$\left[\begin{array}{l}{[\text { Area of a triangle (using two sides and the angle between them).] }} \\ {[\quad \text { Law of Sines }}\end{array}\right]$

In triangle $A B C$, it is known that $\angle B A C=\alpha, \angle B C A=\gamma, A B=c$. Find the area of triangle $A B C$.
Apply the Law of Sines.

## Solution

By the Law of Sines

$$
\frac{c}{\sin \gamma}=\frac{A C}{\sin \left(180^{\circ}-\alpha-\gamma\right)} \Rightarrow A C=\frac{c \sin (\alpha+\gamma)}{\sin \gamma} .
$$

Therefore,

$$
\begin{gathered}
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot A C \sin \alpha= \\
=\frac{1}{2} c \cdot \frac{c \sin (\alpha+\gamma)}{\sin \gamma} \cdot \sin \alpha=\frac{c^{2} \sin \alpha \sin (\alpha+\gamma)}{2 \sin \gamma}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-32.jpg?height=346&width=648&top_left_y=2137&top_left_x=705)

Answer

$\frac{c^{2} \sin \alpha \sin (\alpha+\gamma)}{2 \sin \gamma}$

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