91640029823644051645720435317842392159581760* Then, by division and cancellation, we have 5577836661177128362617195077849548802; therefore the tangent No. 6 has been divided into 5577836661177128362617195077849548802 parts, each of which is equal to tangent No. 7. By Case 6, the polygon was carried to 87651718961354874269698779794778624034 sides, each of which contained two tangents of 1 1972063063734639263984455073299118880* 2 1972063063734639263984455073299118880 Then X 876517189613 17530343792270974853939755958 54874269698779794778624034= 1972063063734639263984455073 By Case 7, the secant No. 7 is 39686625468648065798177627126514337489817601 91640029823644051645720435317842392159581760 155561309093858075352247798426109998456550523587637197163135 Then, by ARTICLES 1 and 2, dividing the assumed circumference by the assumed diameter, we have by cancellation 9778139428756793310999845655052358 694780218565561760 53028674979635202 9778139428756793307855575901087803021311122618187161507044955968527937325 72315021278742593705095186927931392062 22 = 3.142857, or 34, 0937296131596355254253028674979635202 = 7 the true ratio between the circumference and diameter of the given circle. 1 By Case 7, the sine of the given arc 15556130909385807535224779 8426396866254686480657981776271265143374898176012. Then, by ARTICLE 7, multiplying the radius by the sine, we have 2 1555613090938580753552247798426396866254686480657981776271area of the inscribed double triangle for half 26514337489817601 the number of sides. 2 By ARTICLE 10, 155561309093858075352247798426396866254686 or 34, which is the true ratio of the circumference to the diameter of 68648065798177627126514337489817601 23644051645720435317842392159581760 NOTE. The difference between the inscribed and circumscribed polygons for Case 7 has been omitted, for the purpose of allowing the student to make the calculation for himself; the number of squares contained in the given polygon, together with the unit of comparison, has also been left out, as the calculation is made in the same manner as in the previous cases. The decimals of the Summary have also been omitted, as in Case 6, and for the reason mentioned therein, viz. that the decimals for the cosine, the radius, and the tangent will be found in Part Second of this volume. Indeed, the whole of Cases 6 and 7 would have been omitted had it not interfered with the original design of the author, as the other cases are sufficient to enable any one to get a thorough and comprehensive knowledge of the subject, though the curious and the adept are at liberty to extend the work ad infinitum. By Case 1, the 1 50 PROOFS. th of the square described upon the radius was deducted for the purpose of finding the value of the given arc, and th so deducted is the sine of the given arc; the square root of the 1 50 and the square root of the remaining 49 ths is the cosine of the given 1 are; for, substituting the numbers, we have (1/2)2 = 2, and the sine of the given arc. of 2: 50 49 = and 25 25 25 25 25 5 Now, by Case 1, there are 22 double triangles contained in the given Now, by Case 1, there are 22 double triangles contained in the given polygon. 1100 Again, = 175 175 1100 1100 22 175 difference between the inscribed and circumscribed polygons is th of 22 175 22 1 50 therefore the 1 50 Deducting the cosine from the secant of the given arc, we have therefore the difference between the cosine and secant is equivalent to 1 radius is equivalent to deducting theth from the secant of the given arc. 50 Again, by Case 1, the circumscribed polygon is 1078 scribed polygon is 175 ; then, by division and cancellation, 1078 |