough]] it [[insertion, over a scribble]] is even[[insertion, over a scribble]] easy to approximate the value of x; [[strikethrough]][[?]][[/strikethrough]] but, putting d/b - m/bt = -γ; and c/a = τ; x =[[scribble]][[?]][[/scribble]]  √( 1/γ+1 - 1/γ+1 mτ [[superscript]]2[[/superscript]] + [[square root]](1/γ+1 mγτ)² [[/square root]] ) [[/square root]] [[strikethrough]][[/square root]][[/strikethrough]] [[scribble]][[?]][[/scribble]] -1/γ+1 mγτ = =[[scribble]][[?]][[/scribble]] [[/strikethrough]]

[[left margin, written in red ink]]To be taken in at the [[black ink]](*) [[/black ink]] on the foregoing page [[/left margin, written in red ink]]

(*) The numbers given on p.112 are most of them wrong. the true ones are therefore here repeated

   Now. there is given PE = 23°29' the distance of the pole of the ecliptic from the pole of the world; NE = nE = 23°55'49", AE = aE = 35°39'44" the colat.; NEP = 4°42'49"; AEP = 65°33'55" the long. of N and A from σ9 in 1765, by D. [[superscript]] r [[/superscript]] [[underline]] Halley [[/underline]]'s Tables. which being on different sides of σ9, their will give NEA = 70°16'44". It is plain, at the end of the time required, this NEA = nEa, because the longitudes of both stars always increase by the same invariable quantity in the same time.

[[fancy section dividing line]] 

ad/bcm x - a/bct x; wherin y [[superscript]]2[[/superscript]] + x [[superscript]] 2 [[/superscript]] = 1(Rad.) [[scribble]] thence [[insertion]] by reduction and [[/insertion]] putting d/b - m/bt = ±γ, and c/a = τ = cot.EP; x is had = ∓ [[numerator]] γmτ [[/numerator]] / [[denominator]] γ²+1 [[/denominator]]+ [[square root]] √ [[numerator]]1-m²τ² [[/numerator]] / [[denominator]] γ²+1 [[/denominator]] + [[square root]] [[numerator]] γmτ [[/numerator]] / [[denominator]] γ²+1 [[/denominator]] [[/square root]] [[/square root]]; the upper or lower sign taking place according as d/b is greater or less than m/bt; Solved x =

 [[centered]] Or thus  [[/centered]]
Put a, k, b, p, x for the Sines, c, l, d, q, y for the cosines of PE, nE, nEa, aE, PEn. Then by prop. V.B.I. of [[underline]] Em. [[/underline]] Trig. dx + by = [[scribble]] Sine, and by cor. I. to the same prop dy - bx = Cos. of PEA: whence, by Spherics [[numerator]] kx [[/numerator]] / [[denominator]] al-kcy [[/denominator]] = tang. nPE = aPE, & ∵ = [[numerator]] pdx+pby [[/numerator]] / [[denominator]] aq-pcdy+pcbx [[/denominator]] : whence by reduction, ald/bck x - aq/pcb x = y² + x² - al/kc, wherein l/k = 1/m, and q/p = 1/t, as substitude above, which used [[scribble]] in the process as before, the value of x comes out exactly [[strikethrough]]as before [[/strikethrough]] the same, as there found.

     [[larger print]] But best of all thus. [[/larger print]]
Fig. 39 [[insertion]] & 40 are [[insertion]] only a repetition of the 26. [[superscript]] th [[/superscript]] [[insertion]] and 25. [[superscript]] th [[/superscript]] [[insertion]] because the Northpole star at N is on the contrary side the pole P to that of [[underlined]] Alioth [[underlined]] at A, and therefore is apparently better. --- From [[bold]] a [[/bold]] let fall the perpendicular ac upon nE: Then per Spherics, (in fig. 40. nE [[strikethrough]][[written above strikethrough]]P [[written above strikethrough]] is greater than NEP, because N [[strikethrough]] being[[/strikethrough]] now having a greater than that of A [[underline]]minus [[/underline]] 180°.)

Transcription Notes:
τ => Greek letter Tau γ => Greek letter Gamma