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Question
Find the magnitude, in radians and degrees, of the interior angle of a regular duodecagon.
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Solution
\[\text{ Sum of the interior angles of the polygon }= \left( n - 2 \right)\pi\]
Number of sides in the duodecagon = 12
\[ \therefore \text{ Sum of the interior angles of the duodecagon }= \left( 12 - 2 \right)\pi = 10\pi\]
\[\text{ Each angle of the duodecagon }= \frac{\text{ Sum of the interior angles of the polygon }}{\text{ Number of sides }} = \frac{10\pi}{12} = \frac{5\pi}{6}\text{ rad }\]
\[\text{ Each angle of duodecagon }= \left( \frac{5\pi}{6} \times \frac{180}{\pi} \right)^\circ= {150}^\circ\]
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