of that arc.*

The tangent of 45° is equal to the radius,† and an arc of 90° has no tangent.

(M) The secant of an arc is a straight line drawn from the centre of the circle, and produced till it meets the tangent. Thus CT is the secant of the arc BF, or of the angle BCF.

The secant of any arc is equal to the secant of the supplement of that arc. The secant of 45° is equal to double the sine of 45°,§ and an arc of 90° has no secant.

(N) The versed sine of an arc is that part of the diameter which is contained between the sine and the arc. Thus BG is the versed sine of the arc BF, and be is the versed sine of the supplemental arc FAнb.

The versed sine of the supplement of an arc, is the difference between the versed sine of that arc and the diameter.

For bGbB GB; or, GBbBbG.

(0) The co-sine of an arc is the sine of the complement of that arc; or it is that part of the diameter contained between the centre of the circle and the sine. Thus FE is the cosine of the arc BF, being the sine of AF, which is the complement of BF; or CG is the cosine of BF, because CG = FE.

The cosine of an arc is equal to the cosine of its supplement. The cosine of an arc, less than a quadrant, is equal to the radius diminished by the versed sine. For CG = CB

GB.

(P) The co-tangent of an arc is the tangent of the complement of that arc. Thus AK is the co-tangent of the arc BF, being the tangent of the arc AF, which is the complement of BF.

The co-tangent of an arc is equal to the co-tangent of its supplement.

(Q) The co-secant of an arc is the secant of the complement of that arc. Thus CK is the co-secant of the arc BF, being the

secant of the arc AF.

* The supplemental tangent will fall on the contrary side of the diameter. Let BF be any arc, (Plate I. fig. 1.) its supplement is rafb, bt is the tangent of the supplement, and it is equal to BT. For the right angled triangles CBT and cbt are equi-angular, and have the side cв equal to co.

For in this case the angle BCT, measured by the arc BF (Plate I. fig. 1.,) will be half a right angle; consequently the angle BTC will be half a right angle; the triangle is therefore isosceles, and BC= BT.

Though the secant of an arc, and the secant of its supplement, be the same in quantity, they have contrary directions like the tangents. For bt (Plate I. fig. 1.) has been shewn to be the tangent of the supplemental arc Fafb, and equal to TB; it is plain therefore that cr=cl.

For if the arc BF = 45°, BC = BT; and (Euclid 47. of I.) CB 2+ BT2 = CT 2 the square of the secant, CB2 + AC2=AB2; therefore AB=CT, that is, the chord of 90° is equal to the secant of 45°. But the sine of 45° is equal to half the chord of 90° (K. 31. and Note), therefore the sine of 45° is equal to half its

secant.

The co-secant of an arc, and the co-secant of its supplement, are of the same magnitude.

(R) The co-versed sine of an arc is the versed sine of the complement of that arc. Thus, AE is the co-versed sine of the arc BF, being the versed sine of the arc AF.

(S) The co-chord of an arc is the chord of the complement of that arc.

(T) From the preceding definitions it appears that the sine and co-sine of any arc can never exceed the radius; the secant and co-secant always exceed the radius, and the tangent and co-tangent admit of all possible degrees of magnitude. Also, that any side of a triangle may be considered as the radius of a circle, and that the other sides will necessarily become either sines, tangents, or secants.

(U) Thus, 1st. If the hypothenuse ac be made the radius of a circle, it is evident that the perpendicular BC is the sine of the angle A, and that AB is the cosine thereof But the sine of either of the acute angles of a right angled triangle is the cosine of the other, and the contrary; therefore BC is the cosine of c, and AB is the sine thereof.

(W) 2dly. If the base AB be considered as the radius of a circle, BC is evidently the tangent of the angle A, and AC is the secant thereof. But the tangent of one angle is the co-tangent of the other; also the secant of one angle is the co-secant of the other: therefore BC is the co-tangent of c, and ac is the co-secant of c.

Rad

Sinte

A Cos B
A

Sec A

A Rad

Tang

(X) 3dly. If the perpendicular BC be considered as the radius of a circle, it is evident that the base AB will be the tangent of the angle c, and the hypothenuse AC will be the secant thereof. But because the tangent, secant, &c. of one angle of a right angled triangle is the co-tangent, co-secant, &c.' of the other; the base AB will be the co-tangent A Tang B of the angle ▲, and the hypothenuse ac will be

its co-secant.

Ꭰ

INVESTIGATION OF RULES FOR CALCULATING THE SIDES AND ANGLES OF PLANE TRIANGLES.

PROPOSITION I. (Plate I. Fig. 4.)

(Y) In any right angled plane triangle, if any of the three sides be made the radius of a circle; the other sides will be either sines, tangents, or secants, of the respective angle, correspondent to that radius; and will be similar to the sines, tangents, or secants in the tables, when compared with the tabular radius.

