Sum of all angles in an n-sided polygon \(= (n-2)(180°)\)

If the n-sided polygon is REGULAR, each interior angle in the n-sided polygon \(= \frac{(n-2)(180°)}{n}\)

The question tells us that, in polygon X, each angle is an INTEGER.
Since each angle in a regular polygon must be LESS THAN 180°, 179° must be the biggest possible angle in polygon X

So we can write: \(\frac{(n-2)(180)}{n} = 179\) [we'll solve this equation for n]

Multiply both sides by n to get: \((n-2)(180) = 179n\)

Expand the left side to get: \(180n-360 = 179n\)

Solve: \(n = 360\)

So, when polygon X has 360 sides, each interior angle is 179° (the greatest integer value for an interior angle in a regular polygon)

Now that we've maximized the measurement of each angle in polygon X, we can focus our attention on polygon Q

We're told that the number of sides in polygon Q is 1/4 the number of sides in polygon X

So, the number of sides in polygon Q = 360/4 = 90

When we plug 90 into our formula above, we get...

Each angle in a 90-sided polygon \(= \frac{(90-2)(180)}{90}= \frac{(88)(2)}{1}=176\)

Answer: C

Cheers,
Brent