fecant, of any arch which is the measure of any given angle
ABC, is to the fine, verfed fine, tangent, and fecant, of any
other arch which is the meafure of the fame angle, as the ra
dius of the firft is to the radius of the second.

Let AC, MN be mcafures of the angles ABC, according to
def. 1. CD the fine, DA the verfed fine, AE the tangent, and
BE the fecant of the arch AC, according to def. 4. 5. 6. 7.
and NO the fine, OM the verfed fine, MP the tangent, and
BP the fecant of the arch MN, according to the fame defi-
nitions. Since CD, NO, AE, MP are parallel, CD is to NO
as the radius CB to the radius NB, and AE, to MP as AB to
BM, and BC or BA to BD as BN or BM to BO; and, by
converfion, DA to MO as AP to MB. Hence the corollary
is manifeft; therefore, if the radius be fuppofed to be divided
into any given number of equal parts, the fine, verfed fine,
tangent, and fecant of any given angle, will each contain a
given number of thefe parts; and, by trigonometrical tables,
the length of the fine, verfed fine, tangent, and fecant of any
angle may be found in parts of which the radius contains a
given number: and, vice verfa, a number expreffing the length
of the fine, verfed fine, tangent, and fecant being given, the
angle of which it is the fine, verfed fine, tangent, and fecant
may be found.

VIII.

The difference of an angle from a right angle is called the Complement of that angle. Thus, if BH be drawn perpendicular to AB, the angle CBH will be the complement of the angle ABC, or of CBF.

IX.

Let HK be the tangent, CL or DB, which is equal to it, the fine, and BK the fecant of CBH, the complement of ABC, ac cording to def. 4. 6. 7, HK is called the co-tangent, BD the co-fine, and BK the co-fecant of the angle ABC.

COR. 1. The radius is a mean proportional between the tangent and co-tangent.

For, fince HK, BA are parallel, the angles HKB, ABC will be

equal, and the angles KHB, BAE are right; therefore the tri-
angles 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 co-fine
and fecant of any angle ABC.

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

Fig. 3.

IN

PROP. I. FIG. 5.

Na a right angled plain triangle, if the hypothenufe be made radius, the fides become the fines of the angles oppofite to them; and if either fide be made radius, the remaining fide is the tangent of the angle oppofite to it, and the hypothenuse the fecant of the fame angle.

Let ABC be a right angled triangle; if the hypothenufe BC be made radius, either of the fides AC will be the fine of the angle ABC oppofite to it; and if either fide BA be made radias, the other fide AC will be the tangent of the angle ABC oppofite to it, and the hypothenufe BC the fecant of the fame angle.

About B as a center, with BC, BA for diftances, let r circles CD, EA be defcribed, meeting BA, BC in D, E: fince CAB is a right angle, BC being radius, AC is the fine of the angle ABC by def. 4. and BA being radius, AC is the tangent, and BC the fecant of the angle ABC, by def. 6. 7.

COR. 1. Of the hypothenufe a fide and an angle of a right angled triangle, any two being given, the third is alfo given. COR. 2. Of the two fides and an angle of a right angled tri angle, any two being given, the third is also given.

THE

PROP. II. FIG. 6. 7.

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

In right angled triangles this prop. is manifeft from prop. 1. for if the hypothenufe be made radius, the fides are the fines of the angles oppofite to them, and the radius is the fine of a righ | angle (cor. to def. 4.) which is oppofite to the hypothenufe.

In any oblique angled triangle ABC, any two fides AB, AC will be to one another as the fines of the angles ACB, ABC which are oppofite to them.

From C, B draw CE, BD perpendicular upon the oppoft fides AB, AC produced, if need be. Since CEB, CDB are right angles, BC being radius, CE is the fine of the angle CBA, and BD the fine of the angle ACB; but the two triangles CAE, DAB have each a right angle at D and E; and likewife the common angle CAB; therefore they are fimilar, and confequendy,

CA is to AB, as CE to DB; that is, the fides are as the fines of the angles oppofite to them.

COR. Hence of two fides, and two angles oppofite to them, in a plain triangle, any three being given, the fourth is alfo given.

I

PROP. III. FIG. 8.

Na plain triangle, the fum of any two fides is to their difference, as the tangent of half the sum of the angles at the base, to the tangent of half their difference.

Let ABC be a plain triangle, the fum of any two fides AB, AC will be to their difference as the tangent of half the fum of the angles at the base ABC, ACB to the tangent of half their difference.

About A as a center, with AB the greater fide for a distance, let a circle be described, meeting AC produced in E, F, and BC in D; join DA, EB, FB; and draw FG parallel to BC, meeting EB in G.

