ANSWERS TO PROBLEMS
Published December 28. 13. By John G. Smith, Invorcargill :— 18/5? 3 and2/6|S. 14. By Sigma, Shag Valley :—: — Chord of half the arc=4 '472163. Then Vrß2-4"47216V r B 2 -4"472163 2 = 6-633231 and 6633231+8 = 14-633231, hence 4 -472163 3 -M4 •638231=1.886768 =heightof arc /l -366768^+4 472L63 2 =4 -676355 = chord of half the arc 4 676355 x 8 -8 944326 -t-3=9-488838= length of arc neirly. 15. By John Whitty, Gore :— £49. CORRECT ANSWERS. From Sigma, Shag Valley, to 13. G. M. W., Kyeburn Diggings, 13. Jos. Hall, Dunedin, 13.
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Bibliographic details
Otago Witness, Issue 1418, 25 January 1879, Page 21
Word Count
91ANSWERS TO PROBLEMS Otago Witness, Issue 1418, 25 January 1879, Page 21
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