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

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