Given the following flat structure where $AB$ and $BC$ are different lengths of the base of a triangular pyramid ($AC$ is unknown) and $\alpha$ and $\beta$ are different angle cuts. $AD$, $BD$ and $CD$ are known edge lengths.
How can I calculate the angle between $BD$ and the ground when building an irregular pyramid by creating an angle of $\delta$ between $AB$ and $BC$ as shown in the following diagram?
Let $h$ be the (perpendicular) distance from $D$ to the plane of $\triangle ABC$ in the pyramid. Then $$\sin\theta = \frac{h}{|BD|}\qquad\to\qquad \theta = \arcsin\frac{h}{|BD|} \tag{1}$$ Writing $V$ for the volume of the tetrahedron, we also have $$V = \frac13 h |\triangle ABC| = \frac13 h\cdot \frac12 |AB||BC|\sin\delta \quad\to\quad h = \frac{6V}{|AB||BC|\sin\delta} \tag{2}$$ Finally, as seen in Wikipedia's "Tetrahedron" entry, $$6V = |AB||BC||BD|\cdot\sqrt{1+2\cos\alpha^\prime\cos\beta^\prime\cos\delta-\cos^2\alpha^\prime-\cos^2\beta^\prime-\cos^2\delta} \tag{3}$$ where $\alpha^\prime := \angle ABD$ and $\beta^\prime:=\angle CBD$. (I can't tell from your picture, but it may be that $\alpha^\prime=90^\circ-\alpha$ and $\beta^\prime=90^\circ-\beta$.)
Combining $(1)$, $(2)$, $(3)$, we have