Solve the isosceles triangle if the area $t=256 \mathrm{~cm}^{2}$, and the angle at the vertex is $\alpha=48^{\circ} 15^{\prime} 20^{\prime \prime}$.
Let one side of the triangle be denoted by \( l \). Then,

\[
l^{2} \sin \alpha=2 t
\]

or

\[
l=\sqrt{\frac{2 t}{\sin \alpha}}
\]

The base of the triangle is

\[
\begin{gathered}
a=2 l \sin \frac{\alpha}{2}=2 \sin \frac{\alpha}{2} \sqrt{\frac{2 t}{\sin \alpha}}=2 \sqrt{\frac{2 \sin ^{2} \frac{\alpha}{2}}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}} t= \\
=2 \sqrt{t \cdot \operatorname{tg} \frac{\alpha}{2}}
\end{gathered}
\]

Numerical results in this problem:

\[
l=26.19 \mathrm{~cm}, a=21.41 \mathrm{~cm}
\]

and

\[
\beta=90^{\circ}-\frac{\alpha}{2}=65^{\circ} 52^{\prime} 20^{\prime \prime}
\]

(Bauer Jakab, Zenta.)