[ [quadrilateral: calculations, metric relations.]

Specify all convex quadrilaterals for which the sums of the sines of opposite angles are equal.

#
Let's denote the consecutive angles of the quadrilateral as $\alpha, \beta, \gamma$, and $\delta$. According to the problem, $\sin \alpha + \sin \gamma = \sin \beta + \sin \delta$. Since $\alpha + \beta + \gamma + \delta = 360^\circ$, we have $\sin \delta = \sin (360^\circ - \alpha - \beta - \gamma) = -\sin (\alpha + \beta + \gamma)$. Therefore, $\sin \alpha + \sin \gamma = \sin \beta - \sin (\alpha + \beta + \gamma)$. Let's transform this equation using trigonometric formulas.

$$
2 \sin \frac{\alpha + \gamma}{2} \cos \frac{\alpha - \gamma}{2} = -2 \sin \frac{\alpha + \gamma}{2} \cos \left(\beta + \frac{\alpha + \gamma}{2}\right)
$$

Notice that $\alpha + \gamma \neq 360^\circ$, so $\sin \frac{\alpha + \gamma}{2} \neq 0$. Therefore,

$$
\cos \frac{\alpha - \gamma}{2} = -\cos \left(\beta + \frac{\alpha + \gamma}{2}\right), \cos \frac{\alpha - \gamma}{2} + \cos \left(\beta + \frac{\alpha + \gamma}{2}\right) = 0
$$

$$
2 \cos \frac{\frac{\alpha - \gamma}{2} + \beta + \frac{\alpha + \gamma}{2}}{2} \cos \frac{\frac{\alpha - \gamma}{2} - \beta - \frac{\alpha + \gamma}{2}}{2}, \cos \frac{\alpha + \beta}{2} \cos \frac{\gamma + \beta}{2} = 0
$$

From this, we find that $\cos \frac{\alpha + \beta}{2} = 0$ or $\cos \frac{\gamma + \beta}{2} = 0$, meaning $\alpha + \beta = 180^\circ$ or $\gamma + \beta = 180^\circ$. Since angles $\alpha$ and $\beta$ are adjacent to one side of the quadrilateral, in the first case, two opposite sides are parallel, and in the second case, the other two sides are parallel. Therefore, the quadrilateral is either a trapezoid or a parallelogram.

## Answer

Parallelogram or trapezoid.