Solve the triangle if we know one side ($a=51 \mathrm{~m}$), the angle opposite to it $\left(\alpha=59^{\circ} 4^{\prime} 39^{\prime \prime}\right)$, and the area $\left(t=1020 \mathrm{~m}^{2}\right)$.
First solution. It is known that the area of the triangle is

$$
t=\sqrt{s(s-a)(s-b)(s-c)}
$$

and

$$
\operatorname{tg} \frac{\alpha}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}
$$

thus

$$
\frac{t}{\operatorname{tg} \frac{\alpha}{2}}=s(s-a)
$$

or

$$
s^{2}-a s-\frac{t}{\operatorname{tg} \frac{\alpha}{2}}=0
$$

from which

$$
s=75 \text{ m}
$$

We know that

$$
\cos \frac{\alpha}{2}=\sqrt{\frac{s(s-a)}{b c}}
$$

thus

from which

$$
b c=2378
$$

However, if

$$
2 s=a+b+c
$$

then

$$
b+c=2 s-a=99
$$

From $\left(1^{\circ}\right)$ and $\left(2^{\circ}\right)$

$$
b=41 \text{ m}, \quad c=58 \text{ m}.
$$

Furthermore, we obtain that

$$
\beta=43^{\circ} 36^{\prime} 10^{\prime \prime}
$$

and

$$
\gamma=77^{\circ} 19^{\prime} 11^{\prime \prime}
$$

Second solution.

$$
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha
$$

but since

$$
2 b c=\frac{4 t}{\sin \alpha}
$$

therefore

$$
b^{2}+c^{2}=a^{2}+4 t \frac{\cos \alpha}{\sin \alpha}
$$

Adding (2) to (1) and subtracting (1) from (2)

$$
\begin{aligned}
& (b+c)^{2}=a^{2}+\frac{4 t}{\sin \alpha}(1+\cos \alpha)=a^{2}+4 t \operatorname{ctg} \frac{\alpha}{2} \\
& (b-c)^{2}=a^{2}-\frac{4 t}{\sin \alpha}(1-\cos \alpha)=a^{2}-4 t \operatorname{tg} \frac{\alpha}{2}
\end{aligned}
$$

from which it again follows that

$$
b=58 \text{ m} \text { and } c=41 \text{ m}.
$$