[ Linear dependence of vectors ] [ Angles between lines and planes ]

The height of a regular hexagonal pyramid is equal to the side of the base. Find the angle between a lateral face and the plane of the base.
Let ABCDEFP be a given regular hexagonal pyramid with vertex $P$; $M$ - the center of the regular hexagon $A B C D E F$, $K$ - the midpoint of $A B$. Denote $A B=B C=C D=D E=E F=A F=a$. Since the pyramid is regular, $P M$ is its height. In addition, $P K \perp A B$ and $M K \perp A B$. Therefore, the angle between a lateral face and the base plane is the angle $\angle P K M$, and since $M$ is the height of the equilateral triangle $A B M$ with side $a$, then $M K=\frac{a \sqrt{3}}{2}$. From the right triangle $P K M$ we find that

$$
\operatorname{tg} \beta=\operatorname{tg} \angle P K M=\frac{P M}{M K}=\frac{a}{\frac{a \sqrt{3}}{2}}=\frac{2}{\sqrt{3}}
$$

## Answer

$\operatorname{arctg} \frac{2}{\sqrt{3}}$

The height of the regular hexagonal pyramid is equal to the side of the base. Find the angle between adjacent lateral faces.

## Solution

Let ABCDEFP be the given regular hexagonal pyramid with vertex $P$; $M$ - the center of the regular hexagon $A B C D E F$. Denote $A B=B C=C D=D E=E F=A F=a$. Since the pyramid is regular, $P M$ is its height (from the condition it follows that $P M=a$). Therefore, the angle $\angle P D M$ is the angle between the lateral edge and the base plane. In the right triangle $D P M$, it is known that $P M=D M=a$, so $\angle P D M=45^{\circ}$. The line $D M$ is the orthogonal projection of the lateral edge $D P$ onto the base plane of the pyramid. Since $D M \perp C E$, by the theorem of three perpendiculars, $D P \perp C E$. Drop the perpendicular $C H$ from vertex $C$ to the lateral edge $D P$. The line $D P$ is perpendicular to two intersecting lines $C E$ and $C H$ in the plane $C E H$, so the line $D P$ is perpendicular to this plane. Therefore, the angle $\angle C H E$ is the linear angle of the dihedral angle between the planes of the lateral faces $C D P$ and $E D P$. Denote $\angle C H E=\gamma$. Let $N$ be the point of intersection of the diagonals $A D$ and $F C$ of the regular hexagon $A B C D E F$. Then $N$ is the midpoint of the segment $M D$,

$$
N H \perp D P, N H \perp C E, N D=\frac{a}{2}, C N=\frac{a \sqrt{3}}{2} .
$$

From the right triangle $D N H$ we find that

$$
N H=D N \sin \angle N D H=D N \sin \angle P D M=\frac{a}{2} \cdot \sin 45^{\circ}=\frac{a \sqrt{2}}{4}
$$

Therefore,

$$
\operatorname{tg} \frac{\gamma}{2}=\operatorname{tg} \angle N H C=\frac{C N}{N H}=\frac{\frac{a \sqrt{3}}{2}}{\frac{a \sqrt{2}}{4}}=\sqrt{6}
$$

By a known trigonometric formula, we can also find the cosine of the angle $\gamma$:

$$
\cos \gamma=\frac{1-\operatorname{tg}^2 \frac{\gamma}{2}}{1+\operatorname{tg}^2 \frac{\gamma}{2}}=\frac{1-6}{1+6}=-\frac{5}{7}
$$

## Answer

$2 \operatorname{arctg} \sqrt{6}=\arccos \left(-\frac{5}{7}\right)$