Solve the triangle if $a=15 \mathrm{~cm}, \alpha=32^{\circ} 15^{\prime} 25^{\prime \prime}, t=156 \mathrm{~cm}^{2}$.
It is known that

$$
2 t=b c \sin \alpha \quad \text { or } \quad b c=\frac{2 t}{\sin \alpha}
$$

According to Carnot's theorem,

$$
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha=(b+c)^{2}-2 b c(\cos \alpha+1)=(b+c)^{2}-4 b c \cos ^{2} \frac{\alpha}{2}
$$

or

$$
(b+c)^{2}=a^{2}+\frac{8 t}{\sin \alpha} \cos ^{2} \frac{\alpha}{2}=a^{2}+4 t \operatorname{ctg} \frac{\alpha}{2}
$$

Similarly,

$$
a^{2}=(b-c)^{2}=2 b c(\cos \alpha-1)=(b-c)^{2}+4 b c \sin ^{2} \frac{\alpha}{2}
$$

or

$$
(b-c)^{2}=a^{2}+\frac{8 t}{\sin \alpha} \sin ^{2} \frac{\alpha}{2}=a^{2}-4 t \operatorname{tg} \frac{\alpha}{2}
$$

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

$$
b=\frac{\sqrt{a^{2}+4 t \operatorname{ctg} \frac{\alpha}{2}}+\sqrt{a^{2}-4 t \operatorname{tg} \frac{\alpha}{2}}}{2}
$$

and

$$
c=\frac{\sqrt{a^{2}+4 t \operatorname{ctg} \frac{\alpha}{2}}-\sqrt{a^{2}-4 t \operatorname{tg} \frac{\alpha}{2}}}{2}
$$

Now we can calculate the angles using the sine theorem.

Substituting the given numerical values, we get

$$
b=27.92 \mathrm{~cm}, \quad c=21.26 \mathrm{~cm}
$$

and

$$
\beta=99^{\circ} 17^{\prime} 16^{\prime \prime}, \quad \gamma=48^{\circ} 27^{\prime} 21^{\prime \prime}
$$