Angle between two surfaces!

Let the sides of the square be $l$ and let the radius of the circle (and the height of the pyramid) be $h$ .

Since the circle's circumference is equal to the perimeter of the square, $4l = 2\pi h$ , or $h = \frac{2l}{\pi}$ .

The acute dihedral angle of a square pyramid with a base side of $l$ and a height of $h$ is:

$\theta = \cos^{-1}\bigg(\cfrac{l^2}{l^2 + 4h^2}\bigg) = \cos^{-1}\bigg(\cfrac{l^2}{l^2 + 4(\frac{2l}{\pi})^2}\bigg) = \cos^{-1}\bigg(\cfrac{\pi^2}{\pi^2 + 16}\bigg) \approx \boxed{67.57253352°}$

Let point $Q$ on $PD$ be such that $AQ$ and $CQ$ are perpendicular to $PD$ . Then $\angle AQC = \theta$ is the angle between the two faces of the pyramid.

Let the common perimeter be $2\pi$ . Then the radius or the height $OP = 1$ and the side length the square $ABCD$ is $\dfrac \pi 2$ , $BD = \dfrac \pi{\sqrt 2}$ , $OD = \dfrac \pi{2\sqrt 2}$ , and $PD = \sqrt{OP^2+OD^2} = \sqrt{1+\dfrac {\pi^2}8}$ . $\cos \angle PDA = \dfrac {\frac 12 AD}{PD} = \dfrac \pi{\sqrt{16+2\pi^2}} \implies \sin \angle PDA = \sqrt{\dfrac {16+\pi^2}{16+2\pi^2}}$ . And we have:

$\begin{aligned} \sin \frac \theta 2 & = \frac {AO}{AQ} = \frac {AO}{AD \sin \angle PDA} = \sqrt{\frac {16+2\pi^2}{32+2\pi}} \\ \implies \theta & = 2 \sin^{-1} \sqrt{\frac {16+2\pi^2}{32+2\pi}} \approx 112.427466477^\circ \end{aligned}$

Therefore the acute angle between the two faces is $180^\circ - 112.427466477^\circ \approx \boxed{67.6}^\circ$ .

Let $a$ be side of square base $ABCD$ and $r$ radius of the circle.

Let $O(0,0,0), A:(-\dfrac{a}{2},-\dfrac{a}{2},0), B:(-\dfrac{a}{2},\dfrac{a}{2},0), C:(\dfrac{a}{2},\dfrac{a}{2},0), D:(\dfrac{a}{2},-\dfrac{a}{2},0)$ and $P(0,0,r)$

$\vec{BA} = 0\vec{i} + a\vec{j} + 0\vec{k}, \vec{BP} = \dfrac{a}{2}\vec{i} + -\dfrac{a}{2}\vec{j} + r\vec{k}$ and $\vec{BC} = a\vec{i} + 0\vec{j} + 0\vec{k}$

$\implies \vec{u} =\vec{BP} X \vec{BA} = -ar\vec{i} + 0\vec{j} + \dfrac{a^2}{2}\vec{k}$ and $\vec{v} = \vec{BP} X \vec{BC} = 0\vec{i} + a\vec{j} + \dfrac{a^2}{2}\vec{k}$

$\implies \vec{u} \cdot \vec{v} = \dfrac{a^4}{4} = \sqrt{\dfrac{4a^2r^2 + a^4}{4}} \sqrt{\dfrac{4a^2r^2 + a^4}{4}} \cos(\theta) = \dfrac{4a^2r^2 + a^4}{4} \cos(\theta) \implies$

$\dfrac{a^4}{4} = \dfrac{a^2}{4}(4r^2 + a^2)\cos(\theta) \implies \cos(\theta) = \dfrac{a^2}{4r^2 + a^2}$

and $4a = 2\pi r \implies r = \dfrac{2a}{\pi} \implies \cos(\theta) = \dfrac{\pi^2}{16 + \pi^2} \implies \theta = \arccos(\dfrac{\pi^2}{16 + \pi^2}) \approx \boxed{67.57253352^{\circ}}$ .

Let E is mid point of AD, F at lateral edge PD and $AF \perp PD, CF \perp PD$ . Then $\theta = \angle AFC$ is dihedral angles of adjacent faces. Side length of square is $a$ , radius of circle is $r$ :

$\begin{aligned} 2 \pi r &= 4a \implies r = \dfrac 2{\pi} a \\ PE &= \sqrt {r^2 + (\dfrac a2)^2 } = \sqrt {r^2 + \dfrac {a^2}4} \\ PD &= \sqrt {r^2 + (\dfrac {\sqrt 2}2a)^2} = \sqrt {r^2 + \dfrac {a^2}2} \\ S_{\triangle PAD} &= \dfrac 12 AD \times PE = \dfrac 12 AF \times PD \\ \therefore AF &= \dfrac {AD \times PE}{PD} = \dfrac {a \sqrt {r^2 + \dfrac {a^2}4}}{\sqrt {r^2 + \dfrac {a^2}2}} \\ CF &= AF \\ AC^2 &= AF^2 + CF^2 - 2 \cdot AF \cdot CF \cdot \cos {\theta} \\ &= 2 AF^2 - 2 AF^2 \cos {\theta} \\ \therefore \cos \theta &= \dfrac {AC^2 - 2 AF^2}{2 AF^2} = \dfrac {AC^2}{2 AF^2} - 1 \\ &= \dfrac {2a^2}{\dfrac {2a^2 (r^2 + \dfrac {a^2}4)}{r^2 + \dfrac {a^2}2}} - 1 \\ &= \dfrac {4r^2 + 2a^2}{4r^2 + a^2} - 1 = \dfrac {a^2}{4r^2 + a^2} \\ &= \dfrac {a^2}{4(\dfrac {2}{\pi})^2 a^2 + a^2} = \dfrac {\pi ^2}{16 + \pi ^2} \\ \theta &= \cos^{-1} {\dfrac {\pi ^2}{16 + \pi ^2}} \approx \boxed {67.57 \degree} \end{aligned}$