How many special triangles?

Let $\gamma$ be the angle opposite $c$ . The sine rule tells us that $\frac{5}{\sin{30}}=\frac{11}{\sin{\gamma}}$ , or that $\sin{\gamma}=\frac{11}{10}$ , which is impossible as the range of the sine function is $[-1,1]$ .

With the third side length being $b$ , the cosine rule $a^{2} = b^{2} + c^{2} - 2bc\cos(\alpha)$ gives us that

$5^{2} = b^{2} + 11^{2} - 22b\cos(30^{\circ}) \Longrightarrow b^{2} - 11\sqrt{3}b + 96 = 0$ .

By the quadratic formula this has solutions $b = \dfrac{11\sqrt{3} \pm \sqrt{(11\sqrt{3})^{2} - 4 \times 96}}{2} = \dfrac{11\sqrt{3} \pm i\sqrt{21}}{2}$ ,

neither of which is real, i.e., the number of triangles is $\boxed{0}$ .