The sum of all but one of the interior angles of a convex polygon equals 2525° . find the measure of the exterior angle adjacent to the remaining interior angle. can you please help if possible with diagram


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Question Number 55333 by Otchere Abdullai last updated on 21/Feb/19

$T h e s u m o f a l l b u t o n e o f t h e i n t e r i o r$ $a n g l e s o f a c o n v e x p o l y g o n e q u a l s$ $2525 ° . f i n d t h e m e a s u r e o f t h e$ $e x t e r i o r a n g l e a d j a c e n t t o t h e$ $r e m a i n i n g i n t e r i o r a n g l e .$ $c a n y o u p l e a s e h e l p i f p o s s i b l e w i t h$ $d i a g r a m$
	Answered by mr W last updated on 21/Feb/19

	${l e t}^{'} s s a y t h e p o l y g o n h a s n s i d e s .$ $α_{n} i s t h e r e m a i n i n g i n t e r i o r a n g l e .$ $\underset{k = 1}{\sum^{n - 1}} α_{k} + α_{n} = (n - 2) \times 180 °$ $2525 ° + α_{n} = (n - 2) \times 180 °$ $\Rightarrow α_{n} = (n - 2) \times 180 ° - 2525 °$ $c o n v e x p o l y g o n \Rightarrow 0 ° < α_{n} < 180 °$ $i f n = 16 : α_{n} = - 5 ° < 0 °$ $i f n = 18 : α_{n} = 355 ° > 180 °$ $\Rightarrow n = 17$ $\Rightarrow α_{n} = (17 - 2) \times 180 ° - 2525 ° = 175 ° o k!$ $e x t e r i o r a n g l e β_{n} = 180 ° - α_{n} = 5 °$
	Commented by mr W last updated on 21/Feb/19

	Commented by Otchere Abdullai last updated on 21/Feb/19

	$Y e s t h a t s w h y i s a i d y o u a r e p r o f$ $G o d b l e s s y o u a l o t$
	Commented by Otchere Abdullai last updated on 21/Feb/19

	$O! y e s P r o f o f P r o f s$
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