Find the area of triangle $ABC$, if it is known that $AB=a, \angle A=\alpha, \angle B=\beta$.

#
Use the Law of Sines.

## Solution

By the Law of Sines, $\frac{A C}{\sin \angle B}=\frac{A B}{\sin \angle C}$, hence

$$
A C=\frac{A B \sin \angle B}{\sin \angle C}=\frac{a \sin \beta}{\sin \left(180^{\circ}-(\alpha+\beta)\right)}=\frac{a \sin \beta}{\sin (\alpha+\beta)}
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot A C \sin \angle A=\frac{1}{2} \cdot \frac{a^{2} \sin \alpha \sin \beta}{\sin (\alpha+\beta)}
$$

## Answer

$\frac{a^{2} \sin \alpha \sin \beta}{2 \sin (\alpha+\beta)}$.