If a regular polygon has 6 sides then the measure of its interior angle is greater than the measure of its exterior angle by how many degrees?
Sum of all interior angles of regular polygon with $$n$$ sides = $$(n-2)\times180^\circ$$
=> Sum of interior angles of a regular hexagon = $$(6-2)\times180^\circ=720^\circ$$
Also, sum of exterior angle of an polygon = $$360^\circ$$
$$\therefore$$ Required difference = $$\frac{(720-360)}{6}$$
= $$\frac{360^\circ}{6}=60^\circ$$
=> Ans - (C)
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