COR. I.
The radius is a mean proportional between the
tangent and the cotangent of any angle ABC; or tan
ABC x cot ABC R2.

For, fince HK, BA are parallel, the angles HKB, ABC are equal, and KHB, BAE are right angles; therefore the triangles BAE, KHB are fimilar, and therefore AE is to AB, as BH or BA to HK.

COR. 2. The radius is a mean proportional between the cofine and fecant of any angle ABC; or cof. ABC x fec. ABC= R2.

Since CD, AE are parallel, BD is to BC or BA, as BA to BE.

IN

PRO P. I.

N a right angled plane triangle, as the hypotenuse to either of the fides, fo the radius to the fine of the angle oppofite to that fide; and as either of the fides is to the other fide, fo is the radius to the tangent of the angle oppofite to that fide.

Let ABC be a right angled plane triangle, of which BC is the hypotenufe. From the centre C, with any radius CD, defcribe the arch DE; draw DF at right angles to CE, and from E draw EG touching the circle in E, and meeting CB in G; DF is the fine, and EG the tangent of the arch DE, or of the angle C.

The two triangles DFC, BAC are equiangular, because the angles DFC, BAC are

right angles, and the angle at C is common. Therefore, CB: BA:: CD: DF; but CD is the radius, and DF the fine of the angle C, (Def. 4.); therefore CB: BA::R : fin. C.

Alfo, because EG touch

FE

B

A

es the circle in E, CEG is

a right angle, and therefore equal to the angle BAC; and

fince the angle at C is common to the triangles CBA, CGE, thefe triangles are equiangular, wherefore CA: AB:: CE: EG; but CE is the radius, and EG the tangent of the angle C; therefore, CA: AB:: R: tan. C.

COR. 1. As the radius to the fecant of the angle C, fo the fide adjacent to that angle to the hypotenufe. For CG is the fecant of the angle Č (def. 7.), and the triangles CGE, CBA being equiangular, CA: CB:: CE: CG, that is, CA: CB:: R: fec. C.

COR. 2. If the analogies in this propofition,and in the above corollary be arithmetically expreffed, making the radius = 1,

fince fin. C cof. B, because B is the complement of C,

cof. B=

and for the fame reason, cof. C= BC'

BC

A

COR. 3. In every triangle, if a perpendicular be drawn, from any of the angles on the oppofite fide, the fegments of that fide are to one another, as the tangents of the parts into which the oppofite angle is divided by the perpendicular. For, if in the triangle ABC, AD be drawn perpendicular to the base BC,

B

each of the triangles CAD, ABD being right angled, AD: DC::R: tan. CAD, and AD: DB: R: tan. DAB; therefore, ex æquo, DC : DB : : tan. CAD : tan. BAD.

SCHO.

SCHOLIUM.

The propofition, juft demonftrated, is most easily remembered, by ftating it thus: If in a right angled triangle the hypotenuse be made the radius, the fides become the fines of the oppofite angles; and if one of the fides be made the radius, the other fide becomes the tangent of the opposite angle, and the hypotenuse the fecant of it.

PROP. II.

HE fides of a plane triangle are to one another as the fines of the oppofite angles.

TH

From A any angle in the triangle ABC, let AD be drawn perpendicular to BC. And because the triangle ABD is right angled at D, AB: AD:: Ř: fin. B; and, for the fame reason, AC: AD::R: fin. C, and inverfely, AD: AC:: fin. C: R; therefore, ex æquo inversely, AB: AC::fin. C: fin. B. In the fame manner, it may be demonftrated, that AB: BC:: fin. C: fin. A. Therefore, &c. Q. E. D.

B

D

PROP.

TH

HE fum of the fines of any two arches of a circle, is to the difference of their fines, as the tangent of half the sum of the arches to the tangent of half their difference.

Let AB, AC be two arches of a circle ABCD; let E be the centre, and AEG the diameter which paffes through A: fin. AC + fin. AB: fin. AC-fin. AB : : tan. (AC+AB): tan.

(AC—AB.)

Draw BF parallel to AG, meeting the circle again in F. Draw BH and CL perpendicular to AE, and they will be the fines of the arches AB and AC; produce CL till it meet the circle again in D; join DF, FC, DE, EB, EC, DB.

Now, fince EL from the centre is perpendicular to CD, it bifects the line CD in L, and

DK: KC :: tan. DFK: tan CFK; but tan. DFK tan.

BD,

because the angle DFK (20. 3.) is the half of DEB, and is therefore measured by half the arch DB. For the fame rea

fines of the arches AB and AC; and KC is the difference of their fines; alfo BD is the fum of the arches AB and AC, and BC the difference of thofe arches. Therefore, &c. Q. E. D.

14

COR. I. Because EL is the cofine of AC, and EH of AB, FK is the fum of these cofines, and KB their difference; for FK FB+EL EH+EL, and KB LH-EH-EL. Now, FK: KB: tan. FDK: tan. BDK; and tan FDK co tan. DFK, because DFK is the complement of FDK; therefore, FK: KB:: co-tan. DFK : tan. BDK;

that is, FK:KB:: co-tan.

DB: tan. BC. The sum of

the cofines of two arches is therefore to the difference of the fame cofines, as the co-tangent of half the fum of the arches to the tangent of half their difference.

COR. 2. In the right-angled triangle FKD, FK: KD::R: tan DFK: Now FK cof AB + cof AC, KD = fin AB + fin AC, and tan DFK = tan 1⁄2 (AB+ AC), therefore cof AB+ cof AC: fin AB + fin AC :: R: tan (AB+ AC.).

In the fame manner, by help of the triangle FKC, it may fhown that cof AB + cof AC : fin AC — fin AB :: R:

tan 1⁄2 (AC — AB.).

COR. 3. If the two arches AB and AC be together equal to 90°, the tangent of half their fum, that is, of 45°, is equal to the radius. And the arch BC being the excess of DC above DB, or above 90°, the half of the arch BC will be equal to the excess of the half of DC above the half of DB, that is, to the excess of AC above 45°; therefore, when the fum of two arches is 90°, the fum of the fines of those arches is to their difference as the radius to the tangent of the difference between either of them and 45°.

PROP.