A triangle has one angle $\alpha=43^{\circ}$. Determine the angles of the triangle if the area $t$ of the triangle satisfies the relation

$$
2 t=a b \sqrt{\sin ^{2} \alpha+\sin ^{2} \beta+\sin \alpha \sin \beta}
$$
The essence of the solution was to notice that from the conditional equation, the value of the angle $\gamma$ is $120^{\circ}$. This can be achieved in two main ways: either by eliminating angles using distances and then using the relationships between the parts of the triangle, or by trying to transform the given relationship into a trigonometric relationship and performing trigonometric transformations. The former can be followed in the following way.

![](https://cdn.mathpix.com/cropped/2024_05_02_47463d5ba6d77886ca59g-1.jpg?height=219&width=523&top_left_y=308&top_left_x=779)

Solution I: The labeling is shown in the figure. Substitute the expressions $\sin \alpha=\frac{m_{c}}{b}$, $\sin \beta=\frac{m_{c}}{a}$, $2 t=c m_{c}$ into the given relationship and square both sides. Since both sides are positive, we obtain an equivalent relationship to the original one:

$$
c^{2} m_{c}^{2}=a^{2} b^{2}\left(\frac{m_{c}^{2}}{a^{2}}+\frac{m_{c}^{2}}{b^{2}}+\frac{m_{c}^{2}}{a b}\right)=\left(a^{2}+b^{2}+a b\right) m_{c}^{2}
$$

From this, since $m_{c}$ cannot be $0$ (due to $\alpha=43^{\circ}$, the triangle cannot degenerate into a straight line),

$$
c^{2}=a^{2}+b^{2}+a b
$$

On the other hand, by the cosine rule,

$$
c^{2}=a^{2}+b^{2}-2 a b \cos \gamma
$$

Comparing the two expressions, $\cos \gamma=-\frac{1}{2}$, and since $\gamma<180^{\circ}$, $\gamma=120^{\circ}$ is the only possible value. In this case, $\alpha+\beta=60^{\circ}$, so the angles of the triangle are $\alpha=43^{\circ}, \beta=17^{\circ}, \gamma=120^{\circ}$.

Solution II: Using the relationship $2 t=a b \sin \gamma$ for determining the area, dividing by the non-zero value of $a b$ and squaring, we get

$$
\sin ^{2} \gamma=\sin ^{2} \alpha+\sin ^{2} \beta+\sin \alpha \sin \beta
$$

or

$$
\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma+\sin \alpha \sin \beta=0
$$

First, transform the middle difference using the product-to-sum identities for trigonometric functions [we could also achieve the goal by expanding $\sin \gamma=\sin (\alpha+\beta)$]:

$$
\begin{gathered}
\sin ^{2} \beta-\sin ^{2} \gamma=(\sin \beta+\sin \gamma)(\sin \beta-\sin \gamma)= \\
=2 \sin \frac{\beta+\gamma}{2} \cos \frac{\beta-\gamma}{2} \cdot 2 \cos \frac{\beta+\gamma}{2} \sin \frac{\beta-\gamma}{2}= \\
=2 \sin \frac{\beta+\gamma}{2} \cos \frac{\beta+\gamma}{2} \cdot 2 \sin \frac{\beta-\gamma}{2} \cos \frac{\beta-\gamma}{2}= \\
=\sin (\beta+\gamma) \sin (\beta-\gamma)=\sin \alpha \sin (\beta-\gamma)
\end{gathered}
$$

Using this,

$$
\begin{gathered}
\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma+\sin \alpha \sin \beta=\sin \alpha[\sin \alpha+\sin (\beta-\gamma)]+\sin \alpha \sin \beta= \\
=2 \sin \alpha \sin \frac{\alpha+\beta-\gamma}{2} \cos \frac{\alpha-\beta+\gamma}{2}+\sin \alpha \sin \beta= \\
=2 \sin \alpha \sin \frac{180^{\circ}-2 \gamma}{2} \cos \frac{180^{\circ}-2 \beta}{2}+\sin \alpha \sin \beta= \\
=2 \sin \alpha \sin \left(90^{\circ}-\gamma\right) \cdot \cos \left(90^{\circ}-\beta\right)+\sin \alpha \sin \beta= \\
=2 \sin \alpha \sin \beta \cos \gamma+\sin \alpha \sin \beta=2 \sin \alpha \sin \beta\left(\cos \gamma+\frac{1}{2}\right)
\end{gathered}
$$

This product can only be 0 if $\cos \gamma=-\frac{1}{2}$, i.e., $\gamma=120^{\circ}$.

# Remarks: 1. In Solution II, we proved and used the identity

$$
\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma=2 \sin \alpha \sin \beta \cos \gamma, \text { if } \alpha+\beta+\gamma=180^{\circ}
$$

which appeared in problem 560 of our journal (Vol. VIII, p. 82). Several students also used this identity directly in their solutions.

2. Some students derived (1) from (2) by using the sine rule in the form $\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}=\lambda$, or more precisely the relationships $a=2 r \sin \alpha$ etc., where $r$ is the radius of the circumcircle of the triangle. Using the same method, the cosine rule can also be transformed into a trigonometric relationship, precisely into the identity (3), and comparing it with (2) gives the desired result.