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in point O with radius 6, m arc AB=60.find the area of the shaded sector.
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\[{(n+1)^{2}(n+3)!}/(n+1)!(3^{2n+2}) * {n!3^{2n}}/(n^2(n+2)!)\]
The area is in proportion to the degree of the arc. So,\[\frac{60^o}{360^o}=\frac{l}{2\pi r}\]where l is your arc length here, r the radius. Then\[l=\frac{2 \pi r}{6}=2 \pi\] since your radius is 6.
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