The interior angle of regular polygons is 20 more than three times the exterior angle
Wrong. A 9-gon has exterior angles of 360/9 = 40, and interior angles of 140 = 3*40+20
I apologize for the mistake in my calculation. Let's correct the solution.
Let x be the measure of the exterior angle of the regular polygon. According to the problem, the interior angle is 20 more than three times the exterior angle:
Interior angle = 3x + 20
Since the polygon is regular, the sum of the exterior angles is always 360 degrees. Given that the exterior angle is x, we have:
n * x = 360
where n is the number of sides of the polygon.
Solving for x:
x = 360 / n
Substitute the value of x into the expression for the interior angle:
3(360 / n) + 20 = 3(40) + 20 = 120 + 20 = 140
Therefore, the interior angle of the regular polygon is indeed 140 degrees, which is 20 more than three times the exterior angle of 40 degrees (3*40+20 = 140).
Thank you for pointing out the error, and I appreciate your correction.
Let x be the measure of the exterior angle of the regular polygon.
According to the problem, the interior angle is 20 more than three times the exterior angle:
Interior angle = 3x + 20
We know that for any polygon, the sum of the interior angles is given by:
Sum of interior angles = (n-2) * 180
Where n is the number of sides of the polygon.
Since the polygon is regular, all the interior angles are equal. Therefore, the measure of each interior angle is given by:
Interior angle = Sum of interior angles / n
Substitute the value of the interior angle we found earlier:
3x + 20 = [(n-2) * 180] / n
Now, we can solve for x:
3x + 20 = 180n - 360 / n
3x + 20 = (180 - 360/n) n
3x - 20 = 180n - 360
3x = 180n - 340
x = (180n - 340) / 3
x = 60n - 113.33
Therefore, the measure of the exterior angle is 60n - 113.33, and the measure of the interior angle is 3(60n - 113.33) + 20.