The SECANTS are deduced by cor. 4. to the definitions, since they are the reciprocals of the cosines. From the lower tangents and secants, the tangents of arcs that exceed 45° are most easily derived; for (cor. 2. Prop. 4. T.) T,45°+a = Sec,2a + T,2a, Thus, T,46° = Sec,2°+T,2°, or 1.0355303 = 1.0006095+.0349208. PROP. VII. THEOR. In a right angled triangle, the radius is to the sine of an oblique angle, as the hypotenuse to the opposite side. Let the triangle ABC be right angled at B; then R: S,CAB:: AC: CB. For assume AR equal to the given radius, describe the arc RD, and draw the perpendicular RS. The triangles ARS and ACB are evidently similar, and therefore AR: RS ::AC: CB. But, AR being the radius, RS is the sine of the arc RD which measures the angle RAD or CAB; and consequently R: S,A :: AC : CB. S D B PROP. VIII. THEOR. In a right angled triangle, the radius is to the tangent of an oblique angle, as the adjacent side to the opposite side. Let the triangle ABC be right angled at B; then R:T,BAC:: AB: BC. For, assuming AR equal to the given radius, describe the arc RD, and draw the perpendicular RT. The triangles ART and ABC being similar, AR: RT::AB: BC. But, AR being the radius, RT is the tangent of the arc RD which measures the angle at A; and therefore R: T,A::AB: BC. A D R B Cor. Hence the radius is to the secant of an angle, as the adjacent side to the hypotenuse. For AT is the secant of the arc RD, or of the angle at A; and, from similar triangles, AR: AT::AB: AC. PROP. IX. THEOR. The sides of any triangle are as the sines of their opposite angles. In the triangle ABC, the side AB is to BC, as the sine of the angle at C to the sine of that at A. For let a circle be described about the triangle; and the sides AB and BC, being chords of the intercepted arcs or of the angles at the centre, are (cor. def. T.) equal to twice the sines of the halves of those angles, or the angles ACB and CAB at the circumference. But, of the same angles, the chords or sines (VI. 35. El.) are propor tional to the radius; and consequently AB: BC :: S,C: S,A. A B C PROP. X. THEOR. In any triangle, the sum of two sides, is to the difference, as the tangent of half the sum of the angles at the base, to the tangent of half their difference. 2 In the triangle ABC, AB + AC: AB-AC::T,C+B: TC-B 2 For, by the last proposition, AB : AC :: S,C: S,B, and consequently (V. 12. El.) AB+ AC: AB-AC:: S,C+S,B: S,CS,B. But, by Prop. 4, S,C + S,B : ; wherefore S,C — S,B :: T, C+B : T,C-B 2 AB+AC:AB—AC : : T,C+B.T,C—B. Ꭲ Otherwise thus: From the vertex A, and with a distance equal to the greater side AB, describe the semicircle FBD, meeting the other side AC extended both ways to F and D, join BD and BF, which produce to meet straight line DE drawn parallel to CB. Because the isosceles triangle DAB, has the same vertical angle with the triangle CAB, each of its remaining angles ADB and ABD is (I. 34. El.) F E B equal to half the sum of the angles ACB and ABC; and therefore (II. 13. cor.) the defect of ABC from that mean, that is the angle CBD, or its alternate angle BDE, must be equal to half the difference of those angles. Now FBD being (III. 26. El.) a right angle, BF and BE are tangents of the angles BDF and BDE, to the radius DB, and hence are proportional to the tangents of those angles with any other radius. But since CB and DE are parallel, CF, or AB + AC: CD, or AB-AC:: BF: BE; consequently AB + AC: AB—AC ::T, ACB + ABC 2 : T, ACB-ABC ,or AB+AC:AB—AC :: Cot. A Cot. B+A,or-Cot. C+A. : Cor. Suppose another triangle abc to have the sides ab and ac equal to AB and BC, but containing that is, RT, 45°-b:: Cot. B: Cot. B+ A, or, - Cot. C+A. Now, in the right angled triangle abc, ab, or AB, is to ac, or AC, as the radius, to the tangent of the angle at b. PROP. XI. THEOR. In any triangle, as twice the rectangle under two sides, is to the difference between their squares and the square of the base, so is the radius, to the cosine of the contained angle. In the triangle ABC, 2AB × BC : AB2 + AC2 — BC* R: Cos, BAC. |