1 with CE or CB, which, when the arch DB is very small, will therefore be very nearly equal to each other; consequently the sine ED will also be very nearly equal to the tangent BG. To illustrate this farther, draw DF parallel to CB; the triangles GDF, GCB are equiangular; therefore DF: CB:: GF:GB; whence if DF or EB is less than 10,000,000th part of the radius CB, GF, the difference between the sine and tangent will also be less than the 10,000,000 th part of the tangent GB. Cor. Since an arch is greater than its sine, and less than its tangent, and the sine and tangent of a very small arch are nearly equal, the arch which lies between them must be nearly equal to either of them. Wherefore in very small arches, as arch is to arch so is sine to sine. 1 PROP. VIII. PROB. To find the sine of an arch of one minute. The sine of 30° being (equal to half the chord of 60°, that is, equal to half the radius or) equal to the sine of 15° may be found by Prop. 3. In like manner, from the sine of 15° we may obtain the sine of its half 7° 30', and so on by continually bisecting the arcs, and finding the sines, till after eleven bisections of the arc of 30°, we at last get the sine of 52' 54' 3'' 45'' = .000,255,663,462; the cosine of which viz. 999,999,967,318, is very nearly equal to the radius; in which case, as is plain from Prop. 7, the sines are proportional to the arcs. Therefore, as the arch 52" 44" 3'' 45' is to an arch of 1', so is the sine of the former arch=.000,255,763,462 to the sine of l'=.000,290,888,204. When the sine of l'has been thus found, the sine and cosine of 2' may be found, by Prop. 4. and 2, to be .000,581,776,1 and .999,999,830,8 respectively. If any angle (BAC) at the circumference of a circle be bisected by a straight line (AD), and AC one of the sides of the given angle be produced till it be cut by DE, which is made equal to AD, then will CE, the part produced, be equal to the chord AB. In the quadrilateral figure ABDC inscribed in the circle, the angles B, and ACD, are together equal to two right angles, (22. 3,) and therefore equal to the two angles ACD, DCE, (13. 1,) consequently B is=DCE. But (5. 1.) the angle E is=DAC, because by construction DE is =DA, and by the supposition DAC is = DAB; therefore E=DAB. And the arches CD, DB and (29.3.) their chords are equal. Therefore (26. 1.) the triangles DAB, DCE are in every respect equal, and CE=BA. PROP. X. THEOR. Fig. 6. If there be any number of equal arches, the chord of the first will be to the chord of the second, as the chord of the second is to the sum of the chords of the first and third ; or as any one of the chords is to the sum of the preceding and following chords. Let the arches AB, BC, CD, DE, EF, &c. be equal, and let the chords AB, AC, AD, AE, &c. be drawn, then will AB: AC:: AC: AB+AD:: AD: AC+AE :: AE: AD+ AF:: AF: AE+AG. For produce AD to H, AE to I, AF to K, AG to L, SO that each of the triangles ABC, ACH, ADI, AEK, AFL may be an isosceles triangle; then since the arch BC is equal to the arch CD, the angle BAD is bisected by AC, therefore (by Prop. 9.) DH=AĎ. For the same reason, EI=AC; FK = AD; and GL=AE. . And because the angles at A, being at the circumference, and standing on equal arches AB, BC, CD, DE, &c. are equal, the isosceles triangles ABC, ACH, ADI, AEK, AFL, are equiangular, and therefore similar to one another,(4.6.); whence AB: AC:: AC:(AH=)AB+AD:: AD:(AIN)AC+ AE :: AE:(AK=)AD+AF:: AF: (AL=)AE+AG; and so on. Cor. Since (by Prop. 4. Cor. 1.) rad : 2 cos LAB :: chord AB: chord AC, (fig. 6,) therefore, rad: cos LAB:: AB: AC :: AC: AB+AD:: AD: AC+AE, &c., or (halving these chords,) rad: 2 cos LAB:: JAB: JAC:: JAC: JAB+JAD:: AD: AAC+AE, &c. Now, let each of the arches AB, BC, CD, &c. be 2'; then will JAB be the sine of 1', JAC the sine of 2', JAD the sine of 3', AE the sine of 4', and so on. Whence, having found the sine of l' by Prop. 8, and its cosine, viz. 999,999,957,692, by Prop. 2, the sine of 2' is found by this analogy, rad : 2cos l': : sin 1': sin 2'; and from these, the sines of all the following arcs are found for every minute of the quadrant, thus : Sin 3' or cos 89° 57' = 2 cos l'X sin 2'-sinl'= .000,872,664,5. Sin 4 or cos 89 56 = 2 cos 1 X sin 3 —sin2 = .001,163,552,6. Sin 5 or cos 89 55 = 2 cos 1 X sin 4 —sin: = .001,454,440,5. Sin 6 or cos 89 54 =2 cos 1 x sin 5 - sin4 = .001,745,328,4; and so on. In this manner, may the sines of every minute of the quadrant be easily found; but when the sines are found to 60°, the remaining sines may be still more easily found by addition only, by Prop. 6. Cor. 2. When the sine of an arch is given, its tangent and secant may be found by the following analogies; (fig. 3. Plane Trigonometry.) Cos: sin :: rad : tan ; and tan : rad :: rad : cotan ; and cos : rad :: rad: sec; and sin: rad :: rad: cosec. For the triangles DBC, BAE, BHK, having the angles at D, A, and H right angles, and the alternate angles HKB, EBA equal, are equiangular, therefore BD: DC::BA: AE; that is, cos: sin :: rad: tan. To find the length of the circumference. When the radius of a circle is 1, it appears from Prop. 8. that an arch of 52" 44'" ziv 45', that is, of zoth part of the circumference is= .000,255,663,46 nearly: consequently when the radius is I, the whole circumference is =.000,255,663,46 x 24576=6-283,1852 nearly: and therefore when the diameter is 1, the circumference is=3.141,592,6 nearly. |