or 34, which is the true ratio of the circumference to the diameter of 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. 1 By Case 1, the 50 PROOFS. th of the square described upon the radius was deducted for the purpose of finding the value of the given arc, and the square root of the th so deducted is the sine of the given arc; 1 50 and the square root of the remaining 49 ths is the cosine of the given arc; for, substituting the numbers, we have (1/2)2 Now, by Case 1, there are 22 double triangles contained in the given By trigonometry, we have cos. : sin. :: rad.: tang. 1 Then, by division and cancellation, we have 1/2 ÷ 1 7 2 Then, by ARTICLE 6, 1/2 × √ 2 = = area of the circumscribed double triangle. X 1/2 Now, by Case 1, there are 22 double triangles contained in the given Again, 22 175 1100 the inscribed and circumscribed polygons. 22 175 22 1 Х 50; therefore the 1 difference between the inscribed and circumscribed polygons is -th of 50 1 square described upon the radius is equivalent to deducting th from the area of the circumscribed polygon. By trigonometry, R2+ tan.2 sec.2. Substituting the numbers, we have (1/2)2; 50 1 2 2 and (=√/2)2 = then 2 + 2 100 49 49 and 100 10 secant of the given arc. 49; Deducting the cosine from the secant of the given arc, we have therefore the difference between the cosine and secant is equivalent to then, dividing this differ1 10 1 35 7 1 50 1 Consequently, deducting th from the square described upon the 50 1 radius is equivalent to deducting the 5th from the secant of the given arc. Again, by Case 1, the circumscribed polygon is 1078 scribed polygon is 175 is of the circumscribed polygon. 50 1 By Case 2, the sine of the given arc is 12; and the radius is √2; 2XV2= there are 3114 sides; then X 3111 area of 99 the inscribed polygon. 154 But, by Case 1, the area of the inscribed polygon is 25 therefore the area of the inscribed polygon for of the area of the inscribed polygon of Case 2; but 49 the area of the inscribed polygon for Case 1 is of the area of the circumscribed polygon for Case 1; therefore, by (Axiom 1), the area of the inscribed polygon of Case 2 is the same as the area of the circumscribed polygon of Case 1. 1 50 But, by Case 1, the th of the area of the circumscribed polygon was deducted; therefore, by Case 2, the same again. Again, by Case 1, the sine is 5' the given polygon; and, by Case 2, each of these sines is divided into 14 parts, each equal to tangent No. 2; therefore there are 44 X 14 = 1 616 tangents contained in the polygon for Case 1, each equal to 70° must be added to complete the polygon; 1 490V/2. Again, by Case 1, the tangent is 12; and there are 22 sides, each 1 of the tangents for Case 1 has become the sum of the sines for Case 2. Again, by Case 1, the sine is 5' and there are 22 sides, each of two complete the polygon; thus + the sines for Case 1 has become the sum of the tangents for Case 2. 7 Again, by Case 1, the cosine is 5; and, by Case 2, the secant is 99 70° 70 70 ; therefore the sum of |