3.66. The area of an isosceles trapezoid is $S$, the angle between its diagonals, opposite the lateral side, is $\alpha$. Find the height of the trapezoid.
3.66. By the condition, $A B \| D C$; $A D=B C ; \angle B O C=\alpha ; S_{A B C D}=S$ (Fig.3.68). Since $\angle B O C$ is the exterior angle of $\triangle D O C$, then $\angle B O C=$

$=2 \angle O D C$, i.e., $\angle O D C=\frac{\alpha}{2}$. Draw $B N \perp C D, E F \| B N$; from $\triangle O F D$ and $\triangle O E B$ we find $O F=D F \operatorname{tg} \frac{\alpha}{2}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-542.jpg?height=319&width=495&top_left_y=919&top_left_x=660)

Fig. 3.68 $O E=B E \operatorname{tg} \frac{\alpha}{2}$. Let $B N=h$; then $h=O F+O E=(D F+B E) \operatorname{tg} \frac{\alpha}{2}$, $D F+B E=h \operatorname{ctg} \frac{\alpha}{2}$. But $S_{A B C D}=(D F+B E) h$, i.e., $S=h^{2} \operatorname{ctg} \frac{\alpha}{2}$, from which $h=\sqrt{S \operatorname{tg} \frac{\alpha}{2}}$.