And, by ARTICLE 6, the rectangle contained by the radius and the tangent, is equal to the area of the circumscribed double triangle; thus Then, by ARTICLE 7, the rectangle contained by the radius and the sine, is a mean proportional between the inscribed and circumscribed double triangles of half the number of sides; thus, by cancellation 27720 ; extracting the square 3841992011 2X 1 139601 2, we have 4 384199201 triangle for half the number of sides. By ARTICLE 10, the rectangle contained by the rectangle of the radius and the sine, and the number of sides contained in the given polygon; is equal to the area of the entire inscribed polygon of double the number of sides. For, by Case 3, the number of sides contained by the given polygon is 616034, and the rectangle of the radius and sine is 2 ; then 19601 2 = 22 7 7 = 3.142857, or 34, which is the true ratio of the circum ference to the diameter of the given circle. Again, by ARTICLE 8, the rectangle contained by the inscribed double triangle and the number of sides, is equal to the area of the entire in And, by ARTICLE 9, the rectangle contained by the circumscribed double triangle and the number of sides, is equal to the area of the entire circumscribed polygon; thus 1 431222 1386012616031= 97020 V2. 431222 Now, the area of the circumscribed polygon is 97020 165675147853622 37275006481020* Now, by hypothesis, the circumference of a circle is to the diameter 2 as 3 is to 1; therefore 34 times tangent of the given arc is 1 1 or 13860' must be and, by Case 4, the 543339720; then, dividing the tangent No. 3 by the tangent No. 4, we have by cancellation 1 13860 543339720 1 been divided into 39202 equal parts, each of which is equal to the By Case 3, the given polygon was carried to 616034 sides, each of which contained two tangents, and each tangent was 1 13860 ; then By ARTICLE 1, the double of the secant is the constantly assumed diameter; and, by ARTICLE 2, the sum of the tangents is the constantly assumed circumference. Then, dividing the assumed circumference by the assumed diameter, the true ratio between the circumference and diameter of the given circle. Now, by trigonometry, sec.2: tang.2 R2 S2, and, by Case 4, the : 8400 6801 2 2 29521805132967295218051329678400 59043610265935 2 295218051329678400 Х ; extracting the square root, 1 768398401 √2== sine No. 4. 295218051329678400 2 590436102659356801 V590436102659356801' we have Again, by trigonometry, R2 sin.2 = cos.2. Now, by ARTICLE 5, the rectangle contained by the sine and the cosine 1086679440 is equal to the area of the inscribed double triangle; thus 768398401 1 X 7683984011 2 1086679440 5904361026593568011 2. Now, by hypothesis, the circumference of a circle is to the diameter 1 1 3 by the tangent No. 4, we have by cancellation 13860 543339720 been divided into 39202 equal parts, each of which is equal to the By Case 3, the given polygon was carried to 616034 sides, each of which contained two tangents, and each tangent was 1 ; then ; and, by hypothesis, 13860 By ARTICLE 1, the double of the secant is the constantly assumed diameter; and, by ARTICLE 2, the sum of the tangents is the constantly Then, dividing the assumed circumference by the assumed diameter, we have by cancellation 48299328062 543339720 1536796802 48299328063 543339720 543339720 |