## [ Trigonometric relationships in a right triangle ]

In triangle $A B C$, the height $A H$ is equal to $h, \angle B A C=\alpha, \angle B C A=\gamma$. Find the area of triangle $A B C$.

#
Apply the Law of Sines.

## Solution

From the right triangle $C H A$, we find that

$$
A C=\frac{A H}{\sin \gamma}=\frac{h}{\sin \gamma}
$$

By the Law of Sines,

$$
B C=\frac{A C \sin \alpha}{\sin (\alpha+\gamma)}=\frac{h \sin \alpha}{\sin \gamma \sin (\alpha+\gamma)}
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} B C \cdot A H=\frac{h^{2} \sin \alpha}{2 \sin \gamma \sin (\alpha+\gamma)}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-33.jpg?height=378&width=786&top_left_y=1120&top_left_x=636)

Answer

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