Solve the triangle if $a^{2}-b^{2}=19 \mathrm{~m}^{2}, \gamma=126^{\circ} 52^{\prime} 12^{\prime \prime}$ and the diameter of the circumscribed circle $d=21.25 \mathrm{~m}$.
The formula $c=d \sin \gamma$ gives $c$. Furthermore,

$$
a=d \sin \alpha \quad \text { and } \quad b=d \sin \beta,
$$

therefore

$$
\begin{gathered}
a^{2}-b^{2}=d^{2}(\sin \alpha+\sin \beta)(\sin \alpha-\sin \beta)= \\
=d^{2} \cdot 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2} \cdot 2 \sin \frac{\alpha-\beta}{2} \cos \frac{\alpha-\beta}{2}= \\
=d^{2} \sin \gamma \sin (\alpha-\beta)
\end{gathered}
$$

thus

$$
\sin (\alpha-\beta)=\frac{a^{2}-b^{2}}{d^{2} \sin \gamma}
$$

From this formula, we can calculate $\sin (\alpha-\beta)$, and since $\alpha+\beta=180^{\circ}-\gamma$, $\alpha$ and $\beta$ can be determined. Substituting the given values, we get:

$$
\alpha=28^{\circ} 4^{\prime} 21^{\prime \prime}, \beta=25^{\circ} 3^{\prime} 27^{\prime \prime}, a=10 \mathrm{~m}, b=9 \mathrm{~m}, c=17 \mathrm{~m}
$$

(Nándor Ehrenfeld, Nyitra.)