AMC 12 Mock Exam 5 Problem 15

Let $AB=c$. The law of cosines at $A$ gives $13^2=15^2+c^2-2\cdot15\cdot c\cos60^\circ$, i.e., $c^2-15c+56=0$, so $c=7$ or $c=8$. Both work: $7,13,15$ and $8,13,15$ satisfy the triangle inequality, and in each case the angle opposite the side of length $13$ is $60^\circ$ (this is exactly the equation we solved). The sum is $7+8=\boxed{15}$. (The quick route: by Vieta the two roots sum to $15$, but one must check that both roots give genuine triangles.)