BC, FI, KN, respectively, are also the sines of the angle at A. The lines BL, FO, KP, which are the tangents of the same arcs, are likewise the tangents of the angle at A. ART. 26. It appears from cor. to 15.4, that the side of a regular hexagon, inscribed in a circle, is equal to the radius of the circle. But the side of a regular hexagon, inscribed in a circle, subtends an arc of 60°; hence the chord of 60° is equal to the radius of the circle. Again, since a quadrant subtends a right angle at the centre of the circle (Art. 22), it is evident that the sine of a quadrant, or 90°, is the radius of the circle (see Fig. p. 34); thus HC the sine of AH, or ACH is the radius of the circle. Further, if we suppose CG to bisect the right angle ACH, we shall have CGA (which HCG, by 29.1) = ACG; whence AG CA; that is, the tangent of 45°, or half a right angle, the radius. Thus it = = = appears that the chord of 60°, the sine of 90°, and the tangent of 45°, are respectively equal to the radius of the circle. Trigonometrical Propositions. ART. 27. The sines of two angles adapted to any radius have to each other the same ratio as the sines of the same angles adapted to any other radius. the triangles AEC and AKH are similar; as are also AFD Cor. If we substitute the word tangent or secant in place of sine, the proposition will still be true; and the demonstration will be made out in the same manner by drawing tangents to the circles at B and G, and using those tangents, or their secants, instead of the sines. ART. 28. In any right angled plane triangle, as the hypothenuse is to the perpendicular, so is radius to the sine of the angle at the base; as the hypothenuse is to the base, so is radius to the cosine of the angle at the base; and as the base is to the perpendicular, so is radius to the tangent of the angle at the base. D that is, and AB As AC: BC :: radius: the sine of A; ART. 29. In any right lined triangle, the sides have to each other the same ratio as the sines of the opposite angles. the radius BC or AE. Now, from the similar triangles ACD, AEF; As AC CD:: AE: EF (4.6); and alternately, : : AC: AE :: CD: EF, :: sine of B sine of A; these sines being suited to any radius whatever (Art. 27). Q. E. D. ART. 30. In any right lined triangle, the sum of any two sides is, to their difference, as the tangent of half the sum of the angles, opposite to those sides, to the tangent of half their difference. Then it is evident that DB is the sum, and BE the difference, of AC and AB. The outward angle CAD of the triangle ABC, is equal to the two inward and opposite angles, ABC and ACB (32.1). But AEC, at the circumference, is equal to half the angle CAD at the centre (20.3); that is, AEC = half the sum of ABC and ACB. Again, AC = AF; therefore, AFB ACB (5.1). But, ABC = AFB + BAF (32.1) = ACB + BAF; consequently, BAF and therefore ECF = = the difference between ABC and ACB; half that difference (20.3). But EG being parallel to BC, the angle CEG ECF. Furthermore, the angle DCE in a semicircle being a right one (31.3), ECG is also a right angle. Now, because BC is parallel to EG; DB: BE :: DC: CG (2.6). But CD is the tangent of CED, and CG the tangent of CEG, suited to the radius EC; and these tangents have to each other the same ratio as the tangents of the same angles adapted to any other radius (Art. 27). Hence, AC+ AB: AC-AB:: tang of (ABC + ACB): tang of (ABC — ACB). Q. E. D. ART. 31. In any right lined triangle, having two unequal sides; as the less of those sides is to the greater, so is radius to the tangent of an angle; and as radius is to the tangent of the excess of that angle above half a right angle, so is the tangent of half the sum of the angles opposite to those sides, to the tangent of half their difference. B A E = Now, because AE AD, the angle ADE = AED; hence each of those angles is half a right angle. Since the triangles |