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 —sin3 = .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'" Zi45', that is, of 137th part of the circumference is=.000,255,663,46 nearly: consequently when the radius is 1, 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. |