Each interior angle of a regular polygon is $^{'} a^{'}$ times of its exterior angle. Find, in terms of $a$ , the number of sides in the polygon.

(a n + 1)

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2 (a + 1)

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(a - 1)

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2 (a n - 1)

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Solution

The correct option is C $2 (a + 1)$
$L e t e a c h e x t e r n a l a n g l e o f a n - s i d e d r e g u l a r p o l y g o n b e^{'} e^{'} . T o t a l s u m o f e x t e r n a l a n g l e s o f n - s i d e d r e g u l a r p o l y g o n = n \times e B u t t o t a l s u m o f e x t e r n a l a n g l e s o f n - s i d e d r e g u l a r p o l y g o n = 360^{o} S o n \times e = 360 - - - - - - - (1) N o w E a c h A n g l e o f a n - s i d e d R e g u l a r P o l y g o n = \frac{(n - 2) 180}{n}, G i v e n e a c h a n g l e o f a r e u l a r p o l y g o n w i t h n s i d e s =^{'} a^{'} t i m e s t h e e x t e r n a l a n g l e = a \times e S o a \times e = \frac{(n - 2) 180}{n} \Rightarrow n \times e = \frac{(n - 2) 180}{a} - - - - - - - - (2) E q u a t i n g 1 & 2, w e g e t \frac{(n - 2) 180}{a} = 360 \Rightarrow n - 2 = \frac{360 a}{180} = 2 a \Rightarrow n = 2 a + 2 \Rightarrow n = 2 (a + 1)$

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