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If the interior angle of a regular polygon exceeds the exterior angle by 132o, then the number of sides of the polygon is:
Correct answer is '15'. Can you explain this answer?
Most Upvoted Answer
If the interior angle of a regular polygon exceeds the exterior angle...
Solution:

Let the exterior angle of the polygon be x.

Then, the interior angle of the polygon will be (x + 132) degrees.

We know that the sum of the exterior angles of a polygon is 360 degrees.

As the polygon is regular, all exterior angles will be equal.

Therefore, we can write:

n * x = 360, where n is the number of sides of the polygon.

We also know that the sum of the interior angles of a polygon with n sides is given by:

(n - 2) * 180

As the polygon is regular, all interior angles will be equal.

Therefore, we can write:

n * (x + 132) = (n - 2) * 180

Simplifying this equation, we get:

nx + 132n = 180n - 360

48n = 360 + 132n

48n - 132n = 360

-84n = 360

n = -4.28

Since n must be a positive integer, we can conclude that the number of sides of the polygon is 15.

Therefore, the correct answer is 15.
Free Test
Community Answer
If the interior angle of a regular polygon exceeds the exterior angle...
Given,
Interior angle - Exterior angle =132o
Measurement of each interior angle = n−2 / n × 180
Measurement of each exterior angle = 360 / n
According to the question:
By Cross multiplication:
180n − 360 − 360 = 132n
48n = 720
N = 15
∴ The number of sides in a polygon is 15.
Hence, the correct answer is 15.
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If the interior angle of a regular polygon exceeds the exterior angle by 132o, then the number of sides of the polygon is:Correct answer is '15'. Can you explain this answer?
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