Find the measure of each exterior angle of a regular polygon of 15 sides

24^{0}

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30^{0}

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36^{0}

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42^{0}

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Solution

The correct option is A $24^{0}$
$H e r e, l e t t h e n u m b e r o f s i d e s o f t h e p o l y g o n b e n A l s o i t i s a r e g u l a r o n e . T h e t o t a l n u m b e r o f e x t e r i o r a n g l e s = n & t h e y a r e e q u a l t o e a c h o t h e r ∴ T h e s u m o f t h e e x t e r i o r a n g l e s = 360^{o} ∴ E a c h a n g l e = θ = \frac{360^{o}}{n} ⟹ n = \frac{360^{o}}{θ} N o w θ = 24^{o} ∴ n = \frac{360^{o}}{24^{o}} = 15. I t i s a 15 s i d e d r e g u l a r p o l y g o n . A n s - 15 s i d e s$

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