#how to do this polygons
45 messages · Page 1 of 1 (latest)
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Use [180(n-2)]/n, where n is the number of sides of the polygon
It should be +close
I take that back
It should just be 180°(n-2)
The extra n divisor is for an individual angle
well the two angles are the same
So you can find the sum of the missing two angles
then half it
hows this in any way helpful
sum of interior angles - angles known
ok
bro
this is some low effort responses
Maybe ur not putting enough effort into solving the question
What u want
im not giving u the answer
im not asking for the answer
Sure
ok
@wraith anvil
Hello quan_dingle1, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.
@frail citrus @fresh trench The user still needs help with this help request.
What do you still need help with
well there is this formula to find the total sum of interior angle for any given polygon: (2n-4)90
where "n" is the number of angles in a polygon
for urs its 6, so plug in 6
You can also do 180(n-2)