<PRE><B>Question 1147469</B>
To find the measure of an exterior angle of a polygon, use these formula in this order.

{{{s = ((n - 2) 180)/(n)}}}
{{{180 - s = e}}}

In this case, &lt;I&gt;s&lt;/I&gt; is the sum of the interior-angle measurements of the polygon, and &lt;I&gt;e&lt;/I&gt; is the measure of an exterior angle of the polygon.
Let &lt;I&gt;n&lt;/I&gt; be &lt;B&gt;72&lt;/B&gt;.
Then, solve (72 - 2).

72 - 2 = &lt;B&gt;70&lt;/B&gt;

Next, multiply the difference by 180.

70 * 180 = &lt;B&gt;12,600&lt;/B&gt;

Now, divide the product by 72.

12,600 ÷ 72 = &lt;B&gt;175&lt;/B&gt;

Finally, subtract the quotient from 180.

180 - 175 = &lt;B&gt;5&lt;/B&gt;

So, the measure of each exterior angle of a regular 72-gon is &lt;B&gt;5°&lt;/B&gt;.</PRE>