What is the area of a triangle if the radius $R$ of the circumscribed circle, one side $a$, and one angle $\beta$ are given?
Let's draw the altitude $C C_{1}$ in triangle $A B C$. Then

$$
A B=c=B C_{1}+C_{1} A=a \cos \beta+\sqrt{\overline{A C}^{2}-{\overline{C C_{1}}}^{2}}=a \cos \beta+\sqrt{b^{2}-a^{2} \sin ^{2} \beta}
$$

If $C D$ is the diameter of the circumcircle of the triangle, then triangle $C A D$ is a right triangle and

$$
C D A \angle=C B A \angle=\beta,
$$

thus

$$
A C=b=C D \sin \beta=2 R \sin \beta
$$

and therefore

$$
c=a \cos \beta+\sin \beta \sqrt{4 R^{2}-a^{2}}
$$

and

$$
t=\frac{a c \sin \beta}{2}=\frac{a^{2} \sin 2 \beta}{4}+\frac{a \sin ^{2} \beta}{2} \sqrt{4 R^{2}-a^{2}}
$$

(Domokos György, Keszthely.)