AC, as the tangent KD to the radius DC; and that the fecant Pl. Trig. CG is to the radius CA, as the fecant CK to the radius CD.

COR. 1. Hence, if the radius AC be divided into any number of equal parts, and the fine BF be a multiple of one of them; the number of parts in BF is to the number in AC, as * BF to k AC; that is, as EH to CD: Wherefore the numbers denoting the parts in BF and AC, exprefs the ratio, which the fine of the angle ACB, in any circle, has to the radius of that circle.

In the fame manner, the ratios which the fines, verfed fines, tangents, and fecants of any angles, have to the radius, may be expreffed in numbers, if they be multiples of any parts of the radius.

In the Trigonometrical Tables, the radius is fuppofed to be divided into 10,000,000, &c. equal parts: and the fines, tangents, fecants, and verfed fines, are reprefented by the numbers, which shew what multiples of one of these parts are most nearly equal to them.

3. LEM.

COR. 2. Hence, the fines of two arches in any circle are to one another, as the numerical expreffions of the fines of the angles measured by them, in the tables. For the fine of the first arch is to the radius of the circle, as the numerical expreffion of the fine of the angle measured by that arch to the radius of the tables; and the radius of the circle is to the fine of the other arch, as the radius of the tables to the numerical expreffion of that other fine: Therefore, by equality, the firft 1 22. §i fine is to the other fine, as the numerical expreffion of the first to the numerical expreffion of the other fine.

In the fame manner, it may be proved, that the tangents, fecants, and verfed fines, of arches of the fame circle, are to one another, as their numerical expreffions.

COR. 3. From this and the preceding propofitions, it is manifeil, that in a right angled triangle, the hypothenufe is to a fide, as the radius in the tables to the numercial expreffion of the fine of the angle oppofite to that fide: and that one of the fides about the right angle is to the other, as the radius to the tangent of the angle oppofite to that other; as alfo, that a fide is to the hypothenule, as the radius to the fecant of the angle contained by them.

Pl. Trig.

SOLUTION of the CASES of RIGHT-ANGLED

TRIANGLES.

a right angled triangle, of the three fides, and either of the acute angles, any two being given, the other two and the remaining acute angle may be found.

But when the two acute angles are given, the fides cannot be found from them; but only their ratios, which are the fame with those of the fines of their oppofite angles.

When one of the acute angles is given, the other is also given, for it is the complement of the former; and therefore the fine of one of them is the cofine of the other.

C

The feveral cafes of this propofition may be refolved by the help of the first and fecond propofitions, and their corollaries, as in the following table. Or, if two fides be given, the third may be found by the 47th of the 1ft of the Elem. For the fquare of BC is equal to the fquares of BA, AC; and therefore the fquare of AB is equal to the excess of the square of BC above the fquare of AC.

B

A

In the Tables, R. fignifies the radius, fin. a fine, tan. a tangent, fec. a fecant, cos. a cofine, cot. a cotangent; and R2. fignifies the fquare of the radius, fin 2. the fquare of the fine, A the half of A. Alfo R: fin. B:: BC: CA: as radius to the fine of the angle at B, fo is the fide BC to the fide CA; and 38° 42′ 12′′ 38 degrees, 42 minutes, 12 feconds.

IN

[N any triangle, the fides are to one another, as the fines of the angles oppofite to them.

a

Pl. Trig.

If one of the angles ABC be a right angle; by making AC the radius, the fides AB, BC are the fines of the angles at C, a 1. P. T. A oppofite to them; and the radius AC is the fine of the right b Co. 7. angle ABC.

C

b

But, if none of them be a right angle; draw CD, pendiculars to AB, BC: therefore, in the right angled triangles

A

D

E

Def.

ADC, AEC, if the hypothenufe AC be made the radius, the fide CD is the fine a of the angle BAC, and AE is the fine of the angle ACB and becaufe the angles at D, E are right angles B and ABC is common to the triangles ABE, CBD; they are equiangular d; and therefore CB d 32. 1. is to CD, as BA to AE; and, alternately, CB is to BA, as f CD to AE; that is, as the fine of the angle BAC to the fine of the angle ACB.

e

PROP. IV.

IN any triangle, the fum of two fides is to their difference, as the tangent of half the fum of the angles at the bafe, to the tangent of half their diffe

rence.

