12.235. The angle between a lateral edge of a regular quadrilateral pyramid and the plane of the base is equal to the magnitude of the plane angle at the vertex of the pyramid. Find the angle between a lateral face and the plane of the base.
Solution.

Let $SO$ be the height of the regular pyramid $SABCD$ (Fig. 12.102), $\angle SDO = \angle CSD$, and $E$ be the midpoint of $CD$. Then $OE \perp CD$, $SE \perp CD$, and $\angle SEO$ is the angle between the lateral face and the base plane of the pyramid.

If $OE = 1$, $SO = H$, then $CD = 2$, $OC = \sqrt{2}$.

Let $DF$ be the height of $\triangle DSC$, then $\triangle SOD = \triangle DFS$ - by hypotenuse and acute angle, and $DF = SO = H$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0984.jpg?height=658&width=615&top_left_y=77&top_left_x=80)

Fig. 12.101

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0984.jpg?height=652&width=606&top_left_y=78&top_left_x=658)

Fig. 12.102

In $\triangle SOE (\angle SOE = 90^\circ): SE = \sqrt{OE^2 + SO^2} = \sqrt{1 + H^2}$.

From $\triangle SOC (\angle SOC = 90^\circ): SC = \sqrt{OC^2 + SO^2} = \sqrt{2 + H^2}$. $S_{\triangle CSD} = \frac{1}{2} SC \cdot DF = \frac{1}{2} CD \cdot SE \Rightarrow SC \cdot DF = CD \cdot SE \Rightarrow H \sqrt{2 + H^2} = 2 \sqrt{1 + H^2} \Leftrightarrow H^2 (2 + H^2) = 4 + 4H^2 \Leftrightarrow H^4 - 2H^2 - 4 = 0; H^2 = 1 \pm \sqrt{5}$.

Since $H_2 > 0$, then $H^2 = 1 + \sqrt{5}; H = \sqrt{1 + \sqrt{5}}$.

In $\triangle SOE (\angle SOE = 90^\circ)$:

$$
\tan \angle SEO = \frac{SO}{OE} = H = \sqrt{1 + \sqrt{5}} \Rightarrow \angle SEO = \arctan \sqrt{1 + \sqrt{5}}
$$

Answer: $\arctan \sqrt{1 + \sqrt{5}}$.