# Number of sides of a polygon

• Aug 28th 2010, 09:53 PM
Punch
Number of sides of a polygon
http://i952.photobucket.com/albums/a...g/P8260034.jpg

$A, B$ and $C$ are three vertices of a polygon as shown above. The reflex angle $ABC =$ $222^o$ and the size of the remaining exterior angles are equal to $53^o$ each. Calculate the number of sides this polygon has.
• Aug 28th 2010, 09:59 PM
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Quote:

Originally Posted by Punch
$A, B$ and $C$ are three vertices of a polygon as shown above. The reflex angle $ABC =$ $222^o$ and the size of the remaining exterior angles are equal to $53^o$ each. Calculate the number of sides this polygon has.

Sum of measures of exterior angles of a (simple) polygon is 360 degrees. So...
• Aug 28th 2010, 10:04 PM
Punch
note that not all the exterior angles are equal...

360-222=138

360/53 will not work
• Aug 28th 2010, 10:49 PM
Educated
To calculate the total angle of all the interior angles added, we use:

$a=180(n-2)$

a = sum of total internal angles
n = number of sides

Now given the information:
- One internal angle is 138 degrees
- The rest of the internal angles are 127 degrees

Therefore, the total internal angle can be calculated by:

$a = 127(n-1) + 138$

Now use simultaneous equation to solve for n.
• Aug 28th 2010, 10:51 PM
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Quote:

Originally Posted by Punch
note that not all the exterior angles are equal...

Yes I know.

Quote:

Originally Posted by Punch
360-222=138

No, exterior angle is 222-180 = 42

Quote:

Originally Posted by Punch
360/53 will not work

And that's not what I was suggesting.

42 + (n-1)53 = 360. Solve for n.
• Aug 28th 2010, 10:55 PM
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Accidental double post. (I pressed submit once, and then it told me it didn't post because I had to wait 30 seconds between posts, but actually it had posted...)