The angles of triangle $\mathrm{ABC}$ are in the ratio $5: 7: 11$. Solve the triangle if $a=50 \mathrm{~cm}$.
If the angles of a triangle are $\alpha, \beta, \gamma$, then:

$$
\alpha: \beta: \gamma=5: 7: 11
$$

thus

$$
\alpha=39^{\circ} 7^{\prime} 50^{\prime \prime}, \quad \beta=54^{\circ} 46^{\prime} 57^{\prime \prime} \quad \text { and } \quad \gamma=86^{\circ} 5^{\prime} 13^{\prime \prime}
$$

We calculate the sides using the Law of Sines. Specifically:

$$
b=\frac{a \sin \beta}{\sin \alpha}, \quad c=\frac{a \sin \gamma}{\sin \alpha}
$$

Omitting the detailed calculations, we find that:

$$
b=64.72 \mathrm{~cm}
$$

and

$$
c=79.04 \mathrm{~cm}
$$