Let ABC be a triangle, of which the fides AC is greater than AB; the fum of CA, AB is to their difference, as the tangent of half the fum of the angles ABC, ACB to the tangent of half their difference.

b

e 4. 6. f 16. 5.

From the centre A, at the distance AB, describe the femicircle DBE, meeting AC in E, and CA produced in D; therefore DC is the fum of CA, AB, and CE is their difference: join BD, BE; and through E draw a EF parallel to CB; therefore a 31. 1. the angle FEB is equal to EBC and because the angle BAD b29. 1. is equal to the angles ABC, ACB, and that the angle BED at c 32 1. the circumference is half of the angle BAD; the angle BED d 20. 3• is half the fum of the angles ABC, ACB: and becaufe AEB is equal to the angles EBC, ECB, and ABE is equal to AEB, for AE is equal to AB; the angle ABE is equal to the angles

Gga

e

Pl Trig. ECB, EBC: to each of thefe equals add the angle EBC; and the whole angle ABC is eq 1 to ACB and twice EBC; there

f 31. 3.

fore twice EBC is the difference
between the angles ABC, ACB;
and EBC or BEF is half their
difference but because the angle
EBD in a femicircle is a right
angle f; if BE be made the radius,

85. Def. BD is the tangent of BED, and BF the tangent of BEF: and be-. caufe FE is parallel to BC; CD

h2. 6. is to CE, as BD to BF: Wherefore, as the fum of the fides CA, AB to their difference, fo is the tangent of half the fum of the angles ABC, ACB to the tangent of half their difference.

IN

N any triangle, if a perpendicular be drawn to the bafe from the oppofite angle; the bafe is to the fum of the fides, as the difference of the fides to the excefs of twice the greater fegment above the base.

Let ABC be a triangle, of which the fide AC is greater than AB; and from A, draw a AD perpendicular to BC; and from the centre A, at the diftance AC, defcribe the circle CEF, meeting AB produced in E, F, and CB produced in G: and becaufe AD, which paffes through the centre, cuts CG at right angles in D; GD is equal to DC; therefore DC adjacent to the giater fide AC is the greater fegment: as the baie CB to the fum of the fides CA, AB, fo is the difference of CA, AB to the excefs of twice CD above the bafe CB.

D

E

Becaufe EA, AF are each of them equal to AC; BE is the fum of CA, AB, and BF their difference: and be caufe CG is double of CD, GB is the excefs of twice CD above the base CB: But because CG, EF in the circle, cut one another in B, the rectangle EB, BF is equal to the rectangle CB, BG; wherefore CB is to BE, as BF to BG. Therefore, &c.

*

PROP.

*N. B. The foregoing Propofitions are fufficient for folving the cases;

but the following are often ufed.

PROP.

VI.

Pl. Trig.

IN

N any triangle, if any angle be found fuch, that the radius is to its tangent, as one of the fides to the other; then the radius is to the tangent of the difference between this angle and half a right angle, as the tangent of half the fum of the angles at the bafe, to the tangent of half their difference.

Let ABC be a triangle of which the fides AB, AC are unequal; and drawa AD at right angles to AB, and make it equal a II. 1. to AC, and join BD; therefore, BA is to AD or AC, as the b 1. P. T. radius to the tangent of the angle ABD. It is to be proved, that the radius is to the tangent of the difference between the angle ABD and the half of a right angle, as the tangent of half the fum of the angles ACB, ABC to the tangent of half their difference.

Make AE, AF each equal

to AB, and join BE, BF, and E
draw DG para lel to BE:
and because LA is equal to

AB, the angle EBA is equal d to BEA; and EAB is a right angle; therefore each

of the angles BEA, ABE is

B

half a right angle For

e 32. I.

the fime reafon, each of the angles ABF, AFB is half of a right angle; therefore EBF is a right angle: and because DG is parallel to BE; the angle DGE is a right angle f, and each f 29. 1. of the angles GDF, GFD half of a right angle; therefore DG is equal to GF: alfo, BG being the radius, DG is the tangent b of DBG, which is the difference between the angle DBA and FBA half a right angle: and DE is the fum of the fides CA, AB, and DF their difference: but becaufe DG is parallel to BF, BG is to GF or GD, as ED to DF; that is, as the tangent of half the fum of the angles ABC, ACB, to the tangent of half their difference ". Wherefore, &c.

g 2.6.

h 3. P. Te

PROP.