A certain triangle has two sides of $8 \mathrm{dm}$ and $5 \mathrm{dm}$; the angle opposite the first is twice as large as the one opposite the second. What is the length of the third side of the triangle?
According to the Law of Sines:

$$
5: 8=\sin \alpha: \sin 2 \alpha=\sin \alpha: 2 \sin \alpha \cos \alpha=1: 2 \cos \alpha
$$

thus,

$$
\cos \alpha=\frac{4}{5} ; \sin \alpha=\frac{3}{5}, \cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha=\frac{7}{25}
$$

and hence the third side

$$
x=5 \cos 2 \alpha+8 \cos \alpha=\frac{7}{5}+\frac{32}{5}=\frac{39}{5}=7.8 \mathrm{dm}
$$

Number of solutions. 20.