The angle EAB (32. 1.) is equal to the fum of the angles at the bafe, and the angle EFB at the circumference is equal to the half of EAB at the center (20. 3.); therefore EFB is half the fum of the angles at the bafe; but the angle ACB (32. 1.) is equal to the angles CAD and ADC, or ABC together; there fore FAD is the difference of the angles at the base, and FBD at the circumference, or BFG, on account of the parallels FG, BD, is the half of that difference; but fince the angle EBF in a femicircle is a right angle (1. of this) FB being radius, BE, BG, are the tangents of the angles EFB, BFG; but it is manifeft that EC is the fum of the fides BA, AC, and CF their difference; and fince BC, FG are parallel (2. 6.) EC is to CF, as EB to BG; that is, the fum of the fides is to their difference, as the tangent of half the fum of the angles at the bafe to the tangent of half their difference.

IN:

PROP. IV. FIG. 18.

N any plain triangle BAC, whofe two fides are BA, AC and base BC, the lefs of the two fides, which let be BA, is to the greater AC as the radius is to the tangent of an angle; and the radius is to the

a 3

tangent of the excess of this angle above half a right angle as the tangent of half the fum of the angles B and C at the bafe, is to the tangent of half their difference.

At the point A, draw the straight line EAD perpendicular to BA; make AE, AF, each equal to AB, and AD to AC; join BE, BF, BD, and from D, draw DG perpendicular upon BF. And becaufe BA is at right angles to EF, and EA, AB, AF are equal, each of the angles EBA, ABF is half a right angle, and the whole EBF is a right angle; alfo (4. 1. El.) EB is equal to BF. And fince EBF, FGD are right angles, EB is parallel to GD, and the triangles EBF, FGD are fimilar; therefore EB is to BF as DG to GF, and EB being equal to BF, FG must be equal to GD. And because BAD is a right angle, BA the less fide is to AD or AC the greater, as the radius is to the tangent of the angle ABD; and because BGD is a right angle, BG is to GD or GF as the radius is to the tangent of GBD, which is the excess of the angle ABD above ABF half a right angle. But because EB is parallel to GD, BG is to GF as ED is to DF, that is, fince ED is the fum of the fides BA, AC and FD their difference, (3. of this,) as the tangent of half the fum of the angles B, C, at the bafe to the tangent of half their difference. There fore, in any plain triangle, &c. Q. E. D,

I

PROP. V. FIG. 9. and 10.

N any triangle, twice the rectangle contained by any two fides is to the difference of the fum of the fquares of these two fides, and the fquare of the bafe, as the radius is to the co-fine of the angle included by the two fides.

Let ABC be a plain triangle, twice the rectangle ABC contained by any two fides BA, BC is to the difference of the fum of the fquares of BA, BC, and the fquare of the base AC, as the radius to the co-fine of the angle ABC.

From A, draw AD perpendicular upon the opposite side BC; then (by 12, and 13. 2. El.) the difference of the fum of the fquares of AB, BC, and the fquare of the base AC, is equal to twice the rectangle CBD; but twice the rectangle CBA is so

twice the rectangle CBD, that is, to the difference of the fum of the fquares of AB, BC, and the fquare of AC, (1. 6.) as AB to BD; that is, by prop. 1. as radius to the fine of BAD, which is the complement of the angle ABC, that is, as radius to the cofine of ABC.

PROP. VI. FIG. 11.

IN any triangle ABC, whofe two fides are AB, AC,

and bafe BC; the rectangle contained by half the perimeter, and the excess of it above the base BC, is to the rectangle contained by the ftraight lines, by which the half of the perimeter exceeds the other two fides AB, AC, as the fquare of the radius is to the fquare of the tangent of half the angle BAC oppofite to the base.

Let the angles BAC, ABC be bifected by the ftraight lines AG, BG; and, producing the fide AB, let the exterior angle CBH be bifected by the ftraight line BK, meeting AG in K; and from the points G, K, let there be drawn perpendicular upon the fides the ftraight lines GD, GE, GF, KH, KL, KM. Since therefore (4. 4.) G is the center of the circle inscribed in the triangle ABC, GD, GF, GE will be equal, and AD will be equal to AE, BD to BF, and CE to CF. In like manner KH, KL, KM will be equal, and BH will be equal to BM, and AH to AL, because the angles HBM, HAL are bifected by the ftraight lines BK, KA: and because in the triangles KCL, KCM, the fides LK, KM are equal, KC is common and KLC, KMC are right angles, CL will be equal to CM: fince therefore BM is equal to BH, and CM to CL; BC will be equal to BH and CL together; and, adding AB and AC together, AB, AC, and BC will together be equal to AH and AL together: but AH, AL are equal wherefore each of them is equal to half the perimeter of the triangle ABC: but fince AD, AE are equal, and BD, BF, and alfo CE, CF, AB together with FC, will be equal to half the perimeter of the triangle to which AH or AL was fhewn to be equal; taking away therefore the common AB, the remainder FC will be equal to the remainder BH: in the fame manner is it demonftrated, that BF is equal to CL and