3.16. The height of an isosceles trapezoid is equal to \( h \). 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 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 \( h=2, \alpha=15^{\circ}, \beta=75^{\circ} \).
3.16. Given that in trapezoid $ABCD$ we have: $AB = CD, BC \parallel AD$, $M \in AD, AM = MD, N \in BC, BN = NC, BE \perp AD, BE = h, \angle BMC = 2\alpha$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-509.jpg?height=312&width=713&top_left_y=1259&top_left_x=224)

Fig. 3.18

$\angle AND = 2\beta$ (Fig. 3.18). Since $\triangle ABM = \triangle CMD$ and $\triangle ABN = \triangle CDN$ (by two sides and the included angle), we have $BM = MC, AN = ND$. From $\triangle AMN$ and $\triangle BMN$ we find $AM = h \operatorname{tg} \beta, BN = h \operatorname{tg} \alpha$. Therefore,

$$
S_{ABCD} = (AM + BN) h = h^2 (\operatorname{tg} \alpha + \operatorname{tg} \beta) = \frac{h^2 \sin (\alpha + \beta)}{\cos \alpha \cos \beta}
$$

For $h = 2, \alpha = 15^\circ$ and $\beta = 75^\circ$ we get

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