\(5\frac{1}{2}cm\)
\(4\frac{2}{3}cm\)
\(3\frac{1}{2}cm\)
\(2\frac{1}{3}cm\)
Correct answer is D
\(r=\frac{\theta \times l}{360^{\circ }}\\
r=?, \theta =60^{\circ }\), l = radius of the orginal sector of a circle = 14cm
\(r=\frac{60\times 14}{360}=\frac{7}{3}=2\frac{1}{3}cm\)
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