12.137. 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.
## Solution.

Let $O$ be the point of intersection of the diagonals of the given trapezoid $ABCD, AB=CD$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0898.jpg?height=400&width=671&top_left_y=1195&top_left_x=593)

Fig. 12.10 (Fig. 12.10), $\angle AOB = \alpha$.

Since $\angle AOB$ is the external angle of $\triangle AOD$, and $AO = OD$, then $\angle CAD = \frac{\alpha}{2}$.

Let $CK = H$ be the height of the trapezoid.

From $\triangle AKC \left(\angle AKC = 90^\circ\right): AK = H \operatorname{ctg} \frac{\alpha}{2}$.

Then the area of the trapezoid $S = \frac{AD + BC}{2} \cdot CK = AK \cdot CK = H^2 \operatorname{ctg} \frac{\alpha}{2}$.

Finally, $H = \sqrt{S \operatorname{tg} \frac{\alpha}{2}}$.

Answer: $\sqrt{S \operatorname{tg} \frac{\alpha}{2}}$.