The base of a pyramid is a regular pentagon, and three consecutive side faces are inclined to the base plane at angles of $30^{\circ}$, $90^{\circ}$, and $60^{\circ}$, respectively. What are the angles of inclination of the remaining two side faces to the base plane?
Let the base of the pyramid be $ABCDE$, and the lateral faces inclined at $30^\circ$, $90^\circ$, and $60^\circ$ to the base be $FAB$, $FBC$, and $FCD$, respectively. Thus, the projection of the apex $F$ onto the base plane $S$ is at point $F_1$ on segment $BC$. Furthermore, let the projections of $F$ onto lines $AB$, $CD$, $DE$, and $EA$ be $F_2$, $F_3$, $F_4$, and $F_5$, respectively.
![](https://cdn.mathpix.com/cropped/2024_05_02_fd3d97c676ffb5d81162g-1.jpg?height=490&width=1224&top_left_y=308&top_left_x=446)

Thus, the plane of the right triangle $FF_1F_i$ ($i=2,3,4,5$) is perpendicular to the respective line in the base plane, so the angle between the lateral face planes $FAB$, $FCD$, $FDE$, and $FEA$ and the base plane $S$ is measured by $\angle F_1F_iF$ and

$$
\operatorname{ctg} \angle F_1F_iF = \frac{F_1F_i}{F_1F}.
$$

For $i=2$ and $3$, the inclination angles are $30^\circ$ and $60^\circ$, respectively, so

$$
\frac{F_1F_2}{F_1F_3} = \frac{F_1F \operatorname{ctg} 30^\circ}{F_1F \operatorname{ctg} 60^\circ} = \frac{\sqrt{3}}{1 / \sqrt{3}} = 3.
$$

Based on this, from the similar right triangles $F_1F_2B$ and $F_1F_3C$ (where the angles at vertices $B$ and $C$ are the exterior angles of the regular pentagon, $72^\circ$),

$$
\frac{F_1B}{F_1C} = 3,
$$

so $F_1$ is the point dividing segment $BC$ in the ratio $1:3$ closer to $C$. Let $CF_1 = 1$, then the height of the pyramid is

$$
F_1F = F_1F_3 \operatorname{tg} 60^\circ = CF_1 \sin 72^\circ \operatorname{tg} 60^\circ = \sqrt{3} \cos 18^\circ.
$$

Let $F_1F_5$ intersect the diagonal $BD$ of the pentagon at $G$, then $GF_5$ is the distance from the diagonal to $AE$, so, using the right triangle $F_1BG$,

$$
\begin{aligned}
& F_1F_5 = F_1G + GF_5 = F_1B \sin 36^\circ + AB \sin 72^\circ = \\
& \quad = 3 \sin 36^\circ + 4 \sin 72^\circ \\
& \operatorname{ctg} \angle F_1F_5F = \frac{F_1F_5}{F_1F} = \frac{3 \sin 36^\circ}{\cos 18^\circ} + \frac{4}{\sqrt{3}} = \\
& = 2 \sqrt{3} \sin 18^\circ + \frac{4 \sqrt{3}}{3} = 1.070 + 2.309 = 3.379
\end{aligned}
$$

and thus the inclination angle of the face $FAE$ is $16^\circ 29'$.

The calculation of $\operatorname{ctg} \angle F_1F_4F$ differs only in that the first term should be one-third of (1), $0.357 + 2.309 = 2.666$, so the inclination angle of the face $FDE$ is $20^\circ 34'$.

András Vajnági (Budapest, Fazekas M. Gyak. Gimn., III. o. t.)

Remark. The angles in question can also be determined by construction from the projection of the pyramid onto the base plane $S$ and the appropriately rotated images of the lines $FF_i$ in $S$.