VIL tremity of the tangent AE, is called the Secant of the arch AC or angle ABC. Cor. to def. 4. 6. 7. The sine, tangent, and secant of any angle ABC, are likewise the sine, tangent, and secant of its supplement CBF. It is manifest from def. 4. that CD is the sine of the angle CBF. Let CB be produced till it meet the circle again in G; and it is manifest that AE is the tangent, and BE the secant, of the angle ABG or EBF, from def. 6. 7. Cor. to def. 4. 5. 6. 7. The sine, versed sine, tangent, Fig. 4. and secant, of any arch, which is the measure of any given angle ABC, is to the sine, versed sine, tangent, and secant, 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. ing to def. 1.; CD the sine, DA the versed sine, VIII. 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 cotangent, BD the cosine, and BK the cosecant, of the angle ABC. Cor. 1. The radius is a mean proportional between the tangent and cotangent. 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 cosine 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 Cor. 2. Of the two sides and an angle of a right angled triangle, any two being given, the third is also given. 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, AČ, 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 C at 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. 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 and 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 cosine 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 cosine 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 A BC; that is, as radius to the cosine 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, AČ, 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 is 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 |