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Unformatted text preview: with hypotenuse of length 10 and leg BC of length 5, so the length of OC is 5Ö3, and the angle BOC is p/6. This implies that the angle AOB is p/12, since angle AOC is clearly p/4. The area of a circular segment of radius 10 and central angle p/6 is 25p /3. From this it’s easy to compute that triangle AOB has base length 10 and height 5Ö2 sin(p/12) = 5(Ö3  1)/2, giving an area of (25Ö3  1) /2. Put this all together and you get an answer of 4(25p/3  25(Ö3  1)) @ 31.5146 square ft. William Webb also computed the area of regions where the professor is in the range of two or of three students. (Notice that the wet lecturer is never in the range of only one shooter.) Anyone care to compute these?...
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 Spring '10
 AlbertoDelgado
 Geometry, Combinatorics, Pythagorean Theorem, Hypotenuse, triangle, Felice Kelly

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