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AC, as the tangent KD to the radius DC; and that the fecant Pl. Trig. CG is to the radius CA, as the secant 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 k BF tok 3. LEM. AC; that is, as EH to CD: Wherefore the numbers denoting the parts in BF and AC, express the ratio, which the fine of the angle ACB, in any circle, has to the radius of that circle.

In the same manner, the ratios which the fines, versed fines, tangerts, and secants of any angles, have to the radius, may be expressed in numbers, if they be multiples of any parts of the radius.

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

Cor. 2. Hence, the fines of two arches in any circle are to one another, as the numerical expressions of the fincs of the angles measured by them, in the tables. For the fine of the firit arch is to the radius of the circle, as the numerical ex. pression 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 expression of that other fine: Therefore, by equality', the first 122. S. fine is to the other fine, as the numerical expression of the first to the numerical expression of the other fine.

In the same manner, it may be proved, that the tangents, fecants, and versed sines, of arches of the same circle, are to one another, as their numerical expressions.

CoR. 3. From this and the preceding propofitions, it is mania feit, that in a right angled triangle, the hypothen use is to a fide, as the radius in the tables to the numercial expression of the fine of the angle opposite to that fide: and that one of the kides about the righe angle is to the other, as the radius to the tangent of the angle opposite to that other; as also, that a side is to the hypothenule, as the radius to the fecant of the angle cóna tained by then,

SOLU

Pl. Trig.

SOLUTION of the CASES OF RIGHT-ANGLED

TRIANGLES.

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GENERAL PROPOSITION.
N 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 same with those of the fines of their opposite 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.
The several cases of this proposition may be

С
resolved by the help of the first and second
propofitions, and their corollaries, as in the
following table. Or, if two sides be given,
the third may be found by the 47th of the ift
of the Elem. For the square of BC is equal

B

A to the squares of BA, AC; and therefore the square of AB is equal to the excess of the square of BC above the {quare of AC.

In the Tables, R. fignifies the radius, fin. a fine, tan. a tan- gent, sec. a fecant, cos. a cofine, cot. a cotangent; and R2, fignifies the square of the radius, fin 2. the square of the fine, ÇA the half of A. Allo R: sin. 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 seconds.

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PRO P. III.

Pl. Trig.

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CI2. I.

N any triangle, the sides are to one another, as the

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

Def. angle ABC.

But, if none of them be a right angle; draw (CD, AE per. pendiculars to AB, BC: therefo:e, in the right angled triangles ADC, AEC, if the hypothenuse

А AC be made the radius, the side CD is the fine a of the angle BAC, and AE is the fine of the angle ACB: and because 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. I. is to CD, as e BA to AE; and, alternately, CB is to BA, asf e 4. 6. CD to AE; that is, as the fine of the angle BAC to the fine of the angle ACB.

f 16. 5.

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IN any triangle, the sum of 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 diffe

rence.

Let ABC be a triangle, of which the sides AC is greater than AB; the sum 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.

From the centre A, at the distance AB, describe the semicircle DBE, meeting AC in E, and CA produced in D; therefore DC is the sum of CA, AB, and CE is their difference : join BD, BE; and through E draw a EF parallel to CB; therefore the angle FEB is equal to EBC: and because the angle BAD 629. 1. is equal to the angles ABC, ACB, and that the angle BED at c 34. s. the circumference is half of the angle BAD; the angle BED d 20. 3. is half the sum of the angles ABC, ACB : and because AEB is equal to the angles EBC, ECB, and ABE is equal to AEB, e 5.1, for AE is equal to AB; the angle ABE is equal to the angles

Gg

ECB,

a

31. 1.

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Pl Trig. ECB, EBC: to each of these equals add the angle EBC; and V the whole angle ABC is equal to ACB and twice EBC; there

fore twice EBC is the differece

between the angles ABC, ACB; cand EBC or BEF is half their

A difference : but because the angle

EBD in a semicircle is a right f 31. j. angle f; if BE be made the radius, 8 5. Def. BD is the tangent 8 of BED, and BF the tangent of BEF: and be

B cause FE is parallel to BC; CD h 2.6.

is to CE, as BD to BF 1: Wherefore, as the sum of the fides CA, AB to their difference, so is the tangent of half the luin of the angles ABC, ACB to the tangent of half their difference.

IN

PROP. V.
N any trianglé, if a perpendicular be drawn to the

base from the oppofie angle; the base is to the Sum of the fides, as the difference of the fides to the excess of twice the greater segment above the base.

Let ABC be a trianole, of which the fide AC is greater than AB; and from A, draw a AD perpendicular to BC; ard from the centre A, at the dittance AC, describe the circle CEF, meeting AB produced in E, F, and CB produced in G: and becaufe AD, which passes through the centre, cuts CG at right argles in D; GD is equal to DC; therefore DC adjacent to the giater fide AC is the greater segment: as the bale CB to the sum of the fides CA, AB, fo is the difference of CA, AB to the excess of twice CD above the base CB.

Because EA, AF are each of them
equal to AC; BE is the sum of CA,

B.
AB, and BF their difference : and be-

C caule CG is double of CD, GB is the

D excess of twice CD above the base CB: But because CG, EF in the circle, cut one another in B,

the D 35. 3. rectangle EB, BF is equal to the rectangle CB, KG; wherefore a 16.6. CB is to BE, as B to BG. Therefore, &c. *

G

3

PROP.

* N.B. The foregoing Propofitions are fufficient for folying the cases; but the following are often used.

PROP.

VI.

Pla Trig.

IN

N any triangle, if any angle be found fuch, that

the radius is to its tangent, as one of the sides 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 sum of the angles at the tase, to the tangent of half their difference.

Let ABC be a triangle of which the sides AB, AC are un. equal; and draw a AD at right angles to AB, and make it equal a 11. 8. 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 sum of the angles ACB, ABC to the tangent of half their difference. Make AE, AF each equal

G to AB, and join BE, BF, and E А F draw C DG para lel to BE:

DC31. It and because EA is equal to AB, the angle EBA is equal to BEA; and EAB is a

d 5.1. right angle; therefore each of the angles BEA, ABE is

B

C half a right angle For

e 32. . the sime reason, 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: also, BG being the radius, DG is the tangent o of DBG, which is the difference between the angle DBA and FBA half a right angle: and DE is the sum of the sides CA, AB, and DF their difference: but because DG is parallel to BF, BG is to GF or GD, as & ED to DF; that is, as the tan- g 2.6. gent of half the sum of the angles ABC, ACB, to the tangent of half their difference h. Wherefore, &c.

h 3. P. T.

PROP.

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