given angle, will each contain a given number of these parts; and, by trigonometrical tables, the length of the sine, versed sine, tangent, and secant, of any angle, may be found in parts of which the radius contains a given number; and, vice versa, a number expressing the length of the sine, versed sine, tangent, and secant, being given, the angle of which it is the sine, versed sine, tangent, and secant, may be found.

VIII.

Fig. 3. 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 sine and BK the secant of CBH, the complement of ABC, according to def. 4, 6, 7, HK is called the co-tangent, BD the co-sine, and BK the co-secant, of the angle ABC.

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

For, since HK, BA are parallel, the angles HKB, ABC will be equal, and the angles KHB, BAE are right: therefore the triangles BAE, KHB are similar, and therefore AE is to AB, as BH or BA to HK. COR. 2. The radius is a mean proportional between the co-sine and secant of any angle ABC.

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

PROP. I. FIG. 5.

In a right angled plane triangle, if the hypothenuse be made radius, the sides become the sines 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 same angle.

Let ABC be a right angled triangle; if the hypothenuse BC be made radius, either of the sides AC will be the sine of the angle ABC opposite to it; and if either side BA be made radius, the other side AC will be the tangent of the angle ABC opposite to it, and the hypothenuse 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 BA, BC in D, E: since CAB is a right angle, BC being radius, AC is the sine of the angle ABC by def. 4, and BA being radius, AC is the tangent, and BC the secant 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 two sides and an angle of a right angled triangle, any two being given, the third is also given.

PROP. II. FIG. 6, 7.

THE sides of a plane triangle are to one another as the sines 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 sides are the sines of the angles opposite to them, and the radius is the sine of a right angle (cor. to def. 4.) which is opposite to the hypothenuse.

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

From C, B draw CE, BD perpendicular upon the opposite sides AB, AC produced, if need be. Since CEB, CDB, are right angles, BC being radius, CE is the sine of the angle CBA, and BD the sine of the angle ACB: but the two triangles 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 sides are as the sines of the angles opposite to them.

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

PROP. III. FIG. 8.

In a plane triangle, the sum of any two sides 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 plane 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 centre, 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 centre (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 together: 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 sides is to their difference, as the tangent of half the sum of the angles at the base to the tangent of half their difference.

PROP. IV. FIG. 18.

In any plane triangle BAC, whose 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 sum of the angles B and Cat the base 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 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; also (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 side 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, since ED is the sum of the sides BA, AC, and FD their difference (3. of this), as the tangent of half the sum of the angles B, C, at the base, to the tangent of half their difference. Therefore, in any plane triangle, &c. Q. E. D.

PROP. V. FIG. 9. 10.

IN any triangle, twice the rectangle contained by any two sides is to the difference of the sum of the squares of these two sides, and the square of the base, as the radius is to the co-sine of the angle included by the two sides.

Let ABC be a plane triangle; twice the rectangle ABC contained by any two sides BA, BC, is to the difference of the sum of the squares of BA, BC, and the square of the base AC, as the radius to the co-sine 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 sum of the squares of AB, BC, and the square of the base AC, is equal to twice the rectangle CBD; but twice 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.), as AB to BD; that is, by prop. 1, as radius to the sine of BAD, which is the complement of the angle ABC, that is, as radius to the co-sine of ABC.

PROP. VI. FIG. 11.

In any triangle ABC, whose two sides are AB, AC, and base BC, the rectangle contained by half the 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 side 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 in the centre 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 bisected by the straight lines BK, KA: and because in the triangles 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 shown to be equal; taking away therefore the common AB, the remainder FC will be equal to the remainder BH; in the same manner it is demonstrated, that BF is equal to CL; and since the points B, D, G, F, are in a circle, the angle DGF will be equal to the exterior and opposite angle FBH (22. 3.); wherefore their halves BGD, HBK will be equal to one another: the right angled triangles BGD, HBK will therefore be equiangular, and GD will be to BD, as BH to HK, and the rectangle contained by GD, HK will be equal to the rectangle DBH or BFC; but since AH is to HK, as AD to DG, the rectangle HAD (22. 6.) will be to the rectangle contained by HK, DG, or the rectangle BFC, (as the square of AD is to the square of DG, that is) as the square of the radius to the square of the tangent of the angle DAG, that is, the half of BAC: but HA is half the perimeter of the triangle ABC, and AD is the excess of the same above HD, that is, above the base BC: but BF or CL is the excess of HA or AL above the side AC; and FC, or HB, is the excess of the same HA above the side AB; therefore the rectangle contained by half the perimeter, and the excess of the same above the base, viz. the rectangle HAD, is to the rectangle contained by the straight lines by which the half of the perimeter exceeds the other two sides, that is, the rectangle BFC, às

the square of the radius is to the square of the tangent of half the angle BAC opposite to the base. Q. E. D.

PROP. VII. FIG. 12. 13.

In a plane triangle, the base is to the sum of the sides, as the difference of the sides is to the sum or difference of the segments of the base made by the perpendicular upon it from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base.

Let ABC be a plane triangle; if from A the vertex be drawn a straight line AD perpendicular upon the base BC, the base BC will be to the sum of the sides BA, AC, as the difference of the same sides is to the sum or difference of the segments CD, BD, according as the square of AC the greater side is greater or less than the sum of the squares of the lesser side AB, and the base BC.

About A as a centre, with AC the greater side for a distance, let a circle be described meeting AB produced in E, F, and CB in G: it is manifest, that FB is the sum, and BE the difference of the sides; and since AD is perpendicular to GC, GD, CD will be equal; consequently GB will be equal to the sum or difference of the segments CD, BD, according as the perpendicular AD meets the base, or the base produced; that is, (by conv. 12. 13. 2.) according as the square of AC is greater or less than the sum of the squares of AB, BC: but (by 35. 3.) the rectangle CBG is equal to the rectangle EBF; that is, (16. 6.) BC is to BF, as BE is to BG, that is, the base is to the sum of the sides, as the difference of the sides is the sum or difference of the segments of the base made by the perpendicular from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base. Q. E. D.

PROP. VIII. PROB. FIG. 14.

THE sum and difference of two magnitudes being given, to find them.

Half the given sum added to half the given difference, will be the greater, and half the difference subtracted from half the sum, will be the less.

For, let AB be the given sum, AC the greater, and BC the less. Let AD be half the given sum; and to AD, DB, which are equal, let DC be added; then AC will be equal to BD and DC together; that is, to BC, and twice DC; consequently twice DC is the difference, and DC half that difference; but AC the greater is equal to AD, DC; that is, to half the sum added to half the difference, and BC the less is equal to the excess of BD, half the sum above DC half the difference. Q. E. D.