In an isosceles triangle with angles $\alpha, \beta, \gamma$. What are these angles if $\sin ^{2} \alpha+\sin ^{2} \beta=\sin \gamma$?
Solution. We distinguish two cases.

1. case: $\alpha=\beta$ and $\gamma=180^{\circ}-2 \alpha$. Then the equation is:

$$
\begin{aligned}
& 2 \sin ^{2} \alpha=\sin \gamma=\sin \left(180^{\circ}-2 \alpha\right)=\sin 2 \alpha, \quad \text { that is } \\
& 2 \sin ^{2} \alpha=2 \sin \alpha \cos \alpha
\end{aligned}
$$

Simplifying the equation by $2 \sin \alpha \neq 0$, we get $\sin \alpha=\cos \alpha$. Since $0<\alpha<180^{\circ}$, this only holds if $\alpha=\beta=45^{\circ}$ and $\gamma=90^{\circ}$. In this case, the triangle is right-angled and isosceles.

2. case: $\alpha=\gamma$ and $\beta=180^{\circ}-2 \alpha$. Then the equation is:

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

Substituting the value of $\beta$, we get:

$$
\begin{aligned}
\sin ^{2} \alpha+(\sin 2 \alpha)^{2} & =\sin \alpha \\
\sin ^{2} \alpha+4 \sin ^{2} \alpha \cos ^{2} \alpha & =\sin \alpha
\end{aligned}
$$

Simplifying by $\sin \alpha \neq 0$, and using $\cos ^{2} \alpha=1-\sin ^{2} \alpha$:

$$
4 \sin ^{3} \alpha-4 \sin \alpha-\sin \alpha+1=0
$$

From here, $4 \sin \alpha \cdot\left(\sin ^{2} \alpha-1\right)-(\sin \alpha-1)=0$. Let's factorize:

$$
(\sin \alpha-1) \cdot[4 \sin \alpha \cdot(\sin \alpha+1)-1]=0
$$

A product is zero if and only if one of its factors is zero. If $\sin \alpha-1=0$, then $\sin \alpha=1, \alpha=90^{\circ}$. Since $\alpha=\gamma$, this is not possible. If $4 \sin \alpha \cdot(\sin \alpha+1)-1=0$, then we get the following quadratic equation:

$$
4 \sin ^{2} \alpha+4 \sin \alpha-1=0
$$

From here, $\sin \alpha_{1}=\frac{-1-\sqrt{2}}{2}<-1$, which cannot be a solution; $\sin \alpha_{2}=\frac{-1+\sqrt{2}}{2} \approx 0.2071, \alpha_{2} \approx 11.95^{\circ}$. The angles of the triangle in the second case are $\alpha=\gamma \approx 11.95^{\circ}, \beta \approx 156.1^{\circ}$.