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Cor. to Def 4. 5. 6. 7. The fine, versed fine, tangent, and Fig. 4.

secant, of any arch which is the measure of any given angle ABC, is to the fine, versed fine, tangent, and tecant, of any other arch which is the measure of the same angle, as

the radius of the first is to the radius of the second.
Let AC, MN be measures of the angles ABC, according to

def. 1. CD the line, DA the versed sine, AE the tangent,
and BE the fecant of the arch AC, according to def. 4. 5.
6. 7. and NO the fine, OM the versed fine, MP the tan-
gent, and BP the fecani of the arch MN, according to the
same definitions. 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 BD as BN or BM
to BO; and, by conversion, DA to MO as AB to MB.
Hence the corollary is manifest; therefore, if the radius
be supposed to be divided into any given number of equal
parts, the fine, versed fine, tangent, and secant of any given
angle, will each contain a given number of these parts; and,
by trigonometrical tables, the length of the fine, versed fine,
tangent, and fecant of any angle may be found in parts of
which the radius contains a given number : and, vice versa,
a number exprefling the length of the fine, verfed fine,
tangent, and fecant being given, the angle of which it is the
fine, versed fine, tangent, and fecant may be found.
VIII.

Fig. $: 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, according to def. 4. 6. 7, HK is called the co-tan-
gent, BD the co-fine, and BK the co-fecant of the angle

ABC.
Cor. 1. The radius is a mean proportional between the tan-

gent and co-cangent.
For, fince HK, BA are parallel, the angles HKB, ABC will
be equal, and the angles KHP, BAE are right; therefore
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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 co-line

and fecant of any angle ABC. Since CD, AE are parallel, BD is to BC or BA, as BA to

BE.

IN

PRO P. I. Fig. 5. N a right angled plain triangle, if the hypothenuse

be made radius, the sides become the lines of the angles opposite to them; and if either side be made radius, the remaining side is the tangent of the angle opposite to it, and the hypothenuse the secant of the fame angle.

Let ABC be a right angled triangle ; if the hypothenuse BC be made radius, either of the sides AC will be the fine of the angle ABC oppofite to it; and if either side BA be made ra. dius, the other fide AC will be the tangent of the angle ABC opposite to it, and the hypothenufe BC the secant of the same angle.

About B as a centre, with BC, BA for distances, let two circles CD, EA be described, meeting Bil, BC in D, E: Since 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 hypothenuse a side and an angle of a right angled triangle, any two being given, the third is also given.

Cor. 2. Of the iwo sides and an angle of a right angled tri. angle, any two being given, the third is also given.

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THE sides of a plain triangle are to one another, as

the fines of the angles opposite to them.

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

In

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

From C, B draw CE, BD perpendicular upon the opposite 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 iwo iri. aogles CAE, DAB have each a right angle at. D and E; and likewise the common angle CAB; therefore they are similar, and consequently, CA is to AB, as CE to DB; that is, the fides are as the lines of the angles oppofire to them.

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

ven,

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IN a plain triangle, the sum of any two files 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 sum of any two sides AB; AC will be to their difference as the tangent of half the sum of the angles at the base ABC, ACB to the tangent of half their difference.

About A as a center, with AB the greater side 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 sum of the angles at the base, and the angle EFB at the circumference is equal to the half of EAB at the center (20. 3.); therefore EFB is half the sum of the angles at the base; but the angle ACB (32. 1.) is equal to the angles CAD and ADC, or ABC to gether; therefore 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 since the angle EBF in a semicircle is a right angle (1. of this) FB being radius, BE, BG, are the tangents of the angles EFB, BFG; but it is manifest that EC is the sum of the sides BA, AC, and CF their difference; and since BC, FG are parallel (2. 6.) EC is to CF, as EB to BG; that is, the sum of the

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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.

PRO P. IV. FIG. 18. IN any plain triangle BAC, whote two sides are BA,

AC and base BC, the less of the two sides, 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 tangent of the excess of this angle above half a right angle as the tangent of half the sun of the angles B and C at the bale, 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; joio BE, BF, BD, and from D, draw DG perpendicular upon BF. And because 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; allo (4. 1. El.) EB is equal to BF. And since EBF, FGD are right angles, EB is parallel to GD, and the triangles EBF, FGD are similar; 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 live is to AD or AC the greater, as the radius is to the rangent of the angle ABD; and because BGD is a right angle, BG is to GD or GF as the radius is to the can. gent 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 sum of the fides BA, AC and FD their difference, (3. of this), as the tan. gent of half the sum of the angles B, C, at the base to the tangent of half their difference. Therefore, in any plain triangle, &c. e E. D.

PRO P. V. Fig. 9. and 10. IN any triangle, twice the rectangle contained by any

two sides is to the difference of the sum of the {quares of these two sides, and the square of the base, as the radius is to the co-line of the angle included by the two sides.

Let ABC be a plain triangle, twice the rectangle ABC COD gained by any iwo lides BA, BC is to the difference of the fum

ct

of the squares of BA, BC, and the square of the base AC, as the radius to the co-fine of the angle ABC.

From A, draw AD perpendicular upon the opposite fide BC; then (by 12. and 13. 2. El.). The difference of the sum of the squares of AB, BC, and the square of the base AC, is equal to twice the rectangle CBD; but i wice the rectangle CBA is to twice the rectangle CBD, that is, to the difference of the sum of the squares of AB, BC, and the square of AC, (1. 6.) aś AB 10 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 co-fine of ABC.

PRO P. VI. Fig. U.

IN any triangle ABC, whose two sides are AB, AC,

and base BC; the rectangle contained by half, je perimeter, and the excess of it above the base BC, is to the rectangle contained by the straight lines, by which the half of the perimeter exceeds the other two sides AB, AC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base.

Let the angles BAC, ABC be bisected by the straight lines AG, BG; and, producing the fide AB, let the exterior angle CBH be bisected by the straight line BK, meeting AG in K; and from the points G, K, let there be drawn perpendicular upon the sides the straight lines GD, GE, GF, KH, KL, KM. Since therefore (4.4.) G is the centre of the circle in. scribed 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 straight lines BK, KA: And because in the tri. angles KCL, KCM, the sides LK, KM are equal, KC is common and KLC, KMC are right angles, CL will be equal to CM: Since 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 since AD, AE are equal, and BD, BF, and also CE, CF, AB together with FC, will be equal to half the perimeter of the triangle to which AH or AL was shewn to bę equal; taking away therefore the common AB, the remain.

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