Dihedral Angle

Let the center of rectangle $ABCD$ be the origin, so that $A$ is at $(18, -28, 0)$ , $B$ is at $(18, 28, 0)$ , $C$ is at $(-18, 28, 0)$ , and $F$ is at $(12, 16, 24)$ .

The equation of the plane $a_1x + b_1y + c_1z = d_1$ containing $A$ , $B$ , and $F$ must fulfill $18a_1 - 28b_1 = d_1$ , $18a_1 + 28b_1 = d_1$ , and $12a_1 + 16b_1 + 24c_1 = d_1$ , which solves to $b_1 = 0$ , $c_1 = \frac{1}{4}a_1$ , and $d_1 = 18a_1$ , so that the equation when $a_1 = 4$ is $4x + z = 72$ , which has a normal vector of $(4, 0, 1)$ .

The equation of the plane $a_2x + b_2y + c_2z = d_2$ containing $B$ , $C$ , and $F$ must fulfill $18a_2 + 28b_2 = d_2$ , $-18a_2 + 28b_2 = d_2$ , and $12a_2 + 16b_2 + 24c_2 = d_2$ , which solves to $a_2 = 0$ , $c_2 = \frac{1}{2}b_2$ , and $d_2 = 28b_2$ , so that the equation when $b_2 = 2$ is $2y + z = 56$ , which has a normal vector of $(0, 2, 1)$ .

The dihedral angle between face $ABFE$ and $BCGF$ is equivalent to the obtuse angle between these two normal vectors, which is:

$\theta = \cos^{-1} \bigg( -\cfrac{4 \cdot 0 + 0 \cdot 2 + 1 \cdot 1}{\sqrt{4^2 + 0^2 + 1^2} \cdot \sqrt{0^2 + 2^2 + 1^2}} \bigg) = \cos^{-1} \bigg(\cfrac{-1}{\sqrt{85}} \bigg) \approx \boxed{96.2268°}$ .

Noting that $AB$ runs parallel to the $x$ -axis and $BC$ runs parallel to the $y$ -axis, the normal to face $ABFE$ is given by:

$\mathbf{n}_1 = ( 0 , - \sin \theta_1 , \cos \theta_1 )$

where $\theta_1$ is the angle between face $ABFE$ and the horizontal plane, and is given by

$\theta_1 = \tan^{-1} \dfrac{h}{ \frac{1}{2} (36 - 24) } = \tan^{-1}( 4)$

From which it follows that $\cos \theta_1 = \dfrac{1}{\sqrt{17}}$ and $\sin \theta_1 = \dfrac{4}{\sqrt{17} }$

Similarly, the normal to face $BCGF$ is given by:

$\mathbf{n}_2 = ( \sin \theta_2 , 0, \cos \theta_2)$

where $\theta_2$ is the angle between face $BCGF$ are the horizontal plane, and is given by

$\theta_2 = \tan^{-1} \dfrac{h}{ \frac{1}{2} (56 - 32) } = \tan^{-1}( 2 )$

From which it follows that $\cos \theta_2 = \dfrac{1}{\sqrt{5}}$ and $\sin \theta_2 = \dfrac{2}{\sqrt{5} }$

Hence, the two normals to the two planes are

$\mathbf{n}_1 = (0, -\dfrac{4}{\sqrt{17}} , \dfrac{1}{\sqrt{17}} )$ , and,

$\mathbf{n}_2 = (\dfrac{2}{\sqrt{5}} , 0 , \dfrac{1}{\sqrt{5}} )$

Noting that $\mathbf{n}_1$ and $\mathbf{n}_2$ are unit vectors, the angle between them is

$\phi = \cos^{-1} \mathbf{n}_1 \cdot \mathbf{n}_2 = \cos^{-1} \dfrac{1}{\sqrt{85} } = 83.7731497028^{\circ}$

Hence the angle between the two planes is $180^{\circ} - \phi = 96.2268^{\circ}$