1 By Case 4, the tangent is 543339720; and, by Case 5, the tangent is 1 835002744095575440 1 Then, dividing tangent No. 4 by tangent No. 5, we have 543339720 1 835002744095575440 1536796802; therefore the tangent No. 4 is 1536796802 times as large as tangent No. 5; consequently the difference between the tangents Nos. 4 and 5 is exactly the same as the difference between the differences of the inscribed and circumscribed polygons of Cases 3 and 4. By Case 4, the difference between the inscribed and circumscribed polygons is 1 1328413456156882012; and, by Case 5, the difference be 1 tween the inscribed and circumscribed polygons is 313737305594147 1555633890579843041640 2. Then, dividing the difference of Case 4 by the difference of Case 5, the difference between the inscribed and circumscribed polygons of Case 4 is 2361744410637427202 times as large as the difference between the inscribed and circumscribed polygons of Case 5. By Case 5, the tangent is tangent is 1 83500274409557440; and, by Case 6, the 1 1972063063734639263984455073299118880° Then, dividing tangent No. 5 by tangent No. 6, we have by cancel1972063063734639263984455073299X lation 18880 1 835002744095575440 1 =2361744410637427202; therefore the tangent No. 5 is 2361744410637427202 times as large as tangent No. 6; consequently the difference between the tangents Nos. 5 and 6 is exactly the same as the difference between the differences of the inscribed and circumscribed polygons of Cases 4 and 5; and, consequently, the difference between the differences of the inscribed and circumscribed polygons of Cases 5 and 6 will be the same as the differences between the tangents Nos. 6 1 and 7; thus the tangent No. 6 is 197206306373463926398445507329 1 109998456550523587637197163135 91640029823644051645720435317842392159581760° Then, dividing tangent No. 6 by tangent No. 7, we have by cancellation 5577836661177128362617195077849548802; therefore the tangent No. 6 is 5577836661177128362617195077849548802 times as large as tangent No. 7. Again, the difference between the inscribed and circumscribed poly 1 gons of Case 5 is 3137373055941471556338905798430416401/2; then, dividing this by the difference between tangents Nos. 6 and 7, we have 1 3137373055941471556338905798430416402) × (5577836661177128362617195077849548802); which equals the difference between the areas of the inscribed and circumscribed polygons for Case 6. Secant No. 6 is 2788918330588564181308597538924774401 NOTE.-It will no doubt be observed by the student that, in the Summary for Cases 6 and 7, the sine and tangent are not extended in decimals, for the reason that the space could be better occupied with matter more important to the subject. The student will find it a very agreeable pastime to verify the result in these Cases for himself, and it was partly with that view that they were omitted. The sine and tangent in Case 6 should agree together for 72, and in Case 7 for 144, decimal places. The cosine, radius, and secant have also been omitted, partly for the same reason, and partly because the radius with which they should both agree will be found carried out in the Second Part of this book; for Case 6 they should agree like the sine and tangent with each other to 72 decimal places, and for Case 7 to 144 decimal places. 1 By Case 6, the tangent is 1972063063734639263984455073299118 880' and the square which forms the unit of comparison is the square of the tangent, which is, 1 38890327273464518838061949606599216563 67162016449544406781628584372454400 is Again, by Case 6, the rectangle contained by the radius and the sine 2 2788918330588564181308597538924774401 87651718961354874269698779794778624034. Multiplying these together, we have and the number of sides 1753034379227097485393975278891833058856418130859 7 388903272734645188380619496065992165 6367162016449544406781628584372454400 1 07554307875531968564842627379673198284800; therefore there are exactly 24445348571891983269638939752719507554307875531968564842627379673198284800 squares in the given polygon, each of which is expressed by 1 388903272734645188380619496065992165636 7162016449544406781628584372454400° Then, multiplying the number of squares by the value of each square, 244453485718919832696389397527195075543078753196856388903272734645188380619496065992165636716201644954 we have 4842627379673198284800 44 = 62 = area of the inscribed poly 4406781628584372454400 7 CASE 7. In which the Inscribed and Circumscribed Polygons are Carried to 4889069714378396653927787950543901510861575106393712968525 Then, by ARTICLE 3 and (PROPOSITIONS 4 and 5), tangent2 1 By (PROPOSITION 6) and ARTICLE 4, and by cancellation, 5577836661177128362617195077849548802 1972063063734639263984455073299118880 109998456550523587 X = Then (2788918330588564181308597538924774401)2 × 2 — 1 1555613090938580753522477984263968662546864806579817762712 6514337489817601. 1555613090938580753 Multiplying this by tangent No. 7 we have 1099984565505235876 5224779842639686625468648065798177627126514337489817601 3719716313591640029823644051645720435317842392159581760 secant No. 7. By Case 7, the numerator of the secant is 15556130909385807535224779842639686625468648065798177627126514337489817601, and 278891833058856418130859753892 By Case 6, the secant No. 6 is 197206306373463926398445507329 radius, and twice that amount from the assumed diameter. 8 |