A vectorial approach would be quite lean and effective.
Given two faces of the pyramid, sharing the common edge $V P_n$,
and containing the contiguous base points $P_{n-1}$ and $P_{n+1}$,
the dihedral angle between these two faces would be the angle
made by the two vectors ($t_m, t_p$), normal to the common edge and
lying on the respective face.
Clearly, that will be also the angle made by the normal vectors to the faces,
provided that one is taken in the inward, and the other in the outward
direction.
That is, by the right-hand rule,
$$
{\bf n}_{\,m} = \mathop {P_{\,n} P_{\,n - 1} }\limits^ \to \; \times \;\mathop {P_{\,n} V}\limits^ \to \quad \quad {\bf n}_{\,p} = \mathop {P_{\,n} P_{\,n + 1} }\limits^ \to \; \times \;\mathop {P_{\,n} V}\limits^ \to
$$
Then the dihedral angle $\alpha$ will be simply computed from the dot product
$$
\cos \alpha = {{{\bf n}_{\,m} \; \cdot \;{\bf n}_{\,p} } \over {\left| {{\bf n}_{\,m} } \right|\;\;\left| {{\bf n}_{\,p} } \right|}}
$$
Let $K$ be a middle point of $AC$.
Just $$\measuredangle SBK=\measuredangle SAK=\arccos\frac{\sqrt3}{2}=30^{\circ}.$$
BK is a median of $\Delta ABC$ and is not perpendicular to $AC$, otherwise $AB=BC$, which is a contradiction.
By the way, $BK\perp SK$, but we said about it in the first line.
Let $SK'$ be an altitude of the pyramid.
Thus, since $SA=SB=SC$, we obtain: $\Delta SAK'\cong\Delta SBK'$ and $\Delta SAK'\cong\Delta SCK'$, which gives $$AK'=BK'=CK',$$ which says $K'$ is a center of the circumcircle for $\Delta ABC$.
Thus, $K'\equiv K$.
Best Answer
Let the vertex $A$ be origin. So, the equation of diagonal $AC$ will be $y=\frac{x}{3}$. Given that $CF=6$, we get $C=(18,6)$.
Now, we can find equation of $BC$ which is perpendicular to $AC$. From this, we get $B=(20,0)$. And from $A, B, C$, we get $D=(2,18)$.
It is given that all the lateral sides make equal angle with the base. This is possible only when the distance between the vertices and the projection of apex $(E)$ are equal i.e. $AE=BE=CE=DE$.
So, for $AE=DE$, $E$ should lie on perpendicular bisector of $AD$. Same for $BC$. On solving we get $E=(10,0)$.
To find the height of the pyramid, we've $$\tan\beta=\frac{1}{6}=\frac{h}{10}\\ \implies h=\frac{5}{3}$$