The interior angle of regular polygons is 20 more than three times the exterior angle

Answer 1

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.

Answer 2

Wrong. A 9-gon has exterior angles of 360/9 = 40, and interior angles of 140 = 3*40+20

Answer 3

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.