# Thread: Number of sides of a polygon

1. ## Number of sides of a polygon

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

2. Originally Posted by Punch
$\displaystyle A, B$ and $\displaystyle C$ are three vertices of a polygon as shown above. The reflex angle $\displaystyle ABC =$ $\displaystyle 222^o$ and the size of the remaining exterior angles are equal to $\displaystyle 53^o$ each. Calculate the number of sides this polygon has.
Sum of measures of exterior angles of a (simple) polygon is 360 degrees. So...

3. note that not all the exterior angles are equal...

360-222=138

360/53 will not work

4. To calculate the total angle of all the interior angles added, we use:

$\displaystyle a=180(n-2)$

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:

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

Now use simultaneous equation to solve for n.

5. Originally Posted by Punch
note that not all the exterior angles are equal...
Yes I know.

Originally Posted by Punch
360-222=138
No, exterior angle is 222-180 = 42

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.

6. 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...)