10,11

The lower base of a truncated quadrilateral pyramid is a rhombus $ABCD$, where $AB=4$ and $\angle BAD=60^{\circ}$. $AA1, BB1, CC1, DD1$ are the lateral edges of the truncated pyramid, edge $A1B1=2$, edge $CC1$ is perpendicular to the base plane and equals 2. A point $M$ is taken on edge $BC$ such that $BM=3$, and a plane is drawn through points $B1$, $M$, and the center of the rhombus $ABCD$. Find the dihedral angle between this plane and the plane $AA1C1C$.
Let $O$ be the center of the rhombus $ABCD$. From the triangle $OMB$ we find that

$$
O M=\sqrt{B O^{2}+B M^{2}-2 B O \cdot B M \cos 60^{\circ}}=\sqrt{4+9-6}=\sqrt{7}
$$

Construct the line of intersection of the planes $AA_1C_1C$ and $B_1MO$. For this, extend the segments $C_1C$ and $B_1M$ until they intersect at point $E$. Each of the points $O$ and $E$ lies in both the plane $AA_1C_1C$ and the plane $B_1MO$, so $OE$ is the line of intersection of these planes. Construct the linear angle of the dihedral angle between these planes. For this, drop a perpendicular $C_1K$ from point $C_1$ to the edge $OE$ of this dihedral angle, and from point $K$ draw a perpendicular to $OE$ in the plane $B_1MO$. Let this perpendicular intersect the line $B_1E$ at point $L$. Then $C_1KL$ is the desired linear angle. Denote $\angle C_1KL=\gamma$. To find it, we will compute all the sides of the triangle $C_1KL$. Since $MC=1=\frac{1}{2}B_1C_1$ and $MC \parallel B_1C_1$, the segment $MC$ is the midline of the triangle $B_1C_1E$, so $C$ is the midpoint of the segment $EC_1$, and since $OC \perp EC_1$, the median $OC$ of the triangle $EOC_1$ is its height. Therefore, the triangle $EOC_1$ is isosceles. Moreover, from the right triangle $SOC_1$ we find that

$$
OC_1=\sqrt{OC^2+CC_1^2}=\sqrt{\left(\frac{4\sqrt{3}}{2}\right)^2+2^2}=\sqrt{12+4}=4=EC_1
$$

so the triangle $EOC_1$ is equilateral. Then its height $C_1K$ is equal to the height $OC_1$, i.e., $C_1K=\frac{4\sqrt{3}}{2}=2\sqrt{3}$. In the triangle $OME$ it is known that

$$
ME=\sqrt{MC^2+CE^2}=\sqrt{1+4}=\sqrt{5}, OE=4, OM=\sqrt{7}.
$$

Denote $\angle OME=\phi$. Then

$$
\cos \phi=\frac{OE^2+ME^2-OM^2}{2OE \cdot ME}=\frac{16+5-7}{2 \cdot 4 \cdot \sqrt{5}}=\frac{7}{4\sqrt{5}}, \operatorname{tg} \phi=\sqrt{\frac{1}{\cos^2 \phi}-1}=\frac{\sqrt{31}}{7}
$$

From the right triangle $LKE$ we find that

$$
KL=KE \operatorname{tg} \phi=\frac{1}{2}OE \operatorname{tg} \phi=2 \cdot \frac{\sqrt{31}}{7}=\frac{2\sqrt{31}}{7}, LE=\frac{KE}{\cos \phi}=\frac{\frac{2}{7}}{4\sqrt{5}}=\frac{8\sqrt{5}}{7}.
$$

Consider the face $BB_1C_1C$. Denote $\angle B_1EC_1=\beta$. Then

$$
\operatorname{tg} \beta=\frac{B_1C_1}{EC_1}=\frac{1}{2}, \cos \beta=\frac{1}{\sqrt{1+\operatorname{tg}^2 \beta}}=\frac{2}{\sqrt{5}}
$$

By the cosine rule

$$
LC_1=\sqrt{C_1E^2+LE^2-2C_1E \cdot LE \cos \beta}=\frac{4\sqrt{31}}{7}
$$

Therefore,

$$
\cos \gamma=\frac{KL^2+KC_1^2-LC_1^2}{2KL \cdot KC_1}=\frac{\left(\frac{2\sqrt{31}}{7}\right)^2+(2\sqrt{3})^2-\left(\frac{4\sqrt{31}}{7}\right)^2}{2 \cdot \frac{2\sqrt{31}}{7} \cdot 2\sqrt{3}}=\frac{9}{\sqrt{93}}
$$

## Answer

$\arccos \frac{9}{\sqrt{93}}=\operatorname{arctg} \frac{2}{3\sqrt{3}}$