The sines, tangents, &c. of all angles whatever, are calculated to the logarithmical radius of ten thousand millions, and arranged in tables for the convenience of calculation. Now as no triangle can be formed, but another may be formed similar to it, and that the sides about the equal angles of similar triangles are proportional; if ad, ce, ab, and cỏ, each represent the radius in the tables, then will AF and CG be the secants, br and Gb the tangents, and Dd and Ee the sines in the tables. Then,

I. If the hypothenuse be the radius of a circle.

Because the triangles e EC and ABC are similar,

viz. radius : hypoth, :: sine of angle o cosine of ▲ : base. Again, because the triangles and and ABC are similar,

Ad :

AC

viz. radius : hypoth.

::

Dd

:: sine of angle a=cosine of c: perp.

(Z) THEREFORE, the sine of any angle is to the side opposite to it, as the sine of any other angle is to its opposite side.

II. If the base be the radius of a circle.

Because the triangles aby and abc are similar,

radius :

base,

viz. secant of angle a=cosec. of c: hypoth. ::

and, secant of angle a=cosec. of c: hypoth, tang, of angle a=cotan, of c : perp.

(A) HENCE. Radius: base:: tangent of the angle a, or cotangent of the angle c: perpendicular.

III. If the perpendicular be the radius of a circle,

Because the triangles abc and ABC are similar,

CG : AC :: ba : AB
And, co : AC : cb : .BC

viz. secant of angle c=cosec. of a and, secant of angle c=cosec. of a

hypoth.::tang. of angle c=cotan. of ▲ base. hypoth.:: radius : perpend.

(B) HENCE. Tangent of angle c, or co-tangent of A: base:: radius: perpendicular.

Any of the foregoing quantities are proportional by inver sion, &c.

PROPOSITION II.

(C) The excess of the greater of two given magnitudes above half their sum, is equal to half the difference between those magnitudes or half their sum increased by half their difference, gives the greater magnitude; and being diminished by half their difference, leaves the less magnitude.

:

Let Ac and CB be two unequal

magnitudes, of which ac is the A EDC greater.

B

Bisect AB in D and make AECB, then will AD-half the sum of AC and CB, and DC half the difference; for, since AB is bisected in D, and AECB, it follows that EC is bisected in D, and EC is the difference between ac and CB.

Hence AC-AD≈DC; or AD+DCAC, and AD-(ED=DC) = AE (BC). Q. E. D.

PROPOSITION HI.

(D) In any plane triangle, the sine of any angle, is to the side opposite to it, as the sine of any other angle, is to the side opposite to it, and the contrary.

Since any triangle whatever may be inscribed in a circle, let ABC be the given triangle and o the centre of the circle. Draw OD and OE per- A pendicular to the sides of the triangle, then (by Euclid 3 of III. and 30 of III.) the sides of the triangle

E

D

B

will be bisected, and also the arcs which these sides subtend. Now an angle at the centre of a circle is double to an angle at the circumference when they stand on the same arc, (Euclid 20 of III.) therefore an angle at the centre is equal to an angle at the circumference standing on the double arc; or in other words, an angle at the circumference of a circle is measured by half the arc on which it stands, and an angle at the centre by the whole arc on which it stands: hence the angle cop=the angle CAB, and COE=CBA, &c. Now CI is the sine of the angle COD, and cm is the sine of the angle COE. Draw mi, which will be parallel to AB, then per similar triangles ct: CB: cm:

AC, viz. the sine of the angle at A, is to the side BC, as the sine of the angle at B, is to the side Ac, and the contrary by inversion.

COROL. The sides of triangles are to each other as the chords of double their opposite angles.

PROPOSITION IV.

(E) 1. In any plane triangle, the sum of any two sides is to their difference, as the tangent of half the sum of their opposite angles, is to the tangent of half their difference.

Let ABC be any triangle; make BEBC and join CE, then the angle CBE, being the exterior angle of the triangle ABC, is equal to CAB+BCA. (Euclid 32

of I.)

Bisect AE in D and CE in G, then join DG and it will be parallel to AC: draw BF parallel to DG, join BG, and

with the centre в and radius BG describe a circle.

The angle CBG the angle EBG equal to half the sum of the angles CAB and BCA; for the triangles CBG and EBG have the two sides BC and CG, equal to the two sides BE and EG, and the side BG common to both, therefore they are equal in all respects. (Euclid 8 of I.)

BF being parallel to AC, the angle FBE the angle CAB, and the angle CBF the angle ACB. (Euclid 29 of I.)

=

The angle GBF half the difference between the angles CBF and FBE, for it is the excess of the greater above half their sum GBE (C. 35); hence GE tangent of the half sum, and gr= tangent of the half difference between the two angles CBF and FBE. For the same reason, DB being the excess of the greater of the two given lines AB and BE or BC, above half their sum AD, it is equal to half their difference.

By similar triangles, DE (AB+BC): DB (AB-BC):: GE tangent (ACB+CAB): GF tangent (ACB-CAB) Q. E. D. Half the sum of the angles ACB and CAB added to half their difference gives the angle opposite to the greater side (C. 35) for the greater angle is opposite to the greater side: and half the sun diminished by half the difference, gives the angle opposite to the less side; for the less angle is opposite to the less side. (Euclid 18 of I.)

(F) NOTE. Instead of the tangent of half the sum of the opposite angles, the tangent of half the supplement of the con