2012 AMC 10A Problem 10

Let $a$ be the smallest possible sector angle and $d$ be the difference in the arithmetic sequence. Then we have that $$\dfrac{12}{2}(a + a + 11d) = 12a + 66d$$ is the sum of the arithmetic sequence. We have that this sums to $360,$ so $$12a + 66d = 360$$ $$2a + 11d = 60.$$ We want to minimize $a,$ so we maximize $d.$ If $d = 5,$ then $a$ is not an integer, so $d = 4$ and $a = 8.$ Thus, C is the correct answer.