as BN or BM to BO; and, by conversion and alternation, 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 sine, versed sine, tangent, and secant of any 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. IX. The difference between any angle and 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 acute angle ABC, or of the obtuse angle CBF. In like manner, the difference between any arch and a quadrant is called the complement of that arch. Thus HC is the complement of the arch AC, or of the arch FC. X. 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. 5. 7, 8, 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 tan gent and cotangent of any angle ABC. For, since HK, BA are parallel, the angles HKB, ABC are equal, and KHB, BAE are right angles; 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. Note 1.-For the sake of brevity, certain signs and characters, borrowed from arithmetic, and some obvious contractions are often used in trigonometrical investigations. Thus, if a and b denote any two numbers, their sum is denoted by a+b; their difference by a-b, or a b; their product by axb, or a.b; their quotient by; their squares by a2 and b b2; their square roots by a and b; the square root of the sum of their squares by (a+b); the product of their sum into the sum of any other numbers c and d, by (a+b) x (c+d), or (a+b).(c+d). The mark=denotes the equality of the quantities between which it is written: thus, a=b denotes that a is equal to b; and in the statement of analogies, abc:d, or a: bc:d, denotes that a is to b as c is to d, or that the ratio of a to b is the same with that of c to d. Thus also rad or R is used for radius, sin for sine, tan for tangent, sec for secant, cos for cosine, cot for cotangent, cosec for cosecant; and sin2, cos2, tan2, rad2, &c. for the squares of the sine, cosine, tangent, radius, &c. respectively. NOTE 2. In a right angled triangle, the side subtending the right angle is called the hypotenuse; and the other two sides which contain the right angle are called the legs; one of the legs is also called the perpendicular, and the other the base, according to their position. *PROP. I. FIG. 5. In a right angled plane triangle, the hypotenuse is to either of the legs as the radius to the sine of the angle opposite to that leg, and either of the legs is to the other leg as the radius to the tangent of the angle adjacent to the former leg. Let ABC be a right angled plane triangle, of which AC is the hypotenuse; assume AG as the tabular radius; from the centre A with the radius AG describe the arch DG, draw DE perpendicular to AG, and from G draw GF touching the circle in G and meeting AC in F; then is DE the sine, and FG the tangent of the arch DE, or of the angle A. The triangles AED, ABC are equiangular, because the angles AED, ABC are right angles, and the angle A is common; therefore AC is to CB as AD to DE; but AD is the radius, and DE the sine of the angle A; consequently AC: CB: rad: sin A. Again, because FG touches the circle in G, AGF is a right angle, and therefore equal to the angle B, and the angle A is common to the two triangles ABC, AGF; these triangles are therefore equiangular; consequently, AB is to BC as AG to GF; but AG is the radius, and FG the tangent of the angle A; therefore AB: BC:: rad : tan A. COR. 1. Since AF is the secant of the angle A, (def. 8.), and the triangles AFG, ACB are equiangular, BA is to AC as GA to AK; that is, BA: AC:: rad: sec A. COR. 2. In a right angled plane triangle, if the hypotenuse be made radius, the sides become the sines of their opposite angles; and if either leg be made radius, the other leg becomes the tangent of its opposite angle, and the hypotenuse the secant of the same angle. 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 Proposition is manifest from Prop. 1; for if the hypotenuse 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 hypotenuse. 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. From A, draw and AD AC: : *Otherwise. Fig. 16, 17. AD perpendicular to BC; then, by Prop. 1. sin C: rad; therefore, ex æquo inversely, BA: AC: sin C: sin B. LEMMA III. If there be two unequal magnitudes, half their difference added to half their sum is equal to the greater, and half their difference taken from half their sum is equal to the less. Let AC and CB be two unequal magnitudes, of which AC is the greater, and AB the sum. Bisect AB in D; 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. Therefore, &c. Q. E. D. COR. Hence, if the sum and difference of two magnitudes be given, the magnitudes themselves may be found; for to half the sum add half the difference, and it will give the greater; from half the sum subtract half the difference, and it will give the less. 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 diffe rence. 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, (def. 7.) 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. 8. In a plane triangle, the cosine of half the difference of any two angles is to the cosine of half their sum, as the sum of the opposite sides to the third side; and the sine of half the difference of any two angles is to the sine of half their sum, as the difference of the opposite sides to the third side. Let ABC be a plane triangle, then, cos (CB): cos (C+B):: BA+AC: BC, and sin (CB): sin (C+B) :: BA AC: BC. For, in the preceding proposition, it was shown, that EFB is equal to (C+B), and that CBF is equal to (CB); and since EBF is a right angle, CBE is the complement of CBF, and E the complement of BFE. Now, in the triangle CBE, sin CBE sin E:: CE: BC; that is, cos (CB): cos (C+B) :: AB+ AC: BC. Again, in the triangle CBF, sin CBF: sin CFB:: CF: BC; that is, sin (CB) sin (C+B):: AB-AC: BC. Therefore, &c. Q. E. D. PROP. V. 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 to 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 |