4A. Let $S$ be the intersection of the diagonals of a convex quadrilateral $ABCD$. Determine the angle between the diagonals of the quadrilateral, given that

$$
\measuredangle SAB = \measuredangle SBC = 30^{\circ} \text{ and } \measuredangle SCD = \measuredangle SDA = 45^{\circ}.
$$
Solution. Applying the sine theorem to each of the triangles $A C D$, $A B C$, and $A B D$ given in the diagram, we obtain the following equalities

$$
\begin{aligned}
& \frac{\overline{A C}}{\overline{A D}}=\frac{\sin \left(180^{\circ}-x\right)}{\sin 45^{\circ}}=\frac{\sin x}{\sin 45^{\circ}}, \\
& \overline{\overline{A B}}=\frac{\sin \left(x-30^{\circ}\right)}{\sin \left(180^{\circ}-x\right)}=\frac{\sin \left(x-30^{\circ}\right)}{\sin x}
\end{aligned}
$$

$$
\frac{\overline{A B}}{\overline{A D}}=\frac{\sin 45^{\circ}}{\sin \left(150^{\circ}-x\right)}
$$

By substitution in the equation

$$
\frac{\overline{A B}}{\overline{A D}}=\frac{\overline{A B}}{\overline{A C}} \cdot \overline{\overline{A C}}
$$

we obtain the equation

$$
\frac{\sin 45^{\circ}}{\sin \left(150^{\circ}-x\right)}=\frac{\sin \left(x-30^{\circ}\right)}{\sin x} \cdot \frac{\sin x}{\sin 45^{\circ}}
$$

equivalent to

$$
\sin ^{2} 45^{\circ}=\sin \left(150^{\circ}-x\right) \sin \left(x-30^{\circ}\right)
$$

or with the equation

$$
\frac{1}{2}=-\frac{1}{2}\left(\cos 120^{\circ}-\cos \left(180^{\circ}-2 x\right)\right)
$$

The last equation reduces to $\cos 2 x=-\frac{1}{2}$, whose solutions are $x=60^{\circ}$ or $x=120^{\circ}$. Therefore, the smaller angle between the diagonals is $x=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_5cbf21d72fe8685988dag-09.jpg?height=500&width=615&top_left_y=286&top_left_x=865)