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and BH parallel to AO, meeting GD in v and H:
then it is plain, because Dm = Bm, that Du is =
Hv, and mv = nG = En; and that the triangles
OCF, Omn and mDv are similar ; whence we have
the following proportions,

OC: Om :: CF:mn mn X OC=Om x CF.
OC: OF:: Dm: Dv Dvx OC = Dm XOF.
OC:OF :: Om: On On ~ OC=Om OF.

OC: CF :: Dm: mv my x OC=Dm x CF.
Now, by adding the two first of these equations
together, we have mn + Du ~ OC (DG OC)=
Om x CF + Dm X OF; whence DG is known.
Moreover, by taking the latter from the former,
we get mn Dv x OC (BE XOC)= Om RCF
Dm X OF; whence BE is known.

In like manner, by adding the third and fourth equations together, we have On + mv x OC(OEX OC) = Om X OF + Dm X CF; and, by subtracting the latter from the former, we have On-MU X OC (OGXOC)=Om X OF-Dm X CF; whence OE and OG are also known. 2. E. I.

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Hence, if the fines of two arches be denoted
by S and s; their co-fines by C and c; and radius
by R; then will
the fine of their sum

Sc + SC

SC SC the sine of their difference =

R the co-fine of their sum =


R the co-fine of their difference =

Cc + Ss


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fine =

Hence, the line of the double of either arch

2CS (when they are equal) will be =

and its co

C’ - SP

: whence it appears, that the fine of

R the double of any arch, is equal to twice the rečtangle of the fine and co-fine of the single arch, divided by radius; and that its co-fine is equal to the difference of the squares of the fine and co-line of the single arch, also, divided by radius.

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Moreover, because Dm X CF = OC X mv (LOC REG= 40C OE-OG); and Om * OF = OC X On (LOC X 2On=OC X OE + OG), it follows, that the rectangle of the fines of any two arches (AC, CD (BC) is equal to a rectangle under half the radius, and the difference of the co-fines, of the sum and difference of those arches; and that the rectangle of their co-fines is equal to a retangle under half the radius, and the sum of the co-fines, of the sum and difference of the same arches.


The tangents of two arches being given, to find the tangents of the sum, and difference, of those arches.


Let AN ar.

E AM be the two arches, and AB B



с Α. and AC their N


tangents; also

N let NE be the tangent of their sum, in the first

DS case, and the

D tangent of their difference, in the second, and let CF, perpendicular to the radius DN, be drawn : then, because of the equiangular triangles BAD and BFC, we shall have S X CF = X

x BF = X BCS Take each of the last equal quantities from BD", and there will remain BD -- BD x BF (BD x DF) – BD’ — BA X BC: now BD X DF (BD’ - BA X BC): BD X CF (DA X BC):: DF:CF:: DN (DA): NE :: DA?: DA X NE; whence, alternately, BD’ — BAX BC:DA? (::DA X BC:DA X NE):: BC: NE. But the first term, of this proporcion, because BD’=DA + BA’, will also be expressed by DA’ + BA — BA X BC, or by DA’ + BA’ – BAX AB + AC; or, lastly, by, DA? IBA X AC: therefore, the three first terms of the proportion being known, the fourth NE will likewise be known. 2. E. I.

COROLLARY. Hence, if radius be supposed unity, and the tangents of two arches be denoted by T and t, it follows, that the tangent of their sum will be = T + t

and the tangent of their difference = I - T' T t 1FIT


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But, it will be proper to take notice here (once for all) that, if in these, or any other theorems, the tangent, secant, co-fine, co-tangent, &c. of an arch greater than 90 degrees be concerned ; then, instead thereof, the tangent, secant, co-fine, &c. of an arch, as much below go degrees, is to be taken, with a negative sign ; according to the observation in page 5.

Thus, for instance, let BA be an arch greater than 90°, and let the

tangent of the sum of D

AB and AC be required;

B supposing T to represent the tangent of AD (the supplement of AB) and t the tangent of AC: then, by writing T instead of T, in the first of the foregoing theorems, we

- Ttt

-T+t shall have tang. of BC=


I + iT

TI and therefore tang. DC (-tang. BC) = which is the very theorem demonstrated in the ad cafe.

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As the sum of the tangents of any two angles BAC, BAD, is to their difference, so is the fine of the fum of those angles, to the fine of their difference

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Let BC and BD be the two pro

A posed tangents, to

E the radius AB ; take Bd = BD, join Ad, and draw DE and dF per

F pendicular to AC. It is manifest, be- o d


D cause Bd = BD, that Ad = AD, and DAB = DAB, and, consea quently, that CAD is the difference of the two angles BAC and BAD.

Now, by reason of the similar triangles CDE and CdF, it will be, CD (CB + BD) : Cd (CB BD) :: DE : DF; but DE and dF are fines of DAE and DAF to the equal radii AD and Ad: whence the truth of the proposition is manifeft.


Hence it also appears, that the base (CD) of a plane triangle, is to (Cd) the difference of its two segments (made by letting fall a perpendicular), as the fine of the angle (CAD) at the vertex, to the fine of the difference of the angles at the base.


In any plane triangle ABC, it will be, as the sun of the two sides plus the base, is to the sum of the two fides minus the base, so is the co-tangent of half either angle at the base, so the tangent of half the other angle at the base.

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