12.040. The height of an isosceles trapezoid is $h$. The upper base of the trapezoid is seen from the midpoint of the lower base at an angle of $2 \alpha$, and the lower base is seen from the midpoint of the upper base at an angle of $2 \beta$. Find the area of the trapezoid in this general case and calculate it without tables if $h=2, \alpha=15^{\circ}, \beta=75^{\circ}$.
## Solution.

In trapezoid $ABCD$, we have: $AB=CD, BC \| AD, M \in AD, AM=MD$, $N \in BC, BN=NC, BE \perp AD, BE=h, \angle BMC=2\alpha, \angle AND=2\beta$ (Fig. 12.42). Since $\triangle ABM = \triangle CMD$ and $\triangle ABN = \triangle NCD$ (by two sides and the included angle), we have $BM=MC$ and $AN=ND$. From $\triangle AMN$ and $\triangle BMN$, we find $AM = h \tan \beta$ and $BN = h \tan \alpha$. Therefore, $S_{ABCD} = (AM + BN)h = h^2 (\tan \alpha + \tan \beta) = \frac{h^2 \sin (\alpha + \beta)}{\cos \alpha \cos \beta}$. For $h=2, \alpha=15^\circ$ and $\beta=75^\circ$, we have

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-744.jpg?height=352&width=498&top_left_y=112&top_left_x=139)

Fig. 12.43

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-744.jpg?height=372&width=372&top_left_y=94&top_left_x=766)

Fig. 12.44

$$
S_{ABCD} = \frac{4 \sin 90^\circ}{\cos 15^\circ \cos 75^\circ} = \frac{4}{\cos 15^\circ \sin 15^\circ} = \frac{8}{\sin 30^\circ} = 16
$$

Answer: $\frac{h^2 \sin (\alpha + \beta)}{\cos \alpha \cos \beta} ; 16$.