RULE. Multiply the area of the generating circle by 3, and the product is the area of the cycloid. EXAMPLE I. Required the area of the cycloid, when the diameter of the generating circle is 4 feet. 4 4 16 .7854 47124 7854 12.5664 area of the generating circle. 3 37.6992 fq. feet, area of the cycloid. Ex. 2. Required the area of the cycloid, 25.1328. whose base is Anf. 150.79. Ex. 3. Required the area of the cycloid, whose length is 400 feet. Anf. 23562 fq. feet. PROBLEM XXV. To find the fine and cofine of a very small arch, fuch as 1' A small arch fuch as 1', may be confidered nearly equal to its fine. Suppofe, then, the radius of a circle to be 100000, in which cafe the circumference will be 628318.52; therefore. 628318.52 will quote 29.08, the natural fine of 1'. Since the 60 x 360 fquare of the hypothenuse of a right angled triangle is equal to the fum of the fquares of the legs, therefore from the fquare of Y 2 the the radius fubtract the square of the fine of any arch, and the fquare root of the remainder will be the cofine of that arch. Thus 10000000000-845.64-99999.9 the cofine of 1' The verfed fine x B may be found by fubtracting the cofine. from the radius. PROBLEM XXVI. The fine and cofine of any arch being given, to find the fine and cofine of its double. RULE. As the radius is to the cofine of any arch, fo is twice the fine. of that arch to the fine of its double. EXAMPLE Required the fine and cofine of two degrees, the fine of 1a being 1745, and cofine 99985. And Rad Co-fine 1° 10000099985:3490: 3489.47 natural fine of 2 3489.472= If three arches differ equally, the radius is to the cofine of the middle arch as twice the fine of the difference is to the difference of the fines of the greatest and leaft arches. Ex. 2. Required the fine and cofine of 3°, the fine and cofine of 1° and 2° being given. 100000: 99939: 3490 3490 8994510 399756 299817 = 3487.87110 AL the diff of the extreme arches, 1745 5232.87110 the fine of 3°-AF The cofine of which is EA2-AF2-99863 the cofine of 3a Ex. 3. The fine and cofine 2° and 3° being given, required the fine and cofine of 4°. (Sine 6976 Anf.Co-fine 99756 Ex. 4. Required the fine and cofine of 5°, the fine and cofine of 3° and 4° being given. Anf{ Sine 8715-4844 Ex. red the fine and cofine of 6° 5. The fine and cofine of 4° and 5° being given, requi Anf. Sine 10452 In like manner, the fine and cofine of every minute and degree of the quadrant may be found; but when the calculations. are carried on the length of 60°, the fines of the remaining. arches may be found by the following rule: Take the fine of an arch as much below. 60° as the arch whofe fine is required is above (0°, to which add the fine of the number of degrees that the propofed arch exceeds 60°; the fum will be the fine required. Ex. 6. What is the fine of 80°? The fine of no°—20°=40° is 64279. The fine of 80°-60°-20° is 34202 The verfed fine is found by fubtracting the cofine from the ra of 89° 45' 99999 dius. PROBLEM XXVII. To find the tangent and cotangent of every minute and degree of the quadrant, the fines and cofines being given. The tangent and cotangent of any arch may be found by cither of the following proportions: Because the triangles CED and CBA are fimilar, CD : DE ::. ÇA: AB, therefore DEXCA CD AB, the tangent of the arch, EA; that is, the rectangle contained by the fine and radius of any arch, is equal to the rectangle contained by the cofine and tangent of that arch. Hence, RULE I. To find the tangent, multiply the fine of any arch by the radius, and divide the product by the cofine, the quotient will be the tangent of that arch. The cotangent of any arch may be found upon the fame prin ciples: Thus, CL: LE:: CF: FK; therefore, hence, RULE 2. Divide the product of the cofine and radius of any arch by its fine, and the quotient will be the cotangent: or (which is the fame thing) fay, As the fine of any arch is to its co-fine: fo is the radius to the cotangent of that arch. It is alfo obvious, that AB: AC:: AC: FK, therefore AC2 --FK, that is to fay, the radius is a mean proportional be AB tween the tangent and co-tangent of any arch. Hence, the co-tangent may be found by the following rule RULE 3. Divide the fquare of the radius by the tangent of any arch, and the quotient will give the co-tangent of that arch. EXAMPLE I. Required the tangent and co-tangent of 60°, the co-fine being 50000, and fine 86603. Cofine. Sine. Rad. Tang. of 60°. By Rule 1. 50000 : 86603. 100000: 173206 Anf. By Rule 2. to find the cotangent. Sine. Co-fine. 86603 : 50000: 100000: 57734 co-tan. of 60°, or tan. of 3c". Rad. fquared, 10000000000 By RULE 3. 57734, as by Rule 2. tan. 60° 173206 Ex. 2. Required the tangent and co-tangent of 40° 30′ Anf. { STang. 85407 co-tang. II7D85 Ex. 3. Required the tangent and co-tangent of 15° 32′ Anf. STang. 27795 co-tang- 359772 Ex. 4. Required the tangent and co-tangent of 20° 45′ 37886 (co-tang. 263949 Ex. 5. Required the tangent and co-tangent of 80° o' Anf. {Tang. 567123 co-tang. 17633 PROBLEM |