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

OC: Om:: CF: mn

OC: OF:: Dm: Dv

OC: OF:: Om: On

whence

Om xOF.

mv x OC

Dm x CF.

OC: CF :: Dm: mv Now, by adding the two firft of thefe equations together, we have mn + Dv x 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 x OC) = Om x CF 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 × OF + Dm × CF; and, by subtracting the latter from the former, we have On-mu x OC (OGX OC)=Om × OF-Dmx CF; whence OE and OG are also known. Q, E. I.

COROLLARY I.

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

COROLLARY II.

Hence, the fine of the double of either arch

(when they are equal) will be = fine=

R

: whence it appears, that the fine of the double of any arch, is equal to twice the rectangle of the fine and co-fine of the fingle arch, divided by radius; and that its co-fine is equal to the difference of the Squares of the fine and co-fine of the single arch, alfo, divided by radius.

COROLLARY III.

Moreover, becaufe Dm x CF = OC X mv (OC xEGOCXOE-OG); and Om × OF

=

OC x On (OC 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 fum and difference of thofe arches; and that the rectangle of their co-fines is equal to a rectangle under half the radius, and the fum of the co-fines, of the fum and difference of the fame arches.

PROP. III.

The tangents of two arches being given, to find the tangents of the fum, and difference, of thofe arches.

Let

Let AN ar.d

AM be the two
arches, and AB B

and AC their N
tangents; alfo
let NE be the
tangent of their

fum, in the first
cafe, and the

tangent of their difference, in the fecond, and let CF, perpendicular to the radius DN, be drawn then, because of the equiangular triangles BAD and BFC, we fhall have

{{

BD x CF = DA BC?

BD X BF BAX BC by 24. 3.
= S

Take each of the laft equal quantities from BD, and there will remain BD BD × BF (BD × DF) = BD2 — BA × BC: now BD × DF (BD2 — BA x BC): BD x CF (DA x BC) :: DF: CF :: DN (DA): NE :: DA2 : DA × NE; whence, alternately, BD-BA× BC : DA2 (:: DA × BC: DA × NE) :: BC : NE. But the firft term, of this proportion, because BD DA + BA, will alfo be expreffed by DA2 + BA-BA x BC, or by DA2 + BABAX AB+ AC; or, laftly, by, DA BAX AC: therefore, the three firft terms of the proportion being known, the fourth NE will likewife be known. 2. E. I.

2.

COROLLARY.

Hence, if radius be fuppofed unity, and the tangents of two arches be denoted by T and t, it follows, that the tangent of their fum will be = T+, and the tangent of their difference

t

1-tT'

T t

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

D

B

the tangent of AD (the

Thus, for inftance, let BA be an arch greater than 90°, and let the tangent of the fum of AB and AC be required; fuppofing T to reprefent fupplement of AB) and t

the tangent of AC: then, by writing -T instead

of T, in the firft of the

fhall have tang. of BC=

foregoing theorems, we
-T+t
T+t

=

I-tx-T I + tT

and therefore tang. DC (-tang. BC) =

T

I+T

which is the very theorem demonstrated in the 2d

cafe.

PROP. IV. .

As the fum of the tangents of any two angles BAC, BAD, is to their difference, fo is the fine of the fum of those angles, to the fine of their dif ference.

Let

that Ad = AD, and dAB

DAB, and, confequently, that CAd is the difference of the two angles BAC and BAD.

-

Now, by reafon of the fimilar triangles CDE and CdF, it will be, CD (CB+BD): Čd (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 propofition is manifeft.

COROLLARY.

Hence it also appears, that the base (CD) of a plane triangle, is to (Cd) the difference of its two fegments (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 bafe.

PROP. V.

In any plane triangle ABC, it will be, as the fum of the two fides plus the bafe, is to the fum of the two fides minus the bafe, fo is the co-tangent of half either angle at the bafe, to the tangent of half the other angle at the bafe.

In